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WISKUNDIGE GELETTERDHEID
Graad 12
NOG OEFENINGE
Waarskynlikheid.
  
MATHEMATICAL LITERACY
Grade 12
MORE EXERCISES
Probability.
  
  
  
Antwoorde / Answers  1
  
     
      1.1  Daar is 14 pynappel gegeurde
  
             lekkers in die blik,
  
             n(E) = n(pynappel) = 14
  
             Daar is 14 + 8 + 2 + 6 = 40
  
             lekkers in die blik.
  
             Dus n(S) = n(aantal lekkers) = 40
  
$$ \hspace*{13 mm}\mathrm{P(pynappel)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(pynappel)}{n(aantal\kern1mmlekkers)}  } $$



$$ \hspace*{34 mm}=\kern2mm\frac{14}{40}\kern2mm=\kern1mm\frac{7}{20} $$



[ Vraag 1 ]
  
  
     
      1.2  Daar is 6 appelkoos gegeurde
  
             lekkers in die blik,
  
             dus n(E) = n(appelkoos) = 6
  
             Daar is 14 + 8 + 2 + 6 = 40
  
             lekkers in die blik.
  
             Dus, n(S) = n(aantal lekkers) = 40
  
$$ \hspace*{13 mm}\mathrm{P(appelkoos)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(appelkoos)}{n(aantal\kern1mmlekkers)}  } $$


$$ \hspace*{34 mm}=\kern1mm\frac{6}{40}\kern2mm=\kern1mm\frac{3}{20} $$



[ Vraag 1 ]
  
  
     
      1.3  n(koejawel) = 12
  
             n(aantal lekkers) = 40
$$ \hspace*{13 mm}\mathrm{P(koejawel)\kern1mm=\kern1mm\frac{n(E)}{n(S)}  } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(koejawel)}{n(aantal\kern1mmlekkers)}  } $$


$$ \hspace*{34 mm}=\kern1mm\frac{12}{40}\kern2mm=\kern1mm\frac{3}{10} $$



[ V 1 ]
  
  
     
      1.4  Nie 'n koejawel geur nie, d.i.
  
             enige geur behalwe koejawel kan
  
             gekies word. Die aantal nie
  
             koejawel gegeurde
  
             lekkers = 14 + 8 + 6 = 28
  
                                         OF
  
             Die aantal nie koejawel gegeurde
  
             lekkers = aantal lekkers ─ aantal
  
             koejawel gegeurde lekkers  
  
                                     =  40 ─ 12  =  28
  
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmkoejawel)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(nie\kern1mmkoejawel)}{n(aantal\kern1mmlekkers)}  } $$


$$ \hspace*{34 mm}=\kern1mm\frac{28}{40}\kern2mm=\kern1mm\frac{7}{10} $$



  
                                       OF
  
             NB  
                                      3
             P(koejawel) = ------          {1.3}
                                     10
  
$$ \hspace*{13 mm}\mathrm{P(koejawel)\kern1mm+\kern1mmP(nie\kern1mmkoejawel)\kern1mm=\kern1mm1 } $$
$$ \hspace*{13 mm}\mathrm{\frac {3}{10}\kern1mm+\kern1mmP(nie\kern1mmkoejawel)\kern1mm=\kern1mm1 } $$
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmkoejawel)\kern1mm=\kern2mm1\kern1mm─\kern1mm\frac {3}{10} } $$


$$ \hspace*{40 mm}\mathrm{=\kern1mm\frac {7}{10} } $$


  
[ V 1 ]
  
  
     
      1.5  n(nie pynappel) = 8 + 12 + 6 = 26
  
             n(lekkers) = 40
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmpynappel)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(nie\kern1mmpynappel)}{n(aantal\kern1mmlekkers)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{26}{40}\kern2mm=\kern1mm\frac{13}{20} $$



                                       OF
  
             NB  
  
             P(pynappel) + P(nie pynappel) = 1
  
                                         7
             P(pynappel)  =  --------      (1.1)
                                        20
  
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmpynappel)\kern1mm=\kern1mm1\kern1mm─\kern1mm\frac {7}{20} } $$


$$ \hspace*{40 mm}\mathrm{=\kern1mm\frac {13}{20} } $$


  
[ V 1 ]
  
  
     
      1.6  n(nie kersie) = 40 - 8 = 32
  
             n(lekkers) = 40
  
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmkersie)\kern1mm=\kern1mm\frac{n(nie\kern1mmkersie)}{n(aantal\kern1mmlekkers)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{32}{40}\kern2mm=\kern1mm\frac{4}{5} $$



  
                                       OF
  
             n(kersie) = 8 en n(lekkers) = 40
$$ \hspace*{13 mm}\mathrm{P(kersie)\kern1mm=\kern1mm\frac{8}{40}\kern2mm=\kern1mm\frac{1}{5} } $$


$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmkersie)\kern1mm=\kern1mm1\kern1mm─\kern1mm\frac{1}{5}\kern1mm=\kern1mm\frac{4}{5} } $$


  
  
[ V 1 ]
  
  
     
      1.7  n(lemoen) = 0 en n(lekkers) = 40
  
$$ \hspace*{13 mm}\mathrm{P(lemoen)\kern1mm=\kern1mm\frac{n(lemoen)}{n(aantal\kern1mmlekkers)} } $$


$$ \hspace*{31 mm}=\kern1mm\frac{0}{40}\kern2mm=\kern1mm0 $$



  
[ V 1 ]
  
  
     
      1.8  n(nie lemoen) = 40 en
  
             n(lekkers) = 40
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmlemoen)\kern1mm=\kern1mm\frac{n(nie\kern1mmlemoen)}{n(aantal\kern1mmlekkers)} } $$


$$ \hspace*{37 mm}=\kern1mm\frac{40}{40}\kern2mm=\kern1mm1 $$



  
  
                                       OF
  
  
  
             P(nie lemoen)  =  1  ─  P(lemoen)
  
             P(nie lemoen)  =  1  ─  0  =  1
  
  
[ V 1 ]
  
  
     
      1.1  There are 14 pineapple flavoured
  
             sweets in the tin, thus
  
             n(E) = n(pineapple) = 14
  
             There are 14 + 8 + 2 + 6 = 40
  
             sweets in the tin.
  
             n(S) = n(number of sweets) = 40
  
$$ \hspace*{13 mm}\mathrm{P(pineapple)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(pineapple)}{n(number\kern1mmof\kern1mmsweets)}  } $$



$$ \hspace*{34 mm}=\kern2mm\frac{14}{40}\kern2mm=\kern1mm\frac{7}{20} $$



[ Question 1 ]
  
  
     
      1.2  There are 6 apricot flavoured
  
             sweets in the tin, thus
  
            n(E) = n(apricot) = 6
  
             There are 14 + 8 + 2 + 6 = 40
  
             sweets in the tin.
  
             n(S) = n(number sweets) = 40
  
$$ \hspace*{13 mm}\mathrm{P(apricot)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{32 mm}\mathrm{=\kern1mm\frac{n(apricot)}{n(number\kern1mmof\kern1mmsweets)}  } $$


$$ \hspace*{31 mm}=\kern1mm\frac{6}{40}\kern2mm=\kern2mm\frac{3}{20} $$



[ Question 1 ]
  
  
     
      1.3  n(guava)  =  12
  
             n(number of sweets) = 40
$$ \hspace*{13 mm}\mathrm{P(guava)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(guava)}{n(number\kern1mmof\kern1mmsweets)} } $$


$$ \hspace*{29 mm}=\kern1mm\frac{12}{40}\kern2mm=\kern1mm\frac{3}{10} $$



[ Q 1 ]
  
  
     
      1.4  Not a guava flavour, i.e. any flavour
  
             except guava can be chosen.
  
             The number of not guava
  
             flavoured sweets = 14 + 8 + 6 = 28
  
  
          
OR 
 
             The number of not guava flavoured
  
             sweets = number of sweets ─ 
  
             number of guava flavoured sweets
  
                                              = 40 − 12 = 28
  
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmguava)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{32 mm}\mathrm{=\kern1mm\frac{n(not\kern1mmguava)}{n(number\kern1mmof\kern1mmsweets)} } $$


$$ \hspace*{31 mm}=\kern1mm\frac{28}{40}\kern2mm=\kern1mm\frac{7}{10} $$



  
OR

             NB
  
             P(guava) + P(not guava) = 1
  
                                     3
             P(guava)  =  -------        (1.3)
                                    10
$$ \hspace*{13 mm}\mathrm{\frac{3}{10}\kern1mm+\kern1mmP(not\kern1mmguava)\kern1mm=\kern1mm1 } $$


$$ \hspace*{13 mm}\mathrm{P(not\kern1mmguava)\kern1mm=\kern1mm1\kern1mm─\kern1mm\frac{3}{10}\kern1mm=\kern1mm\frac{7}{10} } $$



                                                            [ Q. 1 ]
  
  
  
  
  
     
      1.5  n(not pineapple) = 8 + 12 + 6 = 26
  
              n(sweets) = 40
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmpineapple)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{32 mm}\mathrm{=\kern1mm\frac{n(not\kern1mmpineapple)}{n(number\kern1mmof\kern1mmsweets)} } $$


$$ \hspace*{31 mm}=\kern1mm\frac{26}{40}\kern2mm=\kern1mm\frac{13}{20} $$



  
  
OR

             NB
  
             P(pineapple)  +  P(Xpineapple)  =  1
  
                                            7
             P(pineapple)  =  -------        (1.1)
                                           20
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmpineapple)\kern1mm=\kern1mm1\kern1mm─\kern1mm\frac{7}{20}\kern1mm=\kern1mm\frac{13}{20} } $$



[ Q 1 ]
  
  



     
      1.6  n(not cherry) = 40 ─ 8 = 32
  
             n(sweets) = 40
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmcherry)\kern1mm=\kern1mm\frac{n(not\kern1mmcherry)}{n(number\kern1mmof\kern1mmsweets)} } $$


$$ \hspace*{35 mm}=\kern1mm\frac{32}{40}\kern2mm=\kern1mm\frac{4}{5} $$



  
  
OR

                n(cherry) = 8 and n(sweets) = 40
  
$$ \hspace*{13 mm}\mathrm{P(cherry)\kern1mm=\kern1mm\frac{8}{40}\kern1mm=\kern1mm\frac{1}{5} } $$


$$ \hspace*{13 mm}\mathrm{P(not\kern1mmcherry)\kern1mm=\kern1mm1\kern1mm─\kern1mm\frac{1}{5}\kern1mm=\kern1mm\frac{4}{5} } $$


  
[ Q 1 ]
  
  
     
      1.7  n(orange) = 0 and n(sweets) = 40
  
$$ \hspace*{13 mm}\mathrm{P(orange)\kern1mm=\kern1mm\frac{n(orange)}{n(number\kern1mmof\kern1mmsweets)} } $$


$$ \hspace*{31 mm}\mathrm{=\kern1mm\frac{0}{40}\kern2mm=\kern1mm0 } $$




[ Q 1 ]
  
  
     
      1.8  n(not orange) = 40 and
  
             n(sweets) = 40
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmorange)\kern1mm=\kern1mm\frac{n(not\kern1mmorange)}{n(number\kern1mmof\kern1mmsweets)} } $$


$$ \hspace*{35 mm}=\kern1mm\frac{40}{40}\kern2mm=\kern1mm1 $$



  
  
OR

$$ \hspace*{13 mm}\mathrm{P(not\kern1mmorangeqw)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(not\kern1mmorange) } $$
$$ \hspace*{43 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm0\kern1mm=\kern1mm1} $$


[ Q 1 ]
  
  
  
  
Antwoorde / Answers  2
  
     
      2.1  Daar is 1 koningin (Q) van skoppens
  
             dus n(E) = n(Q skoppens) = 1 en
  
             daar is 52 kaarte,
  
             n(S) = n(kaarte) = 52
  
$$ \hspace*{13 mm}\mathrm{P(Q\kern1mmskoppens)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(Q\kern1mmskoppens)}{n(aantal\kern1mmkaarte)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{1}{52} $$



[ Vraag 2 ]
  
  
     
      2.2  Daar is 2 pakke van 13 swart
  
             kaarte in die pan en dus
  
             n(E) = n(swart kaarte) = 2 X 13 = 26
  
             en daar is 52 kaarte,
  
             n(S) = n(kaarte) = 52
  
$$ \hspace*{13 mm}\mathrm{P(swart\kern1mmkaart)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(swart kaart)}{n(aantal\kern1mmkaarte)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{26}{52}\kern2mm=\kern1mm\frac{1}{2} $$



[ Vraag 2 ]
  
  
     
      2.3  Daar is 1 pak diamante en dus
  
             is daar 13 kaarte.
  
             n(E) = n(diamant) = 13
  
             en daar is 52 kaarte,
  
             n(S) = n(kaarte) = 52
  
$$ \hspace*{13 mm}\mathrm{P(diamant)\kern1mm=\kern1mm\frac{n(diamant)}{n(aantal\kern1mmkaarte)} } $$


$$ \hspace*{33 mm}=\kern1mm\frac{13}{52}\kern2mm=\kern1mm\frac{1}{4} $$



[ V 2 ]
  
  




     
      2.4  Daar is 2 boere, 2 koninginne
  
             (of vroue), 2 konings en
  
             2 ase wat rooi is.
  
               n(E) = n(rooi prentkaart) = 8
  
               en daar is 52 kaarte,
  
             n(S) = n(kaarte) = 52
  
$$ \hspace*{13 mm}\mathrm{P(rooi\kern1mmprent)\kern1mm=\kern1mm\frac{n(rooi\kern1mmprent)}{n(aantal\kern1mmkaarte)} } $$


$$ \hspace*{35 mm}=\kern1mm\frac{8}{52}\kern2mm=\kern1mm\frac{2}{13} $$



[ V 2 ]
  
  
     
      2.5  Daar is 8 rooi prent kaarte en
  
             dus 52 - 8 = 44 nie
  
             rooi prent kaarte.
  
             n(E) = n(nie rooi prentkaart) = 44
  
             en daar is 52 kaarte,
  
             n(S) = n(kaarte) = 52
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmrooi\kern1mmprent)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{33 mm}\mathrm{=\kern1mm\frac{n(nie\kern1mmrooi\kern1mmprent)}{n(aantal\kern1mmkaarte)} } $$


$$ \hspace*{32 mm}=\kern1mm\frac{44}{52}\kern2mm=\kern1mm\frac{11}{13} $$



  
  
OF

             P(nie rooi prent) = 1 ─ P(rooi prent)
  
  
$$ \hspace*{13 mm}\mathrm{P(rooi\kern1mmprent)\kern1mm=\kern1mm\frac{2}{13}\kern2mm.\kern1mm.\kern1mm.\kern1mm[2.4] } $$


$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmrooi\kern1mmprent)\kern1mm=\kern1mm1\kern1mm─\kern1mm\frac{2}{13} } $$


$$ \hspace*{41 mm}\mathrm{=\kern1mm\frac{11}{13} } $$


[ V 2 ]
  
  
     
      2.6  Daar is 8 rooi en 8 swart prent
  
             kaarte, dus 52 - (2 X 8) = 36
  
             nie prent kaarte.
  
             n(E) = n(nie prent) = 36 en
  
             daar is 52 kaarte,
  
             n(S) = n(aantal kaarte) = 52
  
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmprent)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(nie\kern1mmprent)}{n(aantal\kern1mmkaarte)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{36}{52}\kern2mm=\kern1mm\frac{9}{13} $$



  
  
OF

             Daar is 8 + 8 = 16 prent kaarte
  
             en 52 speelkaarte.
$$ \hspace*{13 mm}\mathrm{P(prent\kern1mmkaart)\kern1mm=\kern1mm\frac{n(prent\kern1mmkaart)}{n(aantal\kern1mmkaarte)} } $$


$$ \hspace*{37 mm}=\kern1mm\frac{16}{52}\kern2mm=\kern1mm\frac{4}{13} $$



  
             P(nie 'n prent) = 1 ─ P('n prent)
  
  
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mm'n\kern1mmprent)\kern1mm=\kern1mm1\kern1mm─\kern1mm\frac{4}{13} } $$


$$ \hspace*{39 mm}\mathrm{=\kern1mm\frac{9}{13} } $$


[ V 2 ]
  
  
     
      2.7  Daar is 2 kleure swart
  
             genommerde kaarte.
  
             Hulle is van 2 tot 10 genommer.
  
             Dus is daar 2 X 9 = 18 swart
  
             genommerde kaarte.
  
             n(E) = n(swart genommerde) = 18
  
             en daar is 52 kaarte,
  
             n(S) = n(kaarte) = 52
  
$$ \hspace*{13 mm}\mathrm{P(swart\kern1mmgenommerd)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(swart\kern1mmgenommerd)}{n(aantal\kern1mmkaarte)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{18}{52}\kern2mm=\kern1mm\frac{9}{26} $$



[ V 2 ]
  
  
     
      2.8  Daar is 4 boere in die pak.
  
             n(E) = n(boer) = 4
  
             en daar is 52 kaarte,
  
             n(S) = n(kaarte) = 52
  
$$ \hspace*{13 mm}\mathrm{P(boer)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{28 mm}\mathrm{=\kern1mm\frac{n(boer)}{n(aantal\kern1mmkaarte)} } $$


$$ \hspace*{27 mm}=\kern1mm\frac{4}{52}\kern2mm=\kern1mm\frac{1}{13} $$



[ V 2 ]
  
  
     
      2.9  Die kaarte met nommers groter
  
             as 4 en kleiner as 8 is
  
             5, 6 en 7 en dus is daar 12
  
             sulke kaart in die pak .
  
             n(E) = n(genommerde) = 12
  
             en daar is 52 kaarte,
  
             n(S) = n(kaarte) = 52
  
$$ \hspace*{13 mm}\mathrm{P(genommerd)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(genommerd)}{n(aantal\kern1mmkaarte)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{12}{52}\kern2mm=\kern1mm\frac{3}{13} $$



[ V 2 ]
  
  
     
      2.10  Die kaarte met nommers kleiner
  
               as 7 is 2, 3, 4, 5 en 6
  
               Dus daar is 10 sulke rooi
  
               kaarte in die pak .
  
               n(E) = n(rooi genommerde) = 10
  
               en daar is 52 kaarte,
  
               n(S) = n(kaarte) = 52
  
$$ \hspace*{13 mm}\mathrm{P(rooi\kern1mmgenommerd)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{33 mm}\mathrm{=\kern1mm\frac{n(rooi\kern1mmgenommerd)}{n(aantal\kern1mmkaarte)} } $$


$$ \hspace*{32 mm}=\kern1mm\frac{10}{52}\kern2mm=\kern1mm\frac{5}{26} $$



[ V 2 ]
  
  
     
      2.11  Daar is 2 boere, 2 vroue,
  
               2 konings en 2 ase wat swart is.
  
               n(E) = n(swart prent) = 8
  
               en daar is 52 kaarte,
  
               n(S) = n(kaarte) = 52
  
$$ \hspace*{13 mm}\mathrm{P(swart\kern1mmprent)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{39 mm}\mathrm{=\kern1mm\frac{n(swart\kern1mmprent)}{n(aantal\kern1mmkaarte)} } $$


$$ \hspace*{38 mm}=\kern1mm\frac{8}{52}\kern2mm=\kern1mm\frac{2}{13} $$



[ V 2 ]
  
  
     
      2.12  Daar is 2 kleure van rooi
  
               genommerde kaarte. Hulle is
  
               van 2 tot 10 genommer.
  
               Dus is daar 2 X 9 = 18 rooi
  
               genommerde kaarte
  
               Dus is daar 52 - 18 = 34 nie
  
               rooi genommerde kaarte.
  
               n(E) = n(nie rooi genommerd) = 34
  
               en daar is 52 kaarte,
  
               n(S) = n(aantal kaarte) = 52
  
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmrooi\kern1mmgenommerd)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{27 mm}\mathrm{=\kern1mm\frac{n(nie\kern1mmrooi\kern1mmgenommerd)}{n(aantal\kern1mmkaarte)} } $$


$$ \hspace*{26 mm}=\kern1mm\frac{34}{52}\kern2mm=\kern1mm\frac{17}{26} $$



[ V 2 ]
  
  
     
      2.13  Daar is 4 ase in die pak
  
               dus 52 - 4 = 48 nie ase.
  
               n(E) = n(nie aas) = 48
  
               en daar is 52 kaarte,
  
               n(S) = n(kaarte) = 52
  
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmaas)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{32 mm}\mathrm{=\kern1mm\frac{n(nie\kern1mmaas)}{n(aantal\kern1mmkaarte)} } $$


$$ \hspace*{31 mm}=\kern1mm\frac{48}{52}\kern2mm=\kern1mm\frac{12}{13} $$



[ V 2 ]
  
  
     
      2.1  There is 1 queen of spades
  
             thus n(E) = n(Q spades) = 1 and
  
             there are 52 playing cards
  
             Thus n(S) = n(number of cards) = 52
  
$$ \hspace*{13 mm}\mathrm{P(Q\kern1mmspades)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{30 mm}\mathrm{=  \frac{n(Q spades)}{n(number of cards)}  } $$


$$ \hspace*{29 mm}=  \frac{1}{52} $$



[ Question 2 ]
  
  
     
      2.2  There are 2 suits of black cards in
  
             the pack and thus
  
             n(E) = n(black card) = 2 X 13 = 26
  
             and there are 52 playing cards
  
             Thus n(S) = n(number of cards) = 52
  
$$ \hspace*{13 mm}\mathrm{P(black\kern1mmcard)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{37 mm}\mathrm{=\kern1mm\frac{n(black\kern1mmcard)}{n(number\kern1mmof\kern1mmcards)} } $$


$$ \hspace*{37 mm}=\kern1mm\frac{26}{52}\kern2mm=\kern1mm\frac{1}{2} $$



[ Question 2 ]
  
  
     
      2.3  There is 1 suit of diamonds in
  
              the pack and thus 13 cards
  
              n(E) = n(diamond) = 13 and
  
              there are 52 playing cards
  
              Thus n(S) = n(number of cards) = 52
  
$$ \hspace*{13 mm}\mathrm{P(diamond)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{34 mm}\mathrm{=\kern1mm\frac{n(diamond)}{n(number\kern1mmof\kern1mmcards)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{13}{52}\kern2mm=\kern1mm\frac{1}{4} $$


[ Q 2 ]
  
  
     
      2.4  There are 2 jacks, 2 queens, 2 kings
  
             and 2 aces that are red
  
             n(E) = n(red picture) = 8 and
  
             there are 52 playing cards
  
             Thus n(S) = n(number of cards) = 52
  
$$ \hspace*{13 mm}\mathrm{P(red\kern1mmpicture)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{34 mm}\mathrm{=\kern1mm\frac{n(red\kern1mmpicture)}{n(number\kern1mmof\kern1mmcards)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{8}{52}\kern2mm=\kern1mm\frac{2}{13} $$


[ Q 2 ]
  
     
      2.5  There are 8 red picture cards and
  
             thus 52 - 8 = 44 not red picture cards
  
             n(E) = n(not red picture) = 44 and
  
             there are 52 playing cards
  
             Thus n(S) = n(number of cards) = 52
  
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmred\kern1mmpicture)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{34 mm}\mathrm{=\kern1mm\frac{n(not\kern1mmred\kern1mmpicture)}{n(number\kern1mmof\kern1mmcards)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{44}{52}\kern2mm=\kern1mm\frac{11}{13} $$


  
  
  
  
OR

             P(not red picture) = 1 ─ P(red picture)
  
$$ \hspace*{13 mm}\mathrm{P(red\kern1mmpicture)\kern1mm=\kern1mm\frac{2}{13}\kern3mm.\kern1mm.\kern1mm.\kern1mm[2.4] } $$


$$ \hspace*{13 mm}\mathrm{P(not\kern1mmred\kern1mmpicture)\kern1mm=\kern1mm1\kern1mm─\kern1mm\frac{2}{13} } $$


$$ \hspace*{44 mm}\mathrm{=\kern1mm\frac{11}{13} } $$


[ Q 2 ]
  
  

     
      2.6  There are 8 red and 8 black picture
  
             cards, thus there are 52 - (2 x 8)
  
             i.e. 36 non picture cards.
  
             n(E) = n(not a picture) = 36 and
  
             there are 52 playing cards
  
             Thus n(S) = n(number of cards)
  
                              = 52
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmpicture)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{34 mm}\mathrm{=\kern1mm\frac{n(not\kern1mmpicture)}{n(number\kern1mmof\kern1mmcards)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{36}{52}\kern2mm=\kern1mm\frac{9}{13} $$


  
  
OR

             There are 8 + 8 = 16 picture cards
  
             and 52 playing cards.
$$ \hspace*{13 mm}\mathrm{P(picture\kern1mmcard)\kern1mm=\kern1mm\frac{n(picture\kern1mmcard)}{n(number\kern1mmof\kern1mmcards)} } $$


$$ \hspace*{39 mm}\mathrm{=\kern1mm\frac{16}{52}\kern2mm=\kern1mm\frac{4}{13} } $$


  
  
             P(not a picture)  =  1  ─  P(a picture)
  
  
$$ \hspace*{13 mm}\mathrm{P(not\kern1mma\kern1mmpicture)\kern1mm=\kern1mm1\kern1mm─\kern1mm\frac{4}{13} } $$


$$ \hspace*{41 mm}\mathrm{=\kern1mm\frac{9}{13} } $$


[ Q 2 ]
  
  
     
      2.7  There are 2 suits of black
  
             numbered cards.
  
             These are numbered 2 to 10.
  
             Thus there are 2 X 9 = 18 black
  
             numberd cards.
  
             n(E) = n(black numbered) = 18 and
  
             there are 52 playing cards
  
             Thus n(S) = n(number of cards) = 52
  
$$ \hspace*{13 mm}\mathrm{P(black\kern1mmnumbered)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{34 mm}\mathrm{=\kern1mm\frac{n(black\kern1mmnumbered)}{n(number\kern1mmof\kern1mmcards)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{18}{52}\kern2mm=\kern1mm\frac{9}{26} $$


  
[ Q 2 ]
  
  
     
      2.8  There are 4 jacks in the pack,
  
             n(E) = n(jack) = 4 and
  
             there are 52 playing cards
  
             Thus n(S) = n(number of cards) = 52
  
$$ \hspace*{13 mm}\mathrm{P(jack)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{27 mm}\mathrm{=\kern1mm\frac{n(jack)}{n(number\kern1mmof\kern1mmcards)} } $$


$$ \hspace*{26 mm}=\kern1mm\frac{4}{52}\kern2mm=\kern1mm\frac{1}{13} $$


  
[ Q 2 ]
  
  
     
      2.9  The cards numbered greater
  
             than 4 and smaller than 8 are
  
             5, 6, and 7 and thus there are
  
             12 such cards in the pack.
  
             n(E) = n(numbered) = 12 and
  
             there are 52 playing cards
  
             Thus n(S) = n(number of cards) = 52
  
$$ \hspace*{13 mm}\mathrm{P(numbered)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{36 mm}\mathrm{=\kern1mm\frac{n(numbered)}{n(number\kern1mmof\kern1mmcards)} } $$


$$ \hspace*{35 mm}=\kern1mm\frac{12}{52}\kern2mm=\kern1mm\frac{3}{13} $$


  
[ Q 2 ]
  
  
     
      2.10  The cards numbered smaller
  
                than 7 are 2, 3, 4, 5, and 6
  
                Thus there are 10 such red cards
  
                in the pack.
  
                n(E) = n(red numbered) = 10 and
  
                there are 52 playing cards
  
                n(S) = n(number of cards) = 52
  
$$ \hspace*{13 mm}\mathrm{P(red\ numbered)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{36 mm}\mathrm{=\kern1mm\frac{n(red\ numbered)}{n(number\kern1mmof\kern1mmcards)} } $$


$$ \hspace*{35 mm}=\kern1mm\frac{10}{52}\kern2mm=\kern1mm\frac{5}{26} $$


  
[ Q 2 ]
  
  
     
      2.11  There are 2 jacks, 2 queens,
  
                2 kings and 2 aces that are black
  
                n(E) = n(black picture) = 8 and
  
                there are 52 playing cards
  
                n(S) = n(number of cards) = 52
  
$$ \hspace*{13 mm}\mathrm{P(black\kern1mmpicture)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{36 mm}\mathrm{=\kern1mm\frac{n(black\kern1mmpicture)}{n(number\kern1mmof\kern1mmcards)} } $$


$$ \hspace*{35 mm}=\kern1mm\frac{8}{52}\kern2mm=\kern1mm\frac{2}{13} $$


  
[ Q 2 ]
  
  
     
      2.12  There are 2 suits of ref
  
                numbered cards.
  
                These are numbered 2 to 10.
  
                Thus there are 2 X 9 = 18 red
  
                numberd cards.
  
                Thus there are 52 - 18 = 34 not
  
                red numbered cards in the pack.
  
                n(E) = n(not red numbered) = 34
  
                and there are 52 playing cards
  
                n(S) = n(number cards) = 52
  
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmred\kern1mmnumbered)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{36 mm}\mathrm{=\kern1mm\frac{n(not\kern1mmred\kern1mmnumbered)}{n(number\kern1mmof\kern1mmcards)} } $$


$$ \hspace*{35 mm}=\kern1mm\frac{34}{52}\kern2mm=\kern1mm\frac{17}{26} $$


  
[ Q 2 ]
  
  
     
      2.13  There are 4 aces in the pack,
  
                thus 52 - 4 = 48 non aces.
  
                n(E) = n(non ace) = 48 and
  
                there are 52 playing cards
  
                n(S) = n(number of cards) = 52
  
$$ \hspace*{13 mm}\mathrm{P(not\kern1mman\kern1mmace)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{36 mm}\mathrm{=\kern1mm\frac{n(not\kern1mman\kern1mmace)}{n(number\kern1mmof\kern1mmcards)} } $$


$$ \hspace*{35 mm}=\kern1mm\frac{48}{52}\kern2mm=\kern1mm\frac{12}{13} $$


  
[ Q 2 ]
  
  
  
  
Antwoorde / Answers  3
  
     
      3.1  Daar is 40 leerlinge in die klas.
  
[ Vraag 3 ]
  
  
     
      3.2.1  Daar is 23 dogters en 40 leerlinge
  
                in die klas.
  
                dus n(E) = n(dogters) = 23 en
  
                n(S) = n(leerlinge) = 40
  
$$ \hspace*{16 mm}\mathrm{P(dogters)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(dogters)}{n(leerlinge)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{23}{40} $$



[ V 3 ]
  
  
     
      3.2.2  Daar is 6 seuns in die klas
  
                wat hulle huiswerk gedoen het
  
                en 40 leerlinge in die klas.
  
                dus n(E) = n(S+huiswerk) = 6 en
  
                n(S) = n(leerlinge) = 40
  
$$ \hspace*{16 mm}\mathrm{P(seun + huiswerk)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(seun + huiswerk)}{n(leerlinge)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{6}{40}\kern2mm=\kern1mm\frac{3}{20} $$



[ V 3 ]
  
  
     
      3.2.3  Daar is 7 dogters in die klas wat
  
                nie hulle huiswerk gedoen het nie
  
                en 40 leerlinge in die klas.
  
                dus n(E) = n(D+0huiswerk) = 7 en
  
                n(S) = n(leerlinge) = 40
  
$$ \hspace*{16 mm}\mathrm{P(D + nie + huiswerk)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(D + nie + huiswerk)}{n(leerlinge)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{7}{40} $$



[ V 3 ]
  
  
     
      3.2.4  Daar is 18 leerlinge wat nie
  
                hulle huiswerk gedoen het nie
  
                en 40 leerlinge in die klas.
  
                dus n(E) = n(nie huiswerk) = 18 en
  
                n(S) = n(leerlinge) = 40
  
$$ \hspace*{16 mm}\mathrm{P(nie + huiswerk)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(nie + huiswerk)}{n(leerlinge)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{18}{40}\kern2mm=\kern1mm\frac{9}{20} $$



[ V 3 ]
  
  
     
      3.2.5  Daar is geen Graad 11 leerlinge in
  
                die klas nie en
  
                40 leerlinge in die klas.
  
                dus n(E) = n(Graad 11) = 0 en
  
                n(S) = n(leerlinge) = 40
  
$$ \hspace*{16 mm}\mathrm{P(Graad 11)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(Graad 11)}{n(leerlinge)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{0}{40}\kern2mm=\kern1mm0 $$



[ V 3 ]
  
  
     
      3.3  Daar is 23 dogters in die klas
  
             en 7 het nie huiswerk gedoen nie.
  
             dus n(E) = n(D+nie huiswerk) = 7
  
             en n(S) = n(dogters) = 23
  
$$ \hspace*{16 mm}\mathrm{P(D + nie huiswerk)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(D + nie huiswerk)}{n(Dogters)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{7}{23} $$



[ V 3 ]
  
  
     
      3.4  Daar is 17 seuns in die klas
  
             en 6 het hulle huiswerk gedoen.
  
             dus n(E) = n(S+huiswerk) = 6
  
             en n(S) = n(seuns) = 17
  
$$ \hspace*{16 mm}\mathrm{P(S + huiswerk)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(S + huiswerk)}{n(Seuns)} } $$


$$ \hspace*{34 mm}=\kern1mm\frac{6}{17} $$



[ V 3 ]
  
  
     
      3.1  There 40 pupils in the class.
  
[ Question 3 ]
  
  
     
      3.2.1  There are 23 girls and 40 pupils
  
                in the class.
  
                Thus n(E) = n(girls) = 23 and
  
                n(S) = n(pupils) = 40
  
$$ \hspace*{13 mm}\mathrm{P(girls)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(girls)}{n(pupils)} } $$


$$ \hspace*{29 mm}=\kern1mm\frac{23}{40} $$



[ Q 3 ]
  
  
     
      3.2.2  There are 6 boys in the class that
  
                have done their homework and
  
                40 pupils in the class.
  
                Thus n(E) = n(B+homework) = 6
  
                and n(S) = n(pupils) = 40
  
$$ \hspace*{13 mm}\mathrm{P(boy + homework)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(boy + homework}{n(pupils)} } $$


$$ \hspace*{29 mm}=\kern1mm\frac{6}{40}\kern2mm=\kern1mm\frac{3}{20} $$



[ Q 3 ]
  
  
     
      3.2.3  There are 7 girls in the class that
  
                have not done their homework and
  
                40 pupils in the class.
  
                n(E) = n(G+nohomework) = 7 and
  
                n(S) = n(pupils) = 40
  
$$ \hspace*{13 mm}\mathrm{P(G+nohomework)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(G+no\kern1mmhomework}{n(pupils)} } $$


$$ \hspace*{29 mm}=\kern1mm\frac{7}{40} $$



[ Q 3 ]
  
  
     
      3.2.4  There are 18 pupils in the class
  
                that have not done their homework
  
                and 40 pupils in the class.
  
                n(E) = n(no homework) = 18 and
  
                n(S) = n(pupils) = 40
  
$$ \hspace*{13 mm}\mathrm{P(no\kern1mmhomework)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(no\kern1mmhomework}{n(pupils)} } $$


$$ \hspace*{29 mm}=\kern1mm\frac{18}{40}\kern2mm=\kern1mm\frac{9}{20} $$



[ Q 3 ]
  
  
     
      3.2.5  There are no Grade 11 pupils in
  
                the class and 40 pupils.
  
                n(E) = n(Grade 11) = 0 and
  
                n(S) = n(pupils) = 40
  
$$ \hspace*{13 mm}\mathrm{P(Grade 11)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(Grade 11}{n(pupils)} } $$


$$ \hspace*{29 mm}=\kern1mm\frac{0}{40}\kern2mm=\kern1mm0 $$



[ Q 3 ]
  
  


     
      3.3  There are 23 girls in the class and 7
  
             have not done their homework.
  
             n(E) = n(G+nohomework) = 7 and
  
             n(S) = n(girls) = 23
  
$$ \hspace*{13 mm}\mathrm{P(G\kern1mm;+\kern1mmno\kern1mmhomework)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(G\kern1mm+\kern1mmno\kern1mmhomework}{n(girls)} } $$


$$ \hspace*{29 mm}=\kern1mm\frac{7}{23} $$



[ Q 3 ]
  
  
     
      3.4  There are 17 boys in the class and
  
             6 have done their homework.
  
             Thus n(E) = n(B+homework) = 6 and
  
             n(S) = n(boys) = 17
  
$$ \hspace*{13 mm}\mathrm{P(B\kern1mm+\kern1mmhomework)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$


$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(B\kern1mm+\kern1mmhomework)}{n(boys)} } $$


$$ \hspace*{29 mm}=\kern1mm\frac{6}{17} $$



[ Q 3 ]
  
  
  
  
Antwoorde / Answers  4
  
     
      4.1  Daar is een D en 11 letters,
  
             dus n(E) = n(D) = 1 en
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(D)\kern1mm=\kern1mm\frac{n(D)}{n(letters)} } $$


$$ \hspace*{23 mm}=\kern1mm\frac{1}{11} $$



[ Vraag 4 ]
  
  
     
      4.2  Daar is een K en 11 letters,
  
             dus n(E) = n(K) = 1 en
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(K)\kern1mm=\kern1mm\frac{n(K)}{n(letters)} } $$


$$ \hspace*{23 mm}=\kern1mm\frac{1}{11} $$



[ V 4 ]
  
  
     
      4.3  Daar is vyf klinkers, U, I, A, I, A, en
  
             11 letters.
  
             dus n(E) = n(klinker) = 5 en
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(klinker)\kern1mm=\kern1mm\frac{n(klinker)}{n(letters)} } $$


$$ \hspace*{30 mm}=\kern1mm\frac{5}{11} $$



[ V 4 ]
  
  
     
      4.4  Daar is vyf konsonante, S, D, F, R, K,
  
             en 11 letters.
  
             dus n(E) = n(konsonant) = 5 en
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(konsonant)\kern1mm=\kern1mm\frac{n(konsonant)}{n(letters)} } $$


$$ \hspace*{36 mm}=\kern1mm\frac{5}{11} $$



[ V 4 ]
  
  
     
      4.5  Daar is twee A's,
  
             en 11 letters.
  
             dus n(E) = n(A) = 2 en
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(A)\kern1mm=\kern1mm\frac{n(A)}{n(letters)} } $$


$$ \hspace*{23 mm}=\kern1mm\frac{2}{11} $$



[ V 4 ]
  
  
     
      4.6  Daar is geen B's,
  
             en 11 letters.
  
             dus n(E) = n(B) = 0 en
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(B)\kern1mm=\kern1mm\frac{n(B)}{n(letters)} } $$


$$ \hspace*{23 mm}=\kern1mm\frac{0}{11}\kern2mm=\kern1mm0 $$



[ V 4 ]
  
  
     
      4.7  Daar is geen O's,
  
             en 11 letters.
  
             dus n(E) = n(nie O) = 11 en
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmO)\kern1mm=\kern1mm\frac{n(nie\kern1mmO)}{n(letters)} } $$


$$ \hspace*{23 mm}=\kern1mm\frac{11}{11}\kern2mm=\kern1mm1 $$



  
  
OF

  
             Daar is geen O;s en 11 letters.
  
             dus n(O) = 0 en
  
             en dus n(E) = n(O) = 0 en
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(O)\kern1mm=\kern1mm\frac{n(O)}{n(letters)} } $$


$$ \hspace*{23 mm}=\kern1mm\frac{0}{11}\kern2mm=\kern1mm0 $$



$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmO)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(O) } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm0\kern2mm=\kern1mm1 } $$


[ V 4 ]
  
  
     
      4.8  Daar is een F en 11 letters
  
             en dus n(nie F) = 11 - 1 = 10
  
             dus n(E) = n(nie F) = 10 en
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmF)\kern1mm=\kern1mm\frac{n(nie\kern1mmF)}{n(letters)} } $$


$$ \hspace*{27 mm}=\kern1mm\frac{10}{11} $$

  
  
OF

  
             Daar is een F en 11 letters.
  
             dus n(F) = 1 en
  
             en dus n(E) = n(F) = 1 en
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(F)\kern1mm=\kern1mm\frac{n(F)}{n(letters)} } $$


$$ \hspace*{23 mm}=\kern1mm\frac{1}{11} $$



$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmF)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(F) } $$
$$ \hspace*{29 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm\frac{1}{11}\kern2mm=\kern1mm\frac{10}{11} } $$


[ V 4 ]
  
  
     
      4.9  Daar is geen E's,
  
             en 11 letters.
  
             dus n(E) = n(nie E) = 11 en
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmE)\kern1mm=\kern1mm\frac{n(nie\kern1mmE)}{n(letters)} } $$


$$ \hspace*{27 mm}=\kern1mm\frac{11}{11}\kern1mm=\kern1mm1 $$

  
  
OF

  
             Daar is geen E's en 11 letters.
  
             dus n(E) = 0 en
  
             en dus n(E) = n(E) = 0 en
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(E)\kern1mm=\kern1mm\frac{n(E)}{n(letters)} } $$


$$ \hspace*{23 mm}=\kern1mm\frac{0}{11}\kern1mm=\kern1mm0 $$



$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmE)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(E) } $$
$$ \hspace*{29 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm0\kern2mm=\kern1mm1 } $$


[ V 4 ]
  
  
     
      4.10  Daar is twee I's en 11 - 2 = 9
  
               nie I letters
  
               en 11 letters
  
               dus n(E) = n(nie I) = 9 en
  
               n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmI)\kern1mm=\kern1mm\frac{n(nie\kern1mmI)}{n(letters)} } $$


$$ \hspace*{27 mm}=\kern1mm\frac{9}{11} $$

  
  
OF

  
               Daar is twee I's en 11 letters.
  
               dus n(I) = 2 en
  
               en dus n(E) = n(I) = 2 en
  
               n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(I)\kern1mm=\kern1mm\frac{n(I)}{n(letters)} } $$


$$ \hspace*{20 mm}=\kern1mm\frac{2}{11} $$



$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmI)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(I) } $$
$$ \hspace*{27 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm\frac{2}{11}\kern2mm=\kern1mm\frac{9}{11} } $$


[ V 4 ]
  
  
     
      4.11  Daar is vyf klinkers, U, I, A, I en A,
  
               en 11 letters
  
               dus is daar 11 - 5 = 6 nie klinkers
  
               dus n(E) = n(nie klinker) = 6 en
  
               n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmklinker)\kern1mm=\kern1mm\frac{n(nie\kern1mmklinker)}{n(letters)} } $$


$$ \hspace*{36 mm}=\kern1mm\frac{6}{11} $$

  
  
OF

  
               Daar is vyf klinkers en 11 letters.
  
               dus n(klinker) = 5 en
  
               en dus n(E) = n(klinker) = 5 en
  
               n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(klinker)\kern1mm=\kern1mm\frac{n(klinker)}{n(letters)} } $$


$$ \hspace*{30 mm}=\kern1mm\frac{5}{11} $$



$$ \hspace*{13 mm}\mathrm{P(nie\kern1mmklinker)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(klinker) } $$
$$ \hspace*{37 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm\frac{5}{11}\kern2mm=\kern1mm\frac{6}{11} } $$


[ V 4 ]
  
  
     
      4.1  There is one D and 11 letters
  
             Thus n(E) = n(D) = 1 and
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(D)\kern1mm=\kern1mm\frac{n(D)}{n(letters)} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{1}{11} } $$


[ Question 4 ]
  
  

     
      4.2  There is one K and 11 letters
  
             Thus n(E) = n(K) = 1 and
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(K)\kern1mm=\kern1mm\frac{n(K)}{n(letters)} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{1}{11} } $$


[ Q 4 ]
  
  

     
      4.3  There are five vowels, U, I, A, I, A,
  
             and 11 letters.
  
             Thus n(E) = n(vowel) = 5 and
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(vowel)\kern1mm=\kern1mm\frac{n(vowel)}{n(letters)} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{5}{11} } $$


[ Q 4 ]
  
  

     
      4.4  There are five consonants,
  
             S, D, F, R, K and 11 letters.
  
             Thus n(E) = n(consonant) = 5 and
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(consonant)\kern1mm=\kern1mm\frac{n(consonant)}{n(letters)} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{5}{11} } $$


[ Q 4 ]
  
  

     
      4.5  There are two A's,
  
             and 11 letters.
  
             Thus n(E) = n(A) = 2 and
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(A)\kern1mm=\kern1mm\frac{n(A)}{n(letters)} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{2}{11} } $$


[ Q 4 ]
  
  

     
      4.6  There are no B's,
  
             and 11 letters.
  
             Thus n(E) = n(B) = 0 and
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(B)\kern1mm=\kern1mm\frac{n(B)}{n(letters)} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{0}{11}\kern1mm=\kern1mm0 } $$



[ Q 4 ]
  
  
     
      4.7  There are no O's,
  
             and 11 letters.
  
             Thus n(E) = n(not O) = 11 and
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmO)\kern1mm=\kern1mm\frac{n(not\kern1mmO)}{n(letters)} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{11}{11}\kern1mm=\kern1mm1 } $$


  
      
OR

  
             There is no O's and 11 letters,
  
             and thus n(O) = 0
  
             Thus n(E) = n(O) = 0 and
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(O)\kern1mm=\kern1mm\frac{n(O)}{n(letters)} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{0}{11}\kern1mm=\kern1mm0 } $$




$$ \hspace*{13 mm}\mathrm{P(not\kern1mmO)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(O) } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm0\kern1mm=\kern1mm1 } $$



[ Q 4 ]
  
     
      4.8  There is one F and 11 letters
  
             and thus n(not F) = 11 - 1 = 10
  
             Thus n(E) = n(not F) = 10 and
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmF)\kern1mm=\kern1mm\frac{n(not\kern1mmF)}{n(letters)} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{10}{11} } $$

  
      
OR

  
             There is one F and 11 letters,
  
             and thus n(F) = 1
  
             Thus n(E) = n(F) = 1 and
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(F)\kern1mm=\kern1mm\frac{n(F)}{n(letters)} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{1}{11} } $$



$$ \hspace*{13 mm}\mathrm{P(not\kern1mmF)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(F) } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm\frac{1}{11}\kern1mm=\kern1mm\frac{10}{11} } $$



[ Q 4 ]
  
     
      4.9  There are no E's,
  
             and 11 letters.
  
             Thus n(E) = n(not E) = 11 and
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmE)\kern1mm=\kern1mm\frac{n(not\kern1mmE)}{n(letters)} } $$


$$ \hspace*{29 mm}\mathrm{=\kern1mm\frac{11}{11}\kern1mm=\kern1mm1 } $$

  
      
OR

  
             There is no E's and 11 letters,
  
             and thus n(E) = 0
  
             Thus n(E) = n(E) = 0 and
  
             n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(E)\kern1mm=\kern1mm\frac{n(E)}{n(letters)} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{0}{11}\kern1mm=\kern1mm0 } $$



$$ \hspace*{13 mm}\mathrm{P(not\kern1mmE)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(E) } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm0\kern1mm=\kern1mm1 } $$


[ Q 4 ]
  
  
     
      4.10  There are two I's and 11 - 2 = 9
  
               not I letters
  
               and 11 letters.
  
               Thus n(E) = n(not I) = 9 and
  
               n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmI)\kern1mm=\kern1mm\frac{n(not\kern1mmI)}{n(letters)} } $$


$$ \hspace*{29 mm}\mathrm{=\kern1mmfrac{9}{11} } $$

  
      
OR

  
               There are two I's and 11 letters,
  
               and thus n(I) = 2
  
               Thus n(E) = n(I) = 2 and
  
               n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(I)\kern1mm=\kern1mm\frac{n(I)}{n(letters)} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{2}{11} } $$



$$ \hspace*{13 mm}\mathrm{P(not\kern1mmI)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(I) } $$
$$ \hspace*{28 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm\frac{2}{11}\kern1mm=\kern1mm\frac{9}{11} } $$


[ Q 4 ]
  
  
     
      4.11  There are five vowels, U, I, A, I
  
               and A and 11 letters
  
               thus there are 11-5 = 6 non vowels.
  
               Thus n(E) = n(not vowel) = 6 and
  
               n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmvowel)\kern1mm=\kern1mm\frac{n(not\kern1mmvowel)}{n(letters)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{6}{11} } $$

  
      
OR

  
               There are five vowels and 11
  
               letters and thus n(vowel) = 5
  
               Thus n(E) = n(vowel) = 5 and
  
               n(S) = n(letters) = 11
  
$$ \hspace*{13 mm}\mathrm{P(vowel)\kern1mm=\kern1mm\frac{n(vowel)}{n(letters)} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{5}{11} } $$



$$ \hspace*{13 mm}\mathrm{P(not\kern1mmvowel)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(vowel) } $$
$$ \hspace*{35 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm\frac{5}{11}\kern1mm=\kern1mm\frac{6}{11} } $$


[ Q 4 ]
  
  
  
  
Antwoorde / Answers  5
  
     
      5.1  Daar is een 5 en 10 balle
  
             dus n(E) = n(5) = 1 en
  
             n(S) = n(balle) = 10
  
$$ \hspace*{13 mm}\mathrm{P(5)\kern1mm=\kern1mm\frac{n(5)}{n(balle)} } $$


$$ \hspace*{21 mm}=\kern1mm\frac{1}{100} $$



[ Vraag 5 ]
  
     
      5.2  Daar is 5 ewe getalle, 2,4,6,8
  
             en 10, en 10 balle.
  
             dus n(E) = n(ewe) = 5 en
  
             n(S) = n(balle) = 10
  
$$ \hspace*{13 mm}\mathrm{P(ewe)\kern1mm=\kern1mm\frac{n(ewe)}{n(balle)}  } $$


$$ \hspace*{26 mm}=\kern1mm\frac{5}{10}\kern1mm\kern1mm  =  \frac{1}{2} $$



[ Vraag 5 ]
  
     
      5.3  Daar is 5 getalle kleiner as 6,
  
             1,2,3,4 en 5, en 10 balle.
  
             dus n(E) = n(< 6) = 5 en
  
             n(S) = n(balle) = 10
  
$$ \hspace*{13 mm}\mathrm{P(<\kern1mm6)\kern1mm=\kern1mm\frac{n(<\kern1mm6)}{n(balle)} } $$


$$ \hspace*{26 mm}=\kern1mm\frac{5}{10}\kern1mm=\kern1mm\frac{1}{2} $$



[ V 5 ]
  
     
      5.4  Daar is 3 onewe getalle groter as 3,
  
             5, 7 en 9 en 10 balle.
  
             dus n(E) = n(> 3) = 3 en
  
             n(S) = n(balle) = 10
  
$$ \hspace*{13 mm}\mathrm{P(>\kern1mm3)\kern1mm=\kern1mm\frac{n(>\kern1mm3)}{n(balle)} } $$


$$ \hspace*{26 mm}=\kern1mm\frac{3}{10} $$



[ V 5 ]
  
     
      5.5  Daar is 3 veelvoude van 3,
  
             3, 6 en 9, en 10 balle.
  
             dus n(E) = n(veelvoud) = 3 en
  
             n(S) = n(balle) = 10
  
$$ \hspace*{13 mm}\mathrm{P(veelvoud\kern1mm3)\kern1mm=\kern1mm\frac{n(veelvoud\kern1mm3)}{n(balle)} } $$


$$ \hspace*{37 mm}=\kern1mm\frac{3}{10} $$



[ V 5 ]
  
     
      5.6  Daar is 4 getalle gelyk aan of
  
             groter as 7, 7, 8, 9, 10 en 10 balle.
  
             dus n(E) = n(>= 7) = 4 en
  
             n(S) = n(balle) = 10
  
$$ \hspace*{13 mm}\mathrm{P(>=\kern1mm7)\kern1mm=\kern1mm\frac{n(>=\kern1mm7)}{n(balle)} } $$


$$ \hspace*{28 mm}=\kern1mm\frac{4}{10}\kern1mm=\kern1mm\frac{2}{5} $$



                                                       [ V 5 ]
  
  
     
      5.7  Daar is 2 getalle deelbaar deur 4
  
             4 en 8, en 10 balle.
  
             dus n(E) = n(/4) = 2 en
  
             n(S) = n(balle) = 10
  
$$ \hspace*{13 mm}\mathrm{P(/\kern1mm4)\kern1mm=\kern1mm\frac{n(/\kern1mm4)}{n(balle)} } $$


$$ \hspace*{24 mm}=\kern1mm\frac{2}{10}\kern1mm=\kern1mm\frac{1}{5} $$



                                                       [ V 5 ]
  
  
  
     
      5.8  Daar is een 2 en 9 balle
  
             dus n(E) = n(2) = 1 en
  
             n(S) = n(balle) = 9
  
$$ \hspace*{13 mm}\mathrm{P(2)\kern1mm=\kern1mm\frac{n(2)}{n(balle)} } $$


$$ \hspace*{24 mm}=\kern1mm\frac{1}{9} $$



                                                       [ V 5 ]
  
  
  
     
      5.9  Daar is 6 getalle groter as 3,
  
             5,6,7,8,9 en 10, en 9 balle.
  
             dus n(E) = n(> 3) = 6 en
  
             n(S) = n(balle) = 9
  
$$ \hspace*{13 mm}\mathrm{P(>\kern1mm3)\kern1mm=\kern1mm\frac{n(>\kern1mm3)}{n(balle)} } $$


$$ \hspace*{26 mm}=\kern1mm\frac{6}{9}\kern1mm=\kern1mm\frac{2}{3} $$



                                                       [ V 5 ]
  
  
  
     
      5.1  There is one 5 and 10 balls
  
             Thus n(E) = n(5) = 1 and
  
             n(S) = n(balls) = 10
  
$$ \hspace*{13 mm}\mathrm{P(5)\kern1mm=\kern1mm\frac{n(5)}{n(balls)} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{1}{10} } $$



[ Question 5 ]
  
     
      5.2  There are 5 even numbers, 2, 4, 6, 8
  
             and 10, and 10 balls
  
             Thus n(E) = n(even) = 5 and
  
             n(S) = n(balls) = 10
  
$$ \hspace*{13 mm}\mathrm{P(even)\kern1mm=\kern1mm\frac{n(even)}{n(balls)} } $$


$$ \hspace*{29 mm}\mathrm{=\kern1mm\frac{5}{10}\kern1mm=\kern1mm\frac{1}{2} } $$



[ Question 5 ]
  
     
      5.3  There are 5 numbers smaller than 6,
  
             1, 2, 3, 4, and 5, and 10 balls
  
             Thus n(E) = n(< 6) = 5 and
  
             n(S) = n(balls) = 10
  
$$ \hspace*{13 mm}\mathrm{P(<\kern1mm6)\kern1mm=\kern1mm\frac{n(<\kern1mm6)}{n(balls)} } $$


$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{5}{10}\kern1mm=\kern1mm\frac{1}{2} } $$



[ Q 5 ]
  
     
      5.4  There are 3 uneven numbers greater
  
             than 3, 5, 7 and 9, and 10 balls
  
             Thus n(E) = n(> 3) = 3 and
  
             n(S) = n(balls) = 10
  
$$ \hspace*{13 mm}\mathrm{P(>\kern1mm3)\kern1mm=\kern1mm\frac{n(>\kern1mm3)}{n(balls)} } $$


$$ \hspace*{26 mm}\mathrm{=\kern1mm\frac{3}{10} } $$



[ Q 5 ]
  
     
      5.5  There are 3 multiples of 3, 3, 6 and 9
  
             and 10 balls
  
             Thus n(E) = n(mult3) = 3 and
  
             n(S) = n(balls) = 10
  
$$ \hspace*{13 mm}\mathrm{P(multiple\kern1mm3)\kern1mm=\kern1mm\frac{n(multiple\kern1mm3)}{n(balls)} } $$


$$ \hspace*{36 mm}\mathrm{=\kern1mm\frac{3}{10} } $$



                                                                [ Q 5 ]
  
  
  
     
      5.6  There are 4 numbers equal to or
  
             greater than 7, 7,8,9, 10 and 10 balls
  
             Thus n(E) = n(>=7) = 4 and
  
             n(S) = n(balls) = 10
  
$$ \hspace*{13 mm}\mathrm{P(>=\kern1mm7)\kern1mm=\kern1mm\frac{n(>=\kern1mm7)}{n(balls)} } $$


$$ \hspace*{28 mm}\mathrm{=\kern1mm\frac{4}{10}\kern1mm=\kern1mm\frac{2}{5} } $$



                                                                [ Q 5 ]
  
  
     
      5.7  There are 2 numbers divisible by 4,
  
             4 and 8, and 10 balls
  
             Thus n(E) = n(div4) = 2 and
  
             n(S) = n(balls) = 10
  
$$ \hspace*{13 mm}\mathrm{P(/\kern1mm4)\kern1mm=\kern1mm\frac{n(/\kern1mm4)}{n(balls)} } $$


$$ \hspace*{25 mm}\mathrm{=\kern1mm\frac{2}{10}\kern1mm=\kern1mm\frac{1}{5} } $$



                                                                [ Q 5 ]
  
  
  
     
      5.8  There is one 2 and 9 balls,
  
             Thus n(E) = n(2) = 1 and
  
             n(S) = n(balls) = 9
  
$$ \hspace*{13 mm}\mathrm{P(2)\kern1mm=\kern1mm\frac{n(2)}{n(balls)} } $$


$$ \hspace*{25 mm}\mathrm{=\kern1mm\frac{1}{9} } $$



                                                                [ Q 5 ]
  
  
  
     
      5.9  There are 6 numbers greater than 3,
  
             5,6,7,8,9 and 10, and 9 balls
  
             Thus n(E) = n(>3) = 6 and
  
             n(S) = n(balls) = 9
  
$$ \hspace*{13 mm}\mathrm{P(>\kern1mm3)\kern1mm=\kern1mm\frac{n(>\kern1mm3)}{n(balls)} } $$


$$ \hspace*{26 mm}\mathrm{=\kern1mm\frac{6}{9}\kern1mm=\kern1mm\frac{2}{3} } $$



                                                                [ Q 5 ]
  
  
  
  
  
Antwoorde / Answers  6
  
     
      6.1  Rangskik die data :
  
             2, 3, 3, 4, 5, 6, 8, 9, 12, 17, 21, 30
  
$$ \hspace*{13 mm}\mathrm{Gemiddelde  =  \frac{Σx}{n} } $$


$$ \hspace*{36 mm}\mathrm{=\kern1mm\frac{120}{12}\kern1mm=\kern1mm10 } $$

  
$$ \hspace*{13 mm}\mathrm{mediaan\kern1mm=\kern1mm\frac{6+8}{2}\kern1mm=\kern1mm7 } $$


  
             modus  =  3  . . .  kom 2 voor
  
[ Vraag 6 ]
  
  
     
      6.2.1  Daar is 2 periodes van 3 ure
  
                en 12 kinders.
  
                dus n(E) = n(3) = 2 en
  
                n(S) = n(kinders) = 12
  
$$ \hspace*{16 mm}\mathrm{P(3)\kern1mm=\kern1mm\frac{n(3)}{n(kinders)} } $$


$$ \hspace*{25 mm}=\kern1mm\frac{2}{12}\kern1mm=\kern1mm\frac{1}{6} $$



[ Vraag 6 ]
  
  
     
      6.2.2  Daar is 7 kinders wat TV
  
                vir 6 of meer ure per week gekyk
  
                het en daar is 12 kinders.
  
                dus n(E) = n(6+) = 7 en
  
                n(S) = n(kinders) = 12
  
$$ \hspace*{16 mm}\mathrm{P(6+)\kern1mm=\kern1mm\frac{n(6+)}{n(kinders)} } $$


$$ \hspace*{27 mm}=\kern1mm\frac{7}{12} $$



[ V 6 ]
  
  
     
      6.2.3  Daar is 8 kinders wat TV vir
  
                10 of minder ure per week gekyk
  
                het en daar is 12 kinders.
  
                dus n(E) = n(< 11) = 8 en
  
                n(S) = n(kinders) = 12
  
$$ \hspace*{16 mm}\mathrm{P(<\kern1mm11)\kern1mm=\kern1mm\frac{n(<\kern1mm11)}{n(kinders)} } $$


$$ \hspace*{31 mm}=\kern1mm\frac{8}{12}\kern1mm=\kern1mm\frac{2}{3} $$



[ V 6 ]
  
  
     
      6.2.4  Minder as 3 ure per dag beteken
  
                minder as 21 ure per week.
  
                Daar is 10 kinders wat TV vir 21
  
                of minder ure per week gekyk
  
                het en daar is 12 kinders.
  
                dus n(E) = n(< 3) = 10 en
  
                n(S) = n(kinders) = 12
  
$$ \hspace*{16 mm}\mathrm{P(<\kern1mm3)\kern1mm=\kern1mm\frac{n(<\kern1mm3)}{n(kinders)} } $$


$$ \hspace*{29 mm}=\kern1mm\frac{10}{12}\kern1mm=\kern1mm\frac{5}{6} $$



[ V 6 ]
  
  
     
      6.2.5  Die gemiddelde is 10 ure per
  
                week. Dus minder as die
  
                gemiddelde beteken minder as
  
                10 ure per week.
  
                Daar is 8 kinders wat TV vir
  
                minder as 10 ure per week gekyk
  
                het en daar is 12 kinders.
  
                dus n(E) = n(< 10) = 8 en
  
                n(S) = n(kinders) = 12
  
$$ \hspace*{16 mm}\mathrm{P(<\kern1mm10)\kern1mm=\kern1mm\frac{n(<\kern1mm10)}{n(kinders)} } $$


$$ \hspace*{31 mm}=\kern1mm\frac{8}{12}\kern1mm=\kern1mm\frac{2}{3} $$



[ V 6 ]
  
  
     
      6.2.6  Die mediaan is 7 ure per week.
  
                Dus meer as die mediaan beteken
  
                meer as 7 ure per week.
  
                Daar is 6 kinders wat TV vir
  
                meer as 7 ure per week gekyk
  
                het en daar is 12 kinders.
  
                dus n(E) = n(> mediaan) = 6 en
  
                n(S) = n(kinders) = 12
  
$$ \hspace*{16 mm}\mathrm{P(>\kern1mmmediaan)\kern1mm=\kern1mm\frac{n(>\kern1mmmediaan)}{n(kinders)} } $$


$$ \hspace*{40 mm}=\kern1mm\frac{6}{12}\kern1mm=\kern1mm\frac{1}{2} $$



[ V 6 ]
  
  
     
      6.2.7  Die modus is 3 ure per week.
  
                Dus gelyk aan die modus beteken
  
                3 ure per week.
  
                Daar is 3 kinders wat TV vir
  
                3 ure per week gekyk
  
                het en daar is 12 kinders.
  
                dus n(E) = n(= modus) = 2 en
  
                n(S) = n(kinders) = 12
  
$$ \hspace*{16 mm}\mathrm{P(>\kern1mmmodus)\kern1mm=\kern1mm\frac{n(>\kern1mmmodus)}{n(kinders)} } $$


$$ \hspace*{37 mm}=\kern1mm\frac{2}{12}\kern1mm=\kern1mm\frac{1}{6} $$



[ V 6 ]
  
  
     
      6.1  Arrange the data :
  
             2, 3, 3, 4, 5, 6, 8, 9, 12, 17, 21, 30
  
$$ \hspace*{13 mm}\mathrm{Mean\kern1mm=\kern1mm\frac{Σx}{n} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{120}{12}\kern1mm=\kern1mm10 } $$

  
$$ \hspace*{13 mm}\mathrm{Median\kern1mm=\kern1mm\frac{6+8}{2}\kern1mm=\kern1mm7 } $$


  
             Mode  =  3  . . . appears twice
  
[ Question 6 ]
  
  
     
      6.2.1  There are two periods of 3 hours
  
                and 12 children.
  
                Thus n(E) = n(3) = 2 and
  
                n(S) = n(children) = 12
  
$$ \hspace*{13 mm}\mathrm{P(3)\kern1mm=\kern1mm\frac{n(3)}{n(children)} } $$


$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{2}{12}\kern1mm=\kern1mm\frac{1}{6} } $$



[ Question 6 ]
  
  
     
      6.2.2  There are 7 children that watched
  
                TV for 6 or more hours per week
  
                and 12 children.
  
                Thus n(E) = n(6+) = 7 and
  
                n(S) = n(children) = 12
  
$$ \hspace*{13 mm}\mathrm{P(6+)\kern1mm=\kern1mm\frac{n(6+)}{n(children)} } $$


$$ \hspace*{24 mm}\mathrm{=\kern1mm\frac{7}{12} } $$



[ Q 6 ]
  
  
     
      6.2.3  There are 8 children who watched
  
                TV for 10 or less hours per week
  
                and 12 children.
  
                Thus n(E) = n(< 11) = 8 and
  
                n(S) = n(children) = 12
  
$$ \hspace*{13 mm}\mathrm{P(<\kern1mm11)\kern1mm=\kern1mm\frac{n(<\kern1mm11)}{n(children)} } $$


$$ \hspace*{29 mm}\mathrm{=\kern1mm\frac{8}{12}\kern1mm=\kern1mm\frac{2}{3} } $$



[ Q 6 ]
  
  
     
      6.2.4  Less than 3 hours per day means
  
                less than 21 hours per week.
  
                There are 10 children who
  
                watched TV for less than 21 hours
  
                per week and 12 children.
  
                Thus n(E) = n(< 3) = 10 and
  
                n(S) = n(children) = 12
  
$$ \hspace*{13 mm}\mathrm{P(<\kern1mm3)\kern1mm=\kern1mm\frac{n(<\kern1mm3)}{n(children)} } $$


$$ \hspace*{26 mm}\mathrm{=\kern1mm\frac{10}{12}\kern1mm=\kern1mm\frac{5}{6} } $$



[ Q 6 ]
  
  
     
      6.2.5  The mean is 10 hours per week.
  
                Thus less than the mean means
  
                less than 10 hours per week.
  
                There are 8 children who watch
  
                TV for less than 10 hours per
  
                week and 12 children.
  
                Thus n(E) = n(< 10) = 8 and
  
                n(S) = n(children) = 12
  
$$ \hspace*{13 mm}\mathrm{P(<\kern1mm10)\kern1mm=\kern1mm\frac{n(<\kern1mm10)}{n(children)} } $$


$$ \hspace*{28 mm}\mathrm{=\kern1mm\frac{8}{12}\kern1mm=\kern1mm\frac{2}{3} } $$



[ Q 6 ]
  
  
     
      6.2.6  The median is 7 hours per week.
  
                Thus more than the median means
  
                more than 7 hours per week.
  
                There are 6 children who watch
  
                TV for more than 7 hours per week
  
                and 12 children.
  
                Thus n(E) = n(> median) = 6 and
  
                n(S) = n(children) = 12
  
$$ \hspace*{13 mm}\mathrm{P(>\kern1mmmedian)\kern1mm=\kern1mm\frac{n(>\kern1mmmedian)}{n(children)} } $$


$$ \hspace*{28 mm}\mathrm{=\kern1mm\frac{6}{12}\kern1mm=\kern1mm\frac{1}{2} } $$



[ Q 6 ]
  
  
     
      6.2.7  The mode is 3 hours per week.
  
                Thus equal to the mode means
  
                equal to 3 hours per week.
  
                There are 2 children who watch
  
                TV for 3 hours per week
  
                and 12 children.
  
                Thus n(E) = n(= mode) = 2 and
  
                n(S) = n(children) = 12
  
$$ \hspace*{13 mm}\mathrm{P(>\kern1mmmode)\kern1mm=\kern1mm\frac{n(>\kern1mmmode)}{n(children)} } $$


$$ \hspace*{28 mm}\mathrm{=\kern1mm\frac{2}{12}\kern1mm=\kern1mm\frac{1}{6} } $$



[ Q 6 ]
  
  
  
  
Antwoorde / Answers  7
     
$$ \hspace*{6 mm}\mathrm{7.1\kern1mmP(slaag)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(druip) } $$
$$ \hspace*{28 mm}=\kern1mm1\kern1mm─\kern1mm0,35 $$
$$ \hspace*{28 mm}=\kern1mm0,65 $$


[ Vraag 7 ]
  
  
     
$$ \hspace*{6 mm}\mathrm{7.2\kern1mmP(druip)\kern1mm=\kern1mm\frac{n(aantal\kern1mmdruipelinge)}{n(aantal\kern1mmgeskryf)}} $$
  
  
                            
  
             P(druip)  =  0,35 en leerlinge
  
             wat geskryf het = 74.
  
$$ \hspace*{15 mm}\mathrm{0,35999887\kern1mm=\kern1mm\frac{n(druipelinge)}{n(geskryf)} } $$


$$ \hspace*{15 mm}\mathrm{n(druipelinge)\kern1mm=\kern1mm0,35\kern1mmX\kern1mm74 } $$
$$ \hspace*{39mm}=\kern1mm26 $$


[ V 7 ]
  
  
     
                                      n(leerlinge wat slaag)
      7.3  P(slaag)  =  ──────────────────
                                   n(leerlinge geskryf)
  
  
             P(slaag) = 0,65 en
  
             n(leerlinge wat geskryf het) = 132
  
$$ \hspace*{15 mm}\mathrm{0,65\kern1mm=\kern1mm\frac{n(leerlinge\ geslaag)}{n(leerlinge\ geskryf)} } $$


$$ \hspace*{15 mm}\mathrm{n(geslaag)\kern1mm=\kern1mm0,65\kern1mmX\kern1mm132 } $$
$$ \hspace*{34 mm}=\kern1mm85,8\kern1mm=\kern1mm86 $$

[ V 7 ]
  
  
     
$$ \hspace*{6 mm}\mathrm{7.1\kern1mmP(pass)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(fail) } $$
$$ \hspace*{28 mm}=\kern1mm1\kern1mm─\kern1mm0,35 $$
$$ \hspace*{28 mm}=\kern1mm0,65 $$


[ Question 7 ]
  
  
     
$$ \hspace*{6 mm}\mathrm{7.2\kern1mmP(failas)\kern1mm=\kern1mm\frac{n(number\kern1mmfailed)}{n(number\kern1mmthat\kern1mmwrote)} } $$
  
  
  
             P(fail)  =  0,35  and pupils
  
             that wrote = 74
  
$$ \hspace*{13 mm}\mathrm{0,35999\kern1mm=\kern1mm\frac{n(failures)}{n(written)} } $$


$$ \hspace*{13 mm}\mathrm{n(failures)\kern1mm=\kern1mm0,35\kern1mmX\kern1mm74 } $$
$$ \hspace*{32 mm}\mathrm{=\kern1mm26 } $$


[ Q 7 ]
  
  
     
                                   n(pupils that pass)
      7.3  P(pass)  =  ─────────────
                                  n(pupils that wrote)
  
  
             P(pass)  =  0,65  and
  
             n(pupils that wrote) = 132
  
                               n(pupils that pass)
               0,65  =  ─────────────
                                        132
  
  
$$ \hspace*{15 mm}\mathrm{n(passed)\kern1mm=\kern1mm0,65\kern1mmX\kern1mm132 } $$
$$ \hspace*{33 mm}=\kern1mm85,8\kern1mm=\kern1mm86 $$

[ Q 7 ]
  
  
  
  
Antwoorde / Answers  8
  
     
      8.1  As die draaibord gebalanseer is,
  
             d.w.s. elke kant het dieselfde
  
             kans op daardie sy te val sal
  
$$ \hspace*{16 mm}\mathrm{P(3)\kern1mm=\kern1mm\frac{1}{4}\kern1mm=\kern1mm0,25 } $$


  
                                                       [ V 8 ]
  
  
     
      8.2  Die stelling is korrek as
  
             die draaibord gebalanseer is,
  
             d.w.s. elke sy het dieselfde
  
             kans op daardie sy te val.
  
                                                       [ V 8 ]
  
     
      8.3.1  As die draaibord gebalanseer is,
  
                dan P(3) = 0,25 en
  
                n(S) = 500
  
                n(3) = 0,25 X 500
  
                        =  125.
  
                Die nommer 3 behoort 125 keer
  
                voor te kom.                      [ V 8 ]
  
  
     
      8.3.2
$$ \hspace*{6 mm}\mathrm{8.3.2\kern1mmP(2)\kern1mm=\kern1mm\frac{n(2)}{n(S)}\kern1mm=\kern1mm\frac{130}{500} } $$


$$ \hspace*{25 mm}=\kern1mm0,26 $$
  
                                                       [ V 8 ]
  
  
     
      8.3.3  P(1)  +  P(2)  +  P(3)  +  P(4)  =  1
  
                0,15  +  0,26  +  0,24  +  P(4)  =  1
  
                                  0,65  +  P(4)  =  1
  
                                               P(4)  =  0,35
  
                                                       [ V 8 ]
  
  
     
      8.3.4  Die bord is ongebalanseerd
  
                want die waarkynlikhede
  
                om die verskillende getalle
  
                te kry verskil baie.
  
                                                       [ V 8 ]
  
  
     
      8.1  If the spinner is balanced, i.e. the
  
             sides have the same chance to
  
             land on its side, then
  
$$ \hspace*{13 mm}\mathrm{P(3)  =  \frac{1}{4}   =  0,25 } $$


  
                                                                [ Q 8 ]
  
  
     
      8.2  The statement is true if the
  
             spinner is balanced, i.e. the
  
             sides have the same chance to
  
             land on its side.
                                                                [ Q 8 ]
  
  
     
      8.3.1  If the spinner is balanced,
  
                then P(3) = 0,25 and
  
                n(S)  =  500.
  
                n(3)  =  0,25  X  500.
  
                        =  125.
  
                The number 3 should appear 125 times..
  
                                                                [ Q 8 ]
  
  
     
      8.3.2
$$ \hspace*{6 mm}\mathrm{8.3.2\kern1mmP(2)\kern1mm=\kern1mm\frac{n(2)}{n(S)}\kern1mm=\kern1mm\frac{130}{500} } $$


$$ \hspace*{25 mm}=\kern1mm0,26 $$
  
                                                                [ Q 8 ]
  
  
     
      8.3.3  P(1)  +  P(2)  +  P(3)  +  P(4)  =  1
  
                0,15  +  0,26  +  0,24  +  P(4)  =  1
  
                                  0,65  +  P(4)  =  1
  
                                                P(4)  =  0,35
  
                                                                [ Q 8 ]
  
  
     
      8.3.4  The spinner is unbalanced
  
                because the probabilities to get the
  
                different numbers differ too much.
  
  
  
                                                                [ Q 8 ]
  
  
  
  
Antwoorde / Answers  9
  
     
      9.1  Aantal leerlinge wat deelgeneem
  
             het is 50.                            [ V 9 ]
  
  
     
      9.2.1  Daar is 18 seuns en 50 leerlinge
  
                n(E) = n(seuns) = 18 en
  
                n(S) = n(leerlinge) = 50.
  
$$ \hspace*{16 mm}\mathrm{P(seun)\kern1mm=\kern1mm\frac{n(seun)}{n(leerlinge)} } $$


$$ \hspace*{32 mm}\mathrm{=\kern1mm\frac{18}{50}\kern1mm=\kern1mm36\kern1mm\% } $$


[ Vraag 9 ]
  
  
     
      9.2.2  Twaalf leerlinge koop niks
  
                en daar is 50 leerlinge.
  
                n(E) = n(koop niks) = 12 en
  
                n(S) = n(leerlinge) = 50.
  
$$ \hspace*{16 mm}\mathrm{P(koop\kern1mmniks)\kern1mm=\kern1mm\frac{n(koop\kern1mmniks)}{n(leerlinge)} } $$


$$ \hspace*{39 mm}\mathrm{=\kern1mm\frac{12}{50}\kern1mm=\kern1mm24\kern1mm\% } $$


[ V 9 ]
  
  
     
      9.2.3  Ses leerlinge koop 'n Smulstafie
  
                en daar is 50 leerlinge.
  
                n(E) = n(Smulstafie) = 6 en
  
                n(S) = n(leerlinge) = 50.
  
$$ \hspace*{16 mm}\mathrm{P(smulstafie)\kern1mm=\kern1mm\frac{n(smulstafie)}{n(leerlinge)} } $$


$$ \hspace*{40mm}\mathrm{=\kern1mm\frac{6}{50}\kern1mm=\kern1mm12\kern1mm\% } $$


[ V 9 ]
  
  
     
      9.2.4  Agt leerlinge koop kooklekkers
  
                en daar is 50 leerlinge.
  
                n(E) = n(kooklekkers) = 8 en
  
                n(S) = n(leerlinge) = 50.
  
$$ \hspace*{16 mm}\mathrm{P(kooklekkers)\kern1mm=\kern1mm\frac{n(kooklekkers)}{n(leerlinge)} } $$


$$ \hspace*{43 mm}\mathrm{=\kern1mm\frac{8}{50}\kern1mm=\kern1mm16\kern1mm\% } $$


[ V 9 ]
  
  
     
      9.2.5  Drie dogters koop 'n Sjokolade stafie
  
                en daar is 32 dogters.
  
                Gebruik "d+SS" vir
  
                dogter + Sjokolade Stafie
  
                n(E) = n(d+SS) = 3 en
  
                n(S) = n(dogters) = 32.
  
$$ \hspace*{16 mm}\mathrm{P(d+SS)\kern1mm=\kern1mm\frac{n(d+SS)}{n(dogters)} } $$


$$ \hspace*{34 mm}\mathrm{=\kern1mm\frac{3}{32}\kern1mm=\kern1mm9,375\kern1mm\% } $$


[ V 9 ]
  
  
     
      9.2.6  Twaalf dogters koop droë vrugte
  
                en daar is 32 dogters.
  
                Gebruik "d+dv" vir
  
                dogter + droë vrugte
  
                n(E) = n(d+dv) = 12 en
  
                n(S) = n(dogters) = 32.
  
$$ \hspace*{16 mm}\mathrm{P(d+dv)\kern1mm=\kern1mm\frac{n(d+dv)}{n(dogters)} } $$


$$ \hspace*{34 mm}\mathrm{=\kern1mm\frac{12}{32}\kern1mm=\kern1mm37,5\kern1mm\% } $$


[ V 9 ]
  
  
     
      9.2.7  Twee seuns koop 'n Smulstafie
  
                en daar is 18 seuns.
  
                Gebruik "s+st" vir
  
                seun + Smulstafie
  
                n(E) = n(s+st) = 2 en
  
                n(S) = n(seuns) = 18.
  
$$ \hspace*{16 mm}\mathrm{P(s+st)\kern1mm=\kern1mm\frac{n(s+st)}{n(seuns)} } $$


$$ \hspace*{32 mm}\mathrm{=\kern1mm\frac{2}{18}\kern1mm=\kern1mm11,111\kern1mm\% } $$


[ V 9 ]
  
  
     
      9.2.8  Vier seuns koop niks
  
                en daar is 18 seuns.
  
                Gebruik "s+niks" vir
  
                seun + niks
  
                n(E) = n(s+niks) = 4 en
  
                n(S) = n(seuns) = 18.
  
$$ \hspace*{16 mm}\mathrm{P(seun+niks)\kern1mm=\kern1mm\frac{n(seun+niks)}{n(seuns)} } $$


$$ \hspace*{42 mm}\mathrm{=\kern1mm\frac{4}{18}\kern1mm=\kern1mm22,222\kern1mm\% } $$


[ V 9 ]
  
  
     
      9.3  Aantal leerlinge wat deelgeneem
  
             het is 50 en daar is 650 leerlinge
  
             in die skool. Dus slegs 7,69 % van
  
             die leerlinge het aan die opname
  
             deelgeneem. Die syfer is te laag
  
             om goeie gevolgtrekkings van te maak.
  
[ V 9 ]
  
  
     
      9.4.1  Volgens 9.2.8 is
  
                P(seun + niks) = 22,222 %
  
                Agt dogters koop niks en
  
                daar is 32 dogters.
  
                dus n(dogter en koop niks) = 8
  
                en n(dogters) = 32.
  
$$ \hspace*{16 mm}\mathrm{P(d+niks)\kern1mm=\kern1mm\frac{n(d+niks)}{n(dogters)} } $$


$$ \hspace*{36 mm}\mathrm{=\kern1mm\frac{8}{32}\kern1mm=\kern1mm25\kern1mm\% } $$



                Dus, P(seun + niks) = 22,222 %
  
                en P(dogter + niks) = 25 %
  
                sodat die waarskynlikheid dat
  
                'n dogter niks sal koop is groter
  
                as die van 'n seun en die stelling
  
                is dus vals.
  
[ V 9 ]
  
  
     
      9.4.2  Volgens 9.2.2 is
  
                P(koop niks) = 24 % en
$$ \hspace*{16 mm}\mathrm{dus\kern1mmP(koop)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(koop\kern1mmniks) } $$
$$ \hspace*{39 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm24\kern1mm\% } $$
$$ \hspace*{39 mm}\mathrm{=\kern1mm76\kern1mm\% } $$

$$ \hspace*{16 mm}\mathrm{76\kern1mm\%\kern1mmvan\kern1mm650\kern1mm=\kern1mm\frac{76\kern1mmx\kern1mm650}{100} } $$

$$ \hspace*{41 mm}\mathrm{=\kern1mm494 } $$


                Die waarskynlikheid dat 494 en
  
                meer leerlinge sal koop is goed
  
                en dus is die stelling waar.
  
[ V 9 ]
  
  
     
      9.4.3  Volgens 9.2.5 is
  
                P(dogter + Sjokolade stafie) = 9,375 %
  
                Vier seuns koop 'n Sjokolade stafie
  
                en daar is 18 seuns.
  
                dus n(seun+Sjokolade stafie) = 4
  
                en n(seuns) = 18.
$$ \hspace*{16 mm}\mathrm{P(s+sjs)\kern1mm=\kern1mm\frac{n(s+sjs)}{n(seuns)} } $$


$$ \hspace*{34 mm}\mathrm{=\kern1mm\frac{4}{18}\kern1mm=\kern1mm22,222\kern1mm\% } $$



                Dus, P(s+Sjokostafie) = 22,222 %
  
                en P(d+Sjokostafie) = 9,375 %
  
                sodat die stelling dus waar is.
  
[ V 9 ]
  
  
     
      9.4.4  Vier dogtes koop 'n Smulstafie en
  
                daar is 32 dogters.
  
                n(d+smuls)  =  4 en n(dogters) = 32
  
$$ \hspace*{16 mm}\mathrm{P(d+sms)\kern1mm=\kern1mm\frac{n(d+sms)}{n(dogters)} } $$


$$ \hspace*{34 mm}\mathrm{=\kern1mm\frac{4}{32}\kern1mm=\kern1mm12,5\kern1mm\% } $$



                Vyf seuns koop droë vrugte
  
                en daar is 18 seuns.
  
                Dus, n(s+dv)  =  5 en n(seuns) = 18
  
$$ \hspace*{16 mm}\mathrm{P(s+dv)\kern1mm=\kern1mm\frac{n(s+dv)}{n(seuns)} } $$


$$ \hspace*{34 mm}\mathrm{=\kern1mm\frac{5}{18}\kern1mm=\kern1mm27,778\kern1mm\% } $$



                Dus, dit is meer waarskynlik dat 'n
  
                seun droë vrugte sal koop as wat
  
                'n dogter 'n Smulstafie sal koop.
  
                As daar meer seuns in die skool as
  
                dogters is sal die stelling waar
  
                wees anders is dit vals.
  
                Aangesien ons nie oor die inligting
  
                beskik nie, kan ons nie met
  
                sekerheid sê of die stelling waar of
  
                vals is nie.
  
[ V 9 ]
  
  
     
      9.1  Number of participating pupils is 50.
  
                                                                  [ Q 9 ]
  
  
     
      9.2.1  There are 18 boys and 50 pupils
  
                n(E) = n(boys) = 18 and
  
                n(S) = n(pupils) = 50
  
$$ \hspace*{16 mm}\mathrm{P(boy)\kern1mm=\kern1mm\frac{n(boy)}{n(pupils)}\kern1mm=\kern1mm0,25 } $$


$$ \hspace*{26 mm}\mathrm{=\kern1mm\frac{18}{50}\kern1mm=\kern1mm36\kern1mm\% } $$


[ Question 9 ]
  
  
     
      9.2.2  Twelve pupils buy nothing and
  
                there are 50 pupils.
  
                n(E) = n(buys nothing) = 12 and
  
                n(S) = n(pupils) = 50
  
$$ \hspace*{16 mm}\mathrm{P(buys\kern1mmnothing)\kern1mm=\kern1mm\frac{n(buys\kern1mmnothing)}{n(pupils)} } $$


$$ \hspace*{41 mm}\mathrm{=\kern1mm\frac{12}{50}\kern1mm=\kern1mm24\kern1mm;\% } $$


[ Q 9 ]
  
  
     
      9.2.3  Six pupils buy a Munch bar and
  
                there are 50 pupils.
  
                n(E) = n(Munch bar) = 6 and
  
                n(S) = n(pupils) = 50
  
$$ \hspace*{16 mm}\mathrm{P(Munch\kern1mmbar)\kern1mm=\kern1mm\frac{n(Munch\kern1mmbar)}{n(pupils)} } $$


$$ \hspace*{41 mm}\mathrm{=\kern1mm\frac{6}{50}\kern1mm=\kern1mm12\kern1mm\% } $$


[ Q 9 ]
  
  
     
      9.2.4  Eight pupils buy boiled sweets and
  
                there are 50 pupils.
  
                n(E) = n(boiled sweets) = 8 and
  
                n(S) = n(pupils) = 50
  
$$ \hspace*{14 mm}\mathrm{P(boiled\kern1mmsweets)\kern1mm=\kern1mm\frac{n(boiled\kern1mmsweets)}{n(pupils)} } $$


$$ \hspace*{43 mm}\mathrm{=\kern1mm\frac{8}{50}\kern1mm=\kern1mm16\kern1mm\% } $$


[ Q 9 ]
  
  
     
      9.2.5  Three girls buy a Chocolate bar and
  
                there are 32 girls.
  
                Use "g+CB" for girl + Chocolate bar
  
  
  
                n(E) = n(g+CB) = 3 and
  
                n(S) = n(girls) = 32
  
$$ \hspace*{16 mm}\mathrm{P(g+CB)\kern1mm=\kern1mm\frac{n(g+CB)}{n(girls)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{3}{32\kern1mm}=\kern1mm9,375\kern1mm\% } $$


[ Q 9 ]
  
  
     
      9.2.6  Twelve girls buy dried fruit and
  
                there are 32 girls.
  
                Use *g+df" for girl + dried fruit
  
  
  
                n(E) = n(g+df) = 12 and
  
                n(S) = n(girls) = 32
  
$$ \hspace*{16 mm}\mathrm{P(g+CB)\kern1mm=\kern1mm\frac{n(g+CB)}{n(girls)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{12}{32}\kern1mm=\kern1mm37,5\kern1mm\% } $$


[ Q 9 ]
  
  
     
      9.2.7  Two boys buy a Munch Bar and
  
                there are 18 boys.
  
                Use *b+MB" for boy + Munch Bar
  
  
  
                n(E) = n(b+MB) = 2 and
  
                n(S) = n(boys) = 18
  
$$ \hspace*{16 mm}\mathrm{P(b+MB)\kern1mm=\kern1mm\frac{n(b+MB)}{n(boys)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{2}{18}\kern1mm=\kern1mm11,111\kern1mm\% } $$


[ Q 9 ]
  
  
     
      9.2.8  Four boys buy nothing and
  
                there are 18 boys.
  
                Use *b+not" for boy + buy nothing
  
  
  
                n(E) = n(b+not) = 4 and
  
                n(S) = n(boys) = 18
  
$$ \hspace*{16 mm}\mathrm{P(b+not)\kern1mm=\kern1mm\frac{n(b+not)}{n(boys)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{4}{18}\kern1mm=\kern1mm22,222\kern1mm\% } $$


[ Q 9 ]
  
  
     
      9.3  Number of participating pupils is 50
  
             and there are 650 pupils in the school.
  
             Thus 7,69 % of the pupils took part
  
             in the survey. This too low for
  
             making good decisions.
  
[ Q 9 ]
  
  
  
  
     
      9.4.1  According to 9.2.8
  
                P(boy+nothing) = 22,22%
  
                Eight girls do not buy anything
  
                and there are 32 girls.
  
                thus, n(girl+nothing) = 8 and
  
                n(girls) = 32
  
$$ \hspace*{16 mm}\mathrm{P(g+nothing)\kern1mm=\kern1mm\frac{n(g+nothing)}{n(boys)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{8}{32}\kern1mm=\kern1mm25\kern1mm\% } $$


  
                Thus, P(boy+nothing) = 22,222 %
  
                and P(girl+nothing) = 25 %
  
                so that the probability that a girl will
  
                buy nothing is greater than that of
  
                a boy and therefore the
  
                statement is false.
  
[ Q 9 ]
  
  
     
      9.4.2  According to 9.2.2
  
                P(buys nothing) = 24 % and
$$ \hspace*{16 mm}\mathrm{thus\kern1mmP(buys)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(buys\kern1mmnothing) } $$
$$ \hspace*{39 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm24\kern1mm\% } $$
$$ \hspace*{39 mm}\mathrm{=\kern1mm76\kern1mm\% } $$
$$ \hspace*{16 mm}\mathrm{76\kern1mm\%\kern1mmof\kern1mm650\kern1mm=\kern1mm\frac{76\kern1mmx\kern1mm650}{100} } $$

$$ \hspace*{39 mm}\mathrm{=\kern1mm494 } $$



                The probability that 494 and more
  
                pupils will buy is good and
  
                the statement is thus true.
  
[ Q 9 ]
  
  
     
      9.4.3  According to 9.2.5
  
                P(girl + Chocolate Bar) = 9,375 %
  
                Four boys buy a Chocolate Bar and
  
                there are 18 boys.
  
                Thus n(Boy + Chocolate Bar) = 4
  
                and n(boys) = 18
$$ \hspace*{16 mm}\mathrm{P(b+CB)\kern1mm=\kern1mm\frac{n(b+CB)}{n(boys)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{4}{18}\kern1mm=\kern1mm22,222\kern1mm\% } $$



                P(b+CB)  =  22,222 % and
  
                P(g+CB) = 9,375 % so that
  
                the statement is true.
  
[ Q 9 ]
  
  
     
      9.4.4  Four girls buy a Munch Bar and
  
                there are 32 girls.
  
                Thus n(g+MB) = 4 and n(girls) = 32
  
$$ \hspace*{16 mm}\mathrm{P(g+MB)\kern1mm=\kern1mm\frac{n(g+MB)}{n(girls)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{4}{32}\kern1mm=\kern1mm12,5\kern1mm\% } $$



                Five boys buy dried fruit and
  
                there are 18 boys.
  
                Thus n(b+df) = 5 and n(boys) = 18
  
$$ \hspace*{16 mm}\mathrm{P(b+df)\kern1mm=\kern1mm\frac{n(b+df)}{n(boys)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{5}{18}\kern1mm=\kern1mm27,778\kern1mm\% } $$



                Thus it is more probable that a boy
  
                will buy dried fruit than it is that
  
                a girl will buy a Munch Bar.
  
                If there are more boys in the school
  
                than girls, the statement will be true,
  
                else it will be false.
  
                As we do not have that information
  
                we can not say whether the
  
                statement is true or false.
  
[ Q 9 ]
  
  
  
  
Antwoorde / Answers  10
  
     
      10.1.1  Daar is 3 rooi lekkers en
  
                  3 + 2 = 5 lekkers in die blik
  
                  n(E) = n(rooi) = 3 en n(S) = 5
  
$$ \hspace*{16 mm}\mathrm{P(rooi)\kern1mm=\kern1mm\frac{n(rooi)}{n(lekkers)} } $$


$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{3}{5} } $$



[ Vraag 10 ]
  
  
     
      10.1.2  Daar is 2 groen lekkers en
  
                  3 + 2 = 5 lekkers in die blik
  
                  n(E) = n(groen) = 2 en
  
                  n(S) = n(lekkers) = 5
  
$$ \hspace*{16 mm}\mathrm{P(groen)\kern1mm=\kern1mm\frac{n(groen)}{n(lekkers)} } $$


$$ \hspace*{33 mm}\mathrm{=\kern1mm\frac{2}{5} } $$



[ V 10 ]
  
  
     
      10.2.1  Daar is 3 - 1 = 2 rooi lekkers en
  
                  2 groen lekkers in die blik
  
                  d.w.s. 2 + 2 = 4 lekkers in die blik
  
                  n(E) = n(rooi) = 2 en
  
                  n(S) = n(lekkers) = 4
$$ \hspace*{18 mm}\mathrm{P(rooi)\kern1mm=\kern1mm\frac{n(rooi)}{n(lekkers)} } $$


$$ \hspace*{32 mm}\mathrm{=\kern1mm\frac{2}{4}\kern1mm=\kern1mm\frac{1}{2} } $$



[ V 10 ]
  
  
     
      10.2.2  Daar is 2 groen lekkers en
  
                  2 + 2 = 2 lekkers in die blik
  
                  n(E) = n(groen) = 2 en
  
                  n(S) = n(lekkers) = 2
  
$$ \hspace*{18 mm}\mathrm{P(groen)\kern1mm=\kern1mm\frac{n(groen)}{n(lekkers)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{2}{4}\kern1mm=\kern1mm\frac{1}{2} } $$



[ V 10 ]
  
  
     
      10.3.1  Daar is 2 rooi en 1 groen
  
                  lekkers in die blik en
  
                  2 + 1 = 3 lekkers in die blik
  
                  n(E) = n(rooi) = 2 en n(S) = 3
  
                  n(S) = n(lekkers) = 3
  
$$ \hspace*{18 mm}\mathrm{P(rooi)\kern1mm=\kern1mm\frac{n(rooi)}{n(lekkers)} } $$


$$ \hspace*{32 mm}\mathrm{=\kern1mm\frac{2}{3} } $$



[ V 10 ]
  
  
     
      10.3.2  Daar is 1 groen en
  
                  2 + 1 = 3 lekkers in die blik
  
                  n(E) = n(groen) = 1 en
  
                  n(S) = n(lekkers) = 3
  
$$ \hspace*{18 mm}\mathrm{P(groen)\kern1mm=\kern1mm\frac{n(groen)}{n(lekkers)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{1}{3} } $$



[ V 10 ]
  
  
     
      10.1.1  There are 3 red and 3 + 2 = 5
  
                  sweets in the tin.
  
                  n(E) = n(red) = 3 and
  
                  n(S) = n(sweets) = 5
$$ \hspace*{18 mm}\mathrm{P(red)\kern1mm=\kern1mm\frac{n(red)}{n(sweets)} } $$


$$ \hspace*{31 mm}\mathrm{=\kern1mm\frac{3}{5} } $$


[ Question 10 ]
  
  
     
      10.1.2  There are 2 green and 3 + 2 = 5
  
                  sweets in the tin.
  
                  n(E) = n(green) = 2 and
  
                  n(S) = n(sweets) = 5
  
$$ \hspace*{18 mm}\mathrm{P(green)\kern1mm=\kern1mm\frac{n(green)}{n(sweets)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{2}{5} } $$



[ Q 10 ]
  
  
     
      10.2.1  There are 3 - 1 = 2 red and 2 green
  
                  sweets in the tin.
  
                  i.e. 2 + 2 = 4 sweets in the tin.
  
                  n(E) = n(red) = 2 and
  
                  n(S) = n(sweets) = 4
  
$$ \hspace*{18 mm}\mathrm{P(red)\kern1mm=\kern1mm\frac{n(red)}{n(sweets)} } $$


$$ \hspace*{31 mm}\mathrm{=\kern1mm\frac{2}{4}\kern1mm=\kern1mm\frac{1}{2} } $$


[ Q 10 ]
  
  
     
      10.2.2  There are 2 green and 2 + 2 = 4
  
                  sweets in the tin.
  
                  n(E) = n(green) = 2 and
  
                  n(S) = n(sweets) = 4
  
$$ \hspace*{18 mm}\mathrm{P(green)\kern1mm=\kern1mm\frac{n(green)}{n(sweets)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{2}{4}\kern1mm=\kern1mm\frac{1}{2} } $$



[ Q 10 ]
  
  
     
      10.3.1  There are 2 red and 1 green
  
                  sweets in the tin and
  
                  2 + 1 = 3 sweets in the tin.
  
                  n(E) = n(red) = 2 and
  
                  n(S) = n(sweets) = 3
  
$$ \hspace*{18 mm}\mathrm{P(red)\kern1mm=\kern1mm\frac{n(red)}{n(sweets)} } $$


$$ \hspace*{31 mm}\mathrm{=\kern1mm\frac{2}{3} } $$


[ Q 10 ]
  
  

     
      10.3.2  There is 1 green and
  
                  2 + 1 = 3 sweets in the tin.
  
                  n(E) = n(green) = 1 and
  
                  n(S) = n(sweets) = 3
  
$$ \hspace*{18 mm}\mathrm{P(green)\kern1mm=\kern1mm\frac{n(green)}{n(sweets)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{1}{3} } $$


[ Q 10 ]
  
  
  
  
Antwoorde / Answers  11
  
  
      11.1  
  
Tweede steen / Second die Eerste steen / First die
  
1
  
2
  
3
  
4
  
5
  
6
  
1
  
11
  
21
  
31
  
41
  
51
  
61
  
2
  
12
  
22
  
32
  
42
  
52
  
62
  
3
  
13
  
23
  
33
  
43
  
53
  
63
  
4
  
14
  
24
  
34
  
44
  
54
  
64
  
5
  
15
  
25
  
35
  
45
  
55
  
65
  
6
  
16
  
26
  
36
  
46
  
56
  
66
[ Vraag 11.1 ]
  
  
     
      11.2  Daar is net een gunstige uitkoms
  
               vir twee viere nl 44 en daar
  
               is 36 moontlike uitkomstes.
  
               Dus, n(E) = n(twee viere) = 1
  
               en n(S) = n(moontlikhede) = 36
  
$$ \hspace*{15 mm}\mathrm{P(twee\kern1mmviere)\kern1mm=\kern1mm\frac{n(twee\kern1mmviere)}{n(moontlikhede)} } $$


$$ \hspace*{39 mm}\mathrm{=\kern1mm\frac{1}{36} } $$



[ V 11 ]
  
  
     
      11.3  Die gunstige uitkomstes is
  
               34 en 43 en daar is
  
               36 moontlike uitkomstes.
  
               Dus, n(E) = n(3 en 4) = 2
  
               en n(S) = n(moontlikhede) = 36
  
$$ \hspace*{15 mm}\mathrm{P(3\kern1mmen\kern1mm4)\kern1mm=\kern1mm\frac{n(3\kern1mmen\kern1mm4)}{n(moontlikhede)} } $$


$$ \hspace*{33 mm}\mathrm{=\kern1mm\frac{2}{36}\kern1mm=\kern1mm\frac{1}{18} } $$



[ V 11 ]
  
  
     
      11.4  Daar is net 1 gunstige uitkomste,
  
               nl 34 want die orde is belangrik
  
               en daar is 36 moontlike uitkomstes.
  
               Dus, n(E) = n(3 en 4) = 1
  
               en n(S) = n(moontlikhede) = 36
  
$$ \hspace*{15 mm}\mathrm{P(3\kern1mmen\kern1mm4)\kern1mm=\kern1mm\frac{n(3\kern1mmen\kern1mm4)}{n(moontlikhede)} } $$


$$ \hspace*{33 mm}\mathrm{=\kern1mm\frac{1}{36} } $$



[ V 11 ]
  
  
     
      11.5  Die onewe getalle is 1, 3 en 5
  
               en dus is daar is 9 gunstige
  
               uitkomste, nl 11, 13, 15, 31, 33,
  
               35, 51, 53 en 55, en daar
  
               is 36 moontlike uitkomstes.
  
               Dus, n(E) = n(onewe) = 9
  
               en n(S) = n(moontlikhede) = 36
  
$$ \hspace*{15 mm}\mathrm{P(onewe)\kern1mm=\kern1mm\frac{n(onewe)}{n(moontlikhede)} } $$


$$ \hspace*{33 mm}\mathrm{=\kern1mm\frac{9}{36}\kern1mm=\kern1mm\frac{1}{4} } $$



[ V 11 ]
  
  
     
      11.6  Die getalle kleiner as 3 is 1 en 2
  
               en dus is daar is 4 gunstige
  
               uitkomste, nl 11, 12, 21, 22,
  
               en daar is 36 moontlike uitkomstes.
  
               Dus, n(E) = n(< 3) = 4
  
               en n(S) = n(moontlikhede) = 36
  
$$ \hspace*{15 mm}\mathrm{P(<\kern1mm3)\kern1mm=\kern1mm\frac{n(< 3)}{n(moontlikhede)} } $$


$$ \hspace*{29 mm}\mathrm{=\kern1mm\frac{4}{36}\kern1mm=\kern1mm\frac{1}{9} } $$



[ V 11 ]
  
  
     
      11.7  Die veelvoude van 3 is 3 en 6
  
               en dus is daar is 4 gunstige
  
               uitkomste, nl 33, 36, 63 en 66,
  
               en daar is 36 moontlike uitkomstes.
  
               Dus, n(E) = n(veelv 3) = 4
  
               en n(S) = n(moontlikhede) = 36
  
$$ \hspace*{15 mm}\mathrm{P(veelvoude\kern1mm3)\kern1mm=\kern1mm\frac{n(veelvoude\kern1mm3)}{n(moontlikhede)} } $$


$$ \hspace*{42 mm}\mathrm{=\kern1mm\frac{4}{36}\kern1mm=\kern1mm\frac{1}{9} } $$



[ V 11 ]
  
  
     
      11.8  Die gunstige uitkomstes is
  
               16, 25, 34, 43, 52, 61 en dus
  
               is daar 6 gunstige uitkomstes
  
               en daar is 36 moontlike uitkomstes.
  
               Dus, n(E) = n(som = 7) = 6
  
               en n(S) = n(moontlikhede) = 36
  
$$ \hspace*{15 mm}\mathrm{P(som\kern1mm=\kern1mm3)\kern1mm=\kern1mm\frac{n(som\kern1mm=\kern1mm3)}{n(moontlikhede)} } $$


$$ \hspace*{38 mm}\mathrm{=\kern1mm\frac{6}{36}\kern1mm=\kern1mm\frac{1}{6} } $$



[ V 11 ]
  
  
     
      11.9  Die gunstige uitkomstes is
  
               46, 55, 56, 64, 65, 66 en dus
  
               is daar 6 gunstige uitkomstes
  
               en daar is 36 moontlike uitkomstes.
  
               Dus, n(E) = n(som > 9) = 6
  
               en n(S) = n(moontlikhede) = 36
  
$$ \hspace*{15 mm}\mathrm{P(som\kern1mm>\kern1mm9)\kern1mm=\kern1mm\frac{n(som\kern1mm>\kern1mm9)}{n(moontlikhede)} } $$


$$ \hspace*{38 mm}\mathrm{=\kern1mm\frac{6}{36}\kern1mm=\kern1mm\frac{1}{6} } $$



[ V 11 ]
  
  
     
      11.10  Die gunstige uitkomstes is
  
                 11, 22, 33, 44, 55 en 66 en dus
  
                 is daar 6 gunstige uitkomstes
  
                 en daar is 36 moontlike uitkomstes.
  
                 Dus, n(E) = n(gelyk) = 6
  
                 en n(S) = n(moontlikhede) = 36
  
$$ \hspace*{17 mm}\mathrm{P(gelyk)\kern1mm=\kern1mm\frac{n(gelyk)}{n(moontlikhede)} } $$


$$ \hspace*{33 mm}\mathrm{=\kern1mm\frac{6}{36}\kern1mm=\kern1mm\frac{1}{6} } $$



[ V 11 ]
  
  
     
      11.11  Die gunstige uitkomstes is
  
                 46, 56, 66, 44, 45, 33, 34, 35, 22
  
                 23, 24, 11, 12, 13, 64, 65, 54, 43,
  
                 53, 32, 42, 21 en 31 en dus
  
                 is daar 23 gunstige uitkomstes
  
                 en daar is 36 moontlike uitkomstes.
  
                 Dus, n(E) = n(verskil) = 23
  
                 en n(S) = n(moontlikhede) = 36
  
$$ \hspace*{17 mm}\mathrm{P(verskil)\kern1mm=\kern1mm\frac{n(verskil)}{n(moontlikhede)} } $$


$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{23}{36} } $$



[ V 11 ]
  
  
     
      11.12  Die gunstige uitkomstes is
  
                 16, 25, 34, 43, 52 en 61 en dus
  
                 is daar 6 gunstige uitkomstes
  
                 en daar is 36 moontlike uitkomstes.
  
                 Dus, n(E) = n(som = 7) = 6
  
                 en n(S) = n(moontlikhede) = 36
  
$$ \hspace*{17 mm}\mathrm{P(som\kern1mm=\kern1mm7)\kern1mm=\kern1mm\frac{n(som\kern1mm=\kern1mm7)}{n(moontlikhede)} } $$


$$ \hspace*{40 mm}\mathrm{=\kern1mm\frac{6}{36}\kern1mm=\kern1mm\frac{1}{6} } $$



[ V 11 ]
  
  
     
      11.13  Daar is 36 gunstige uitkomstes
  
                 en daar is 36 moontlike uitkomstes.
  
                 Dus, n(E) = n(som < 13) = 36
  
                 en n(S) = n(moontlikhede) = 36
  
$$ \hspace*{17 mm}\mathrm{P(som\kern1mm<\kern1mm13)\kern1mm=\kern1mm\frac{n(som\kern1mm<\kern1mm13)}{n(moontlikhede)} } $$


$$ \hspace*{41 mm}\mathrm{=\kern1mm\frac{36}{36}\kern1mm=\kern1mm1 } $$



[ V 11 ]
  
  
     
      11.14  Daar geen gunstige uitkomstes nie
  
                 en daar is 36 moontlike uitkomstes.
  
                 Dus, n(E) = n(som > 15) = 0
  
                 en n(S) = n(moontlikhede) = 36
  
$$ \hspace*{17 mm}\mathrm{P(som\kern1mm>\kern1mm15)\kern1mm=\kern1mm\frac{n(som\kern1mm>\kern1mm15)}{n(moontlikhede)} } $$


$$ \hspace*{41 mm}\mathrm{=\kern1mm\frac{0}{36}\kern1mm=\kern1mm0 } $$



[ V 11 ]
  
  
     
      11.15  Die gunstige uitkomstes is
  
                 45 en 54 en dus
  
                 is daar 2 gunstige uitkomstes
  
                 en daar is 36 moontlike uitkomstes.
  
                 Dus, n(E) = n(aXb = 20) = 2
  
                 en n(S) = n(moontlikhede) = 36
  
$$ \hspace*{17 mm}\mathrm{P(a\ x\ b\ =\ 20)\kern1mm=\kern1mm\frac{n(a\ x\ b\ =\ 20)}{n(moontlikhede)} } $$


$$ \hspace*{42 mm}\mathrm{=\kern1mm\frac{2}{36}\kern1mm=\kern1mm\frac{1}{18} } $$



[ V 11 ]
  
  
     
      11.16  Die gunstige uitkomstes is 45, 46,
  
                 54, 55, 56, 64, 65 en 66 en dus
  
                 is daar 8 gunstige uitkomstes
  
                 en daar is 36 moontlike uitkomstes.
  
                 Dus, n(E) = n(aXb > 18) = 8
  
                 en n(S) = n(moontlikhede) = 36
  
$$ \hspace*{17 mm}\mathrm{P(a\ x\ b\ >\ 18)\kern1mm=\kern1mm\frac{n(a\ x\ b\ >\ 18)}{n(moontlikhede)} } $$


$$ \hspace*{42 mm}\mathrm{=\kern1mm\frac{8}{36}\kern1mm=\kern1mm\frac{2}{9} } $$



[ V 11 ]
  
  
     
      11.2  There is only one favourable outcome
  
               for two fours and there are
  
               36 possible outcomes
  
               Thus, n(E) = n(two fours) = 1 and
  
               n(S) = n(possible) = 36
  
$$ \hspace*{18 mm}\mathrm{P(two\ fours)\kern1mm=\kern1mm\frac{n(two\ fours)}{n(possibilities)} } $$


$$ \hspace*{40 mm}\mathrm{=\kern1mm\frac{1}{36} } $$



[ Question 11 ]
  
  
     
      11.3  The favourable outcomes are
  
               34 and 43 and there are
  
               36 possible outcomes
  
               Thus, n(E) = n(3 and 4) = 2 and
  
               n(S) = n(possible) = 36
  
$$ \hspace*{18 mm}\mathrm{P(3\ and\ 4)\kern1mm=\kern1mm\frac{n(3\ and\ 4)}{n(possibilities)} } $$


$$ \hspace*{38 mm}\mathrm{=\kern1mm\frac{2}{36}\kern1mm=\kern1mm\frac{1}{18} } $$



[ Q 11 ]
  
     
      11.4  There is only one favourable outcome,
  
               viz. 34, because order is important,
  
               and there are 36 possible outcomes
  
               Thus, n(E) = n(3 and 4) = 1 and
  
               n(S) = n(possible) = 36
  
$$ \hspace*{18 mm}\mathrm{P(3\ and\ 4)\kern1mm=\kern1mm\frac{n(3\ and\ 4)}{n(possibilities)} } $$


$$ \hspace*{38 mm}\mathrm{=\kern1mm\frac{1}{36} } $$



[ Q 11 ]
  
  
     
      11.5  The uneven numbers are 1, 3 and 5
  
               and therefore there are 9 favourable
  
               outcomes, viz. 11, 13, 15, 31, 33, 35,
  
               51, 53 and 55 and
  
               there are 36 possible outcomes
  
               Thus, n(E) = n(uneven) = 9 and
  
               n(S) = n(possible) = 36
  
$$ \hspace*{18 mm}\mathrm{P(uneven)\kern1mm=\kern1mm\frac{n(uneven)}{n(possibilities)} } $$


$$ \hspace*{38 mm}\mathrm{=\kern1mm\frac{9}{36}\kern1mm=\kern1mm\frac{1}{4} } $$



[ Q 11 ]
  
  
     
      11.6  The numbers < 3 are 1 and 2
  
               and therefore there are 4 favourable
  
               outcomes, viz. 11, 12, 21 and 22
  
               and there are 36 possible outcomes
  
               Thus, n(E) = n(< 3) = 9 and
  
               n(S) = n(possible) = 36
  
$$ \hspace*{18 mm}\mathrm{P(<\ 3)\kern1mm=\kern1mm\frac{n(<\ 3)}{n(possibilities)} } $$


$$ \hspace*{32 mm}\mathrm{=\kern1mm\frac{4}{36}\kern1mm=\kern1mm\frac{1}{9} } $$



[ Q 11 ]
  
  
     
      11.7  The multiples of 3 are 3 and 6
  
               and therefore there are 4 favourable
  
               outcomes, viz. 33, 36, 63 and 66
  
               and there are 36 possible outcomes
  
               Thus, n(E) = n(mult. 3) = 9 and
  
               n(S) = n(possible) = 36
  
$$ \hspace*{18 mm}\mathrm{P(multiples\kern1mm3)\kern1mm=\kern1mm\frac{n(multiples\ 3)}{n(possibilities)} } $$


$$ \hspace*{43 mm}\mathrm{=\kern1mm\frac{4}{36}\kern1mm=\kern1mm\frac{1}{9} } $$



[ Q 11 ]
  
  
     
      11.8  The favourable outcomes are
  
               16, 25, 34, 43, 52, 61 thus
  
               there are 6 favourable outcomes and
  
               there are 36 possible outcomes
  
               Thus, n(E) = n(sum = 7) = 9 and
  
               n(S) = n(possible) = 36
  
$$ \hspace*{18 mm}\mathrm{P(sum\ =\kern1mm3)\kern1mm=\kern1mm\frac{n(sum\kern1mm=\kern1mm3)}{n(possibilities)} } $$


$$ \hspace*{41 mm}\mathrm{=\kern1mm\frac{6}{36}\kern1mm=\kern1mm\frac{1}{6} } $$



[ Q 11 ]
  
  
     
      11.9  The favourable outcomes are
  
               46, 55, 56, 64, 65 and 66 and thus
  
               there are 6 favourable outcomes and
  
               there are 36 possible outcomes
  
               Thus, n(E) = n(sum > 9) = 6 and
  
               n(S) = n(possible) = 36
  
$$ \hspace*{18 mm}\mathrm{P(sum\ >\ 9)\kern1mm=\kern1mm\frac{n(sum\kern1mm>\kern1mm9)}{n(possibilities)} } $$


$$ \hspace*{41 mm}\mathrm{=\kern1mm\frac{6}{36}\kern1mm=\kern1mm\frac{1}{6} } $$



[ Q 11 ]
  
  
     
      11.10  The favourable outcomes are
  
                 11, 22, 33, 44, 55 and 66 and thus
  
                 there are 6 favourable outcomes and
  
                 there are 36 possible outcomes
  
                 Thus, n(E) = n(equal) = 6 and
  
                 n(S) = n(possible) = 36
  
$$ \hspace*{18 mm}\mathrm{P(equal)\kern1mm=\kern1mm\frac{n(equal)}{n(possibilities)} } $$


$$ \hspace*{34 mm}\mathrm{=\kern1mm\frac{6}{36}\kern1mm=\kern1mm\frac{1}{6} } $$



[ Q 11 ]
  
  
     
      11.11  The favourable outcomes are
  
                 46, 56, 66, 44, 45, 33, 34, 35, 22
  
                 23, 24, 11, 12, 13, 64, 65, 54, 43,
  
                 53, 32, 42, 21 and 31 and thus
  
                 there are 23 favourable outcomes
  
                 and 36 possible outcomes
  
                 Thus, n(E) = n(differ) = 23 and
  
                 n(S) = n(possible) = 36
  
$$ \hspace*{18 mm}\mathrm{P(difference)\kern1mm=\kern1mm\frac{n(difference)}{n(possibilities)} } $$


$$ \hspace*{42 mm}\mathrm{=\kern1mm\frac{23}{36} } $$



[ Q 11 ]
  
  
     
      11.12  The favourable outcomes are
  
                 16, 25, 34, 43, 52 and 61 and thus
  
                 there are 6 favourable outcomes and
  
                 there are 36 possible outcomes
  
                 Thus, n(E) = n(sum = 7) = 6 and
  
                 n(S) = n(possible) = 36
  
$$ \hspace*{18 mm}\mathrm{P(sum\kern1mm=\kern1mm7)\kern1mm=\kern1mm\frac{n(sum\kern1mm=\kern1mm7)}{n(possibilities)} } $$


$$ \hspace*{41 mm}\mathrm{=\kern1mm\frac{6}{36}\kern1mm=\kern1mm\frac{1}{6} } $$



[ Q 11 ]
  
  
     
      11.13  There are 36 favourable outcomes
  
                 and 36 possible outcomes
  
                 Thus, n(E) = n(sum < 13) = 36 and
  
                 n(S) = n(possible) = 36
  
$$ \hspace*{18 mm}\mathrm{P(sum\ <\ 13)\kern1mm=\kern1mm\frac{n(sum\ <\ 13)}{n(possibilities)} } $$


$$ \hspace*{43 mm}\mathrm{=\kern1mm\frac{36}{36}\kern1mm=\kern1mm1 } $$


[ Q 11 ]
  
  

     
      11.14  There are no favourable outcomes
  
                 and there are 36 possible outcomes
  
                 Thus, n(E) = n(sum > 15) = 0 and
  
                 n(S) = n(possible) = 36
  
$$ \hspace*{18 mm}\mathrm{P(sum\ >\ 15)\kern1mm=\kern1mm\frac{n(sum\kern1mm>\kern1mm15)}{n(possibilities)} } $$


$$ \hspace*{43 mm}\mathrm{=\kern1mm\frac{0}{36}\kern1mm=\kern1mm0 } $$



[ Q 11 ]
  
  
     
      11.15  The favourable outcomes are
  
                 45 and 54 and thus
  
                 there are 2 favourable outcomes and
  
                 there are 36 possible outcomes
  
                 Thus, n(E) = n(aXb =20) = 2 and
  
                 n(S) = n(possible) = 36
  
$$ \hspace*{18 mm}\mathrm{P(a\kern1mmx\kern1mmb\kern1mm=\kern1mm20)\kern1mm=\kern1mm\frac{n(a\kern1mmx\kern1mmb\kern1mm=\kern1mm20)}{n(possibilities)} } $$


$$ \hspace*{43 mm}\mathrm{=\kern1mm\frac{2}{36}\kern1mm=\kern1mm\frac{1}{18} } $$



[ Q 11 ]
  
  
     
      11.16  The favourable outcomes are 45, 46,
  
                 54, 55, 56, 64, 65 and 66 and thus
  
                 there are 8 favourable outcomes and
  
                 there are 36 possible outcomes
  
                 Thus, n(E) = n(aXb > 18) = 8 and
  
                 n(S) = n(possible) = 36
  
$$ \hspace*{18 mm}\mathrm{P(a\ x\kern1mmb\kern1mm>\ 18)\kern1mm=\kern1mm\frac{n(a\kern1mmx\ b\kern1mm>\ 18)}{n(possibilities)} } $$


$$ \hspace*{43 mm}\mathrm{=\kern1mm\frac{8}{36}\kern1mm=\kern1mm\frac{2}{9} } $$



[ Q 11 ]
  
  
  
  
Antwoorde / Answers  12
  
      12.1  a = WV ;  b = GV ;  c = VG
  
[ Vraag 12 ]
  
  
       
$$ \hspace*{5 mm}\mathrm{12.2\kern1mmP(verloor\ eerste\ wedstryd)\kern1mm=\kern1mm\frac{1}{3} } $$



[ V 12 ]
  
  
       
     12.3   P(wen 2de)  :  Daar is drie uitkomstes,
  
               nl. WW, GW en VW
$$ \hspace*{15 mm}\mathrm{P(wen\ 2de\ wedstryd)\kern1mm=\kern1mm\frac{1}{3} } $$



[ V 12 ]
  
  
      $$ \hspace*{5 mm}\mathrm{12.4\kern1mmP(wen\ albei)\kern1mm=\kern1mm\frac{2}{9} } $$


[ V 12 ]
  
  
       
     12.5   Daar is 5 uitkomstes,
  
               nl. WV, GV, VW, VG en VV
  
$$ \hspace*{15 mm}\mathrm{P(verloor\ ten\ minste\ een)\kern1mm=\kern1mm\frac{5}{9} } $$



[ V 12 ]
  
  
       
     12.6   Daar is 5 uitkomstes,
  
               nl. WG, GW, GG, GV en VG
  
$$ \hspace*{15 mm}\mathrm{P(ten\ minste\ een\ gelykop)\kern1mm=\kern1mm\frac{5}{9} } $$



[ V 12 ]
  
  
       
     12.7   Daar is 4 uitkomstes,
  
               nl. WW, WG, GW en GG
  
$$ \hspace*{15 mm}\mathrm{P(verloor\ nie\ een\ nie)\kern1mm=\kern1mm\frac{4}{9} } $$



[ V 12 ]
  
  
      12.1  a = WV ;  b = GV ;  c = VG
  
[ Question 12 ]
  
  
       
$$ \hspace*{5 mm}\mathrm{12.2\kern1mmP(lose\ first\ match)\kern1mm=\kern1mm\frac{1}{3} } $$


[ Q 12 ]
  
  

       
     12.3   P(w1n 2nd)  :  There are three
  
               outcomes, viz. WW, DW en LW
  
$$ \hspace*{15 mm}\mathrm{P(win\ 2nd)\kern1mm=\kern1mm\frac{1}{3} } $$


[ Q 12 ]
  
       
$$ \hspace*{5 mm}\mathrm{12.4\kern1mmP(win\ both\ matches)\kern1mm=\kern1mm\frac{2}{9} } $$


[ Q 12 ]
  
  

       
     12.5   There are 5 outcomes,
  
               nl. WL, DL, LW, LD and LL
  
$$ \hspace*{15 mm}\mathrm{P(lose\ at\ least\ one)\kern1mm=\kern1mm\frac{5}{9} } $$



[ Q 12 ]
  
  
       
     12.6   There are 5 outcomes,
  
               nl. WD, DW, DD, DL and LD
  
$$ \hspace*{15 mm}\mathrm{P(at\ least\ one\ draw)\kern1mm=\kern1mm\frac{5}{9} } $$



[ Q 12 ]
  
  
       
     12.7   There are 4 outcomes,
  
               viz. WW, WD, DW and DD
  
$$ \hspace*{15 mm}\mathrm{P(lose\ not\ one)\kern1mm=\kern1mm\frac{4}{9} } $$



[ Q 12 ]
  
  
  
  
Antwoorde  Answers  13
  
     13.1  a = SDS ;  b = DSS ;  c = DDS
  
[ Vraag 13 ]
  
  
       
$$ \hspace*{5 mm}\mathrm{13.2\kern1mmP(1ste\ kind\ 'n\ seun)\kern1mm=\kern1mm\frac{1}{2}} $$



[ V 13 ]
  
  
       
     13.3   Daar is 3 moontlike uitkomste, nl.
  
               SDD, DSD and DDS.
  
$$ \hspace*{15 mm}\mathrm{P(2\ dogters)\kern1mm=\kern1mm\frac{3}{8} } $$




[ V 13 ]
  
  
       
     13.4   Daar is 3 moontlike uitkomstes, nl.
  
               SDD, DSD en DDS.
  
$$ \hspace*{15 mm}\mathrm{P(net\ een\ seun)\kern1mm=\kern1mm\frac{3}{8} } $$




[ V 13 ]
  
  
       
     13.5   Daar is net 1 moontlike uitkoms, nl.
  
               SSS.
  
$$ \hspace*{15 mm}\mathrm{P(geen\ dogters)\kern1mm=\kern1mm\frac{1}{8} } $$




[ V 13 ]
  
  
  
     13.1  a = SDS ;  b = DSS ;  c = DDS
  
[ Question 13 ]
  
  
       
$$ \hspace*{5 mm}\mathrm{13.2\kern1mmP(first\ child\ a\ boy)\kern1mm=\kern1mm\frac{1}{2} } $$



[ Q 13 ]
  
  
       
     13.3   There are 3 possible outcomes, viz.
  
               BGG, GBG and GGB.
  
$$ \hspace*{15 mm}\mathrm{P(2\ daughters)\kern1mm=\kern1mm\frac{3}{8} } $$




[ Q 13 ]
  
  
       
     13.4   There are 3 possible outcomes, viz.
  
               BGG, GBG and GGB.
  
$$ \hspace*{15 mm}\mathrm{P(only\ one\ boy)\kern1mm=\kern1mm\frac{3}{8} } $$




[ Q 13 ]
  
  
       
     13.5   There is only 1 possible outcome, viz.
  
               BBB.
  
$$ \hspace*{15 mm}\mathrm{P(no\ daughters)\kern1mm=\kern1mm\frac{1}{8} } $$




[ Q 13 ]
  
  
  
  
  
Antwoorde / Answers  14
  
     14.1  a = 2M/T ;  b = 3K/H ;
  
               c = 4K/H ;  d = 4M/T
  
[ Vraag 14 ]
  
  
      14.2   Daar is net een moontlikheid, nl. 1K
  
$$ \hspace*{15 mm}\mathrm{P(1\ en\ K)\kern1mm=\kern1mm\frac{1}{12} } $$



[ V 14 ]
  
  


      14.3   Daar is drie moontlikhede, nl.
  
                2K/H, 4K/H en 6K/H
  
$$ \hspace*{15 mm}\mathrm{P(ewe\ getal\ en\ K)\kern1mm=\kern1mm\frac{3}{12}\kern1mm=\kern1mm\frac{1}{4} } $$



[ V 14 ]
  
  
      14.4   Daar is drie moontlikhede, nl.
  
                4M/T, 5M/T en 6M/T
  
$$ \hspace*{15 mm}\mathrm{P(getal\ >\ 3\ en\ M)\kern1mm=\kern1mm\frac{3}{12}  =  \frac{1}{4} } $$



[ V 14 ]
  
  
      14.5   Daar is twee moontlikhede, nl.
  
                1M/T en 3M/T
  
$$ \hspace*{15 mm}\mathrm{P(onewe\ getal\ <\ 5\ en\ M)\kern1mm=\kern1mm\frac{2}{12} } $$


$$ \hspace*{59 mm}\mathrm{=\kern1mm\frac{1}{6} } $$



[ V 14 ]
  
  
      14.6   Daar is ses moontlikhede, nl.
  
                1M/T, 2M/T, 3M/T, 4M/T, 5M/T
  
                en 6M/T
  
$$ \hspace*{15 mm}\mathrm{P(enige\ getal\ en\ ;M)\kern1mm=\kern1mm\frac{6}{12} } $$


$$ \hspace*{49 mm}\mathrm{=\kern1mm\frac{1}{2} } $$



[ V 14 ]
  
  

     14,7   Daar is twee moontlikhede, nl.
  
                3K/H en 3M/T
  
$$ \hspace*{15 mm}\mathrm{P(3\ en\ K\ of\ M)\kern1mm=\kern1mm\frac{2}{12}  =  \frac{1}{6} } $$



[ V 14 ]
  
  
     14.8   Daar is geen moontlikheid nie.
  
$$ \hspace*{15 mm}\mathrm{P(2\ en\ nie\ K\ of\ M)\kern1mm=\kern1mm\frac{0}{12}\kern1mm=\kern1mm0 } $$



[ V 14 ]
  
  
     14.1  a = 2M/T ;  b = 3K/H ;
  
               c = 4K/H ;  d = 4M/T
  
[ Question 14 ]
  
  
     14.2   There is only one favourable
  
                outcome, viz. 1H
  
$$ \hspace*{16 mm}\mathrm{P(1\ and\ H)\kern1mm=\kern1mm\frac{1}{12} } $$



[ Q 14 ]
  
  
     14.3   There are three favourable outcomes,
  
                viz. 2K/H, 4K/H and 6K/H
  
$$ \hspace*{16 mm}\mathrm{P(even\ number\ and\ H)\kern1mm=\kern1mm\frac{3}{12}\kern1mm=\kern1mm\frac{1}{4} } $$



[ Q 14 ]
  
  
     14.4   There are three favourable outcomes,
  
                4M/T, 5M/T en 6M/T
  
$$ \hspace*{16 mm}\mathrm{P(number\ >\ 3\ and\ T)\kern1mm=\kern1mm\frac{3}{12}  =  \frac{1}{4} } $$



[ Q 14 ]
  
  
     14.5   There are two favourable outcomes,
  
                1viz. M/T and 3M/T
  
$$ \hspace*{16 mm}\mathrm{P(uneven\ number\ <\ 5\ and\ T)\kern1mm=\kern1mm\frac{2}{12} } $$


$$ \hspace*{67 mm}\mathrm{=\kern1mm\frac{1}{6} } $$



[ Q 14 ]
  
  
     14.6   There are six favourable outcomes,
  
                1M/T, 2M/T, 3M/T, 4M/T, 5M/T
  
                and 6M/T
  
$$ \hspace*{16 mm}\mathrm{P(any\ number\ and\ M)\kern1mm=\kern1mm\frac{6}{12} } $$



$$ \hspace*{54 mm}\mathrm{=\kern1mm\frac{1}{2} } $$



[ Q 14 ]
  
  
     14.7   There are two favourable outcomes,
  
                3K/H and 3M/T
  
$$ \hspace*{16 mm}\mathrm{P(3\ and\ H\ or\ T)\kern1mm=\kern1mm\frac{2}{12}\kern1mm=\kern1mm\frac{1}{6} } $$



[ Q 14 ]
  
  
     14.8   There are no favourable outcomes,
  
$$ \hspace*{16 mm}\mathrm{P(2\ and\ neither\ H\ or T)\kern1mm=\kern1mm\frac{0}{12}\kern1mm=\kern1mm0 } $$



[ Q 14 ]