WISKUNDIGE GELETTERDHEID
GRAAD 12
NOG OEFENINGE
Waarskynlikheid : antwoorde.
MATHEMATICAL LITERACY
GRADE 12
MORE EXERCISES
Probability : answers
1.1 Daar is 14 pynappel gegeurde lekkers
in die blik, dus n(E) = n(pynappel) = 14
Daar is 14 + 8 + 2 + 6 = 40 lekkers
in die blik.
Dus, n(S) = n(aantal lekkers) = 40
n(E)
P(pynappel) = ────
n(S)
n(pynappel)
= ────────────
n(aantal lekkers)
14 7
= ─── = ───
40 20
[ V 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
n(appelkoos)
P(appelkoos) = ─────────
n(lekkers)
6 3
= ─── = ───
40 20
[ V 1 ]
1.3 n(koejawel) = 12
n(aantal lekkers) = 40
n(koejawel)
P(appelkoos) = ─────────
n(lekkers)
12 3
= ─── = ───
40 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
Dus n(nie koejawel geur) = 28 en
n(aantal lekkers) = 40
n(nie koejawel)
P(nie koejawel) = ───────────
n(lekkers)
28 7
= ─── = ───
40 10
OF
NB P(koejawel) + P(nie koejawel) = 1
3 7
─── +
─── = 1
10 10
Dus P(nie koejawel) = 1 ─ P(koejawel)
3 7
P(nie koejawel) = 1 ─ ─── = ───
10 10
[ V 1 ]
1.5 n(nie pynappel) = 8 + 12 + 6 = 26
n(lekkers) = 40
n(nie pynappel)
P(nie pynappel) = ───────────
n(lekkers)
26 13
= ─── = ───
40 20
OF
NB P(pynappel) + P(nie pynappel) = 1
7 13
─── +
─── = 1
20 20
Dus P(nie pynappel) = 1 ─ P(pynappel)
7 13
P(nie pynappel) = 1 ─ ─── = ───
20 20
[ V 1 ]
1.6 n(nie kersie) = 40 - 8 = 32
n(lekkers) = 40
n(nie kersie)
P(nie kersie) = ───────────
n(lekkers)
32 4
= ─── = ───
40 5
OF
n(kersie) = 8 en n(lekkers) = 40
8 1
P(kersie) = ─── = ──
40 5
Dus P(nie kersie) = 1 ─ P(kersie)
1 4
P(nie kersie) = 1 ─ ─── = ───
5 5
[ V 1 ]
1.7 n(lemoen) = 0 en n(lekkers) = 40
n(lemoen)
P(lemoen) = ────────
n(lekkers)
0
= ─── = 0
40
[ V 1 ]
1.8 n(nie lemoen) = 40 en n(lekkers) = 40
n(nie lemoen)
P(nie lemoen) = ──────────
n(lekkers)
40
= ─── = 1
40
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
in the tin.
Thus n(S) = n(number of sweets) = 40
n(E)
P(pineapple) = ────
n(S)
n(pineapple)
= ──────────────
n(number of sweets)
14 7
= ───
= ───
40 20
[ Q 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
in the tin.
Thus n(S) = n(number of sweets) = 40
n(apricot) 6
P(apricot) = ──────── = ────
n(sweets) 40
3
= ───
20
[ Q. 1 ]
1.3 n(guava) = 12
n(number of sweets) = 40
n(guava) 12
P(guava) = ──────── = ────
n(sweets) 40
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
Thus n(not guava flavoured) = 28
and n(sweets) = 40
n(not guava)
P(not guava) = ─────────
n(sweets)
28 7
= ─── = ───
40 10
OR
NB P(guava) + P(not guava) = 1
3 7
─── +
─── = 1
10 10
Thus P(not guava) = 1 ─ P(guava)
3 7
P(not guava) = 1 ─ ─── = ───
10 10
[ Q. 1 ]
1.5 n(not pineapple) = 8 + 12 + 6 = 26
n(sweets) = 40
n(not pimeapple)
P(not pineapple) = ───────────
n(sweets)
26 13
= ─── = ───
40 20
OR
NB
P(pineapple) + P(not pineapple) = 1
7 13
─── +
─── = 1
20 20
∴ P(not pineapple) = 1 ─ P(pineapple)
7 13
P(not pineapple) = 1 ─ ─── = ───
20 20
[ Q. 1 ]
1.6 n(not cherry) = 40 - 8 = 32
n(sweets) = 40
n(not cherry)
P(not cherry) = ───────────
n(sweets)
32 4
= ─── = ───
40 5
OR
n(cherry) = 8 and n(sweets) = 40
8 1
P(cherry) = ─── = ───
40 5
P(not cherry) = 1 ─ P(cherry)
1 4
P(not cherry) = 1 ─ ─── = ───
5 5
[ Q. 1 ]
1.7 n(orange) = 0 and n(sweets) = 40
n(orange) 0
P(orange) = ──────── = ───
n(sweets) 40
= 0
[ Q. 1 ]
1.8 n(not orange) = 40 and n(sweets) = 40
n(not orange) 40
P(not orange) = ───────── = ───
n(sweets) 40
= 1
OR
P(not orange) = 1 ─ P(orange)
P(not orange) = 1 ─ 0
= 1
[ Q. 1 ]
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
n(E)
P(Q skoppens) = ────
n(S)
n(Q skoppens)
= ────────────
n(aantal kaarte)
1
= ───
52
[ V 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
n(E)
P(swart kaart) = ────
n(S)
n(swart kaart)
= ────────────
n(aantal kaarte)
26 1
= ─── = ──
52 2
[ V 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
n(E)
P(diamant) = ────
n(S)
n(diamant)
= ───────────
n(aantal kaarte)
13 1
= ─── = ──
52 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
n(E)
P(rooi prent) = ────
n(S)
n(rooi prent)
= ───────────
n(aantal kaarte)
8 2
= ─── = ──
52 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
n(E)
P(nie rooi prent) = ────
n(S)
n(nie rooi prent)
= ───────────
n(aantal kaarte)
44 11
= ─── = ──
52 13
OF
P(nie rooi prent) = 1 ─ P(rooi prent)
11 2
P(nie rooi prent) = 1 ─ ─── = ───
13 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 prentkaart) = 44
en daar is 52 kaarte,
n(S) = n(kaarte) = 52
n(E)
P(nie prent) = ────
n(S)
n(nie prent)
= ───────────
n(aantal kaarte)
36 9
= ─── = ──
52 13
OF
Daar is 8 + 8 = 16 prent kaarte
en 52 speelkaarte.
n(prent kaart)
P(prent kaart) = ───────────
n(aantal kaarte)
16 4
= ─── = ───
52 13
P(nie 'n prent) = 1 ─ P('n prent)
4 9
P(nie 'n prent) = 1 ─ ─── = ───
13 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
n(E)
P(swart genommerd) = ────
n(S)
n(swart genommerd)
= ──────────────
n(aantal kaarte)
18 9
= ─── = ──
52 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
n(E)
P(boer) = ────
n(S)
n(boer)
= ────────────
n(aantal kaarte)
4 1
= ─── = ──
52 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
n(E)
P(genommerd) = ────
n(S)
n(genommerd)
= ────────────
n(aantal kaarte)
18 9
= ─── = ──
52 26
[ 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
n(E)
P(rooi genommerd) = ────
n(S)
n(rooi genommerd)
= ─────────────
n(aantal kaarte)
10 5
= ─── = ──
52 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
n(E)
P(swart prent) = ────
n(S)
n(swart prent)
= ────────────
n(aantal kaarte)
8 2
= ─── = ──
52 13
[ V 2 ]
2.12 Daar is 2 kleure van rooi
genommerde kaarte. Hulle is
van 2 tot 10 genommer.
Dus is daar 3 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(kaarte) = 52
n(E)
P(nie rooi genommerd) = ────
n(S)
n(nie rooi genommerd)
= ────────────────
n(aantal kaarte)
34 17
= ─── = ──
52 26
[ V 2 ]
2.12 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
n(E)
P(nie aas) = ────
n(S)
n(nie aas)
= ────────────
n(aantal kaarte)
48 12
= ─── = ──
52 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
n(E)
P(Q spades) = ────
n(S)
n(Q spades)
= ──────────────
n(number of cards)
1
= ───
52
[ Q 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
n(E)
P(black card) = ────
n(S)
n(black card)
= ─────────────
n(number of cards)
26 1
= ─── = ───
52 2
[ Q 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
n(E)
P(diamond) = ────
n(S)
n(diamond)
= ──────────────
n(number of cards)
13 1
= ─── = ───
52 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
n(E)
P(red picture) = ────
n(S)
n(red picture)
= ─────────────
n(number of cards)
8 2
= ─── = ───
52 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
n(E)
P(not red picture) = ────
n(S)
n(not red picture)
= ─────────────
n(number of cards)
44 11
= ─── = ───
52 13
OR
P(not red picture) = 1 ─ P(red picture)
11 2
P(not red picture) = 1 ─ ─── = ───
13 13
[ Q 2 ]
2.6 There are 8 red and 8 black picture cards,
thus not a picture card = 52 - (2 x 8) = 36
n(E) = n(not a picture) = 36 and
there are 52 playing cards
Thus n(S) = n(number of cards) = 52
n(E)
P(not a picture) = ────
n(S)
n(not a picture)
= ─────────────
n(number of cards)
36 9
= ─── = ───
52 13
OR
There are 8 + 8 = 16 picture cards
and 52 playing cards.
n(a picture)
P(picture card) = ─────────────
n(number of cards)
16 4
= ─── = ───
52 13
P(not a picture) = 1 ─ P(a picture)
4 9
P(not a picture) = 1 ─ ─── = ───
13 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
n(E)
P(black numbered) = ────
n(S)
n(black numbered)
= ──────────────
n(number of cards)
18 9
= ─── = ───
52 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
n(E)
P(jack) = ────
n(S)
n(jack)
= ──────────────
n(number of cards)
4 1
= ─── = ───
52 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
n(E)
P(numbered) = ────
n(S)
n(numbered)
= ──────────────
n(number of cards)
18 9
= ─── = ───
52 26
[ 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
Thus n(S) = n(number of cards) = 52
n(E)
P(red numbered) = ────
n(S)
n(red numbered)
= ──────────────
n(number of cards)
10 5
= ─── = ───
52 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
Thus n(S) = n(number of cards) = 52
n(E)
P(black picture) = ────
n(S)
n(black picture)
= ─────────────
n(number of cards)
8 2
= ─── = ───
52 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
Thus n(S) = n(number of cards) = 52
n(E)
P(not red numbered) = ────
n(S)
n(not red numbered)
= ──────────────
n(number of cards)
34 17
= ─── = ───
52 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
Thus n(S) = n(number of cards) = 52
n(E)
P(non ace) = ────
n(S)
n(non ace)
= ─────────────
n(number of cards)
48 12
= ─── = ───
52 13
[ Q 2 ]
3.1 Daar is 40 leerlinge in die klas.
[ V 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
n(E)
P(dogters) = ────
n(S)
n(dogters)
= ─────────
n(leerlinge)
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
n(E)
P(S+huiswerk) = ────
n(S)
n(S+huiswerk)
= ──────────
n(leerlinge)
6 3
= ── = ──
40 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
n(E)
P(D+0huiswerk) = ────
n(S)
n(D+0uiswerk)
= ──────────
n(leerlinge)
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
n(E)
P(nie huiswerk) = ────
n(S)
n(nie huiswerk)
= ──────────
n(leerlinge)
18 9
= ── = ──
40 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) = 18 en
n(S) = n(leerlinge) = 40
n(E)
P(Graad 11) = ────
n(S)
n(Graad 11)
= ──────────
n(leerlinge)
18
= ── = 0
40
[ 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
n(E)
P(D+nie huiswerk) = ────
n(S)
n(D+nie huiswerk)
= ─────────────
n(leerlinge)
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
n(E)
P(S+huiswerk) = ────
n(S)
n(S+huiswerk)
= ───────────
n(leerlinge)
6
= ───
17
[ V 3 ]
3.1 There 40 pupils in the class.
[ Q 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
n(E)
P(girls) = ────
n(S)
n(girls)
= ────────
n(pupils)
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
n(E)
P(B+homework) = ────
n(S)
n(B+homework)
= ───────────
n(pupils)
6 3
= ─── = ───
40 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.
Thus n(E) = n(G+nohomework) = 7 and
n(S) = n(pupils) = 40
n(E)
P(G+nohomework) = ────
n(S)
n(G+nohomework)
= ─────────────
n(pupils)
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.
Thus n(E) = n(no homework) = 18 and
n(S) = n(pupils) = 40
n(E)
P(no homework) = ────
n(S)
n(no homework)
= ────────────
n(pupils)
18 9
= ─── = ───
40 20
[ Q 3 ]
3.2.5 There are no Grade 11 pupils in the class and
40 pupils in the class.
Thus n(E) = n(Grade 11) = 0 and
n(S) = n(pupils) = 40
n(E)
P(Grade 11) = ────
n(S)
n(Grade 11)
= ─────────
n(pupils)
0
= ─── = 0
40
[ Q 3 ]
3.3 There are 23 girls in the class and 7
have not done their homework.
Thus n(E) = n(G+nohomework) = 7 and
n(S) = n(girls) = 23
n(E)
P(G+nohomework) = ────
n(S)
n(G+nohomework)
= ─────────────
n(girls)
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
n(E)
P(B+homework) = ────
n(S)
n(B+homework)
= ───────────
n(pupils)
6
= ───
17
[ Q 3 ]
4.1 Daar is een D en 11 letters,
dus n(E) = n(D) = 1 en
n(S) = n(letters) = 11
n(E)
P(D) = ────
n(S)
n(D) 1
= ─────── = ──
n(letters) 11
[ V 4 ]
4.2 Daar is een K en 11 letters,
dus n(E) = n(K) = 1 en
n(S) = n(letters) = 11
n(E)
P(K) = ────
n(S)
n(K) 1
= ─────── = ──
n(letters) 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
n(E)
P(klinker) = ────
n(S)
n(klinker) 5
= ─────── = ──
n(letters) 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
n(E)
P(konsonant) = ────
n(S)
n(konsonant) 5
= ───────── = ──
n(letters) 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
n(E)
P(A) = ────
n(S)
n(A) 2
= ─────── = ──
n(letters) 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
n(E)
P(B) = ────
n(S)
n(B) 0
= ─────── = ── = 0
n(letters) 11
[ 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
n(E)
P(nie O) = ────
n(S)
n(nie O) 11
= ─────── = ── = 1
n(letters) 11
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
n(E) n(O)
P(O) = ──── = ──────
n(S) n(letters)
0
= ── = 0
11
P(nie O) = 1 ─ P(O)
= 1 ─ 0 = 1
[ 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
n(E)
P(nie F) = ────
n(S)
n(nie F) 10
= ─────── = ──
n(letters) 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
n(E) n(F) 1
P(F) = ──── = ────── = ──
n(S) n(letters) 11
P(nie F) = 1 ─ P(F)
1 10
= 1 ─ ── = ──
11 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
n(E)
P(nie E) = ────
n(S)
n(nie E) 11
= ─────── = ── = 1
n(letters) 11
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
n(E) n(E)
P(E) = ──── = ──────
n(S) n(letters)
0
= ── = 0
11
P(nie E) = 1 ─ P(E)
= 1 ─ 0 = 1
[ 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
n(E)
P(nie I) = ────
n(S)
n(nie I) 9
= ─────── = ──
n(letters) 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
n(E) n(I) 2
P(I) = ──── = ────── = ──
n(S) n(letters) 11
P(nie I) = 1 ─ P(I)
2 9
= 1 ─ ── = ──
11 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
n(E)
P(nie klinker) = ────
n(S)
n(nie klinker) 9
= ────────── = ──
n(letters) 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
n(E) n(I) 2
P(klinker) = ──── = ────── = ──
n(S) n(letters) 11
5
= ────
11
P(nie klinker) = 1 ─ P(klinker)
5 6
= 1 ─ ── = ──
11 11
[ V 4 ]
4.1 There is one D and 11 letters
Thus n(E) = n(D) = 1 and
n(S) = n(letters) = 11
n(E)
P(D) = ────
n(S)
n(D) 1
= ──────── = ──
n(letters) 11
[ Q 4 ]
4.2 There is one K and 11 letters
Thus n(E) = n(K) = 1 and
n(S) = n(letters) = 11
n(E)
P(K) = ────
n(S)
n(K) 1
= ──────── = ──
n(letters) 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
n(E)
P(vowel) = ────
n(S)
n(vowel) 5
= ─────── = ──
n(letters) 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
n(E)
P(consonant) = ────
n(S)
n(consonant) 5
= ────────── = ──
n(letters) 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
n(E)
P(A) = ────
n(S)
n(A) 2
= ──────── = ──
n(letters) 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
n(E)
P(B) = ────
n(S)
n(B) 0
= ─────── = ── = 0
n(letters) 11
[ 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
n(E)
P(not O) = ────
n(S)
n(not O) 11
= ─────── = ── = 1
n(letters) 11
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
n(E)
n(O)
P(O) = ──── = ───────
n(S) n(letters)
0
= ─── = 0
11
P(not O) = 1 ─ P(O)
= 1 ─ 0 = 1
[ Q 4 ]
4.8 There are 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
n(E)
P(not F) = ────
n(S)
n(not F) 10
= ─────── = ──
n(letters) 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
n(E) n(F) 1
P(F) = ──── = ─────── = ──
n(S) n(letters) 11
P(not F) = 1 ─ P(F)
1 10
= 1 ─ ─── = ──
11 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
n(E)
P(not E) = ────
n(S)
n(not E) 11
= ─────── = ── = 1
n(letters) 11
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
n(E)
n(E)
P(E) = ──── = ───────
n(S) n(letters)
0
= ─── = 0
11
P(not E) = 1 ─ P(E)
= 1 ─ 0 = 1
[ 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
n(E)
P(not I) = ────
n(S)
n(not I) 9
= ─────── = ──
n(letters) 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
n(E)
n(I)
P(I) = ──── = ───────
n(S) n(letters)
2
= ───
11
P(not I) = 1 ─ P(I)
2 9
= 1 ─ ─── = ───
11 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
n(E)
P(not vowel) = ────
n(S)
n(not vowel) 6
= ───────── = ──
n(letters) 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
n(E)
n(vowel)
P(vowel) = ──── = ───────
n(S) n(letters)
5
= ───
11
P(not vowel) = 1 ─ P(vowel)
5 6
= 1 ─ ─── = ───
11 11
[ Q 4 ]
5.1 Daar is een 5 en 10 balle
dus n(E) = n(5) = 1 en
n(S) = n(balle) = 10
n(E)
P(5) = ────
n(S)
n(5) 1
= ─────── = ───
n(balle) 10
[ V 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
n(E)
P(5) = ────
n(S)
n(ewe) 5 1
= ────── = ─── = ───
n(balle) 10 2
[ V 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
n(E)
P(5) = ────
n(S)
n(ewe) 5 1
= ────── = ─── = ───
n(balle) 10 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
n(E)
n(> 3)
P(> 3) = ──── = ──────
n(S)
n(balle)
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
n(E)
P(veelvoud) = ────
n(S)
n(veelvoud) 3
= ───────── = ───
n(balle) 10
[ V 5 ]
5.6 Daar is 4 getalle gelyk aan of
groter as 7, 7,8 9 en 10, en 10 balle.
dus n(E) = n(>= 7) = 4 en
n(S) = n(balle) = 10
n(E)
n(>= 7)
P(>= 7) = ──── = ──────
n(S)
n(balle)
4
2
= ─── = ───
10
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
n(E)
n(/ 4)
P(/ 4) = ──── = ──────
n(S)
n(balle)
2
1
= ─── = ───
10
5
[ V 5 ]
5.8 Daar is een 2 en 9 balle
dus n(E) = n(2) = 1 en
n(S) = n(balle) = 9
n(E)
n(2)
P(2) = ──── = ──────
n(S)
n(balle)
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
n(E)
n(> 3)
P(> 3) = ──── = ──────
n(S)
n(balle)
6
2
= ─── = ───
9
3
[ V 5 ]
5.1 There is one 5 and 10 balls
Thus n(E) = n(5) = 1 and
n(S) = n(balls) = 10
n(E)
P(5) = ────
n(S)
n(5) 1
= ────── = ───
n(balls) 10
[ Q 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
n(E)
P(even) = ────
n(S)
n(even) 5 1
= ────── = ─── = ───
n(balls) 10 2
[ Q 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
n(E) n(< 6)
P(< 6) = ──── = ──────
n(S) n(balls)
5 1
= ─── = ───
10 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
n(E) n(> 3)
P(> 3) = ──── = ──────
n(S) n(balls)
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
n(E) n(mult3)
P(mult3) = ──── = ──────
n(S) n(balls)
3
= ───
10
[ Q 5 ]
5.6 There are 4 numbers equal to or
greater than 7, 7,8,9 and 10 and 10 balls
Thus n(E) = n(>=7) = 4 and
n(S) = n(balls) = 10
n(E) n(>=7)
P(>=7) = ──── = ──────
n(S) n(balls)
4 2
= ─── = ───
10 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
n(E) n(div4)
P(div4) = ──── = ──────
n(S) n(balls)
2 1
= ─── = ───
10 5
[ Q 5 ]
5.8 There is one 2 and 9 balls,
Thus n(E) = n(2) = 1 and
n(S) = n(balls) = 9
n(E) n(2)
P(2) = ──── = ──────
n(S) n(balls)
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
n(E) n(>3)
P(>3) = ──── = ──────
n(S) n(balls)
6 2
= ─── = ───
9 3
[ Q 5 ]
6.1 Rangskik die data :
2, 3, 3, 4, 5, 6, 8, 9, 12, 17, 21, 30
Σ x 120
Gemiddelde = ─── = ────
n 12
= 10
6 + 8
mediaan = ───── = 7
2
modus = 3 . . . kom 2 voor
[ V 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
n(E)
n(3)
P(3) = ──── = ───────
n(S)
n(kinders)
2 1
= ─── = ───
12 6
[ V 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
n(E)
n(6+)
P(6+) = ──── = ───────
n(S)
n(kinders)
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
n(E)
n(< 11)
P(< 11) = ──── = ───────
n(S)
n(kinders)
8 2
= ─── = ───
12 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
n(E)
n(< 3)
P(< 3) = ──── = ───────
n(S)
n(kinders)
10 5
= ─── = ───
12 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
n(E)
n(< 10)
P(< 10) = ──── = ───────
n(S)
n(kinders)
8 2
= ─── = ──
12 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
n(E)
P(> mediaan) = ────
n(S)
n(> mediaan)
= ─────────
n(kinders)
6 1
= ─── = ──
12 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
n(E)
P(= modus) = ────
n(S)
n(= modus)
= ─────────
n(kinders)
2 1
= ─── = ──
12 6
[ V 6 ]
6.1 Arrange the data :
2, 3, 3, 4, 5, 6, 8, 9, 12, 17, 21, 30
Σ x 120
Mean = ──── = ───
n 12
= 10
6 + 8
Median = ───── = 7
2
Mode = 3 . . . appears twice
[ Q 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
n(E) n(3)
P(3) = ──── = ────────
n(S) n(children)
2
1
= ─── = ───
12
6
[ Q 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
n(E) n(6+)
P(6+) = ──── = ────────
n(S) n(children)
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
n(E) n(< 11)
P(< 11) = ──── = ────────
n(S) n(children)
8
2
= ─── = ───
12
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
n(E) n(< 3)
P(< 3) = ──── = ────────
n(S) n(children)
10
5
= ─── = ───
12
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
n(E) n(< 10)
P(< 10) = ──── = ────────
n(S) n(children)
8
2
= ─── = ───
12
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
n(E)
P(> median) = ────
n(S)
n(> median)
P(> median) = ──────────
n(children)
6
1
= ─── = ───
12
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
n(E)
P(= mode) = ────
n(S)
n(= mode)
= ─────────
n(children)
2
1
= ─── = ───
12
6
[ Q 6 ]
7.1 P(slaag) = 1 ─ P(druip)
= 1 ─ 0,35
= 0,65
[ V 7 ]
n(leerlinge wat druip)
7.2 P(druip) = ──────────────────
n(leerlinge wat geskryf het)
P(druip) = 0,35 en leerlinge
wat geskryf het = 74.
n(druipelinge)
0,35 = ───────────
74
n(druipelinge) = 0,35 X 74
= 26
[ V 7 ]
n(leerlinge wat slaag)
7.3 P(slaag) = ──────────────────
n(leerlinge wat geskryf het)
P(slaag) = 0,65 en
n(leerlinge wat geskryf het) = 132
n(leerlinge wat slaag)
0,65 = ──────────────────
n(leerlinge wat geskryf het)
n(geslaag) = 0,65 X 132
= 85,8 = 86
[ V 7 ]
7.1 P(pass) = 1 ─ P(fail)
= 1 ─ 0,35
= 0,65
[ Q 7 ]
n(pupils that fail)
7.2 P(fail) = ─────────────
n(pupils that wrote)
P(fail) = 0,35 and pupils
that wrote = 74
n(failures)
0,35 = ─────────
74
n(failures) = 0,35 X 74
= 26
[ 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
n(passed) = 0,65 X 132
= 85,8 = 86
[ Q 7 ]
8.1 As die draaibord gebalanseer is,
d.w.s. elke kant het dieselfde
kans op daardie sy te val sal
1
P(3) = ─── = 0,25
4
[ 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 ]
n(2) 130
8.3.2 P(2) = ──── = ────
n(S) 500
= 0,26
[ 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
1
P(3) = ─── = 0,25
4
[ 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 ]
n(2) 130
8.3.2 P(2) = ──── = ────
n(S) 500
= 0,26
[ 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 ]
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.
n(seun)
P(seun) = ─────────
n(leerlinge)
18
= ─── = 36 %
50
[ V 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.
n(koop niks)
P(koop niks) = ─────────
n(leerlinge)
12
= ─── = 24 %
50
[ 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.
n(Smulstafie)
P(Smulstafie) = ─────────
n(leerlinge)
6
= ─── = 12 %
50
[ 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.
n(kooklekkers)
P(kooklekkers) = ─────────
n(leerlinge)
8
= ─── = 16 %
50
[ 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.
n(d+SS)
P(d+SS) = ─────────
n(dogters)
3
= ─── = 9,375 %
32
[ 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.
n(d+dv)
P(d+dv) = ─────────
n(dogters)
12
= ─── = 37,5 %
32
[ 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.
n(s+st)
P(s+st) = ────────
n(seuns)
2
= ─── = 11,111 %
18
[ 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.
n(s+niks)
P(s+niks) = ────────
n(seuns)
2
= ─── = 22,222 %
18
[ 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.
n(d+niks)
P(d+niks) = ────────
n(dogters)
8
= ─── = 25 %
32
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
dus P(koop) = 1 ─ P(koop niks)
= 1 ─ 24 %
= 76 %
76 X 650
76 % van 650 = ───────
100
= 494
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.
n(s+sjs) 4
P(s+sjs) = ──────── = ───
n(seuuns) 18
= 22,222 %
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
n(d+sms) 4
P(d+sms) = ──────── = ───
n(dogters) 32
= 12,5 %
Vyf seuns koop droë vrugte
en daar is 18 seuns.
Dus, n(s+dv) = 5 en n(seuns) = 18
n(s+dv) 5
P(s+dv) = ──────── = ───
n(seuns) 18
= 27,778 %
Dus, dit mer 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
n(boy)
P(boy) = ───────
n(pupils)
18
= ─── = 36 %
50
[ Q 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
n(buys nothing)
P(buys nothimg) = ────────────
n(pupils)
12
= ─── = 24 %
50
[ 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
n(Munch bar)
P(Munch bar) = ────────────
n(pupils)
6
= ─── = 12 %
50
[ 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
n(boiled sweets)
P(boiled sweets) = ───────────
n(pupils)
8
= ─── = 16 %
50
[ 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
n(g+CB)
P(g+CB) = ───────
n(girls)
3
= ─── = 9,375 %
32
[ 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
n(g+df)
P(g+df) = ───────
n(girls)
12
= ─── = 37,5 %
32
[ 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
n(b+MB)
P(b+MB) = ───────
n(boys)
2
= ─── = 11,111 %
18
[ 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
n(b+not)
P(b+not) = ───────
n(boys)
4
= ─── = 22,222 %
18
[ 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
n(g+nothing)
P(g+nothing) = ──────────
n(girls)
8
= ─── = 25 %
32
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
thus P(buys) = 1 ─ P(buys nothing)
= 1 ─ 24 %
= 76 %
76 X 650
76 % of 650 = ─────────
100
= 494
The probability that 494 and more
pupils will buy is good and
the statement si 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
n(b+CB) 4
P(b+CB) = ─────── = ───
n(boys) 18
= 22,222 %
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
n(g+MB) 4
P(g+MB) = ─────── = ───
n(girls) 32
= 12,5 %
Five boys buy dried fruit and
there are 18 boys.
Thus n(b+df) = 5 and n(boys) = 18
n(b+df) 5
P(b+df) = ─────── = ───
n(boys) 18
= 27,778 %
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 ]
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
n(S) = n(lekkers) = 5
n(rooi) 3
P(rooi) = ──────── = ───
n(lekkers) 5
[ V 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
n(groen) 2
P(groen) = ─────── = ──
n(lekkers) 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
n(rooi)
P(rooi) = ────────
n(lekkers)
2 1
= ─── = ───
4 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
n(groen) 2
P(groen) = ─────── = ──
n(lekkers) 5
2 1
= ─── = ──
4 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
n(rooi) 2
P(rooi) = ──────── = ───
n(lekkers) 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
n(groen) 1
P(groen) = ─────── = ──
n(lekkers) 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
n(red) 3
P(red) = ─────── = ───
n(sweets) 5
[ Q 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
n(green) 2
P(green) = ─────── = ───
n(sweets) 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
n(red)
P(red) = ───────
n(sweets)
2 1
= ─── = ───
4 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
n(green) 2
P(green) = ─────── = ───
n(sweets) 5
2 1
= ─── = ───
4 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
n(red) 2
P(red) = ─────── = ───
n(sweets) 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
n(green) 1
P(green) = ─────── = ───
n(sweets) 3
[ Q 10 ]
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 |
|
[ V / Q 11 ]
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
n(twee viere)
P(twee viere) = ───────────
n(moontlikhede)
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
n(3 en 4)
P(3 en 4) = ───────────
n(moontlikhede)
2 1
= ─── = ───
36 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
n(3 en 4)
P(3 en 4) = ───────────
n(moontlikhede)
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
n(onewe)
P(onewe) = ───────────
n(moontlikhede)
9 1
= ─── = ──
36 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
n(< 3)
P(< 3) = ───────────
n(moontlikhede)
4 1
= ─── = ──
36 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
n(veelv 3)
P(veelv 3) = ───────────
n(moontlikhede)
4 1
= ─── = ──
36 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
n(som = 7)
P(som = 7) = ───────────
n(moontlikhede)
6 1
= ─── = ──
36 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
n(som > 9)
P(som > 9) = ───────────
n(moontlikhede)
6 1
= ─── = ──
36 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
n(gelyk)
P(gelyk) = ───────────
n(moontlikhede)
6 1
= ─── = ──
36 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
n(verskil)
P(verskil) = ───────────
n(moontlikhede)
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
n(som = 7)
P(som = 7) = ───────────
n(moontlikhede)
6 1
= ─── = ──
36 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
n(som < 13)
P(som < 13) = ───────────
n(moontlikhede)
36
= ─── = 1
36
[ 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
n(som > 15)
P(som > 15) = ───────────
n(moontlikhede)
0
= ─── = 0
36
[ 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
n(aXb = 20)
P(aXb = 20) = ───────────
n(moontlikhede)
2 1
= ─── = ──
36 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
n(aXb > 18)
P(aXb > 18) = ───────────
n(moontlikhede)
8 4
= ─── = ──
36 18
[ V 11 ]
11.17 Die volkome vierkante is
1, 4, 9, 16, 25 en 36 sodat
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(vierkant) = 8
en n(S) = n(moontlikhede) = 36
n(vierkant)
P(vierkant) = ───────────
n(moontlikhede)
6 1
= ─── = ──
36 6
[ V 11 ]
11.18 Die veelvoude van 8 is
8, 16, 24 en 32 sodat
die gunstige uitkomstes is
24, 42, 44, 46 en 64 en dus
is daar 5 gunstige uitkomstes
en daar is 36 moontlike uitkomstes.
Dus, n(E) = n(veelv, 8) = 5
en n(S) = n(moontlikhede) = 36
n(veelv, 8)
P(veelv, 8) = ───────────
n(moontlikhede)
5
= ───
36
[ V 11 ]
11.19 Die natuurlike getalle is
1, 2, 3, 4, 5, ens. sodat
die gunstige uitkomstes is
11, 21, 31, 41, 51, 61, 22, 42,
62, 33, 63, 44, 55 en 66 en dus
is daar 14 gunstige uitkomstes
en daar is 36 moontlike uitkomstes.
Dus, n(E) = n(natuurlik) = 14
en n(S) = n(moontlikhede) = 36
n(natuurlik)
P(natuurlik) = ───────────
n(moontlikhede)
14 7
= ─── = ──
36 18
[ 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
n(two fours)
P(two fours) = ─────────
n(possible)
1
= ───
36
[ Q 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
n(3 and 4)
P(3 and 4) = ─────────
n(possible)
2 1
= ─── = ───
36 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
n(3 and 4)
P(3 and 4) = ─────────
n(possible)
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
n(uneven)
P(uneven) = ─────────
n(possible)
9 1
= ─── = ───
36 4
[ Q 11 ]
11.6 The numbers smaller than 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
n(< 3)
P(< 3) = ─────────
n(possible)
4 1
= ─── = ───
36 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
n(mult. 3)
P(mult. 3) = ─────────
n(possible)
4 1
= ─── = ───
36 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
n(sum = 7)
P(sum = 7) = ─────────
n(possible)
6 1
= ─── = ───
36 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
n(sum > 9)
P(sum > 9) = ─────────
n(possible)
6 1
= ─── = ───
36 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
n(equal)
P(equal) = ─────────
n(possible)
6 1
= ─── = ───
36 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
there are 36 possible outcomes
Thus, n(E) = n(differ) = 23 and
n(S) = n(possible) = 36
n(differ)
P(differ) = ─────────
n(possible)
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
n(sum = 7)
P(sum = 7) = ─────────
n(possible)
6 1
= ─── = ───
36 6
[ Q 11 ]
11.13 There are 36 favourable outcomes and
there are 36 possible outcomes
Thus, n(E) = n(sum < 13) = 36 and
n(S) = n(possible) = 36
n(sum < 13)
P(sum < 13) = ─────────
n(possible)
36
= ─── = 1
36
[ 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
n(sum > 15)
P(sum > 15) = ─────────
n(possible)
0
= ─── = 0
36
[ 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
n(aXb =20)
P(aXb =20) = ─────────
n(possible)
2 1
= ─── = ───
36 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
n(aXb > 18)
P(aXb > 18) = ─────────
n(possible)
8 4
= ─── = ───
36 18
[ Q 11 ]
11.17 The favourable outcomes are
1, 4, 9, 16, 25 and 36 so that
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(square) = 6 and
n(S) = n(possible) = 36
n(square)
P(square) = ─────────
n(possible)
6 1
= ─── = ───
36 6
[ Q 11 ]
11.18 The multiples of 8 are
8, 16, 24 and 32 so that
the favourable outcomes are
24, 42, 44, 46 and 64 and thus
there are 5 favourable outcomes and
there are 36 possible outcomes
Thus, n(E) = n(mult. 8) = 5 and
n(S) = n(possible) = 36
n(mult. 8)
P(mult. 8) = ─────────
n(possible)
5
= ───
36
[ Q 11 ]
11.19 The natural numbers are
1, 2, 3, 4, 5 etc. so that
the favourable outcomes are
11, 21, 31, 41, 51, 61, 22, 42,
62, 33, 63, 44, 55 and 66 and thus
there are 14 favourable outcomes and
there are 36 possible outcomes
Thus, n(E) = n(natural) = 6 and
n(S) = n(possible) = 36
n(natural)
P(natural) = ─────────
n(possible)
14 7
= ─── = ───
36 18
[ Q 11 ]
12.1 a
= WV ; b
= GV ; c
= VG
Vr. / Qu. 12
1
12.2
P(verloor eerste wedstryd) = ──
3
Vr. / Qu. 12
12.3
P(wen 2de) : Daar is drie uitkomstes,
nl. WW, GW en VW
1
∴ P(wen 2de) = ──
9
Vr. / Qu. 12
1
12.4
P(wen albei) = ──
9
Vr. / Qu. 12
12.5
Daar is 5 uitkomstes,
nl. WV, GV, VW, VG en VV
5
∴ P(verloor ten minste een) = ──
9
Vr. / Qu. 12
12.6
Daar is 5 uitkomstes,
nl. WG, GW, GG, GV en VG
5
∴ P(ten minste een gelykop) = ──
9
Vr. / Qu. 12
12.7
Daar is 4 uitkomstes,
nl. WW, WG, GW en GG
4
∴ P(verloor nie een nie) = ──
9
Vr. / Qu. 12
12.1 a
= WV ; b
= GV ; c
= VG
Vr. / Qu. 12
1
12.2
P(lose first match) = ──
3
Vr. / Qu. 12
12.3
P(w1n 2nd) : There are three outcomes,
viz. WW, DW en LW
1
∴ P(win 2nd) = ──
9
Vr. / Qu. 12
1
12.4
P(win both) = ──
9
Vr. / Qu. 12
12.5
There are 5 outcomes,
nl. WL, DL, LW, LD and LL
5
∴ P(lose at least one) = ──
9
Vr. / Qu. 12
12.6
There are 5 outcomes,
nl. WD, DW, DD, DL and LD
5
∴ P(at least one draw) = ──
9
Vr. / Qu. 12
12.7
There are 4 outcomes,
viz. WW, WD, DW and DD
4
∴ P(lose not one) = ──
9
Vr. / Qu. 12
13.1 a
= SDS ; b
= DSS ; c
= DDS
Vr. / Qu. 13
1
13.2
P(1ste kind 'n seun) = ──
2
Vr. / Qu. 13
13.3
Daar is 3 moontlike uitkomste, nl.
SDD, DSD and DDS.
3
P(2 dogters) = ──
8
13.4
Daar is 3 moontlike uitkomstes, kl.
SDD, DSD en DDS.
3
P(net 1 seun) = ──
8
Vr. / Qu. 13
13.5
Daar is net 1 moontlike uitkoms, nl.
SSS.
1
P(geen dogters) = ──
8
Vr. / Qu. 13
13.1 a
= SDS ; b
= DSS ; c
= DDS
Vr. / Qu. 13
1
13.2
P(first child a boy) = ──
2
Vr. / Qu. 13
13.3
There are 3 possible outcomes, viz.
BGG, GBG and GGB.
3
P(2 daughters) = ──
8
13.4
There are 3 possible outcomes, viz.
BGG, GBG and GGB.
3
P(only 1 boy) = ──
8
Vr. / Qu. 13
13.5
There is only 1 possible outcome, viz.
BBB.
1
P(no daughters) = ──
8
Vr. / Qu. 13
14.1 a
= 2M/T ; b
= 3K/H ; c
= 4K/H ; d
= 4M/T
Vr. / Qu. 14
14.2
Daar is net een moontlikheid, nl. 1K/H
1
P(1 en K) = ───
12
Vr. / Qu. 14
14.3
Daar is drie moontlikhede, nl.
2K/H, 4K/H en 6K/H
3 1
P(ewe getal en K) = ─── = ──
12 4
Vr. / Qu. 14
14.4
Daar is drie moontlikhede, nl.
4M/T, 5M/T en 6M/T
3 1
P(getal > 3 en M) = ─── = ──
12 4
Vr. / Qu. 14
14.5
Daar is twee moontlikhede, nl.
1M/T en 3M/T
2 1
P(onewe getal < 5 en M) = ─── = ──
12 6
Vr. / Qu. 14
14.6
Daar is ses moontlikhede, nl.
1M/T, 2M/T, 3M/T, 4M/T, 5M/T en 6M/T
6 1
P(enige getal en M) = ─── = ──
12 6
Vr. / Qu. 14
14.7
Daar is twee moontlikhede, nl.
3K/H en 3M/T
2 1
P(3 en k of M) = ─── = ──
12 6
Vr. / Qu. 14
14.8
Daar is geen moontlikheid nie.
0
P(2 en nie kruis of munt nie) = ─── = 0
12
Vr. / Qu. 14
14.1 a
= 2M/T ; b
= 3K/H ; c
= 4K/H ; d
= 4M/T
Vr. / Qu. 14
14.2
There is only one favourable outcome,
viz. 1K/H
1
P(1 and H) = ───
12
Vr. / Qu. 14
14.3
There are three favourable outcomes,
viz. 2K/H, 4K/H and 6K/H
3 1
P(even number and H) = ─── = ──
12 4
Vr. / Qu. 14
14.4
There are three favourable outcomes,
4M/T, 5M/T en 6M/T
3 1
P(number > 3 and T) = ─── = ──
12 4
Vr. / Qu. 14
14.5
There are two favourable outcomes,
1M/T and 3M/T
2 1
P(uneven number < 5 and T) = ─── = ──
12 6
Vr. / Qu. 14
14.6
There are six favourable outcomes,
1M/T, 2M/T, 3M/T, 4M/T, 5M/T and 6M/T
6 1
P(any number and T) = ─── = ──
12 6
Vr. / Qu. 14
14.7
There are two favourable outcomes,
3K/H and 3M/T
2 1
P(3 and H or T) = ─── = ──
12 6
Vr. / Qu. 14
14.8
There are no favourable outcomes,
0
P(2 and neither H or T) = ─── = 0
12
Vr. / Qu. 14