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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
$$ \hspace*{13 mm}\mathrm{P(pineapple)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{n(pineapple)}{n(number\kern1mmof\kern1mmsweets)} } $$
$$ \hspace*{34 mm}=\kern2mm\frac{14}{40}\kern2mm=\kern1mm\frac{7}{20} $$
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
$$ \hspace*{13 mm}\mathrm{P(apricot)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{32 mm}\mathrm{=\kern1mm\frac{n(apricot)}{n(number\kern1mmof\kern1mmsweets)} } $$
$$ \hspace*{31 mm}=\kern1mm\frac{6}{40}\kern2mm=\kern2mm\frac{3}{20} $$
1.3 n(guava) = 12
n(number of sweets) = 40
$$ \hspace*{13 mm}\mathrm{P(guava)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(guava)}{n(number\kern1mmof\kern1mmsweets)} } $$
$$ \hspace*{29 mm}=\kern1mm\frac{12}{40}\kern2mm=\kern1mm\frac{3}{10} $$
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
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmpineapple)\kern1mm=\kern1mm1\kern1mm─\kern1mm\frac{7}{20}\kern1mm=\kern1mm\frac{13}{20} } $$
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
$$ \hspace*{13 mm}\mathrm{P(cherry)\kern1mm=\kern1mm\frac{8}{40}\kern1mm=\kern1mm\frac{1}{5} } $$
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmcherry)\kern1mm=\kern1mm1\kern1mm─\kern1mm\frac{1}{5}\kern1mm=\kern1mm\frac{4}{5} } $$
1.7 n(orange) = 0 and n(sweets) = 40
$$ \hspace*{13 mm}\mathrm{P(orange)\kern1mm=\kern1mm\frac{n(orange)}{n(number\kern1mmof\kern1mmsweets)} } $$
$$ \hspace*{31 mm}\mathrm{=\kern1mm\frac{0}{40}\kern2mm=\kern1mm0 } $$
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
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmorangeqw)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(not\kern1mmorange) } $$
$$ \hspace*{43 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm0\kern1mm=\kern1mm1} $$
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
$$ \hspace*{13 mm}\mathrm{P(Q\kern1mmspades)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{30 mm}\mathrm{= \frac{n(Q spades)}{n(number of cards)} } $$
$$ \hspace*{29 mm}= \frac{1}{52} $$
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
$$ \hspace*{13 mm}\mathrm{P(black\kern1mmcard)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{37 mm}\mathrm{=\kern1mm\frac{n(black\kern1mmcard)}{n(number\kern1mmof\kern1mmcards)} } $$
$$ \hspace*{37 mm}=\kern1mm\frac{26}{52}\kern2mm=\kern1mm\frac{1}{2} $$
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
$$ \hspace*{13 mm}\mathrm{P(diamond)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{34 mm}\mathrm{=\kern1mm\frac{n(diamond)}{n(number\kern1mmof\kern1mmcards)} } $$
$$ \hspace*{34 mm}=\kern1mm\frac{13}{52}\kern2mm=\kern1mm\frac{1}{4} $$
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
$$ \hspace*{13 mm}\mathrm{P(red\kern1mmpicture)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{34 mm}\mathrm{=\kern1mm\frac{n(red\kern1mmpicture)}{n(number\kern1mmof\kern1mmcards)} } $$
$$ \hspace*{34 mm}=\kern1mm\frac{8}{52}\kern2mm=\kern1mm\frac{2}{13} $$
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)
$$ \hspace*{13 mm}\mathrm{P(red\kern1mmpicture)\kern1mm=\kern1mm\frac{2}{13}\kern3mm.\kern1mm.\kern1mm.\kern1mm[2.4] } $$
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmred\kern1mmpicture)\kern1mm=\kern1mm1\kern1mm─\kern1mm\frac{2}{13} } $$
$$ \hspace*{44 mm}\mathrm{=\kern1mm\frac{11}{13} } $$
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)
$$ \hspace*{13 mm}\mathrm{P(not\kern1mma\kern1mmpicture)\kern1mm=\kern1mm1\kern1mm─\kern1mm\frac{4}{13} } $$
$$ \hspace*{41 mm}\mathrm{=\kern1mm\frac{9}{13} } $$
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
$$ \hspace*{13 mm}\mathrm{P(black\kern1mmnumbered)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{34 mm}\mathrm{=\kern1mm\frac{n(black\kern1mmnumbered)}{n(number\kern1mmof\kern1mmcards)} } $$
$$ \hspace*{34 mm}=\kern1mm\frac{18}{52}\kern2mm=\kern1mm\frac{9}{26} $$
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
$$ \hspace*{13 mm}\mathrm{P(jack)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{27 mm}\mathrm{=\kern1mm\frac{n(jack)}{n(number\kern1mmof\kern1mmcards)} } $$
$$ \hspace*{26 mm}=\kern1mm\frac{4}{52}\kern2mm=\kern1mm\frac{1}{13} $$
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
$$ \hspace*{13 mm}\mathrm{P(numbered)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{36 mm}\mathrm{=\kern1mm\frac{n(numbered)}{n(number\kern1mmof\kern1mmcards)} } $$
$$ \hspace*{35 mm}=\kern1mm\frac{12}{52}\kern2mm=\kern1mm\frac{3}{13} $$
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
$$ \hspace*{13 mm}\mathrm{P(red\ numbered)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{36 mm}\mathrm{=\kern1mm\frac{n(red\ numbered)}{n(number\kern1mmof\kern1mmcards)} } $$
$$ \hspace*{35 mm}=\kern1mm\frac{10}{52}\kern2mm=\kern1mm\frac{5}{26} $$
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
$$ \hspace*{13 mm}\mathrm{P(black\kern1mmpicture)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{36 mm}\mathrm{=\kern1mm\frac{n(black\kern1mmpicture)}{n(number\kern1mmof\kern1mmcards)} } $$
$$ \hspace*{35 mm}=\kern1mm\frac{8}{52}\kern2mm=\kern1mm\frac{2}{13} $$
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
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmred\kern1mmnumbered)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{36 mm}\mathrm{=\kern1mm\frac{n(not\kern1mmred\kern1mmnumbered)}{n(number\kern1mmof\kern1mmcards)} } $$
$$ \hspace*{35 mm}=\kern1mm\frac{34}{52}\kern2mm=\kern1mm\frac{17}{26} $$
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
$$ \hspace*{13 mm}\mathrm{P(not\kern1mman\kern1mmace)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{36 mm}\mathrm{=\kern1mm\frac{n(not\kern1mman\kern1mmace)}{n(number\kern1mmof\kern1mmcards)} } $$
$$ \hspace*{35 mm}=\kern1mm\frac{48}{52}\kern2mm=\kern1mm\frac{12}{13} $$
3.1 There 40 pupils in the class.
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
$$ \hspace*{13 mm}\mathrm{P(girls)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(girls)}{n(pupils)} } $$
$$ \hspace*{29 mm}=\kern1mm\frac{23}{40} $$
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
$$ \hspace*{13 mm}\mathrm{P(boy + homework)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(boy + homework}{n(pupils)} } $$
$$ \hspace*{29 mm}=\kern1mm\frac{6}{40}\kern2mm=\kern1mm\frac{3}{20} $$
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
$$ \hspace*{13 mm}\mathrm{P(G+nohomework)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(G+no\kern1mmhomework}{n(pupils)} } $$
$$ \hspace*{29 mm}=\kern1mm\frac{7}{40} $$
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
$$ \hspace*{13 mm}\mathrm{P(no\kern1mmhomework)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(no\kern1mmhomework}{n(pupils)} } $$
$$ \hspace*{29 mm}=\kern1mm\frac{18}{40}\kern2mm=\kern1mm\frac{9}{20} $$
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
$$ \hspace*{13 mm}\mathrm{P(Grade 11)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(Grade 11}{n(pupils)} } $$
$$ \hspace*{29 mm}=\kern1mm\frac{0}{40}\kern2mm=\kern1mm0 $$
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
$$ \hspace*{13 mm}\mathrm{P(G\kern1mm;+\kern1mmno\kern1mmhomework)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(G\kern1mm+\kern1mmno\kern1mmhomework}{n(girls)} } $$
$$ \hspace*{29 mm}=\kern1mm\frac{7}{23} $$
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
$$ \hspace*{13 mm}\mathrm{P(B\kern1mm+\kern1mmhomework)\kern1mm=\kern1mm\frac{n(E)}{n(S)} } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{n(B\kern1mm+\kern1mmhomework)}{n(boys)} } $$
$$ \hspace*{29 mm}=\kern1mm\frac{6}{17} $$
4.1 There is one D and 11 letters
Thus n(E) = n(D) = 1 and
n(S) = n(letters) = 11
$$ \hspace*{13 mm}\mathrm{P(D)\kern1mm=\kern1mm\frac{n(D)}{n(letters)} } $$
$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{1}{11} } $$
4.2 There is one K and 11 letters
Thus n(E) = n(K) = 1 and
n(S) = n(letters) = 11
$$ \hspace*{13 mm}\mathrm{P(K)\kern1mm=\kern1mm\frac{n(K)}{n(letters)} } $$
$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{1}{11} } $$
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
$$ \hspace*{13 mm}\mathrm{P(vowel)\kern1mm=\kern1mm\frac{n(vowel)}{n(letters)} } $$
$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{5}{11} } $$
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
$$ \hspace*{13 mm}\mathrm{P(consonant)\kern1mm=\kern1mm\frac{n(consonant)}{n(letters)} } $$
$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{5}{11} } $$
4.5 There are two A's,
and 11 letters.
Thus n(E) = n(A) = 2 and
n(S) = n(letters) = 11
$$ \hspace*{13 mm}\mathrm{P(A)\kern1mm=\kern1mm\frac{n(A)}{n(letters)} } $$
$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{2}{11} } $$
4.6 There are no B's,
and 11 letters.
Thus n(E) = n(B) = 0 and
n(S) = n(letters) = 11
$$ \hspace*{13 mm}\mathrm{P(B)\kern1mm=\kern1mm\frac{n(B)}{n(letters)} } $$
$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{0}{11}\kern1mm=\kern1mm0 } $$
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
$$ \hspace*{13 mm}\mathrm{P(O)\kern1mm=\kern1mm\frac{n(O)}{n(letters)} } $$
$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{0}{11}\kern1mm=\kern1mm0 } $$
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmO)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(O) } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm0\kern1mm=\kern1mm1 } $$
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
$$ \hspace*{13 mm}\mathrm{P(F)\kern1mm=\kern1mm\frac{n(F)}{n(letters)} } $$
$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{1}{11} } $$
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmF)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(F) } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm\frac{1}{11}\kern1mm=\kern1mm\frac{10}{11} } $$
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
$$ \hspace*{13 mm}\mathrm{P(E)\kern1mm=\kern1mm\frac{n(E)}{n(letters)} } $$
$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{0}{11}\kern1mm=\kern1mm0 } $$
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmE)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(E) } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm0\kern1mm=\kern1mm1 } $$
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
$$ \hspace*{13 mm}\mathrm{P(I)\kern1mm=\kern1mm\frac{n(I)}{n(letters)} } $$
$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{2}{11} } $$
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmI)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(I) } $$
$$ \hspace*{28 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm\frac{2}{11}\kern1mm=\kern1mm\frac{9}{11} } $$
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
$$ \hspace*{13 mm}\mathrm{P(vowel)\kern1mm=\kern1mm\frac{n(vowel)}{n(letters)} } $$
$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{5}{11} } $$
$$ \hspace*{13 mm}\mathrm{P(not\kern1mmvowel)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(vowel) } $$
$$ \hspace*{35 mm}\mathrm{=\kern1mm1\kern1mm─\kern1mm\frac{5}{11}\kern1mm=\kern1mm\frac{6}{11} } $$
5.1 There is one 5 and 10 balls
Thus n(E) = n(5) = 1 and
n(S) = n(balls) = 10
$$ \hspace*{13 mm}\mathrm{P(5)\kern1mm=\kern1mm\frac{n(5)}{n(balls)} } $$
$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{1}{10} } $$
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
$$ \hspace*{13 mm}\mathrm{P(even)\kern1mm=\kern1mm\frac{n(even)}{n(balls)} } $$
$$ \hspace*{29 mm}\mathrm{=\kern1mm\frac{5}{10}\kern1mm=\kern1mm\frac{1}{2} } $$
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
$$ \hspace*{13 mm}\mathrm{P(<\kern1mm6)\kern1mm=\kern1mm\frac{n(<\kern1mm6)}{n(balls)} } $$
$$ \hspace*{30 mm}\mathrm{=\kern1mm\frac{5}{10}\kern1mm=\kern1mm\frac{1}{2} } $$
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
$$ \hspace*{13 mm}\mathrm{P(>\kern1mm3)\kern1mm=\kern1mm\frac{n(>\kern1mm3)}{n(balls)} } $$
$$ \hspace*{26 mm}\mathrm{=\kern1mm\frac{3}{10} } $$
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 ]
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
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
$$ \hspace*{13 mm}\mathrm{P(3)\kern1mm=\kern1mm\frac{n(3)}{n(children)} } $$
$$ \hspace*{23 mm}\mathrm{=\kern1mm\frac{2}{12}\kern1mm=\kern1mm\frac{1}{6} } $$
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
$$ \hspace*{13 mm}\mathrm{P(6+)\kern1mm=\kern1mm\frac{n(6+)}{n(children)} } $$
$$ \hspace*{24 mm}\mathrm{=\kern1mm\frac{7}{12} } $$
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
$$ \hspace*{13 mm}\mathrm{P(<\kern1mm11)\kern1mm=\kern1mm\frac{n(<\kern1mm11)}{n(children)} } $$
$$ \hspace*{29 mm}\mathrm{=\kern1mm\frac{8}{12}\kern1mm=\kern1mm\frac{2}{3} } $$
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
$$ \hspace*{13 mm}\mathrm{P(<\kern1mm3)\kern1mm=\kern1mm\frac{n(<\kern1mm3)}{n(children)} } $$
$$ \hspace*{26 mm}\mathrm{=\kern1mm\frac{10}{12}\kern1mm=\kern1mm\frac{5}{6} } $$
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
$$ \hspace*{13 mm}\mathrm{P(<\kern1mm10)\kern1mm=\kern1mm\frac{n(<\kern1mm10)}{n(children)} } $$
$$ \hspace*{28 mm}\mathrm{=\kern1mm\frac{8}{12}\kern1mm=\kern1mm\frac{2}{3} } $$
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
$$ \hspace*{13 mm}\mathrm{P(>\kern1mmmedian)\kern1mm=\kern1mm\frac{n(>\kern1mmmedian)}{n(children)} } $$
$$ \hspace*{28 mm}\mathrm{=\kern1mm\frac{6}{12}\kern1mm=\kern1mm\frac{1}{2} } $$
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
$$ \hspace*{13 mm}\mathrm{P(>\kern1mmmode)\kern1mm=\kern1mm\frac{n(>\kern1mmmode)}{n(children)} } $$
$$ \hspace*{28 mm}\mathrm{=\kern1mm\frac{2}{12}\kern1mm=\kern1mm\frac{1}{6} } $$
$$ \hspace*{6 mm}\mathrm{7.1\kern1mmP(pass)\kern1mm=\kern1mm1\kern1mm─\kern1mmP(fail) } $$
$$ \hspace*{28 mm}=\kern1mm1\kern1mm─\kern1mm0,35 $$
$$ \hspace*{28 mm}=\kern1mm0,65 $$
$$ \hspace*{6 mm}\mathrm{7.2\kern1mmP(failas)\kern1mm=\kern1mm\frac{n(number\kern1mmfailed)}{n(number\kern1mmthat\kern1mmwrote)} } $$
P(fail) = 0,35 and pupils
that wrote = 74
$$ \hspace*{13 mm}\mathrm{0,35999\kern1mm=\kern1mm\frac{n(failures)}{n(written)} } $$
$$ \hspace*{13 mm}\mathrm{n(failures)\kern1mm=\kern1mm0,35\kern1mmX\kern1mm74 } $$
$$ \hspace*{32 mm}\mathrm{=\kern1mm26 } $$
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
$$ \hspace*{15 mm}\mathrm{n(passed)\kern1mm=\kern1mm0,65\kern1mmX\kern1mm132 } $$
$$ \hspace*{33 mm}=\kern1mm85,8\kern1mm=\kern1mm86 $$
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 ]
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
$$ \hspace*{16 mm}\mathrm{P(boy)\kern1mm=\kern1mm\frac{n(boy)}{n(pupils)}\kern1mm=\kern1mm0,25 } $$
$$ \hspace*{26 mm}\mathrm{=\kern1mm\frac{18}{50}\kern1mm=\kern1mm36\kern1mm\% } $$
9.2.2 Twelve pupils buy nothing and
there are 50 pupils.
n(E) = n(buys nothing) = 12 and
n(S) = n(pupils) = 50
$$ \hspace*{16 mm}\mathrm{P(buys\kern1mmnothing)\kern1mm=\kern1mm\frac{n(buys\kern1mmnothing)}{n(pupils)} } $$
$$ \hspace*{41 mm}\mathrm{=\kern1mm\frac{12}{50}\kern1mm=\kern1mm24\kern1mm;\% } $$
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
$$ \hspace*{16 mm}\mathrm{P(Munch\kern1mmbar)\kern1mm=\kern1mm\frac{n(Munch\kern1mmbar)}{n(pupils)} } $$
$$ \hspace*{41 mm}\mathrm{=\kern1mm\frac{6}{50}\kern1mm=\kern1mm12\kern1mm\% } $$
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
$$ \hspace*{14 mm}\mathrm{P(boiled\kern1mmsweets)\kern1mm=\kern1mm\frac{n(boiled\kern1mmsweets)}{n(pupils)} } $$
$$ \hspace*{43 mm}\mathrm{=\kern1mm\frac{8}{50}\kern1mm=\kern1mm16\kern1mm\% } $$
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
$$ \hspace*{16 mm}\mathrm{P(g+CB)\kern1mm=\kern1mm\frac{n(g+CB)}{n(girls)} } $$
$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{3}{32\kern1mm}=\kern1mm9,375\kern1mm\% } $$
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
$$ \hspace*{16 mm}\mathrm{P(g+CB)\kern1mm=\kern1mm\frac{n(g+CB)}{n(girls)} } $$
$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{12}{32}\kern1mm=\kern1mm37,5\kern1mm\% } $$
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
$$ \hspace*{16 mm}\mathrm{P(b+MB)\kern1mm=\kern1mm\frac{n(b+MB)}{n(boys)} } $$
$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{2}{18}\kern1mm=\kern1mm11,111\kern1mm\% } $$
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
$$ \hspace*{16 mm}\mathrm{P(b+not)\kern1mm=\kern1mm\frac{n(b+not)}{n(boys)} } $$
$$ \hspace*{35 mm}\mathrm{=\kern1mm\frac{4}{18}\kern1mm=\kern1mm22,222\kern1mm\% } $$
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.
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.
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.
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.
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.
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 |
|