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11 More about Probability

11 More about ProbabilityActivity

Activity 11.1 (p. 11.19)

1. (a)

(i) 0

(ii) 0

(iii) 0

(b) Yes No

(c)

2. (a)

(i)

(ii)

(iii)

(b) Yes No

(c)

Activity 11.2 (p. 11.33)1. (a) Yes No

(b

)4 5 7 8

R (R, 4) (R, 5) (R, 7) (R, 8)

Y (Y, 4) (Y, 5) (Y, 7) (Y, 8)

G (G, 4) (G, 5) (G, 7) (G, 8)

2. , ,

3.

Activity 11.3 (p. 11.45)

1. (a) 1 2 3 4

1 --- (1, 2) (1, 3) (1, 4)

2 (2, 1) --- (2, 3) (2, 4)

3 (3, 1) (3, 2) --- (3, 4)

4 (4, 1) (4, 2) (4, 3) ---

(b) 12

2. (a)

(i)

(ii)

(iii)

(b) yes

3.

Classwork

Classwork (p. 11.5)(a) A = {3, 5, 6}, B = {1, 2}, U = {1, 2, 3, 4, 5, 6}

(b) A = {0, 1, 2, 4}, B = {1, 2, 5}, U = {0, 1, 2, 3, 4, 5, 6}

Classwork (p. 11.6)(a) A = {4, 8}, B = {1, 2, 3}, C = {1, 3, 5, 7, 9}

(b) (i) = {1, 2, 3, 4, 8}(ii) = {1, 2, 3, 5, 7, 9}(iii) = (iv) = {1, 3}(v) = {1, 2, 3, 5, 6, 7, 9, 10}(vi) = {2, 4, 6, 8, 10}(vii) = {1, 2, 3, 5, 6, 7, 9, 10}(viii) = {4, 8}

(c)

Classwork (p. 11.18) 1. 2. 3. 4.

Classwork (p. 11.24)1. 2. 3. 4.

Classwork (p. 11.32)1. 2. 3. 4.

Classwork (p. 11.45)1. 2. 3. 4.

Classwork (p. 11.57)

1. or 2. or

117

1

A2

34

56

7

89

10

BU

C

NSS Mathematics in Action (2nd Edition) 5B Full Solutions

3. or 4. or

Quick Practice

Quick Practice 11.1 (p. 11.11)(a) S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(b) (i) X = {2, 4, 6, 8, 10}(ii) Y = {7, 8, 9, 10}

Quick Practice 11.2 (p. 11.12)(a) Event B = {5, 10, 15, 20}

∴

(b)

Quick Practice 11.3 (p. 11.13)Box B

4 5 6

1 5 6 7

Box A 2 6 7 8

3 7 8 9

Let E be the event that the sum of two numbers is less than 7. Then, E′ is the event that the sum of two numbers is not less than 7, i.e. greater than or equal to 7.From the table, we have n(S) = 9, n(E) = 3, n(E′) = 6.(a) P(the sum of two numbers is less than 7)

(b) P(the sum of two numbers is greater than or equal to 7)

Quick Practice 11.4 (p. 11.21)

(a)

(b)

Quick Practice 11.5 (p. 11.22)Let F1 be the event that the chosen candidate fails paper I,and F2 be the event that the chosen candidate fails paper II.∵ F1 and F2 are non-mutually exclusive events.∴ P(fails paper I or paper II)

Quick Practice 11.6 (p. 11.23)Let E1 be the event that two even numbers are obtainedand E2 be the event that the sum is 10.The number of all possible outcomes = 6 × 6 = 36(a) ∵ E1 = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2),

(6, 4), (6, 6)}

∴

(b) ∵ E2 = {(4, 6), (5, 5), (6, 4)}

∴

(c)

(d)

Quick Practice 11.7 (p. 11.26)Let A be the event that the student’s favourite subject is English or Chinese.

118


Then A′ is the event that the student’s favourite subject is neither English nor Chinese.

Quick Practice 11.8 (p. 11.27)By the counting principle,

(a) (i) Let A be the event that the sum of the numbers is less than 6.Then, event A = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1),

(2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

∴

(ii)

(b) (i) Let B be the event that both numbers shown are odd.Then, event B = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3),

(3, 5), (5, 1), (5, 3), (5, 5)}

∴

(ii)


(a)

(b)

Quick Practice 11.10 (p. 11.36) P(Patrick does not win the game)

P(Daisy does not win the game)

(a)

(b)

(c)


119


Let P stand for passing a paper and F stand for failing a paper.(a) P(passes exactly two papers)

(b)

Quick Practice 11.12 (p. 11.43)(a) There are 8 possible outcomes for the condition ‘the person

chosen comes from shop A’. 3 of them satisfy the event ‘he/she is a sales person’.

∴

(b) There are 5 possible outcomes for the condition ‘the personchosen is a sales person’. 3 of them satisfy the event ‘he/she comes from shop A’.

∴

Quick Practice 11.13 (p. 11.44)(a) There are 5 outcomes for the condition ‘sum = 8’.

2 of them satisfy the event ‘product = 12’,i.e. {(2, 6), (6, 2)}.

∴

(b) There are 4 outcomes for the condition ‘product = 12’. 2 of them satisfy the event ‘sum = 8’,i.e. {(2, 6), (6, 2)}.

∴


(a)

(b)

Quick Practice 11.15 (p. 11.49)(a) P(the mango candy is in Billy’s bag)

(b) P(the mango candy is in Tom’s bag)

(c) P(Tom draws an apple candy from Billy’s bag)



120


Let A1 and A2 be the events that a class A student is drawn inthe 1st and the 2nd draw respectively, and B1 and B2 be theevents that a class B student is drawn in the 1st and the 2nddraw respectively.

(a) (i)

(ii)

(b)

Quick Practice 11.18 (p. 11.58)(a) Number of ways of selecting 4 people

Number of ways of selecting 2 particular couples

∴

(b) Number of ways of selecting 2 couples

∴

Quick Practice 11.19 (p. 11.59)(a) Number of ways of drawing 4 cards from 7 cards

When the sum of the four numbers is even, the numbers are

either ‘4 odd numbers’ or ‘2 odd and 2 even numbers’.Number of ways of choosing 4 odd numbers

Number of ways of choosing 2 odd and 2 even numbers

Total number of favourable outcomes

∴

(b)

Quick Practice 11.20 (p. 11.60)Total number of ways of arranging the 8 students = 8!(a) Since the students from school A must stand next to each

other, we can treat them as one unit.

A1 A2 A3 A4 S1 S2 S3 S4

one unit

Number of ways of arranging the 4 other students and 1 unit= (4 + 1)!= 5!Number of ways of arranging the 4 students from school A= 4!

∴

(b) Since the students from school A are separated by a studentfrom other schools, they must stand in the arrangementslike these:

A1 S1 A2 S2 A3 S3 A4 S4

orS1 A1 S2 A2 S3 A3 S4 A4

In each arrangement,Number of ways of arranging the students from school A= 4!Number of ways of arranging the 4 other students= 4!∴ P(each separated by a student from other schools)

Quick Practice 11.21 (p. 11.61)Number of cakes = 8 × 3 = 24Total number of ways of choosing 3 cakes

(a) Number of ways of choosing 1 shop from 8 shops

121


Number of ways of choosing 3 cakes from 1 shop

∴

(b) Number of ways of choosing 3 shops from 8 shops

Number of ways of choosing 1 cake from each of the 3 chosen shops

∴

Further Practice

Further Practice (p. 11.13)1. Let E be the event of drawing a yellow card with a happy

face and F be the event of drawing a red card.We have n(S) = 7, n(E) = 3, n(F) = 0

(a)

(b)

2. Let E be the event that the picnic is held in January andF be the event that the picnic is held on a Friday.n(S) = 31, n(E) = 31, n(F) = 4

(a)

(b)

Further Practice (p. 11.27)1. (a)

(b)

2. Let E be the event that the selected student passes theEnglish test and C be the event that the selected studentpasses the Chinese test.

(a)

(b)

(c) P(passes the English test but fails the Chinese test)P(passes the English test) P(passes both tests)

3. ∵ The probability of getting a red ball is 0.4.

∴ ...... (1)

∵ The probability of getting a red ball or a green ball is 0.6.

∴

i.e. ...... (2)

By substituting x = 2 into (1), we have

Further Practice (p. 11.38)

122


1. (a)

(b)

2. (a)

(b)

(c)

3. Let Y stand for the pointer landing on the yellow regions, R stand for the pointer landing on the purple regions and U stand for the pointer landing on the blue region.

, and

Further Practice (p. 11.52)1. (a) Let H stand for head and T stand for tail.

∴ The possible outcomes are{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

(b) (i)

(ii)

2. (a) P(B)

(b)

123


3. (a) (i) P(Kelvin wins the prize)= P(Kelvin wins in the 1st draw) + P(Kelvin wins in the 2nd draw) + P(Kelvin wins in the 3rd draw)

=

=

(ii) P(Chloe wins the prize)= 1 P(Kelvin wins the prize)

=

=

(b) From (a), the probabilities for them to win the prize are the same.∴ It is fair for Chloe to draw after Kelvin.∴ The claim is not agreed.

Further Practice (p. 11.62)1. Number of ways of selecting 4 students

(a) Number of ways of selecting 4 girls

∴

(b)

2. Total number of ways of arranging the 5 participants = 5!(a) Since Lydia performs immediately after Vincy, we can

treat Lydia and Vincy as a unit.Number of ways of arranging the 3 other participants and 1 unit

∴

(b) Number of ways of arranging 3 other participants= 3!

∴

3. (a) Total number of ways of drawing 5 cards

Number of ways of drawing the letters A, B and C

Number of ways of drawing 2 number cards∴

(b) Total number of orders of drawing 5 cards

Number of ways of drawing the letters A, B and CSince the letters A, B and C are successively drawn in

the order ‘ABC’, we can treat ‘ABC’ as a unit.Number of ways of drawing 2 number cardsNumber of ways of arranging the 2 number cards and

1 unit∴

Exercise

Exercise 11A (p. 11.7)Level 11. (a)

(b)

124


(c)

(d)

2. (a) (i)

(ii)

(b) (i)

(ii)

3. (a) False

(b) True

(c) True

(d) False

4. (a)

(b)

(c)

(d)

5. (a)

(b)

(c)

(d)

6. (a)

(b)

(c)

(d)

(e)

(f)

7. (a)

(b) ∵∴ A and B are not disjoint.

Level 28. (a)

(b)

(c)

(d)

9. (a)

(b)

125

A BU

1216

20

6 10

40 2

Y

U

Z

8

X

14 18


(c)

(d)

10. (a)

(b)

11. (a)

(b)

12. (a) (i)

(ii)

(b)

From (a)(ii),∴

13. (a) (i)

(ii)

(b) ∵∴ A and C are disjoint.

14. (a)

(b) (i)

(ii)

Exercise 11B (p. 11.14)Level 11. (a)

(b) (i)(ii)

2. (a) Let H stand for head and T stand for tail.∴

(b) (i)(ii)

3. (a)

(b)

4. (a)

(b) (i)

(ii)

5. (a) ,

(b) (i)

(ii)

6. (a)

126

A BU


(b)

7. (a)

(b)

8. (a)

(b)

Level 29. Let W stand for a white straw, R stand for a red straw and

G stand for a green straw.

1st draw 2nd draw Possible Outcomes

(a)

(b)

10. (a)

(b) (i)

(ii)

11. Let E be the event that the box taken contains no defective light bulbs. Then, E′ is the event that the box taken contains at least 1 defective light bulb.From the table, we haven(S) = 4 + 15 + 11 + 6 + 4 = 40, n(E) = 4 and n(E′) = 36

(a)

(b)

12. (a)

(b)

13. (a) Second throw1 2 3 4

1 (1, 1) (1, 2) (1, 3) (1, 4)

First throw2 (2, 1) (2, 2) (2, 3) (2, 4)3 (3, 1) (3, 2) (3, 3) (3, 4)4 (4, 1) (4, 2) (4, 3) (4, 4)

(b) (i) P(sum is greater than 5)

(ii)

14. (a) ∵ The probability of drawing a $10-coin is .

127

W

R

G

W ..............WW

W ..............RW

W ..............GW

R ..............WR

R .............. RR

R ..............GR


∴

(b) Total number of coins = 14 + 6 + 8 = 28

(i)

(ii)

15. Let x be the total number of balls.

Then the number of red balls ,

and the number of blue balls .

If we take x = 6, we havetotal number of balls = 6, number of red balls = 2 and number of blue balls = 3.If we take x = 12, we have total number of balls = 12, number of red balls = 4 andnumber of blue balls = 6(or any other answers in the ratio 6 : 2 : 3)

16. (a)

(b) (i)

(ii)

17. (a) S = {3456, 3465, 3546, 3564, 3645, 3654,4356, 4365, 4536, 4563, 4635, 4653,5346, 5364, 5436, 5463, 5634, 5643,6345, 6354, 6435, 6453, 6534, 6543}

(b) (i)

(ii)

(c)

Exercise 11C (p. 11.28)Level 11. (a) Events E and F are both mutually exclusive and

complementary.

(b) Events E and F are neither mutually exclusive nor complementary.

(c) Events E and F are mutually exclusive.

2. (a) The chosen letter is a consonant. / The chosen letter is ‘P’, ‘R’ or ‘M’.

(b) The number obtained is less than or equal to 3. / The number obtained is not greater than 3. / The number obtained is 1, 2 or 3.

(c) At least one head shows up.

3. Total number of pairs of shoes = 8 + 5 + 2 = 15

4.

5. (a) P(lemon candy)

(b) P(apple candy or orange candy)

128


6. P(not defective)

7. Total number of coins in the cash box= 13 + 25 + 18 + 19 + 23 + 8 + 6= 112

(a)

(b)

8. (a)

(b)

(c)

9.

(a)

(b)

(c)

10. Let E be the event that the sum of the numbers drawn is less than 4. Then E′ is the event that the sum of the numbers drawn is not less than 4.Total number of possible outcomes 3 4 12(a) E = {(1, 1), (1, 2), (2, 1)}

∴ P(less than 4)

(b) P(not less than 4)

11. (a) (i) There are 26 red cards in a deck of 52 playing cards.

P(a red card)

(ii) There are 4 Jacks in a deck of 52 playing cards.

P(a Jack)

(b) P(a red card or a Jack) P(a red card) P(a Jack) P(a red Jack)

129


12. (a) ∵ Among the fifteen numbers 11, 12, 13, …, 25, there are 5 prime numbers (11, 13, 17, 19 and 23) and 7 even numbers (12, 14, 16, 18, 20, 22 and 24).

∴

(b) ∵ Among 11, 12, …, 25, there are 5 numbers (12, 15, 18, 21 and 24) which are divisible by 3, 4 numbers (12, 16, 20 and 24) which are divisible by 4 and 2 numbers (12 and 24) which are divisible by both 3 and 4.

∴

13. Let E be the event that the outcomes are the same. Then E′ is the event that the outcomes are not the same, i.e. the outcomes are different.Total number of possible outcomes(a) E = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

P(the outcomes are the same)

(b) P(the outcomes are different)

Level 214. Total number of students 22 20 16 24 82

(a) P(a boy from 5A)

(b) P(a boy)

(c) P(from 5A)

(d) P(a boy or from 5A)

15. (a) From 1 to 100, there are 50 numbers which are divisible by 2.

P(divisible by 2)

(b) From 1 to 100, there are 20 numbers which are divisible by 5.

P(divisible by 5)

(c) From 1 to 100, there are 10 numbers which are divisible by both 2 and 5.P(divisible by 2 or 5)= P(divisible by 2) + P(divisible by 5) P(divisible by both 2 and 5)

=

=

130


16. Total number of possible outcomes = 6 × 6 = 36 (a) E = {(1, 4), (2, 3), (3, 2), (4, 1)}

(b) F = {(1, 6), (2, 3), (3, 2), (6, 1)}

(c)

17. (a)

(b)

(c)

∵ A staff who is a female but not a clerk is the same as a staff who is neither a male nor a clerk.

∴

18. (a) P(likes singer A or singer B)= P(likes singer A) + P(likes singer B) P(likes both singers)

=

=

(b)

Alternative Solution

19. By the counting principle,

Let H stand for head and T stand for tail.(a) Let A be the event of getting no tails.

A = {HHHHH}

∴

(b) Let B be the event of getting exactly one tail.B = {THHHH, HTHHH, HHTHH, HHHTH, HHHHT}

∴

(c)

131


20. (a) P(drama, sports or music)

P(news or cartoon)

(b) (i)

(ii)

21. (a) Number of students read exactly one of the newspapers

(b) (i) P(reads exactly one of the newspapers)

(ii) P(reads either newspaper A or newspaper B)

(iii) P(does not read any of the two newspapers)

22.

By substituting (2) into (1), we have

By substituting (3) into (1), we have

23. (a) Total number of possible outcomes

Let A be the event that the product of the three

numbers obtained is less than or equal to 2. Then A′ is

the event that the product of the three numbers

obtained is greater than 2.

A ={(1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1)}

(b) Let B be the event that the product of the three

numbers obtained is less than 140. Then B′ is the event

that the product of the three numbers obtained is

greater than or equal to 140.

B′ ={(4, 6, 6), (5, 5, 6), (5, 6, 5), (5, 6, 6), (6, 4, 6),

132


(6, 5, 5), (6, 5, 6), (6, 6, 4), (6, 6, 5), (6, 6, 6)}

Exercise 11D (p. 11.38)Level 11. P(taking two cans of coke)

= P(taking 1 can of coke from the first refrigerator and

taking 1 can of coke from the second refrigerator)

= P(taking 1 can of coke from the first refrigerator)

P(taking 1 can of coke from the second refrigerator)

=

=

2.

3.

4.

5.

6. (a) (i)

133


(ii)

(b)

7. (a)

(b)

8. (a)

(b)

(c)

9. (a)

(b)

134


10. (a)

(b)

(c)

11.

12. (a)

(b)

(c)

13. (a) P(Betty wins exactly one prize)

(b) P(Betty wins at most one prize)P(Betty wins exactly one prize)

+ P(Betty gets no prize)

Alternative Solution P(Betty wins at most one prize)

1 P(Betty wins two prizes)

14. (a)

(b)

15.

Level 2

16. (a) (i)

135


(ii)

(iii)

(b)

17. Let S stand for a worker who suffers from the occupational disease and N stand for a worker who does not suffer from the occupational disease.(a)

(b)

18. Let A stand for a person who accepts the interview andN stand for a person who does not accept the interview.

(a)

(b)

(c)

19. Let W stand for Kenneth winning a round and L stand for Kenneth losing a round.

(a)

(b) P(win at least two rounds)

20.

21. P(6 heads)

136


P(3 heads followed by 3 tails)

∵ The probabilities of the two cases are the same.∴ May’s claim is incorrect.

22. Let G stand for a goal and M stand for a missed shoot.

(a)

(b)

23. Let B stand for drawing a blue ball and G stand for drawing a green ball.(a)

(b)

24. Let W stand for Bob winning the game, N stand for Bob

not winning the game.

(a)

(b)

∴ The probability that Bob wins at least once is not the same as the probability that Amy wins at most twice.

25. (a)

(b)

∴ The possible values of p are and .

26. (a)

P(congestion on AFED)

∴ Route AFED has a lower chance of having traffic congestion.

Alternative Solution

137


∴ Route AFED has a lower chance of having traffic congestion.

(b) P(Mr Chan will encounter traffic congestion)= P(select route ABCD and encounter traffic congestion) + P(select route AFED and encounter traffic congestion)

=

=

27. P(loses in 50 m breast stroke)

P(loses in 100 m free style)

P(loses in 200 m relay) 1 x

Exercise 11E (p. 11.53)Level 1

1. (a) Total number of girls

P(wears glasses | girl)

(b) Total number of students wearing glasses

P(boy | wears glasses)

2. (a) There are 7 possible outcomes for the condition ‘the cup is green’. 4 of them satisfy the event ‘the number on the cup is even’.

∴ P(even | green cup)

(b) There are 10 possible outcomes for the condition ‘the number on the cup is odd’. 7 of them satisfy the event ‘the cup is red’.

∴ P(red cup | odd)

3. (a) After the Queen of hearts is drawn, there are 51 cards left and 12 of them are hearts.∴ P(2nd card is a heart | 1st card is Queen of hearts)

(b) After the Queen of hearts is drawn, there are 51 cards left and 40 of them are not face cards.∴ P(2nd card is not a face card |1st card is Queen of

hearts)

4. (a) P(2 daughters)

(b) P(2 daughters | not all boys)

138


5. (a) There are 9 possible outcomes for the condition ‘both numbers are even’. 2 of them satisfy the event ‘the product of the two numbers obtained is 12’, i.e. {(2, 6) and (6, 2)}.

(b) There are 4 possible outcomes for the condition ‘the product of the two numbers obtained is 12’. 2 of them satisfy the event ‘both numbers are even’,i.e. {(2, 6) and (6, 2)}.

6. (a)

(b)

7. (a)

(b)

8. (a)

(b)

9. P(1 red clip and 1 blue clip)= P(1st clip is red and 2nd clip is blue) + P(1st clip is blue and 2nd clip is red)= P(1st clip is red) P(2nd clip is blue | 1st clip is red) + P(1st clip is blue) P(2nd clip is red | 1st clip is blue)

=

=

10. (a) The three successive draws are independent since the buttons are replaced.P(1st button is blue, 2nd button is green and 3rd button

is red)

(b) The three successive draws are dependent since the buttons are not replaced.P(1st button is blue, 2nd button is green and 3rd button

is red)

139


11. (a) (i)

(ii)

(b)

12. (a) P(1 bottle of orange juice and 1 bottle of tea are taken out)= P(1st bottle is orange juice) P(2nd bottle is tea | 1st bottle is orange juice) + P(1st bottle is tea) P(2nd bottle is orange juice | 1st bottle is tea)

=

=

(b) P(same type of drink are taken out)

= P(2 bottles of tea) + P(2 bottles of coffee) + P(2 bottles of orange juice)= P(1st bottle is orange juice) P(2nd bottle is orange juice | 1st bottle is

orange juice) + P(1st bottle is tea)

P(2nd bottle is tea | 1st bottle is tea) + P(1st bottle is coffee)

P(2nd bottle is coffee | 1st bottle is coffee)

=

=

13. (a)

(b) P(opens the door on the 2nd attempt)= P(fails on the 1st attempt) P(succeeds on the 2nd attempt | fails on

the 1st attempt)

=

=

(c) P(opens the door within three attempts)= 1 P(fails to open the door within three attempts)

=

=

Alternative SolutionP(opens the door on the 3rd attempt)= P(fails on the first 2 attempts) P(succeeds on the 3rd attempt | fails on the first

2 attempts)

=

=

P(opens the door within three attempts)= P(opens the door on 1st attempt) + P(opens the door on 2nd attempt) + P(opens the door on 3rd attempt)

=

=

140


14.

15.

Level 216. (a) P(the letters are the same)

= P(both letters are ‘O’) + P(both letters are ‘N’) + P(both letters are ‘G’)= P(1st letter is ‘O’) P(2nd letter is ‘O’ | 1st letter is ‘O’) + P(1st letter is ‘N’) P(2nd letter is ‘N’ | 1st letter is ‘N’) + P(1st letter is ‘G’) P(2nd letter is ‘G’ | 1st letter is ‘G’)

=

=

(b) P(the letters are different)P(the letters are the same)

17. P(a particular pair of socks is drawn)

P(a matching pair)

18.

19. Let P1 ,P2 and P3 be the events that the 1st, the 2nd and the 3rd selected student pass the test respectively, F1 ,F2 and F3 be the events that the 1st , the 2nd and the 3rd selected student fail the test respectively.

20. Let A, B, C and D be the events that Andrew, Betty, Calvin and Doris draw the princess sticker respectively.

P(A) =

P(B)=

=

=

P(C)=

=

=

∴ All of them have equal chance of getting the princess stickers.

21. (a) P(two face cards are drawn)= P(1st card is a face card) P(2nd card is a face card | 1st card is a face card)

=

=

(b) P(two spade cards are drawn | two face cards are drawn)

141


22. (a) P(two gold coins)= P(1st coin is a gold coin) P(2nd coin is a gold coin | 1st coin is a gold coin)

=

=

(b)

(c) P(two gold coins | two coins are of the same type)

23. Let K1 and K2 be the events that the 1st and the 2nd towels drawn are black respectively, Y be the event that the 1st towel drawn is yellow.P(the towel in drawer C is black)= P(K1) P(K2 | K1) + P(Y) P(K2 | Y)

=

=

24. (a) (i)

(ii)

(b)

142


25. Let X and Y stand for drawing a red ball and a black ball respectively.

(a)

(b)

26. (a) (i) P(2 boys are taller than 1.75 m and neither of them plays basketball)

=

=

(ii) P(2 boys are not taller than 1.75 m and neither of them plays basketball)

=

=

P(1 boy is taller than 1.75 m while 1 boy is not, and neither of them plays basketball)

=

=

P(neither of them plays basketball)= P(2 boys are taller than 1.75 m and

neither of them plays basketball) + P(2 boys are not taller than 1.75 m and

neither of them plays basketball) + P(1 boy is taller than 1.75 m while 1 boy is

not, neither of them plays basketball)

=

=

(b)

P(both of them play basketball)= P(2 boys are taller than 1.75 m and both of them play basketball) + P(2 boys are not taller than 1.75 m and both of them play basketball) + P(1 boy is taller than 1.75 m while 1 boy is

not, both of them play basketball)

=

=

∴ The probability of choosing two boys who play basketball is greater than the probability of choosing two boys who do not play basketball.

∴ Natalie’s claim is incorrect.

143


Exercise 11F (p. 11.62)Level 1

1. (a) Number of possible combinations

(b) There are 3 vowels (‘E’, ‘O’ and ‘I’) in the word.

2. (a) Number of possible passwords formed

(b)

3. Number of ways of drawing 5 balls from 18 balls

(a) Number of ways of drawing 5 yellow balls

(b) Number of ways of drawing 3 red balls

Number of ways of drawing 2 yellow balls

4. Number of possible four-letter strings formedThere are 3 vowels (‘A’, ‘I’ and ‘E’) and 5 consonants (‘P’, ‘M’, ‘N’, ‘S’ and ‘T’) in the set.Number of ways of forming a string that 2 consonants for the first 2 letters and 2 vowels for the last 2 letters

P(first two letters are consonants and the last two letters are vowels)

5. Number of ways of taking 3 cartons of milk from 20 cartons of milk

(a) Number of ways of taking 3 non-expired cartons of milk

(b) Number of ways of taking 1 expired carton of milk

Number of ways of taking 2 non-expired cartons of milk

6. Number of ways of drawing 3 cards from 12 cards(a) ∵ There are 4 cards marked with numbers which are

multiples of 3.∴ Number of ways of drawing 3 cards which are not

multiples of 3

(b)

144


7. Number of ways of choosing 4 toys cars from 17 toy cars

(a) Number of ways of choosing 4 black cars

(b) Number of ways of choosing 4 red cars

Number of ways of choosing 4 yellow cars

8. Total number of arrangements of all the 8 peopleSince Jayden and Zoe must sit together, we can treat them as one unit. For example,

J Z F1 F2 F3 F4 F5 F6

Number of ways of arranging the 6 other friends and 1 unit

Number of ways of arranging Jayden and Zoe

∴ P(Jayden and Zoe sat together)

9. Number of ways of selecting 3 people from 11 people

(a) Number of ways of selecting 2 doctors

Number of ways of selecting 1 dentist

(b) Number of ways of selecting 1 doctor

Number of ways of selecting 2 dentists

(c)

10. (a) Number of ways of drawing 4 batteries from 13 batteries

Number of ways of selecting 4 new batteries

P(all the batteries drawn are new batteries)

(b)

145


11. Number of ways of choosing 4 photos from 22 photos

(a) Number of ways of choosing the other 3 photos

P(a particular photo is included)

(b) Number of ways of choosing 2 colour photosNumber of ways of choosing 2 black-and-white photos

P(exactly 2 colour photos are selected)

12. Number of ways of choosing 5 students from 20 students

(a) Number of ways of choosing 2 students (except Mary, Emily and John)

P(Mary, Emily and John are all chosen)

(b) Number of ways of choosing 3 students (except Mary, Emily and John) from 17 students

P(Mary and Emily are chosen but not John)

13. Total number of ways of arranging the 7 people(a) We can treat 3 women as one unit and 4 men as

another unit. For example,

W1W3W2 M1M2M3M4

Number of ways of arranging the 3 womenNumber of ways of arranging the 4 men

∴ P(3 women followed by 4 men)

(b) We can treat the particular man and woman as oneunit. For example,

WpMp M1 M2 M3 W1 W2

Number of ways of arranging 5 people and 1 unit

∴ P(a particular man will speak immediately after a particular woman)

14. Total number of ways of arranging the 6 students(a) Consider Wendy and Cherry seat at the two end of the

row. For example,

W S1 S2 S3 S4 C

Number of ways of arranging Wendy and Cherry

Number of ways of arranging the other 4 students

∴ P(Wendy and Cherry seated at the two end of the row)

(b) Consider Wendy and Cherry are seated next to each other. We can treat Wendy and Cherry as a unit. For example,

W C S1 S2 S3 S4

Number of ways of arranging Wendy and CherryNumber of ways of arranging the other 4 students and 1 unit

∴ P(Wendy and Cherry are not seated next to each other)

Level 215. Number of ways of drawing 5 teapots from 18 teapots

(a) Number of ways of drawing 5 non-cracked teapots

146


(b) Number of ways of drawing 1 cracked teapot

Number of ways of drawing 4 non-cracked teapots

16. Number of ways of selecting 4 songs from 17 songs

(a) Number of ways of selecting 3 old songs

Number of ways of selecting 1 new song

Number of ways of selecting 4 old songs

(b)

17. Total number of ways of arranging the 9 people(a) We can treat the 3 pianists as a unit. For example,

P1P2P3 S1 S2 S3 S4 S5 S6

Number of ways of arranging the 6 singers and 1 unit

Number of ways of arranging the 3 pianists∴ P(all pianists stand next to each other)

(b) Consider the 3 pianists as a unit which is in the middle of the row. For example,

S1 S2 S3 P1P2P3 S4 S5 S6

Number of ways of arranging the 3 pianistsNumber of ways of arranging the 6 singers∴ P(all pianists stand in the middle of the row)

18. Total number of ways of arranging the 9 people(a) We can treat the 5 coaches as a unit and the 4 trainees

as another unite. For example,

C1C2C3C4C5 T1T2T3T4

Number of ways of arranging the 2 unitsNumber of ways of arranging the 5 coachesNumber of ways of arranging the 4 trainees

(b) Since the coaches and the trainees sit alternately, they must sit in an arragement like this:

C1 T1 C2 T2 C3 T3 C4 T4 C5

Number of ways of arranging the 5 coachesNumber of ways of arranging the 4 trainees

19. (a) Total number of possible arrangements

147


We can treat the 3 particular flags as a unit. For example,

P1P2P3 F1 F2 F3 F4 F5

Number of ways of arranging the other 5 flags and 1 unitNumber of ways of arranging the 3 particular flags

∴

Any 2 of the 3 particular flags cannot be hoisted next to each other. In other words, these 3 flags must be separated by the remaining 5 flags.

There are 6 possible positions for these 3 flags.Number of ways of arranging the 3 particular flags

Number of ways of arranging the other 5 flags

∴

(b) ∵

∴ E and F are not complementary events.

20. Number of ways of taking 3 fruits from 17 fruits

(a) Number of ways of taking 3 oranges

Number of ways of taking 3 pears

Number of ways of taking 3 mangoes

(b) Number of ways of taking 1 orange

Number of ways of taking 1 pear

Number of ways of taking 1 mango

(c)

21. Let A, B and C stand for attending lessons for a subject taught in building A, B and C respectively.Number of ways of choosing 4 subjects from 10 subjects

(a) Number of ways of choosing 2A1B1C =

148


Number of ways of choosing 1A2B1C =

Number of ways of choosing 1A1B2C =

(b) Number of ways of choosing 2A1B1C =

Number of ways of choosing 2A2B =

Number of ways of choosing 2A2C =

22. Number of ways of choosing 5 people from 10 people

(a) Number of ways of choosing 5 childrenNumber of ways of choosing 4 children and 1 adult

Number of ways of choosing 3 children and 2 adults

(b) Number of ways of choosing 2 adults and 3 children

23. Number of ways of arranging the 9 booksNo two cookery books are next to each other. In other words, these 3 cookery books must be separated by the remaining 6 music books.

There are 7 possible positions for these 3 cookery books.

Number of ways of arranging the 6 music books

Number of ways of arranging the 3 cookery books

24. Total number of 4-digit numbers formed(a) To form a 4-digit number which is greater than or

equal to 5000, the first digit must be greater than or equal to ‘5’, i.e. the number of ways of choosing the first digit is∴ Number of possible 4-digit numbers which is

greater than or equal to 5000

149


∴ P(greater than or equal to 5000)

(b) There are two cases to form a 4-digit number which is less than 5400.Case (i): The first digit is less than or equal to ‘4’.

∴ The number of possible 4-digit numbers formed under this case

Case (ii): The first digit is ‘5’ and the second digit is less than ‘4’.∴ The number of possible 4-digit numbers

formed under this case

∴ P(less than 5400)

Check Yourself (p. 11.69)

1.

2.

3.

(a)

(b)

(c)

150


4. (a)

(b)

(c)

5. (a)

(b)

6. (a)

(b)

7. Number of ways of choosing 4 toys from 9 toys

Number of ways of choosing 2 teddy bears

Number of ways of choosing 2 robots

151


Revision Exercise 11 (p. 11.70)Level 11. Total number of workers = 18 + 27 + 3 + 24 + 11 + 7 = 90

(a)

(b)

(c)

2. ‘Made in Korea’, ‘made in China’ and ‘made in Hong Kong’ are mutually exclusive events.

(a)

(b)

3. (a)

(b)

(c)

4. (a)

(b)

5. (a) P(passes at least one of the subjects)= P(passes Mathematics) + P(passes English) – P(passes both subjects)

(b) P(fails both subjects)= 1 P(passes at least one of the subjects)

6. (a)

(b) P(at least 1 of the mobile phones malfunction within 1 year)= 1 P(no mobile phones malfunction within 1 year)= 1 0.7225

=

7. (a)

152

Cannot bounce well

Wrong company logos

15

5

10


(b)

(c)

8. (a)

(b) (i)

(ii)

9. In order to successfully book a table at their favourite restaurant and he is not late for the dinner, Tom needs to remember to book a table, their favourite restaurant is not fully booked and he is not late for the dinner.

∴ P(Tom successfully books a table at their favourite restaurant and he is not late for the dinner)

153


10. (a)

(b)

11. (a) P(divisible by 3 | number on the 1st bear is 2)

(b)

12. (a)

(b)

(c)

13. (a)

(b)

154


14. (a) P(Natalie wins in the first round)= P(Michelle cannot get a ‘6’ in the first round and

Natalie gets a ‘6’ in the first round)

=

=

(b)

15.

16.

17. (a) P(mustard | ketchup)

(b) P(ketchup | mustard)

(c) P(neither ketchup nor mustard)

18. (a) Number of ways of selecting 5 staff from 18 staff

Number of ways of selecting 5 female staff

(b) Number of ways of selecting 1 male staff

Number of ways of selecting 4 female staff

(c)

19. (a) Number of ways of arranging the 11 people

Number of ways of arranging the other 10 people

P(a particular boy sits on either end)

155


(b) We can treat the 6 boys as a unit and the 5 girls as another unit. For example,

B1B2B3B4B5B6 G1G2G3G4G5

Number of ways of arranging the 2 unitsNumber of ways of arranging the 6 boysNumber of ways of arranging the 5 girls

∴ P(all the boys are on one side)

20. (a) Number of ways of arranging the 5 lettersNumber of ways of choosing a vowel for the 1st letter

Number of ways of arranging the other 4 letters

∴ P(1st letter is a vowel)

(b) Consider the consonants and vowels form an alternating pattern. Then the 1st, 3rd and 5th letters are consonants and the 2nd and 4th letters are vowels.Number of ways of arranging the 3 consonantsNumber of ways of arranging the 2 vowelsP(the consonants and vowels form an alternating pattern)

21. (a) P(the first prize)

(b) P(the second prize)

Level 222. (a) P(obtains level 5 in all 3 subjects)

(b) P(obtains level 5 in exactly 2 subjects)

(c) P(obtains level 5 in at most one subject)

23. (a) Number of students that have not joined the English club or the art club

P(none of them has joined the English club or the art club)

(b) Number of students that have joined either the English club or the art club

P(only one of them has not joined any club)

24. (a) Let C and V stand for taking a chocolate cupcake and a vanilla cupcake respectively.

(b)

156


25. (a)

(b)

(c) P(team A and team B win alternately)

26. (a) (i)

(ii)

(b)

∴ The principal is correct.

27. (a)

(b) Let F stand for a boy getting the flu and N stand for a boy not getting a flu.

28. The favourable outcomes are listed below:Colour Favourable outcomes

(the sum of the numbers obtained is odd)Blue nil

Yellow (2, 3), (2, 5), (3, 2), (3, 10), (5, 2), (5, 10),(10, 3), (10, 5)

Pink (4, 9), (6, 9), (8, 9), (9, 4), (9, 6), (9, 8)∴ Total number of favourable outcomes = 8 + 6 = 14

Total number of possible outcomes = 10 10 = 100

∴ P(same colour and sum is odd)

29. Number of ways of drawing 2 eggs from 8 eggs

Number of ways of drawing 2 good eggs

Number of ways of drawing 1 good egg

Number of ways of drawing 1 rotten egg

157


∴

30. (a) The possible pairs of numbers with sum equal to 8 are (1, 7), (3, 5), (4, 4), (5, 3) and (7, 1).

(b) The possible pairs of numbers with sum equal to 8 and the first number is greater than second number are (5, 3) and (7, 1).

31. (a) P(Peter wins the game within two draws by him)= P(Peter wins the game in his 1st draw) + P(Peter wins the game in his 2nd draw)

=

=

(b)

Since there are only 5 white balls, John will definitely win the game if Peter cannot win in his 3rd draw.∴ P(Peter wins the game)

∴Peter have a greater chance of winning.

32. Number of ways of drawing 4 glasses from 20 glasses

Number of ways of drawing 4 non-defective glassesThe required probability

33. Let H stand for a head and T stand for a tail.(a)

158


(b)

34. (a)

(b)

(c)

35. (a)

(b)

36. (a) Number of ways of drawing 5 balls from 25 balls

Number of ways of drawing 1 red ball

Number of ways of drawing 4 non-red balls

(b) Number of ways of drawing 2 red balls

Number of ways of drawing 3 non-red balls

159


37. Total number of ways of arranging the 6 people(a) Consider the 2 boys are the first 2 persons in the

queue. For example,

B1 B2 G1 G2 G3 G4

Number of ways of arranging the 2 boysNumber of ways of arranging the 4 girls∴ P(the 2 boys are the first 2 persons in the queue)

(b) Consider the two boys are the first and the last persons in the queue. For example,

B1 G1 G2 G3 G4 B2

Number of ways of arranging the 2 boysNumber of ways of arranging the 4 girls∴ P(the first and the last persons in the queue are

both boy)

(c) We can treat the 2 boys as a unit. For example,

G1 G2 G3 G4 B1B2

Number of ways of arranging the 2 boysNumber of ways of arranging the 4 girls and 1 unit

∴

(d) We can treat the 4 girls as a unit. For example,

G1G2G3G4 B1 B2

Number of ways of arranging the 4 girlsNumber of ways of arranging the 2 boys and 1 unit

∴

38. (a) Number of ways of choosing 4 students from 30 students

Number of ways of choosing 4 universities from 10 universities

Number of ways of choosing 1 student from 3 students of a university

P(all from different universities)

(b) Number of ways of choosing 3 universities from 10 universities

Number of ways of choosing 1 university which has 2 students in the study group

Number of ways of choosing 2 students from 3 students of a university

P(from 3 different universities)

(c) P(from at least 3 different universities)= P(all from different universities) + P(from 3 different universities)

=

=

39. (a)

160


(b) (i)

(ii)

(iii)

40. (a) (i)

(ii)

161


(b) (i) Number of students passing paper I only

Number of students passing paper II only

Number of students passing 1 paper only

Number of ways of choosing 2 students from 45 students

Number of ways of choosing 2 students from 21 students who passed 1 paper only

(ii) Number of ways of choosing 2 students from25 students who passed paper II

Number of ways of choosing 1 student from 9 students who passed paper II only

Number of ways of choosing 1 student from 16 students who passed both papers

41. Let L, R and G stand for turning left, turning right and going straight at a junction respectively, and N, E, S and W

stand for walking due North, East, South and West at a junction respectively.

(a) (i)

(ii)

(iii)

(b) (i) Number of ways of choosing 1 tourist who arrives at park S

(ii)

162


42. (a) (i)

(ii)

(b)

(c) (i)

∴ It is possible that team A gets total scores of 4 points but they are not the only champion.

(ii)

163


Multiple Choice Questions (p. 11.78)1. Answer: A

2. Answer: D Number of white and even number cards = 8

3. Answer: D

4. Answer: C Let C stand for a correct answer and W stand for a wrong

answer.P(only one correct answer) P(CWW) P(WCW) P(WWC)

5. Answer: BP(same entrance)

6. Answer: D

7. Answer: BLet W stand for a white chopstick and B stand for a black chopstick.P(same colour)

8. Answer: ANumber of ways of choosing 2 balls from 5 balls∵ Sum of the numbers of all the balls

= 1 + 2 + 3 + 4 + 5 = 15∴ There are 4 possible outcomes for the sum of the

numbers on Jennifer’s balls is greater than or equal to 8, i.e. (3, 5), (4, 5), (5, 4) and (5, 3).

∴ P(the sum of the numbers on Jennifer’s balls is greater than that on Kimmy’s)= P(the sum of the numbers on Jennifer’s balls is greater than or equal to 8)

=

=

9. Answer: BTotal number of rotten oranges∴ P(comes from box A | rotten orange)

10. Answer: BThere are 3 letters which are the same in the words ‘ACTION’ and ‘EQUALITY’. They are ‘A’, ‘T’ and ‘I’. P(2 letters chosen are the same)

164


P(both ‘A’) + P(both ‘T’) + P(both ‘I’)

P(2 letters chosen are different)= 1 P(2 letters chosen are the same)

=

=

11. Answer: CLet L stand for turning left and R stand for turning right.

12. Answer: B

13. Answer: DNumber of ways of arranging the four students = 4!P(names are called in alphabetical order)

14. Answer: BNumber of ways of arranging the seven people = 7!We can treat Mr and Mrs Cheung as a unit. For example,

F1 F2 F3 F4 F5 Mr Cheung, Mrs Cheung

Number of ways of arranging Mr and Mrs CheungNumber of ways of arranging the other 5 people and 1 unit

∴

15. Answer: C Number of ways of choosing 4 eggs from 12 eggs

Number of ways of choosing 2 rotten eggs

Number of ways of choosing 2 non-rotten eggsP(exactly 2 eggs chosen are rotten)

HKMO (p. 11.79)1. Number of ways of drawing 3 numbers from 6 numbers

2.

165


∵

∴ The value of B is 3.

3.

4. P(white ball = black ball) =

Let P(white ball > black ball) = x.Then P(white ball < black ball) = x.P(white ball > black ball) + P(white ball = black ball) + P(white ball < black ball) = 1

∴ P(white ball > black ball) =

Alternative SolutionBy the counting principle, the total number of possible outcomes is 10 10 100.

White ball drawn

Favourable black ball drawn

Number of favourable outcomes

10 1, 2, 3, 4, 5, 6, 7, 8, 9 99 1, 2, 3, 4, 5, 6, 7, 8 88 1, 2, 3, 4, 5, 6, 7 77 1, 2, 3, 4, 5, 6 66 1, 2, 3, 4, 5 5

5 1, 2, 3, 4 44 1, 2, 3 33 1, 2 22 1 11 – 0

Total 45

∴ P(white ball > black ball)

Exam Focus

Exam-type Questions (p. 11.81)

1. (a) Number of ways of selecting 4 cats from 15 cats

Number of ways of selecting 1 white cat

Number of ways of selecting 1 brown cat

Number of ways of selecting 1 grey cat

Number of ways of selecting 1 black cat

(b) Number of ways of selecting 4 white cats

Number of ways of selecting 4 brown cats

(c)

2. Total number of ways of arranging the 10 people(a) Since John and Mary are next to each other, we can

treat them as a unit. Since Alex and Elaine are next to each other, we can treat them as another unit. For example,

JM AE P1 P2 P3 P4 P5 P6

166


Number of ways of arranging John and MaryNumber of ways of arranging Alex and ElaineNumber of ways of arranging the other 6 people and 2 units

∴

(b) Consider John, Mary, Alex and Elaine are all sit in the front row. For example,

P2 P3 P4 P5 P6

JM AE P1

Number of ways of arranging John and MaryNumber of ways of arranging Alex and ElaineNumber of ways of arranging 1 person and 2 units in the front row

Number of ways of arranging the other 6 people∴

3. (a) Number of ways of forming the 1st group

Number of ways of forming the 2nd group

Number of ways of forming the 3rd group

Number of ways of forming the 4th groupNumber of ways of grouping the students

(b) Consider each group has one boy only.Number of ways of forming the 1st groupNumber of ways of forming the 2nd group

Number of ways of forming the 3rd group

Number of ways of forming the 4th group

4. (a) Mean =

=

167


Standard deviation =

kg11.72840806 kg

=

(b) Let x kg be the weight with standard score greater than 0.8.

∴ There are 5 people whose weights have standard scores greater than 0.8.Number of ways of selecting 2 people from 18 people

Number of ways of selecting 2 people whose weights have standard scores greater than 0.8

(c) (i) ∵ The mean of the weights of the remaining 17 people in the group is higher than 72 kg.

∴ The weight of the withdrawn person must be less than 72 kg.

There are 8 people whose weights are less than 72 kg.The required probability

(ii) To keep the mean weight unchanged after withdrawing 2 people, their mean weight must be equal to 72 kg.There are 4 favourable pairs of people,

i.e. (71, 73), (68, 76), (62, 82) and (62, 82).The required probability

5. Answer: C

168


6. Answer: CLet B stand for drawing a black ball and W stand for drawing a white ball.

7. Answer: C

8. Answer: A

Investigation Corner (p. 11.86)

(a) (i)

(ii)

(b) (i) He can put 1 red ball in the 1st box, 1 red ball in the 2nd box and then the rest in the 3rd box.

(ii)

(c) (i) Case I: 50 boxes should be requested. Put a red ball in each of the first 49 boxes and 1 red ball and all white balls in the last box.

ORCase II: 51 boxes should be requested. Put a red ball

in each of the first 50 boxes and all white balls in the last box.

(ii) Case I: 50 boxes are given.P(win)

Case II: 51 boxes are given.P(win)

169


170

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