5bch15(more About Probability)

15 More about Probability

15 More about Probability • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Activity

4.

Number of combinations = 4 × 3 = 12 1 P(tuna salad and soft drink) = 12

5.

(a)

Activity 15.1 (p. 11) 1.

Experiment Throwing a die once Drawing a card from a deck of 52 cards at random Choosing a letter from the word ‘TRICK’

Event A Obtaining ‘1’ Getting an odd number Drawing a king

Event B Obtaining ‘2’ Getting a prime number Drawing the ace of hearts Drawing a spade

Drawing a queen The letter is ‘T’ The letter is ‘I’

The letter is ‘I’ The letter is a vowel

Are A and B mutually exclusive? (/ )

P(A)

P(B)

P(A or B)

P(A) + P(B)



1 6

1 6

1 3

1 3



1 2

1 2

2 3

1



1 13

1 52

5 52

5 52



1 13

1 4

4 13

17 52



1 5

1 5

2 5

2 5



1 5

1 5

1 5

2 5

2.

P ( A or B ) = P ( A) + P( B )

3.

No

Activity 15.2 (p. 19) 1.

1 P(tuna salad) = 4

2.

1 P(soft drink) = 3

3.

Yes

1 1 1 = × 12 4 3 (b) P ( A and B ) = P ( A) × P ( B )

Follow-up Exercise p. 7 1.

(a) ∵ There are 13 hearts in the deck. ∴ Number of outcomes favourable to the event = 13 13 = ∴ P(heart) 52 1 = 4 (b) ∵ There are 4 queens in the deck. ∴ Number of outcomes favourable to the event = 4 4 = ∴ P(queen) 52 1 = 13

2.

(a) Since there are 4 possible outcomes in choosing a letter from the word ‘LIST’ and 6 possible outcomes in choosing a letter from the word ‘STUPID’, by the counting principle, the total number of possible outcomes in choosing a letter from each of the words is: 4 × 6 = 24 There are 3 favourable outcomes: II, SS and TT 3 = 24 ∴ P(same letter) 1 = 8

1

Certificate Mathematics in Action Full Solutions 5B

(b) There are 3 consonants in the word ‘LIST’ and 4 consonants in the word ‘STUPID’. By the counting principle, the total number of favourable outcomes in choosing a consonant from each of the words is: 3 × 4 = 12 12 = ∴ P(both consonants) 24 1 = 2 3.

Let C stand for a correct answer and W stand for a wrong answer. By using a tree diagram, we have: First Second Third Outcomes question question question C CCC C W CCW C C CWC W W CWW C

WCC

W

WCW

C

WWC

W

WWW

(c) From the table, there are 8 times getting a ‘1’, 33 times getting a ‘3’ and 21 times getting a ‘5’. ∴ Number of times getting an odd number = 8 + 33 + 21 = 62 62 = 100 ∴ P(odd number) 31 = 50 p. 14 1.

(b) There are 2 favourable outcomes: 24 and 28 2 = ∴ P(divisible by 4) 10 1 = 5

C W

(c) P(odd number or divisible by 4) = P (odd number) + P (divisible by 4) 1 1 = + 2 5 7 = 10

W From the tree diagram, there are 8 possible outcomes. (a) There is 1 favourable outcome for getting 3 correct answers, i.e. CCC. 1 ∴ P(3 correct answers) = 8 (b) There are 3 favourable outcomes for getting 2 wrong answers, i.e. CWW, WCW and WWC. 3 ∴ P(2 wrong answers) = 8 4.

(a) From the table, there are 24 times getting a ‘2’. 24 = ∴ P(‘2’) 100 6 = 25 (b)

2

Total number of possible outcomes = 10 (a) There are 5 favourable outcomes: 21, 23, 25, 27 and 29 5 = ∴ P(odd number) 10 1 = 2

2.

Total number of people in the group = 16 + 12 + 12 + 21 + 28 + 16 + 4 + 11 = 120 (a) P(man with blood type ‘AB’ or woman with blood type ‘A’) = P(man with blood type ‘AB’) + = P(woman with blood type ‘A’) 28 12 = + 120 120 =

1 3

(b) P(‘A’ or ‘B’ or ‘AB’) = P(‘A’) + P(‘B’) + P(‘AB’) 16 + 12 12 + 21 28 + 16 = + + 120 120 120 105 = 120 7 = 8

15 More about Probability

3.

Let G stand for a green ball, R stand for a red ball, B stand for a blue ball and W stand for a white ball. By the counting principle, the total number of possible outcomes in drawing two balls is: 3 × 4 = 12 (a) There is 1 favourable outcome: GG 1 ∴ P(both green) = 12 (b) There are 2 favourable outcomes: RR and RR 2 = ∴ P(both red) 12 1 = 6 (c) There are 3 favourable outcomes: GR, GR and RG 3 = ∴ P(one green and one red) 12 1 = 4

p. 24

1.

= P (prime) × P(prime) 3 3 P(both prime) = 6 × 6 1 = 4

2.

= P (odd and even or even and odd) = P (odd and even) + P(even and odd) = P (odd) × P (even) + P (even) × P (odd ) P(sum is odd) 1 1 1 1 = × + × 2 2 2 2 1 = 2

3.

= 1 − P (fail Chinese) × P (fail English)

p. 17 1. (a) A’ = getting an odd number (b) A’ = getting a heart, diamond or club (c) A’ = the man does not have a private car 2.

3.

= 1 − P (wear glasses) P(does not wear glasses) = 1 − 0.4 = 0.6 By the counting principle, the total number of possible outcomes in throwing three dice is: 6 × 6 × 6 = 216 The complementary event of ‘product is greater than 2’ is ‘product equals to 1 or 2’. ∵

∴

∴

There are 4 favourable outcomes that the product of the three numbers equals to 1 or 2, i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1) and (2, 1, 1). 4 = P(product equals to 1 or 2) 216 1 = 54 P(product greater than 2) = 1 − P (product equals to 1 or 2) 1 = 1− 54 53 = 54

P(Amy passes at least one subject) = 1 − P (Amy fails both subjects) = 1 − [1 − P( pass Chinese)] × [1 − P (pass English )] = 1 − (1 − 0.75) × (1 − 0.6) = 0.9

4.

1 1 = 2 2 3 1 P(lose the 200 m race) = 1 − = 4 4 1 7 P(lose the 400 m race) = 1 − = 8 8 Let W stand for winning a race and L stand for losing a race. P(Harry wins exactly one race) = P ( WLL) + P (LWL) + P( LLW) 1 1 7 1 3 7 1 1 1 = × × + × × + × × 2 4 8 2 4 8 2 4 8 7 21 1 = + + 64 64 64 29 = 64 P(lose the 100 m race) = 1 −

p. 29 1. Let C stand for a correct answer and W stand for a wrong answer. Since Anthony answers the first question correctly, the possible outcomes are CCC, CCW, CWC and CWW. Among the 4 possible outcomes, only CCW and CWC are favourable outcomes. ∴ P(2 of 3 questions are correct | 1st question is correct) 2 = 4 1 = 2

3

Certificate Mathematics in Action Full Solutions 5B

2.

(a) ∵ ∴

The first card drawn is the ace of spades.

2.

Only 13 hearts are left after the first draw. Only 51 cards are left after the first draw.

∴ P(heart | the first card drawn is the ace of spades) 13 = 51 (b) ∵

The first card drawn is the ace of spades.

Only 3 aces are left after the first draw. Only 51 cards are left after the first draw. ∴ P(ace | the first card drawn is the ace of spades) ∴

(b) P(same colour) = P (BB or EE or WW ) = P ( BB) + P (EE ) + P ( WW ) 2 = + P (1st one is blue) × P (2nd one is blue | 15 1st one is blue) + P(1st one is white) × P( 2nd one is white | 1st one is white) 2 2 1 4 3 = + × + × 15 10 9 10 9 13 = 45

3 51 1 = 17 =

3.

(a) Total number of students = 425 + 218 + 157 = 800 218 = ∴ P(disagree | student) 800 109 = 400

3.

(b) Total number of people agree = 425 + 92 = 517 92 ∴ P(teacher | agree) = 517 p. 32 1. (a) P(both defective) = P (1st one is defective) × P (2nd one is defective | 1st one is defective) 15 14 = × 100 99 7 = 330 (b) P(both non-defective) = P (1st one is non-defective) × P( 2nd one is non-defective | 1st one is non-defective) 85 84 × 100 99 119 = 165 =

4

Let B denote a black glove, E denote a blue glove and W denote a white glove. (a) P(BB) = P(1st one is black) × P( 2nd one is black | 1st one is black) 4 3 = × 10 9 2 = 15

P(colour-blind) = P (male and colour-blind or female and colour-blind) = P ( male and colour-blind) + P (female and colour-blind) = P ( male) × P (colour-blind | male) + P (female) × P (colour-blind | female) = 0.62 × 0.03 + (1 − 0.62) × 0.02 = 0.0262

Exercise Exercise 15A (p. 8) Level 1 1.

Total number of possible outcomes = 3 + 4 + 5 = 12 (a) ∵ The bag contains 3 red balls. ∴ Number of outcomes favourable to the event = 3 3 = ∴ P(red ball) 12 1 = 4 (b) ∵ The bag contains 4 white balls. ∴ Number of outcomes favourable to the event = 4 4 = ∴ P(white ball) 12 1 = 3

15 More about Probability

(c)

∵ The bag contains 0 green balls.

5.

∴ Number of outcomes favourable to the event = 0 0 ∴ P(green ball) = 12 =0 2.

(b) There are 6 favourable outcomes: 4, 8, 12, 16, 20 and 24 6 = ∴ P(multiple of 4) 24 1 = 4

Total number of possible outcomes = 6 (a) There are 4 favourable outcomes: 1, 2, 3 and 4 4 = ∴ P(not greater than 4) 6 2 = 3

(c) There are 2 favourable outcomes: 12 and 24 2 = ∴ P(common multiple of 3 and 4) 24 1 = 12

(b) There are 3 favourable outcomes: 1, 2 and 3 3 = ∴ P(less than 4) 6 1 = 2 3.

Total number of possible outcomes = 10

6.

(a) ∵ The word contains 2 ‘L’. ∴ Number of outcomes favourable to the event = 2 2 = 10 ∴ P(‘L’) 1 = 5 ∴ Number of outcomes favourable to the event = 5 5 = 10 ∴ P(vowel) 1 = 2

∴ Number of boxes containing less than three defective light bulbs = 4 + 15 + 11 = 30 30 = ∴ P(less than three defective light bulbs) 40 3 = 4

(a) There is only 1 favourable outcome. 1 ∴ P(jack of clubs) = 52 (b) ∵ There are 9 number cards in each suit and there are 4 suits. ∴ Number of outcomes favourable to the event = 9× 4 = 36 36 = ∴ P(number card) 52 9 = 13

(a) From the table, there are 4 boxes containing no defective light bulbs. 4 = 40 ∴ P(no defective light bulbs) 1 = 10 (b) From the table, there are 11 boxes containing two defective light bulbs, 15 boxes containing one defective light bulb and 4 boxes containing no defective light bulbs.

(b) ∵ The word contains 5 vowels.

4.

(a) There are 8 favourable outcomes: 3, 6, 9, 12, 15, 18, 21 and 24 8 = 24 ∴ P(multiple of 3) 1 = 3

7.

(a) From the table, there are 115 times getting a ‘2’. 115 = 1000 ∴ P(‘2’) 23 = 200

5

Certificate Mathematics in Action Full Solutions 5B

(b) From the table, there are 115 times getting a ‘2’, 103 times getting a ‘3’ and 118 times getting a ‘5’. ∴ Number of times getting a prime number = 115 + 103 + 118 = 336 336 = 1000 ∴ P(prime number) 42 = 125 8.

Total number of balls = 6, red ball = 2, blue ball = 3 or total number of balls = 12, red ball = 4, blue ball = 6 (or any other reasonable answers)

9.

It is uncertain. Their probabilities are equal if bag A and bag B contain the same number of balls.

Level 2 10. Number of times the tail shows up = 80 − 32 = 48 48 = ∴ P(getting a tail) 80 3 = 5 11. Let B stand for a boy and G stand for a girl. By using a tree diagram, we have: First child Second child Third child Outcomes B BBB B G BBG B B BGB G G BGG B

GBB

G

GBG

B

GGB

G

GGG

B G G From the tree diagram, there are 8 possible outcomes. (a) There is 1 favourable outcome for all boys, i.e. BBB. 1 ∴ P(all boys) = 8 (b) There are 3 favourable outcomes for two boys and one girl, i.e. BBG, BGB and GBB. 3 ∴ P(two boys and one girl) = 8

6

12.

15 More about Probability

13. Let W stand for a white straw, R stand for a red straw and G stand for a green straw. By using a tree diagram, we have: Bag A Bag B Outcomes W WW W WW W R WR R WR G WG W RW W RW R R RR R RR G RG W GW W GW G R GR R GR G GG W GW W GW G R GR R GR G GG From the tree diagram, there are 20 possible outcomes. (a) There are 6 favourable outcomes for two straws of the same colour, i.e. WW, WW, RR, RR, GG and GG. 6 = ∴ P(same colour) 20 3 = 10 (b) There are 5 favourable outcomes for one red and one green straw, i.e. RG, GR, GR, GR and GR. 5 = 20 ∴ P(one red and one green) 1 = 4 13. Since there are 6 possible outcomes in throwing a die once, by the counting principle, the total number of possible outcomes in throwing two dice is: 6 × 6 = 36 (a) There are 6 favourable outcomes: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) 6 = ∴ P(the sum is 7) 36 1 = 6 (b) There are 4 favourable outcomes: (3, 6), (4, 5), (5, 4) and (6, 3) 4 = 36 ∴ P(the sum is 9) 1 = 9

7

Certificate Mathematics in Action Full Solutions 5B

14. Since there are 4 possible outcomes for urn A and 4 possible outcomes for urn B, by the counting principle, the total number of possible outcomes in drawing one ball from each urn is: 4 × 4 = 16 (a) There are 2 favourable outcomes: (5, 5) and (7, 7) 2 = 16 ∴ P(same number) 1 = 8 (b) There are 2 favourable outcomes: (1, 8) and (7, 2) 2 = 16 ∴ P(sum of 9) 1 = 8 (c) There are 5 favourable outcomes: (5, 7), (5, 8), (7, 5), (7, 7) and (7, 8) 5 ∴ P(sum greater than 11) = 16

Exercise 15B (p. 17) Level 1 1. P(red card or jack of spades) = P ( red card) + P ( jack of spades) 26 1 = + 52 52 27 = 52

2.

= 1 − P (defective) 5 P(non-defective) = 1 − 1000 199 = 200

3.

= P ( rainy) + P (cloudy) 1 2 P(rainy or cloudy) = 4 + 5 13 = 20

15. By the counting principle, the total number of possible outcomes in forming a four-digit number is: 4 × 3 × 2 × 1 = 24 (a) There are 6 favourable outcomes: 3564, 3654, 5364, 5634, 6354 and 6534 6 = ∴ P(ends with a ‘4’) 24 1 = 4 (b) There are 12 favourable outcomes in which 6 outcomes end with ‘4’ and 6 outcomes end with ‘6’. 12 = ∴ P(even number) 24 1 = 2 (c) Since 3 + 4 + 5 + 6 = 18 is a multiple of 3, any numbers formed by 3, 4, 5 and 6 must be a multiple of 3. ∴ There is no prime number formed by 3, 4, 5 and 6. 0 ∴ P(prime number) = 24 =0

4.

(a) There are 4 favourable outcomes: U, E, A and I 4 ∴ P(vowel) = 13 = 1 − P ( vowel) 4 (b) P(consonant) = 1 − 13 9 = 13

5.

Total number of people in the group = 47 + 62 + 33 + 18 + 40 = 200 (a) P(‘Cartoon’ or ‘Popular music’) = P(‘Cartoon’) + P(‘Popular music’) 18 40 = + 200 200 29 = 100 (b) P(‘Drama’ or ‘Sports’ or ‘Popular music’) = 1 – P(‘News’ or ‘Cartoon’) = 1 – [P(‘News’) + P(‘Cartoon’)] 18   47 = 1−  +   200 200  =

8

27 40

15 More about Probability

6.

Let H stand for a head and T stand for a tail. By the counting principle, the total number of possible outcomes in tossing two coins is: 2 × 2 = 4 (a) There are 2 favourable outcomes: HT and TH 2 = ∴ P(one head and one tail) 4 1 = 2 (b) There is 1 favourable outcome: TT 1 ∴ P(two tails) = 4 (c) P(no tails) = 1 − P(either or both of them are tails) = 1 − [ P (one head and one tail) + P( two tails)] 1 1 = 1−  +  2 4 1 = 4

7.

(a) There are 6 favourable outcomes: 1, 3, 5, 7, 9 and 11 6 = 12 ∴ P(odd number) 1 = 2 (b) There are 3 favourable outcomes: 4, 8 and 12 3 = 12 ∴ P(multiple of 4) 1 = 4 (c) P(odd number or multiple of 4) = P (odd number) + P (multiple of 4) 1 1 = + 2 4 3 = 4

8.

By the counting principle, the total number of possible outcomes is: 2 × 2 × 2 = 8 There is 1 favourable outcome for all boys. 1 ∴ P(all boys) = 8 = 1 − P (all boys) 1 ∴ P(at least one girl) = 1 − 8 7 = 8

9.

They are not complementary events because if one even number and one odd number are obtained, neither events occur. 9

Certificate Mathematics in Action Full Solutions 5B

10. No, because getting a ‘2’ is a favourable outcome to both events of ‘getting a prime number’ and ‘getting an even number’. Level 2 11. By the counting principle, the total number of possible outcomes is: 6 × 6 = 36 There are 10 favourable outcomes for sum less than 6: sum equals 2: (1, 1) sum equals 3: (1, 2), (2, 1) sum equals 4: (1, 3), (2, 2), (3, 1) sum equals 5: (1, 4), (2, 3), (3, 2), (4, 1) 10 = ∴ P(sum less than 6) 36 5 = 18 = 1 − P( black or white) = 1 − [ P (black) + P ( white)] 12. P(neither black nor white) = 1 −  4 + 1   15 6  17 = 30 = 1 − P( team A wins or draws) = 1 − [ P ( team A wins) + P(draws)] 13. P(team B wins) = 1 −  3 + 1  8 4 3 = 8 14. (a) P(either team A or team B wins) = P ( team A wins) + P( team B wins) 1 1 = + 3 4 7 = 12 (b) P(neither of the teams win) = 1 − P(either team A or team B wins) 7 = 1− 12 5 = 12 15. By the counting principle, the total number of possible outcomes is: 6 × 6 = 36 (a) There are 3 favourable outcomes for sum less than 4: sum equals 2: (1, 1) sum equals 3: (1, 2), (2, 1)

10

3 36 ∴ P(sum less than 4) 1 = 12 =

(b) There are 4 favourable outcomes for sum equals 5: (1, 4), (4, 1), (2, 3) and (3, 2) 4 = ∴ P(sum equals 5) 36 1 = 9 = 1 − P (sum equals 5) 1 ∴ P(sum not equals 5) = 1 − 9 8 = 9 (c) There are 8 favourable outcomes: sum equals 4: (1, 3), (2, 2), (3, 1) sum equals 5: (1, 4), (2, 3), (3, 2), (4, 1) sum equals 12: (6, 6) 8 = P(sum equals 4 or 5 or 12) 36 2 = 9

16. (a)

(b)

(c)

120° 360° 1 = 3

P (MTR ) =

20° 360° 1 = 18

P ( taxi) =

P ( MTR or taxi) = P( MTR ) + P( taxi) 1 1 = + 3 18 7 = 18

(d) Angle subtended by the sector ‘bus’ = 360° − 120° − 20° − 14° − 33° − 13° = 160° = 1 − P( bus) 160° = 1− 360° ∴ P(not by bus) 200 = 360 5 = 9

15 More about Probability

Exercise 15C (p. 25)

(b) P(a CD of male singer and a CD of female singer) = P(CD of male singer from 1st box and CD of = female singer from 2nd box or CD of female singer = from 1st box and CD of male singer from 2nd box) = P(CD of male singer from 1st box and CD of = female singer from 2nd box) + P(CD of female = singer from 1st box and CD of male singer from = 2nd box) = P(CD of male singer from 1st box) × P(CD of = female singer from 2nd box) + P(CD of female = singer from 1st box) × P(CD of male singer from = 2nd box) 3 3 5 2 = × + × 8 5 8 5 19 = 40

Level 1

1.

= P (1st one is blue) × P (2nd one is blue) 4 3 P(both blue) = 8 × 9 1 = 6

2.

P(uses broadband service and has 4 members) = P (uses broadband service) × P (has 4 members) = 0.8 × 0.6 = 0.48

3.

(a) P(both cards are spade) = P (1st card is spade) × P (2nd card is spade) 13 13 = × 52 52 1 = 16 (b) P(one is ace and one is king) = P(1st card is ace and 2nd card is king or = 1st card is king and 2nd card is ace) = P(1st card is ace and 2nd card is king) + = P(1st card is king and 2nd card is ace) = P(1st card is ace) × P(2nd card is king) + = P(1st card is king) × P(2nd card is ace) 4 4 4 4 = × + × 52 52 52 52 2 = 169

4.

(a) P(product is odd) = P (1st number is odd) × P( 2nd number is odd) 3 3 = × 6 6 1 = 4 = 1 − P (product is odd ) 1 (b) P(product is even) = 1 − 4 3 = 4

5.

(a) P(2CDs of male singers) = P(CD of male singer from 1st box) × = P(CD of male singer from 2nd box) 3 2 = × 3+5 2+3 3 = 20

6.

(a) P(both solve the problem) = P(Terrence solves the problem) × P(Sara solves the = problem) 2 3 = × 3 4 1 = 2 (b) P(exactly one of them solves the problem) = P(Terrence solves the problem and Sara cannot or = Sara solves the problem and Terrence cannot) = P(Terrence solves the problem and Sara cannot) + = P(Sara solves the problem and Terrence cannot) = P(Terrence solves the problem) × P(Sara cannot = solve the problem) + P(Terrence cannot solve the = problem) × P(Sara solves the problem) 2  3  2 3 = × 1 −  + 1 −  × 3  4  3 4 =

5 12

Level 2 7. Let S stand for a worker suffers from occupational illness and N stand for a worker does not suffer from occupational illness. P(exactly two of the three workers suffer from the illness) = P(SSN or SNS or NSS) = P(SSN) + P(SNS) + P(NSS) = 0.15 × 0.15 × (1 – 0.15) + 0.15 × (1 – 0.15) × 0.15 + = (1 – 0.15) × 0.15 × 0.15 = 0.057 375

11

Certificate Mathematics in Action Full Solutions 5B

8.

9.

P(two letters drawn are the same) = P ( HH or EE or II or SS) = P ( HH) + P (EE ) + P ( II) + P (SS) 1 1 1 1 1 1 1 1 = × + × + × + × 11 7 11 7 11 7 11 7 4 = 77 Let S stand for a son and D stand for a daughter. P(both children are of the same sex) = P (SS or DD) = P (SS) + P (DD) 3 2 3 4 = × + × 3+3 2+ 4 3+3 2+ 4 1 = 2

10. (a) P(no one hits the target) = P(Alan, Anthony and Scarlet do not hit the target) = P(Alan not hit the target) × P(Anthony not hit the = target) × P(Scarlet not hit the target)  1  1  1 = 1 −  × 1 −  × 1 −   2  3  4 =

1 4

= 1 − P (no one hits the target ) 1 (b) P(the target is hit) = 1 − 4 3 = 4 11. Let C stand for a correct answer and W stand for a wrong answer. = P (CCCC) 1 1 1 1 (a) P(all answers correct) = 4 × 4 × 4 × 4 1 = 256 (b) P(only 3 answers correct) = P(WCCC or CWCC or CCWC or CCCW) = P(WCCC) + P(CWCC) + P(CCWC) + P(CCCW) 3 1 1 1 1 3 1 1 = × × × + × × × + 4 4 4 4 4 4 4 4 1 1 3 1 1 1 1 3 × × × + × × × 4 4 4 4 4 4 4 4 3 = 64

12

12. Let M stand for a shot is made and N stand for a shot is not made. = P ( MMMM ) 2 2 2 2 (a) P(four shots in four trials) = 5 × 5 × 5 × 5 16 = 625 (b) P(more than 2 shots out of 4 trials) = P (3 shots out of 4 trials or 4 shots out of 4 trials) = P (3 shots out of 4 trials) + P ( 4 shots out of 4 trials) = P ( MMMN or MMNM or MNMM or 16 625 = P ( MMMN) + P ( MMNM) + P (MNMM ) + NMMM) +

16 625 2 2 2  2 2 2  = × × × 1 −  + × × 1 − 5 5 5  5 5 5  P ( NMMM) +

2 × 5

2 2  2 2 2  2 + × 1 −  × × + 1 −  × 5 5  5 5 5  5 2 2 2 16 × × + 5 5 5 625 112 = 625 13. (a) P(B) = P(upper branch and lower branch and upper = branch) = P(upper branch) × P(lower branch) × P(upper = branch) 1 1 1 = × × 2 2 2 1 = 8 (b) P(C or D) = P(C) + P(D) = P(upper branch and lower branch and lower branch) = + P(lower branch and uppest branch) = P(upper branch) × P(lower branch) × = P(lower branch) + P(lower branch) × = P(uppest branch) 1 1 1 1 1 = × × + × 2 2 2 2 3 7 = 24

15 More about Probability

14. (a) P(all the students bring the textbook) = P(1st, 2nd, … , 40th students bring the textbook) = P(1st student brings) × P(2nd student brings) × … = × P(40th student brings) = [1 – P(1st student forgets to bring)] × [1 – P(2nd = student forgets to bring)] × … × [1 – P(40th student = forgets to bring)] = (1 − 0.01) × (1 − 0.01) × ... × (1 − 0.01) = 0.99

(b) Total number of students wear glasses = 10 + 15 = 25 10 = P(boy | wear glasses) 25 2 = 5 4.

40

= 0.6690 (cor. to 4 sig. fig.)

P(first tape is used while the second one is new) = P(first tape is used) × P(second tape is new | first = tape is used) 8 42 = × 50 49 24 = 175

(b) P(at least one student forgets to bring the textbook) = 1 − P(all the students bring the textbook ) = 1 − 0.6690 = 0.3310 (cor. to 4 sig. fig.)

(b) P(both of them are used) = P(first tape is used and second tape is used) = P(first tape is used) × P(second tape is used | first = tape is used) 8 7 = × 50 49 4 = 175

Exercise 15D (p. 33) Level 1 1. There are 3 possible outcomes for an odd number: 1, 3 and 5 Among the possible outcomes, only 3 and 5 are favourable outcomes. 2 ∴ P(prime number | odd number) = 3 5. 2.

(a) There are 12 possible outcomes for a face card. Among the possible outcomes, only the jack, queen and king of spades are favourable outcomes. 3 = 12 ∴ P(spade | face card) 1 = 4

(a) Total number of girls = 15 + 9 = 24 15 P(wear glasses | girl) 24 5 = 8 =

(a) P(first piece is a corner piece while the second is not) = P(first piece is a corner piece) × P(second piece is = not a corner piece | first piece is a corner piece) 4 196 = × 200 199 98 = 4975 (b) P(both are corner pieces) = P(1st piece is a corner piece and 2nd piece is a = corner piece) = P(1st piece is a corner piece) × P(2nd piece is a = corner piece | 1st piece is a corner piece) 4 3 = × 200 199 3 = 9950

(b) There are 39 possible outcomes for not a club. Among the possible outcomes, only 13 cards are favourable outcomes. 13 = ∴ P(spade | not a club) 39 1 = 3 3.

(a) Total number of tapes = 42 + 8 = 50

6.

(a) P(all boys) = P(1st boy and 2nd boy and 3rd boy) = P(1st boy) × P(2nd boy | 1st boy) × = P(3rd boy | 1st and 2nd boy) 22 21 20 = × × 42 41 40 11 = 82

13

Certificate Mathematics in Action Full Solutions 5B

(b) P(all are of the same sex) = P(all boys or all girls) = P(all boys) + P(all girls) 11 = + P (1st girl) × P( 2nd girl | 1st girl) × 82 P (3rd girl | 1st and 2nd girl) 11 20 19 18 + × × 82 42 41 40 67 = 287 =

7.

(a) ∵ The first ball drawn is a multiple of 5. ∴ Only 29 balls are left after the first draw. Only 4 balls are left that is a multiple of 7. ∴ P(2nd ball is a multiple of 7 | 1st ball is a 4 multiple of 5) = 29 (b) ∵ The first ball drawn is a multiple of 5. ∴ Only 29 balls are left after the first draw. Only 5 balls are left that is a multiple of 5. ∴ P(2nd ball is a multiple of 5 | 1st ball is a 5 multiple of 5) = 29

Level 2 8. P(both even) = P(1st card is even and 2nd card is even) = P(1st card is even) × P(2nd card is even | 1st card is even) 5 4 = × 10 9 2 = 9 9.

Let R denote a red ball and Y denote a yellow ball. Total number of balls = 14 + 10 = 24 = P (R ) × P( Y | R ) 14 10 (a) P(RY) = 24 × 23 35 = 138 = P (RR or YY) = P (RR ) + P( YY) = P (R ) × P( R | R ) + P (Y ) × P (Y | Y ) (b) P(same colour) 14 13 10 9 = × + × 24 23 24 23 34 = 69

14

10. Let G denote a green form and W denote a white form. Total number of forms = 50 + 30 = 80 (a) P(three forms are of the same colour) = P(GGG or WWW) = P(GGG) + P(WWW) = P(1st G) × P(2nd G | 1st G) × P(3rd G | 1st and = 2nd G) + P(1st W) × P(2nd W | 1st W) × = P(3rd W | 1st and 2nd W) 50 49 48 30 29 28 = × × + × × 80 79 78 80 79 78 91 = 316 (b) ∵ There are only two colours of forms. ∴ P(three forms are of different colours) = 0 11. (a) Since there is only 1 key can open the door. 1 ∴ P(1st trial) = 3 (b) P(second trial) = P (1st trial fails) × P (2nd trial | 1st trial fails) 2 1 = × 3 2 1 = 3 (c) P(last trial) = P(1st trial fails) × P(2nd trial fails | 1st trial fails) × = P(last trial | 1st and 2nd trials fail) 2 1 1 = × × 3 2 1 1 = 3 12. By the counting principle, the total number of possible outcomes is: 6 × 6 = 36 (a) There are 5 favourable outcomes for the sum of the numbers is 6: (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1) 5 ∴ P(sum of the numbers is 6) = 36 (b) There are 3 odd numbers on a die. By the counting principle, the total number of favourable outcomes is: 3× 3 = 9 9 = ∴ P(both odd) 36 1 = 4

15 More about Probability

(c) Among the 9 possible outcomes that both numbers are odd, only 3 of them are favourable outcomes: (1, 5), (3, 3) and (5, 1) 3 = ∴ P(sum is 6 | both are odd) 9 1 = 3

(ii)

(b) Percent of blood sample shows positive result = 0.098% + 6.993% = 7.091%

13. (a) P(smoker) = P(male and smoker or female and smoker) = P(male and smoker) + P(female and smoker) = P(male) × P(smoker | male) + = P(female) × P(smoker | female) = 0.515 × 0.13 + (1 − 0.515) × 0.07

Percent of blood sample that is false positive = 6.993% ∴ P(false positive | positive result) 6.993% = 7.091% = 0.9862 (cor. to 4 sig. fig.)

= 0.1009 = 1 − P (smoker) (b) P(non-smoker) = 1 − 0.1009 = 0.8991 14. Let R denote a red ball, G denote a green ball and W denote a white ball. Total number of balls = 5 + 3 + 2 = 10 (a) P(one red and one white) = P(RW or WR) = P(RW) + P(WR) = P(R) × P(W | R) + P(W) × P(R | W) 5 2 2 5 = × + × 10 9 10 9 2 = 9 (b) P(same colour) = P(RR or GG or WW) = P(RR) + P(GG) + P(WW) = P(R) × P(R | R) + P(G) × P(G | G) + = P(W) × P(W | W) 5 4 3 2 2 1 = × + × + × 10 9 10 9 10 9 14 = 45 15. (a) (i)

P(true positive result) = P(AIDS and positive result) = P(AIDS) × P(positive result) = 0.1% × 0.98 = 0.098%

P(false positive result) = P(no AIDS and positive result) = P(no AIDS) × P(positive result) = (1 – 0.1%) × 0.07 = 6.993%

(c) Any reasonable answers.

Revision Exercise 15 (p. 37) Level 1 1. (a) There are 8 favourable outcomes: 2, 3, 5, 7, 11, 13, 17 and 19 8 = ∴ P(prime) 20 2 = 5 (b) There are 5 favourable outcomes: 4, 8, 12, 16 and 20 5 = ∴ P(multiple of 4) 20 1 = 4 (c) P(prime or multiple of 4) = P (prime) + P( multiple of 4) 2 1 = + 5 4 13 = 20 2.

Total number of people = 88 + 113 + 78 + 21 = 300 78 (a) P(disagree) 300 13 = 50 =

15

Certificate Mathematics in Action Full Solutions 5B

(b) P(‘strongly agree’ or ‘agree’) = P(‘strongly agree’) + P(‘agree’) 88 113 = + 300 300 67 = 100 3.

Total number of possible outcomes = 52 − 6 = 46 (a) ∵ All the jack of black suits are removed.

(b) P(sum is even) = P(both even or both odd) = P(both even) + P(both odd) = P(even) × P(even) + P(odd) × P(odd) 5 5 5 5 = × + × 10 10 10 10 1 = 2 6.

∴ P(a jack of black suits) = 0 (b) There are 2 favourable outcomes: ace of hearts and ace of diamonds 2 = 46 ∴ P(an ace of red suits) 1 = 23 (c) There are 10 favourable outcomes: A, 2, 3, …, 10 of clubs 10 = ∴ P(a club) 46 5 = 23 4.

5.

16

P(both are members) = P(1st is a member and 2nd is a member) = P(1st is a member) × P(2nd is a member | = 1st is a member) 80 79 = × 1200 1199 79 = 17 985

(a) P(sum is even) = P(1st is even and 2nd is even or 1st is odd and = 2nd is odd) = P(1st is even and 2nd is even) + = P(1st is odd and 2nd is odd) = P(1st is even) × P(2nd is even | 1st is even) + = P(1st is odd) × P(2nd is odd | 1st is odd) 5 4 5 4 = × + × 10 9 10 9 4 = 9

(a) P(both defective) = P(defective from A) × P(defective from B) 16 25 = × 800 1000 1 = 2000 (b) P(both are not defective) = P(not defective from A) × P(not defective from B) 800 − 16 1000 − 25 = × 800 1000 1911 = 2000

7.

Let C denote a correct answer and W denote a wrong answer. (a) P(one answer wrong) = P (CCW or CWC or WCC) = P (CCW) + P (CWC) + P ( WCC) 2 2  2 2  2 2 × × 1 −  + × 1 −  × + 3 3  3 3  3 3  2 2 2 1 −  × ×  3 3 3 4 = 9 =

(b) P(all answers wrong) = P ( WWW )  2  2  = 1 −  × 1 −  × 1 −  3  3  1 = 27

2  3

(c) P(at least one answer correct) = 1 − P(all answers wrong) 1 = 1− 27 26 = 27

15 More about Probability

8.

(a) P(both late) = P(Vincent late) × P(Andrew late) 1 1 = × 3 4 1 = 12 (b) P(at least one of them arrives punctually) = 1 – P(both late) 1 = 1− 12 11 = 12 (c) P(only one of them arrives punctually) = P(Vincent late and Andrew punctual or = Vincent punctual and Andrew late) = P(Vincent late and Andrew punctual) + = P(Vincent punctual and Andrew late) = P(Vincent late) × P(Andrew punctual) + = P(Vincent punctual) × P(Andrew late) 1  1  1 1 = × 1 −  + 1 −  × 3  4  3 4 =

9.

5 12

Total number of batteries = 2 + 4 =6 (a) P(1st is used and 2nd is new) = P (1st is used) × P (2nd is new | 1st is used) 2 4 = × 6 5 4 = 15 (b) P(both used) = P (1st is used) × P( 2nd is used | 1st is used) 2 1 × 6 5 1 = 15 =

10. Let H stand for a head and T stand for a tail. There are 7 possible outcomes for at least one head: HTT, THT, TTH, HHT, HTH, THH and HHH Among the possible outcomes, there are 3 favourable outcomes: HHT, HTH and THH 3 ∴ P(exactly two heads | at least one head) = 7

11. For the 1st time the rumour spreads, Alfred can choose any 40 classmates out of the 40 possible classmates. For the 2nd to the 10th times the rumour spreads, the student can choose any 39 classmates (excluding Alfred) out of the 40 possible classmates. Since every time the rumour spreads are independent events. ∴

P(still has not returned to Alfred in 10 times) =

40  39  ×  40  40 

9

= 0.796 (cor. to 3 sig. fig.) 12. F = a head is obtained when a coin is tossed or F = a queen is drawn from a deck of 52 cards (or any other reasonable answers) 13. G = the number obtained is a prime number G = the number obtained is greater than 3 (or any other reasonable answers) Level 2 14. By the counting principle, the total number of possible outcomes: 6 × 6 = 36 (a) There are 6 favourable outcomes: (1, 1), (2, 1), (1, 2), (2, 2), (1, 3) and (3, 1) 6 = 36 ∴ P(less than 5) 1 = 6 (b) There are 5 favourable outcomes: (2, 6), (6, 2), (3, 5), (5, 3) and (4, 4) 5 ∴ P(equal to 8) = 36 (c) P(less than 5 or equal to 8) = P (less than 5) + P (equal to 8) 1 5 = + 6 36 11 = 36 15. (a) P(all of them fail) = P ( Ivy fails) × P (Grace fails) × P( Winky fails)  1  2  5 = 1 −  × 1 −  ×  1 −   2  5  8 9 = 80

17

Certificate Mathematics in Action Full Solutions 5B

(b) P(at least one of them passes) = 1 − P(all of them fail) 9 = 1− 80 71 = 80 16. (a) P(sum of 2) = P(‘1’ and ‘1’) = P(‘1’) × P(‘1’) 308 308 = × 2000 2000 = 0.0237 (cor. to 3 sig. fig.) (b) Number of times of not getting a ‘4’ = 2000 − 458 = 1542 P(at least one ‘4’) = 1 – P(no ‘4’ in two throws) = 1 – P(no ‘4’) × P(no ‘4’) 1542 1542 = 1− × 2000 2000 = 0.406 (cor. to 3 sig. fig.) (c) P(a sum of 10 or more) = P(‘4’ and ‘6’ or ‘6’ and ‘4’ or ‘5’ and ‘5’ or = ‘5’ and ‘6’ or ‘6’ and ‘5’ or ‘6’ and ‘6’) = P(‘4’ and ‘6’) + P(‘6’ and ‘4’ ) + P(‘5’ and ‘5’ ) + = P(‘5’ and ‘6’) + P(‘6’ and ‘5’) + P(‘6’ and ‘6’) 458 312 312 458 206 206 = × + × + × + 2000 2000 2000 2000 2000 2000 206 312 312 206 312 312 × + × + × 2000 2000 2000 2000 2000 2000 = 0.139 (cor. to 3 sig. fig.) 17. Let R denote a red sock, W denote a white sock and B denote a black sock. Total number of socks = 6 + 4 + 2 = 12 (a) P(one red and one black) = P (RB or BR) = P (RB) + P (BR ) = P (R ) × P (B | R ) + P ( B) × P (R | B) 6 2 2 6 = × + × 12 11 12 11 2 = 11

18

(b) P(same colour) = P(RR or WW or BB) = P(RR) + P(WW) + P(BB) = P(R) + P(R | R) + P(W) × P(W | W) + = P(B) × P(B | B) 6 5 4 3 2 1 = × + × + × 12 11 12 11 12 11 1 = 3 18. Let W denote a white marble and B denote a black marble. Total number of marbles = 3 + 2 =5 2 (a) P(first trial) = 5 = P ( WB) = P ( W ) × P (B | W ) (b) P(second trial) = 3 × 2 5 4 3 = 10 (c) P(fourth trial) = P(WWWB) = P(W) × P(W | 1st W) × P(W | 1st and 2nd W) × = P(B | 1st, 2nd and 3rd W) 3 2 1 2 = × × × 5 4 3 2 1 = 10 19. (a) P(defective) = P(defective from box A or defective from box B) = P(defective from box A) + P(defective from box B) = P(box A) × P(defective | box A) + = P(box B) × P(defective | box B) 1 2 1 6 = × + × 2 8 2 14 19 = 56 (b) Total number of defective bulbs = 2 + 6 =8 6 8 ∴ P(box B | defective) 3 = 4 =

15 More about Probability

20. P(defective) = P(defective from machine A or = defective from machine B) = P(defective from machine A) + = P(defective from machine B) = P(machine A) × P(defective | machine A) + = P(machine B) × P(defective | machine B) = 0.6 × 0.02 + 0.4 × 0.04 = 0.028 60° 360 ° 21. (a) P(land on sector POQ) 1 = 6 =

(b) (i)

∠QOR = 60° × 1.5 = 90°

(b) By using geometric sequence, we have 1 1 1 8 = first term = , common ratio = 1 2 4 2 = ∴ P(Katherine wins in her nth trial)

1 =  2

=

(c) P(one hits sector POQ and one hits sector QOR) = P(1st hits sector POQ and 2nd hits sector QOR or = 1st hits sector QOR and 2nd hits sector POQ) = P(1st hits sector POQ and 2nd hits sector QOR) + = P(1st hits sector QOR and 2nd hits sector POQ) 60° 90° 90° 60° = × + × 360° 360° 360° 360° 1 = 12 22. Let H denote a head and T denote a tail. = P ( H) (a) (i) P(wins in her 1st trial) 1 = 2 = P (TTH ) 1 1 1 (ii) P(wins in her 2nd trial) = 2 × 2 × 2 1 = 8 = P (TTTTH ) 1 1 1 1 1 (iii) P(wins in her 3rd trial) = 2 × 2 × 2 × 2 × 2 1 = 32

n −1

2 n −1

(c) P(Katherine wins the game) = P(wins in her 1st trial or wins in her 2nd trial or … = or wins in her nth trial or …) = P(wins in her 1st trial) + P(wins in her 2nd trial) + = … + P(wins in her nth trial) + … =

60° + 90° 360° (ii) P(land on sector POR) 5 = 12

11   24

=

=

1 1 1 + + ... +   2 8 2 1 2 1−

2 n −1

+ ...

1 4

2 3

23. (a) P(head appears) = P(normal coin and head or = two-headed coin and head) = P(normal coin and head) + = P(two-headed coin and head) 2 1 1 2 = × + × 3 2 3 2 2 = 3 (b) By using a tree diagram, we have: head normal coin tail head normal coin tail head two-headed coin head From the tree diagram, there are 4 possible outcomes. Among the possible outcomes, only 2 are favourable outcomes. 2 = ∴ P(two-headed coin | head) 4 1 = 2

19

Certificate Mathematics in Action Full Solutions 5B

(c) (i)

There are 2 favourable outcomes of getting two heads in two trials: (1) normal coin and head and head; (2) two-headed coin and head and head For (1), P(normal coin and head and head) 2 1 1 = × × 3 2 2 1 = 6 For (2), P(two-headed coin and head and head) 1 2 2 = × × 3 2 2 1 = 3 ∴ P(two heads in two trials) = P(normal coin and head and head) + = P(two-headed coin and head and head) 1 1 = + 6 3 1 = 2

(ii) P(two tails in two trials) = P(normal coin and tail and another normal = coin and tail) = P(normal coin) × P(tail) × = P(another normal coin | normal coin) × P(tail) 2 1 1 1 = × × × 3 2 2 2 1 = 12

3.

= 4.

2.

20

Answer: D Let C denote a correct answer and W denote a wrong answer. P(only one question is correct) = P(CWW or WCW or WWC) = P(CWW) + P(WCW) + P(WWC) 1  1  1  1 = × 1 −  × 1 −  + 1 −  × 4  4  4  4

= 5.

27 64

Answer: B Let B denote a son and G denote a daughter. By using a tree diagram, we have: First child

Second child Outcomes B BB

B G B

BG GB

G

GG

G

Answer: B There are 3 favourable outcomes: 2, 5 and 17 3 = 6 ∴ P(prime number) 1 = 2 Answer: A P(both are late) = P ( John is late) × P (Mary is late) = 0.4 × 0.3 = 0.12

9 10

1  1  1  1 1 × 1 −  + 1 −  × 1 −  × 4  4  4  4 4

Multiple Choice Questions (p. 41) 1.

Answer: D P(the problem is solved) = 1 – P(Paul and Mary cannot solve the problem) = 1 – P(Paul cannot solve the problem) × = P(Mary cannot solve the problem)  2  5 = 1 −  1 −  × 1 −   5  6

From the tree diagram, there are 3 possible outcomes to have a son. Among the possible outcomes, only 1 is a favourable outcome. 1 ∴ P(two sons | has a son) = 3 6.

Answer: B Total number of rotten oranges = 20 + 10 = 30 20 30 ∴ P(comes from box A | rotten orange) 2 = 3 =

15 More about Probability

7.

Answer: D P(at least one dart hits the target) = 1 – P(two darts do not hit the target) = 1 – P(1st dart does not hit the target) × = P(2nd dart does not hit the target)  3  3 = 1 −  1 −  × 1 −   5  5 =

8.

9.

Total number of students in the class = 27 + 13 = 40 P(same sex) = P(BB or GG) = P(BB) + P(GG) = P(B) × P(B | B) + P(G) × P(G | G) 27 26 13 12 = × + × 40 39 40 39 11 = 20

21 25

Answer: C P(at least one of them is absent) = 1 – P(all present) = 1 – P(Kitty presents) × P(Alice presents) × = P(Alan presents)  1  1  1 = 1 −  1 −  ×  1 −  × 1 −   3  3  3 =

12. Answer: B Let B denote a boy and G denote a girl.

19 27

13. Answer: A P(open in at least 3 trials) = 1 – P(open in less than 3 trials) = 1 – P(open in the 1st trial or 2nd trial) = 1 – [P(open in the 1st trial) + P(open in the 2nd trial)] 2 = 1 −  + P (fail in the 1st trial) × P (open in the 2nd trial | 10 fail in the 1st trial)]

Answer: A There are 3 possible seats for Tracy after Lily takes a seat. Among the possible seats, 2 seats are next to Lily. 2 ∴ P(Tracy sits next to Lily) = 3

10. Answer: B Let W denote a white chopstick and B denote a black chopstick. P(same colour) = P(WW or BB) = P(WW) + P(BB) = P(W) × P(W | W) + P(B) × P(B | B) 2 1 2 1 = × + × 4 3 4 3 1 = 3 11. Answer: D P(greater than or equal to 5) = 1 – P(less than 5) = 1 – P((1, 2) or (1, 3) or (2, 1) or (3, 1)) = 1 – [P(1, 2) + P(1, 3) + P(2, 1) + P(3, 1)] = 1 – [P(1) × P(2 | 1) + P(1) × P(3 | 1) + P(2) × P(1 | 2) + = P(3) × P(1 | 3)] 1 1 1 1 1 1 1 1 = 1−  × + × + × + ×   4 3 4 3 4 3 4 3 =

2 3

8 2  2 = 1−  + ×   10 10 9  17 = 1− 45 28 = 45

HKMO (p. 43) 1.

P(all odd) = P(1st is odd and 2nd is odd and 3rd is odd) = P(1st is odd) × P(2nd is odd | 1st is odd) × P(3rd is odd | = 1st and 2nd are odd) 5 4 3 = × × 9 8 7 5 = 42

2.

By 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 9 8

1, 2, 3, 4, 5,6, 7, 8, 9 1, 2, 3, 4, 5, 6, 7, 8 1, 2, 3, 4, 5, 6, 7

9 8 7

21

Certificate Mathematics in Action Full Solutions 5B

7 6 5 4 3 2 1

1, 2, 3, 4, 5, 6 1, 2, 3, 4, 5 1, 2, 3, 4 1, 2, 3 1, 2 1 – Total

45 100 ∴ P(white ball > black ball) 9 = 20 =

22

6 5 4 3 2 1 0 45