Chapter 4 Probability Distributions for Discrete Variables Chapter …bartlettstp.weebly.com/uploads/2/3/0/2/23023680/textbook... · 2018. 9. 9. · The probability that the marble

MHR Data Management 12 Solutions 1

Chapter 4 Probability Distributions for Discrete Variables

Chapter 4 Prerequisite Skills

Chapter 4 Prerequisite Skills Question 1 Page 142

a) Four of the 52 cards in a standard deck are kings. The probability of selecting a king is 4

52,

or 1

13.

b) Twenty-six of the 52 cards in a standard deck are red cards. The probability of selecting a red

card is 26

52, or

1

2.

c) Thirteen of the 52 cards in a standard deck are spades. The probability of selecting a spade is

13

52, or

1

4.


a) If you were to list all 216 possible outcomes, the possible sums are any natural from 3 to 18.

When three different numbers form a grouping, there are 3! different ways of arranging these

numbers. When two different numbers form the grouping, there are three different ways of

arranging these numbers.

Sum Possible Groupings

Number of

Outcomes Probability

3 (1,1,1) 1 1

216

4 (1,2,1) 3 3

216

5 (1,3,1), (1,2,2) 6 6

216

6 (1,4,1), (1,3,2), (2,2,2) 10 10

216

7 (1,4,2), (1,3,3), (5,1,1), (3,2,2) 15 15

216

8 (1,4,3), (1,2,5), (1,1,6), (4,2,2), (3,3,2) 21 21

216

9 (6,2,1), (5,3,1), (5,2,2), (4,4,1), (4,3,2), (3,3,3) 25 25

216

10 (6,3,1), (6,2,2), (5,3,2), (5,4,1), (4,4,2), (4,3,3) 27 27

216

11 (6,4,1), (6,3,2), (5,5,1), (5,4,2), (5,3,3), (4,4,3) 27 27

216

12 (6,5,1), (6,4,2), (6,3,3), (5,5,2), (5,4,3), (4,4,4) 25 25

216

13 (6,6,1), (6,5,2), (6,4,3), (5,5,3), (5,4,4) 21 21

216

14 (6,4,4), (6,5,3), (5,5,4), (6,6,2) 15 15

216

15 (6,6,3), (6,4,5), (5,5,5) 10 10

216

16 (6,6,4), (6,5,5) 6 6

216

17 (6,6,5) 3 3

216

18 (6,6,6) 1 1

216

2 MHR Data Management 12 Solutions

b) The sum of the probabilities is 1.


a) Since three of the eight doors are red, the probability that the food was placed behind a red

door is 3

8.

b) Since the food is behind only one of the are eight doors, the probability that the rat will select

the correct door is 1

8.


a) There are 14 marbles with 6 red and 1 green.

For n(S) = 14 and n(A) = 7,

( )( )

( )

1

2

7

14

n AP A

n S

The probability that the marble is red or green is 1

2.

b) There are 14 marbles with 1 green and 3 yellow (neither red nor blue).

For n(S) = 14 and n(A) = 4,

( )( )

( )

2

7

4

14

n AP A

n S

The probability that the marble is neither red nor blue is 2

7.


a) Of the 12 face cards, 4 are queens and 4 are kings. The probability of selecting a queen or a

king is 8

12, or

2

3.

b) Of the 12 face cards, 6 are red cards and 1 queen of spades. The probability of selecting a red

card or the queen of spades is 7

12.

c) There are no cards that are red and a spade. The probability of selecting a red card and a

spade is 0.



a) Since the outcome of rolling a die does not affect the outcome of rolling another die, these are

independent events.

b) Since the outcome of dealing a first card affects the second card dealt, these are dependent

events.

c) Since the outcome of flipping a coin does not affect the outcome of flipping another coin,

these are independent events.

d) Since the outcome of selecting a first male affects the selecting of a second male, these are

dependent events.


a)

b) Yes. The events (head, tail, tail) are independent. The outcome of flipping a coin does not

affect the outcome of flipping another coin.


a) There are 13 hearts. The probability that a player will select a queen followed by a king is

1 1

13 12 , or

1

156.

b) Selecting a queen and a king can be done in two ways: queen then king or king then queen.

The probability that a player will select a queen and a king is 1

2156

, or 1

78.

c) There are 13 hearts with 10 non-face cards. The probability that a player will not select a face

card on either draw is 10 9

13 12 , or

15

26.



a) There are 12 men and 10 women. Three men and three women are selected.

12 3 10 3

12! 10!

9!3! 7!3!

26 400

C C

Three men and three women can be selected in 26 400 ways.

b) For more men than women out of six names, select 4, 5, or 6 men.

12 4 10 2

12! 10!

8!4! 8!2!

22 275

C C

12 5 10 1

12! 10!

7!5! 9!1!

7920

C C

12 6 10 0

12! 10!

6!6! 10!0!

924

C C

More men than women can be selected in 22 275 + 7920 + 924, or 31 119 ways.


a) There are 3 chocolate chip, 4 peanut butter, and 6 butterscotch cookies.

Select 2 chocolate chip, 2 peanut butter, and 1 butterscotch cookie.

3 2 4 2 6 1

3! 4! 6!

1!2! 2!2! 5!1!

108

C C C

Two chocolate chips, two peanut butter, and one butterscotch cookie can be selected in 108 ways.

b) There are 3 chocolate chip, 4 peanut butter, and 6 butterscotch cookies.

Select 0 chocolate chips. There are 10 non-chocolate chip.

No chocolate chips cookies can be selected in 10C5, or 252 ways.

c) There are 3 chocolate chip, 4 peanut butter, and 6 butterscotch cookies.

Select at least 1 of each type of cookie.

Case 1: 1 chocolate chip, 1 peanut butter, 3 butterscotch: 3C1×4C1×6C3






3C1×4C1×6C3 + 3C1×4C2×6C2 + 3C1×4C3×6C1 + 3C2×4C1×6C2 + 3C3×4C1×6C1 + 3C2×4C2×6C1 = 894

At least one of each type of cookie can be selected in 894 ways.


a) There are 15! ways to assign the books. Order of assignment does not matter. For example,

ABC is the same as BAC, ACB, BCA, CBA, and CAB. So, there are 3! ways for the books to be

assigned to each person.

The number of ways that 15 books can be assigned to five people is 15!

3!3!3!3!3!, or 168 168 000.

This is the same as 15 3 12 3 9 3 6 3 3 3C C C C C .


b) The number of ways that 15 books can be assigned to three people is 15 5 10 5 5 5C C C , or

756 756.


a) 12 5 8 3

12! 8!

7!5! 5!3!

12 11 10 9 8 8 7 6

5! 3!

44 352

C C

b) 8 5

12 7

8!

3!5!12!

5!7!

8! 5!7!

3!5! 12!

8 7 6 5 4

12 11 10 9 8

0.0707...

C

C

c) 7 3 9 5

16 8

7! 9!

4!3! 4!5!16!

8!8!

7! 9! 8!8!

4!3! 4!5! 16!

0.3426...

C C

C

d) 2 3 2 3

5 2

2 3

5!(0.6) (0.4) (0.6) (0.4)

3!2!

5 4(0.6) (0.4)

2 1

0.2304

C

e)

4 2 4 2

6 4

2

6

1 2 6! 1 2

3 3 2!4! 3 3

6 5 2

2 1 3

0.0823...

C


a) For n = 4, 4 4 0 3 1 2 2 1 3 0 4

4 0 4 1 4 2 4 3 4 4

4 0 3 1 2 2 1 3 0 4

4 3 2 2 3 4

( )

1 4 6 4 1

4 6 4

x y C x y C x y C x y C x y C x y

x y x y x y x y x y

x x y x y xy y

b) Use 4x and 3y. 5 5 0 4 1 3 2 2 3 1 4 0 5

5 0 5 1 5 2 5 3 5 4 5 5

5 0 4 1 3 2 2 3 1 4 0 5

5 4 3 2 2 3

(4 3 ) (4 ) (3 ) (4 ) (3 ) (4 ) (3 ) (4 ) (3 ) (4 ) (3 ) (4 ) (3 )

1(4 ) (3 ) 5(4 ) (3 ) 10(4 ) (3 ) 10(4 ) (3 ) 5(4 ) (3 ) 1(4 ) (3 )

1024 3840 5760 4320

x y C x y C x y C x y C x y C x y C x y

x y x y x y x y x y x y

x x y x y x y

4 5

1620 243xy y


c) Use 0.3 and 0.7. 6 6 0 5 1 4 2 3 3 2 4

6 0 6 1 6 2 6 3 6 4

1 5 0 6

6 5 6 6

6 0 5 1 4 2 3 3 2 4

(0.3 0.7) (0.3) (0.7) (0.3) (0.7) (0.3) (0.7) (0.3) (0.7) (0.3) (0.7)

(0.3) (0.7) (0.3) (0.7)

1(0.3) (0.7) 6(0.3) (0.7) 15(0.3) (0.7) 20(0.3) (0.7) 15(0.3) (0.7) 6(0.

C C C C C

C C

1 5 0 63) (0.7) 1(0.3) (0.7)

0.000 729 .010 206 0.059 535 0.185 22 0.324135 0.302 526 0.117 649

1

d) Use 1

4 and

3

4.

5 5 0 4 1 3 2 2 3 1 4 0 5

5 0 5 1 5 2 5 3 5 4 5 5

5 0 4 1 3

1 3 1 3 1 3 1 3 1 3 1 3 1 3

4 4 4 4 4 4 4 4 4 4 4 4 4 4

1 3 1 3 1 31 5 10

4 4 4 4 4 4

C C C C C C

2 2 3 1 4 0 51 3 1 3 1 3

10 5 14 4 4 4 4 4

1 15 90 270 405 243

1024 1024 1024 1024 1024 1024

1


Chapter 4 Section 1 Probability Distributions

Chapter 4 Section 1 Example 1 Your Turn Page 147

a) The random variable, X, is the number of rooms in apartments in a particular complex.

b)

c) The area of each bar represents its probability, as a percent. The width of each bar is 1, so the

probability, as a percent is shown on the vertical axis.


d) Four-room apartments occur most frequently, and the probability decreases as the room value

increases or decreases from four.

e) Sum = 100%. So the sum of the probabilities is 1. This confirms the results in the example.


a)


Number of Blue, x Distribution of Blue Frequency Probability, P(x)

0 RRRR 1 1

16

1 BRRR RBRR RRBR RRRB 4 1

4

2 RRBB RBRB RBBR

BRRB BRBR BBRR 6

3

8

3 RBBB BRBB BBRB BBBR 4 1

4

4 BBBB 1 1

16

b)

c) 4

0

1 1 3 1 1( ) 0 1 2 3 4

16 4 8 4 16

1 6 3 40

4 8 4 16

4 12 12 4

16

2

x

x P x

The expected number of times the spinner lands on blue is 2.

d) On average, the spinner would land on blue 2 out of the 4 spins.


a) total value of all prizes

( ) price

10 000 000

per ticketnumbe

500 0

r of tickets sold

2.75 5

2.25

00 10 50 0005

4 000 000

E X

The expected value per ticket is –$2.25.

b) Answers may vary. The results of both Example 3 and this question show that you lose

money if you buy a ticket. They are not a good investment.


c) Answers may vary. To make buying a ticket more attractive, the price could be reduced or

more prizes could be given away.

Chapter 4 Section 1 R1 Page 151

While it is impossible to have 1.8 children, expected values are predicted average values and

should not be rounded.


Answers may vary. Two examples of a discrete probability distribution are the number of email

messages you receive each day of the week and the number of students in each mathematics class

at your school. These examples are discrete because each must be a whole number.


Answers may vary.

Create a table showing all possible sums of the two 12-sided dice.

Determine the frequency of each sum and its probability.

Then, construct a histogram to illustrate the probability distribution.

Chapter 4 Section 1 Question 1 Page 151

a) Discrete: The number of points scored in a basketball game must be a whole number.

b) Continuous: The time spent playing a basketball game can be measured in fractions of a

second.

c) Discrete: While mass in general is continuous, the mass of weights in a weight room are

typically whole numbers of kilograms, not fractions.

d) Discrete: The number of windows in a classroom will be a whole number.

e) Continuous: The area of a window in a classroom can be measured in fractions of square

centimetres.


Expectation can be found using 0

( ) ( )n

i i

i

E x x P x

.

It is the predicted average of all possible outcomes.

It does equal the mean of the outcomes weighted according to their respective frequencies.

The sum in the expected value calculations does not equal 1. Answer A.


In Example 2, the discrete random variable is the number of girls in a family of three children.

Answer C.



a)

b) For ease of graphing, convert probabilities to decimal values.


a) 4

1

( ) 1(0.3) 2(0.2) 3(0.1) 4(0.4)

2.6

x

x P x

The expected value of x is 2.6.

b) 1 3 1 1 1 1

( ) 0 2 4 6 8 105 10 5 10 10 10

6 4 6 8 10

10 5 10 10 10

6 8 6 8 10

10

3.8

E x



a) The random variable, X, is the diameter of the marbles in a bag.


b) Discrete: The number of marbles in a bag is a whole number.

c) First calculate the probabilities.

Diameter (mm) Frequency Probability, P(x)

12.0 5 1

12

13.0 11 11

60

14.0 24 2

5

20.0 15 1

4

25.0 5 1

12

d) The area of each bar represents its probability. The width of each bar is 1, so the probability

is shown on the vertical axis.

e) 12.0(5) 13.0(11) 14.0(24) 20.0(15) 25.0(5)

weighted mean60

16.0666...

The weighted mean of the outcomes equals the expectation.


a) Create a table showing all possible sums of the two 8-sided dice.

Determine the frequency of each sum.

First Die

Second Die 1 2 3 4 5 6 7 8

1 2 3 4 5 6 7 8 9

2 3 4 5 6 7 8 9 10

3 4 5 6 7 8 9 10 11

4 5 6 7 8 9 10 11 12

5 6 7 8 9 10 11 12 13

6 7 8 9 10 11 12 13 14

7 8 9 10 11 12 13 14 15

8 9 10 11 12 13 14 15 16


Sum, x Frequency

2 1

3 2

4 3

5 4

6 5

7 6

8 7

9 8

10 7

11 6

12 5

13 4

14 3

15 2

16 1

b) Use a graphing calculator.

c) 1 1 3 1 5 3 7 1 7 3 5 1

( ) 2 3 4 5 6 7 8 9 10 11 12 1364 32 64 16 64 32 64 8 64 32 64 16

3 1 114 15 16

64 32 64

2 3 12 5 30 21

64 32 64 16 64

E x

56 9 70 33 60 13 42 15 16

32 64 8 64 32 64 16 64 32 64

2 6 12 20 30 42 56 72 70 66 60 52 42 30 16

64

9

On average, the expected sum of two dice is 9.



Answers may vary. The rectangle has a perimeter of 24 cm.

Rectangle Width

(cm), x

Distribution of

Dimensions

Frequency Probability,

P(x)

1 1 by 11 1 1

6

2 2 by 10 1 1

6

3 3 by 9 1 1

6

4 4 by 8 1 1

6

5 5 by 7 1 1

6

6 6 by 6 1 1

6


a) A total of 2000 tickets are old, n(S) = 2000. There are nine prizes, n(A) = 9. The probability

of winning a prize is 9

2000.

b) total value of all prizes

Expected payoutnumbe

500

r of tickets sold

0.52

3(100) 5(50)

2000

5

The expected payout per ticket is $0.525.

c) Determine the difference between the ticket price and payout per ticket.

$5 – $0.525 = $4.475

The expected profit per ticket is $4.475.

d) A 90% profit per ticket would mean $4.50 per ticket.

Solving x – 0.525 = 4.5, gives x = 5.025. This means the price of a ticket would have to be $5.03.



a) Let F represent a face card and NF represent not a face card.

Number of

Face Cards, x Distribution of Face Cards Frequency

Probability,

P(x)

0 NF,NF,NF 1 1

8

1 F,NF,NF NF,F,NF NF,NF,F 3 3

8

2 F,F,NF F,NF,F NF,F,F 3 3

8

3 F,F,F 1 1

8

b)

c) Discrete: The number of face cards must be a whole number.

d) 1 3 3 1

( ) 0 1 2 38 8 8 8

3 6 3

8

12

8

1.5

E x

The expected value is 1.5. On average, the expected number of face cards in three trials is 1.5.


a) Let D represent rolling doubles and ND represent not rolling doubles. The probability of

rolling doubles on two standard dice is 1

6. The probability of not rolling doubles on two standard

dice is 5

6.


b) To calculate the probability of each outcome, multiply the probabilities along each branch.

Number of

Doubles Distribution of Doubles Probability

0 ND,ND,ND 125

216

1 D,ND,ND ND,D,ND ND,ND,D 75

216

2 D,D,ND D,ND,D ND,D,D 15

216

3 D,D,D 1

216

c) 125 75 15 1

( ) 0 1 2 3216 216 216 216

75 30 3

216

108

216

0.5

E x

The expected number of doubles in three rolls is 0.5.


Refer to question 2 of Prerequisite Skills.



Answers may vary.


a) Answers may vary. There are many more values in these data sets that start with one

compared to other digits.

b) 1(30.1) 2(17.6) 3(12.5) 4(9.7) 5(7.9) 6(6.7) 7(5.8) 8(5.1) 9(4.6)

( )100

344.1

100

3.441

E x

The expected value is 3.441. On average, the expected first digit is 3.441.


a) The probability of cutting the first ace on the first try is 4

52, or

1

13.

b) If the first ace appears on the second try, the card on the first try was anything other than an

ace.

48 4 192

52 52 2704

12

169

The probability of cutting the first ace on the second try is 12

169.

c) If the first ace appears on the third try, the card on the first and second try was anything other

than an ace.

48 48 4 9216

52 52 52 140 608

144

2197

The probability of cutting the first ace on the third try is 144

2197.

d) If the first ace appears on the n try, the card on the first, second, third, …(n – 1)th try was

anything other than an ace.

The probability of cutting the first ace on the nth try is

148 4

52 52

n

.



Divide the circle into a total of 15 equal sectors of 24°. Assign the number 1 to a sector of 24°.

Assign the number 2 to a sector size of 48°, 3 to sector of 72°, 4 to a sector of 96°, and 5 to sector

of 120°. Then, the probability of each sector is 1 2 3 4 5

, , , , and15 15 15 15 15

, respectively.


Let the probability of rolling a 1, 2, 3, 4, and 6 be 1

7. Then, the probability of rolling a 5 is

2

7.

To calculate the probability of each outcome, multiply the probabilities of rolling each value on

the dice. Then, add results for each sum. For example, probability of a sum of 6 is

1 1 1 23 2

7 7 7 7

.

First Die

Second Die 1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12


Sum Frequency Probability

2 1 1

49

3 2 2

49

4 3 3

49

5 4 4

49

6 5 7

49

7 6 8

49

8 5 7

49

9 4 6

49

10 3 4

49

11 2 4

49

12 1 1

49

1 2 3 4 7 8 7 6 4 4 1( ) 2 3 4 5 6 7 8 9 10 11 12

49 49 49 49 49 49 49 49 49 49 49

2 6 12 20 42 56 56 54 40 44 12

49

7.020...

E x

The expected sum of the two weighted dice is approximately 7.02.

Chapter 4 Section 2 Uniform Distributions


a) Yes, the probability distribution of areas is uniform. Each randomly generated radius is

equally likely and there is a single trial.


b)

Random Number, x P(x) x • P(x)

1 1

8

1

8

2 1

8

2

8

3 1

8

3

8

4 1

8

4

8

5 1

8

5

8

6 1

8

6

8

7 1

8

7

8

8 1

8

8

8

c)

d) Use the sum of the values of x ∙ P(x).

1 2 3 4 5 6 7 8( )

8 8 8 8 8 8 8 8

36

8

4.5

E x

The expectation is 4.5. On average, the expected radius length will be 4.5.



Calculate the expectation. Points received indicate a positive value for the random variable, x.

Points lost indicate a negative value for x.

Spin Point Value, x P(x) x • P(x)

1 –1 1

8 –

1

8

2 4 1

8

4

8

3 –3 1

8 –

3

8

4 8 1

8

8

8

5 –5 1

8 –

5

8

6 −6 1

8 –

6

8

7 –7 1

8 –

7

8

8 16 1

8

16

8

Use the sum of the values of x ∙ P(x).

1 4 3 8 5 6 7 16( )

8 8 8 8 8 8 8 8

6

8

0.75

E x

A fair game will have an expectation equal to 0. This is not a fair game because the player will

win 0.75 point on each turn, on average.


Yes, randomly selecting students by their student number is uniform. Each randomly generated

student number is equally likely in a single trial.


a) A spinner with a uniform distribution will have equal sectors.

b) A spinner with a non-uniform distribution will have unequal sectors.



total value of all prizesExpected payout

number of tickets sold

1500

2

.7

000

0 5


Determine the difference between the ticket price and payout per ticket.

$2 – $0.75 = $1.25

The expected profit per ticket is $1.25 (or loss to the customer of $1.25). This gives an advantage

to the school.


a) Since there are different probabilities for different sums in a trial, this is not a uniform

distribution.

b) Since there is an equal probability of selecting each card in a trial, this is a uniform

distribution.

c) Since there is an equal probability of each song being randomly selected in a trial, this is a

uniform distribution.

d) Since there are different probabilities for different numbers of boys in a family of five, this is

not a uniform distribution.

e) Since there is an equal probability of each person being randomly selected in a trial, this is a



Uniform distribution: In the case of political parties using robo-callers to telephone all

constituents in a riding, the probability is 1 for each person.

Uniform distribution: In the case of three people being selected at random from a group of four

girls and five boys, there is equal probability of each person being randomly selected.

Not a uniform distribution: In the case of dealing one card, face down, to each of five players, the

probability changes with each card dealt.

Uniform distribution: In the case of a school randomly selecting a student to attend a conference,

there is equal probability of each student being randomly selected.

Answer C.



The probability of each number is 1

10. Use this common fraction to determine the expected value.

1( ) (1 2 3 4 5 6 7 8 9 10)

10

1 10(10 1)

10 2

5.5

E x

The expected outcome is 5.5. Answer D.



Points lost indicate a negative value for x. Let n represent the number of green balls, n > 4.

Ball

Colour

Number of

Balls

Point Value, x P(x) x • P(x)

red 4 3 4

4 + n

12

4 + n

green n –2 n

4 + n

–2n

4 + n

Use the sum of the values of x ∙ P(x). Set E(X) = 0 for a fair game.

1 1 2 2

12 20

4 4

( ) ( ) ( )

12 20

4

0 12 2

2 12

6

E x x P x x P x

n

n n

n

n

n

n

n

For a fair game, there must be 6 green balls.


a) The probability of each value is 1

5. Use this common fraction to determine the expected

value.

1( ) (5 10 15 20 25)

5

1(75)

5

15

E x

The expected value is 15.

b) The probability of each value is 12.5%, or 0.125. Use this common value to determine the

expected value.


( ) 0.125(0 1 2 3 4 5 6 7)

7(7 1)0.125

2

3.5

E x

The expected value is 3.5.


a)


1 1

12

1

12

2 1

12

2

12

3 1

12

3

12

4 1

12

4

12

5 1

12

5

12

6 1

12

6

12

7 1

12

7

12

8 1

12

8

12

9 1

12

9

12

10 1

12

10

12

11 1

12

11

12

12 1

12

12

12

Use this common fraction to determine the expected value.

1( ) (1 2 3 4 5 6 7 8 9 10 11 12)

12

1 12(12 1)

12 2

6.5

E x

The expected outcome is 6.5.

b) This does not mean that the time represented by the expectation is most likely. The

expectation is simply the predicted average of all outcomes. Each number between 1 and 12 is

equally likely.


a) If you assume that ‘any specific card” means a card, the probability of any specific card in the

deck is 1

52.


b) Since there is an equal probability of selecting any specific card in a trial, this is a uniform

distribution.

c) The probability of any specific card on the second selection is 1

51.

d) Since the probability of any specific card changes after a card is removed, this is not a



Let A = 1, B = 2, C = 3, D = 4, and E = 5.

Random Answer, x P(x)

1 1

5

2 1

5

3 1

5

4 1

5

5 1

5


a) There are four possible outcomes: P confesses but Q denies, P and Q deny, P and Q confess,

and Q confesses but P denies, corresponding to 0, 1, 2, and 3 years, respectively.

Number of Years in

Prison, x

P(x)

0 1

4

1 1

4

2 1

4

3 1

4


b) Answers may vary. The only chance of going free is if prisoner P confesses. If Q confesses,

then P should also confess and he will get prison time of 2 years, whereas if P denies, P will go to

jail for 3 years. If Q denies, then P should confess because he will be set free; otherwise P will go

to jail for 1 year.


a) Answers may vary. My predictions for the expected outcomes of the dice are 2.5, 3.5, 4.5,

6.5, and 10.5 (the average of the face values for each platonic solid).

b) Use this common fraction to determine the expected value.

For 4 faces,

1( ) (1 2 3 4)

4

10

4

2.5

E x


For 6 faces,

1( ) (1 2 3 4 5 6)

6

1 6(6 1)

6 2

3.5

E x


For 8 faces,

1( ) (1 2 3 4 5 6 7 8)

8

1 8(8 1)

8 2

4.5

E x


For 12 faces,

1( ) (1 2 3 4 5 6 7 8 9 10 11 12)

12

1 12(12 1)

12 2

6.5

E x


c) Answers may vary. My findings in part b) confirm my prediction for the icosahedron.


Calculate the expectation. Each sector is equally likely with a probability of 1

10. The total of the

winning options is $127 000. The total of the losing options is –$120 000.

Use the sum of the values of x ∙ P(x).


1( ) (127 000 120 000)

10

7000

10

700

E x

The expected outcome is winning $700.

The amount of risk money that would make this a fair game is $700.


a) Find the area of each ring.

For region A, 2 2π(20 12 )

256π

A

For region B, 2 2π(12 8 )

80π

A

For region C, 2π8

64π

A

Since the areas of the three regions are not equal, this is not a uniform distribution.

b) Answers may vary. To make this a fair game, let region A be worth 5 points, region B be

worth 16 points, and region C be worth 20 points.

c) Answers may vary. A similar target with a uniform distribution would have regions of equal

area. Choose radii of 8 cm, 128 cm, and 192 cm.


The total prizes are worth $1.50(50 000) + $1200(2) + $15 000 + $35 000, or $127 400.

total value of all prizes( ) price per coffe

127 4001.50

1000 000

enumber of coffees sold

0.1274 1.5

1.3726

E X

The expected value of a cup of coffee is –$1.3726.


Determine the expected payout per ticket for total prizes worth $1 110 000.

total value of all prizesExpected payout price per ticket

number of

1 11

tickets sol

0 000

500 000

d

2.22


Let c represent the cost of a ticket. The desired profit is 0.6c. Solve for c.

2.22 0.6

0.4 2.22

5.55

c c

c

c

A ticket should cost $5.55 in order for the charity to make a 60% profit.



Answers may vary.

a) A spinner with 9 equal sectors, labelled 1 to 9, can be used in a uniform distribution that has

P(x) = 1

9.

b) A spinner with 15 equal sectors, labelled 1 to 15, can be used in a uniform distribution that

has an expected outcome of 8. The expected outcome is the average of the values on the spinner.

c) A scenario similar to the game presented in Example 2 will provide an outcome for an unfair

game. The game involves rolling a die. A player who rolls an even number receives points equal

to three times the face value of the die. If the player rolls an odd number, the player loses five

times the face value of the die.

Roll Point Value, x P(x) x • P(x)

1 –5 1

6 –

5

6

2 6 1

6

6

6

3 –15 1

6 –

15

6

4 12 1

6

12

6

5 –25 1

6 –

25

6

6 18 1

6

18

6

Sum –9

6

9( )

6

1.5

E x

In this case, the player will lose 1.5 points on each turn, on average.


Answers may vary.


A uniform distribution has possible outcomes from 1 to n. The probability of each outcome is 1

n.

Use this common fraction to determine the expected value.

1( ) (1 2 3 )

1 ( 1)

2

1

2

E x nn

n n

n

n

The expected outcome is 1

2

n .



Answers may vary. A table of possible winnings is shown.

Number Card 1 Winnings ($) Card 2 Winnings ($)

Midpoint

(Expectation)

1 40 + 2(1) = 42 100 – 12 = 99 70.5

2 40 + 2(2) = 44 100 – 22 = 96 70.0

3 40 + 2(3) = 46 100 – 32 = 91 68.5

4 40 + 2(4) = 48 100 – 42 = 84 66.0

5 40 + 2(5) = 50 100 – 52 = 75 62.5

6 40 + 2(6) = 52 100 – 62 = 64 58.0

7 40 + 2(7) = 54 100 – 72 = 51 52.5

8 40 + 2(8) = 56 100 – 82 = 36 46.0

9 40 + 2(9) = 58 100 – 92 = 19 38.5

10 40 + 2(10) = 60 100 – 102 = 0 30.0

If a number from 1 to 6 is chosen, more money is won with card 2. If a number from 7 to 10 is

chosen, more money is won with card 1. Choosing 1 gives the greatest expected outcome at

$70.50.

Chapter 4 Section 3 Binomial Distributions


a) The probability of success (an ace) on any individual cut of a deck of cards1

13.

The probability of failure is 12

13. There will be one success and nine failures in 10 tires. So, there

is some combination of

91 12

13 13

. The ace can occur on any of the 10 tries, in 10C1 ways. The

nine non-aces can occur in the remaining 9C9 ways. The probability of success on one try and

failure on the other nine is

9

10 1 9 9

1 12

13 13C C

, or approximately 0.3743.

b) There will be three successes and seven failures in 10 tires. So, there is some combination of 3 7

1 12

13 13

. The ace can occur on any three of the 10 tries, in 10C3 ways. The seven non-aces

can occur in the remaining 7C7 ways. The probability of success on three tries and failure on the

other seven is

3 7

10 3 7 7

1 12

13 13C C

, or approximately 0.0312.


a) X = the number of occurrences of a girl

b) Each trial is independent, and on each trial the probability of a girl is 0.5.


Number of Girls, x Probability, P(x) x • P(x)

0 0 6

6 0 (0.5) (0.5) 0.015 625C 0

1 1 5

6 1(0.5) (0.5) 0.093 75C 0.093 75

2 2 4

6 2 (0.5) (0.5) 0.234 375C 0.468 75

3 3 3

6 3(0.5) (0.5) 0.3125C 0.9375

4 4 2

6 4 (0.5) (0.5) 0.234 375C 0.9375

5 5 1

6 5(0.5) (0.5) 0.093 75C 0.468 75

6 6 0

6 6 (0.5) (0.5) 0.015 625C 0.093 75

c) The probabilities sum to 1.

d) Use a graphing calculator.

Compared to the shape of the histogram in Example 2, the probability is more closely bell-shaped

with the mode at x = 3 girls. It is slightly skewed to the left.

e) Use the values from the table in part b). 6

0

( ) 0 0.093 75 0.468 75 0.9375 0.093 75 0.468 75 0.015 625

3

x

x P x

On average, you can expect 3 girls in a family of six children.


a) In this case, the probability of success means conditions for rain, so

p = 40% q = 1 – 0.4

= 0.4 = 0.6

Use the indirect method to determine P(x < 6) = 1 – P(x ≥ 6).

6 2 7 1 8 0

8 6 8 7 8 8

( 6) 1 (6) (7) (8)

1 (0.4) (0.6) (0.4) (0.6) (0.4) (0.6)

0.9502

95.02%

P x P P P

C C C

There is about 95.02% chance that it rained on fewer than six days.

b) ( )

(8 0.4)

3.2

E X np

On average, there will be 3.2 days of rain out of eight days with the correct atmospheric

conditions.



In the binomial distribution, the coefficient nCx represents the number of ways each of the

number of successes can happen.


Answers may vary. In a binomial distribution, p and q represent the probability of success and

failure, respectively. They are related by p = 1 – q. For example, the probability of drawing a

heart (success) from a deck of cards is 0.25, while the probability of not drawing a heart (failure)

is 0.75.


Answers may vary. The statement that about 11% of Canadians are left-handed means that in a

group of 25 people, the expected number of left-handed people is 2.75. However, the probability

of one person in 25 being left-handed is 1 24

25 1(0.11) (0.89)C , or about 16.78%.


The probability of three successes in seven independent trials in a binomial distribution can be

represented by 3 4

7 3C p q . Answer A.


For n = 8 and p = 0.5,

8

)

( )

4

0.5

(E X np

Answer B.


Look for independent trials whose outcomes are either success or failure. This is not the scenario

in the case of the number of queens in a five-card hand, the sum when two dice are rolled, or each

lane for a 100-m race. An example of a binomial distribution is the probabilities of the number of

times a 5 occurs when spinning a spinner six times. Answer D.



a) and b)


c) Assume a success is a rolling a one. Use a graphing calculator to determine all

P(x) = nCxpxq

n−x , where n = 4, x = 0, 1, 2, 3, 4, p = 0.75, and q = 0.25.

d) Use p = 0.75 and q = 0.25. 4 4 0 3 1 2 2 1 3

4 0 4 1 4 2 4 3

0 4

4 4

4 0 3 1 2 2 1 3 0 4

(0.75 0.25) (0.75) (0.25) (0.75) (0.25) (0.75) (0.25) (0.75) (0.25)

(0.75) (0.25)

1(0.75) (0.25) 4(0.75) (0.25) 6(0.75) (0.25) 4(0.75) (0.25) 1(0.75) (0.25)

C C C C

C

The expansion contains the same terms as determined using the binomial distribution formula, but

in reverse order.


a) Use a graphing calculator to determine all P(x) = nCxpxq

n−x , where n = 6, x = 0, 1, 2, 3, 4, 5, 6,

p = 0.3, and q = 0.7.

b) Use a graphing calculator to determine all P(x) = nCxpxq

n−x , where n = 8, x = 0, 1, 2, 3, 4, 5, 6,

7, 8, p = 1

9, and q =

8

9.



For n = 2000 and p = 1

6,

12000

( )

333.333...

6

E X np

The expected number of times a 6 appears when rolling a die 2000 times is about 333.33.


a) The probability of a family of five children having exactly two girls is 2 3

5 2 (0.5) (0.5)C , or

0.3125.

b) The probability of a family of five children having exactly three boys is 3 2

5 3(0.5) (0.5)C , or

0.3125. This is the same as part a).


a) The probability that two people out of six choose the number 9 is 2 4

6 2 (.05) (0.95)C , or about

0.0305.

b) Use the indirect method to determine P(x ≥ 2) = 1 – P(x < 2).

0 6 1 5

6 0 6 1

( 2) 1 (0) (1)

1 (0.05) (0.95) (0.05) (0.95)

0.0328

P x P P

C C

There is about a 0.0328 chance that at least two people out of six choose the number 9.



n−x , where n = 5, x = 0, 1, 2, 3, 4, 5,

p = 1

6, and q =

5

6.



c) Use a graphing calculator to calculate the sum of the x ∙ P(x) values.

The expected number of sums of 7 in five rolls is about 0.8333.

For n = 5 and p = 1

6,

15

6

( )

5

6

E X np

The expected number of of sums of 7 in five rolls is 5

6, or about 0.8333.

The two methods agree.



n−x , where n = 8, x = 0, 1, 2, 3, 4, 5, 6,

7, 8, p = 0.45, and q = 0.55.


b) Use a graphing calculator to calculate the sum of the x ∙ P(x) values.

The expected number of bull’s-eyes in eight attempts is 3.6.

c) Read the value of P(8) from the table created in part b). Since the probability of eight bull’s-

eyes in eight attempts is 0.00168, it is highly unlikely.


a) The probability that you roll two 3s when rolling five dice is

2 3

5 2

1 5

6 6C

, or about 0.1608.

b) Use 1

6 and

5

6.

5 5 0 4 1 3 2 2 3 1 4 0 5

5 5 5 4 5 3 5 2 5 1 5 0

5 0 4 1 3

1 5 1 5 1 5 1 5 1 5 1 5 1 5

6 6 6 6 6 6 6 6 6 6 6 6 6 6

1 5 1 5 1 51 5 10

6 6 6 6 6 6

C C C C C C

2 2 3 1 4 0 51 5 1 5 1 5

10 5 16 6 6 6 6 6

c) The fourth term:

2 3

5 2

1 5

6 6C

.

d) The general term, x n xn xC p q , in the binomial expansion represents the probability of x

successes in n independent trials, where p is the probability of success on an individual trial and q

is the probability of failure on that same individual trial (q = 1 – p).


a) The probability that three bulbs out of 10 do not meet specification is 3 7

10 3(0.06) (0.94)C , or

about 0.0168.

b) The probability that seven bulbs out of 10 do not meet specification is 7 3

10 7 (0.06) (0.94)C , or

about 2.7901 × 10–7

.

c) Interpret between 3 and 7 to mean 3, 4, 5, 6, or 7 do not meet specifications.

The probability that between three and seven bulbs out of 10 do not meet specification is 3 7

10 3(0.06) (0.94)C + 4 6

10 4 (0.06) (0.94)C + 5 5

10 5(0.06) (0.94)C + 6 4

10 6 (0.06) (0.94)C + 7 3

10 7 (0.06) (0.94)C , or about 0.018 84.


d) Answers may vary. I used the direct method in part c). An alternate method is to use the

indirect method: 1 – P(0) – P(1) – P(2) – P(8) – P(9) – P(10).

e) Answers may vary. The expected number of bulbs that do not meet specification out of 10

bulbs is 10(0.06), or 0.6. Yes, the inspector should be concerned if two bulbs do not meet

specifications.



Each point represents a car given away.

Number of Cars

That Start Point Value, x P(x) x • P(x)

0 0 0 5

5 0

1 9

10 10C

0

1 0 1 4

5 1

1 9

10 10C

0

2 2 2 3

5 2

1 9

10 10C

0.1458

3 3 3 2

5 3

1 9

10 10C

0.0243

4 4 4 1

5 4

1 9

10 10C

0.0018

5 5 5 0

5 5

1 9

10 10C

0.000 05

Sum 0.171 95

A fair game will have an expectation equal to 0. This game favours the contestant, on average.


a) The probability that Jamal is successful on eight of ten free-throws is 8 2

10 8 (0.8) (0.2)C , or

about 0.3020. He is 30.2% likely to be successful on 8 of 10 free-throw attempts.

b) Use the direct method to determine P(x ≥ 8).

8 2 9 1 10 0

10 8 10 9 10 10

( 8) (8) (9) (10)

(0.8) (0.2) (0.8) (0.2) (0.8) (0.2)

0.6777...

P x P P P

C C C

He is about 67.8% likely to be successful on at least 8 of 10 free-throw attempts.


a) The probability that Jean will be successful on two out of eight multiple choice questions is 2 6

8 2 (0.25) (0.75)C , or about 0.3115.

b) Assume that a pass means more than four correct questions.

Use the direct method to determine P(x ≥ 4).


4 4 5 3 6 2 7 1 8 0

8 4 8 5 8 6 8 7 8 8

( 4) (4) (5) (6) (7) (8)

(0.25) (0.75) (0.25) (0.75) (0.25) (0.75) (0.25) (0.75) (0.25) (0.75)

0.113807...

P x P P P P P

C C C C C

The probability that Jean will pass is about 0.113 807.

c) The expected number of correct answers on the quiz is 8(.25), or 2.

d) Answers may vary. The probability distribution will be approximately bell-shaped, with the

mode at 2. It will be skewed to the right.

e) Answers may vary. Since the probability of success increases, the probability distribution will

more closely resemble a bell-shape, with the mode at 8(0.4), or about 3.

f) Probability histogram for part d) Probability histogram for part e)


a) The binomial distribution is not a suitable model in the situation because the probability of

success changes with each trial.

b) Answers may vary. Example: A jar contains 12 red balls and eight green balls. Six balls are

removed with replacement. What is the probability that four of the balls are red?

c) Answers may vary. The probability that four of the six balls are red is 4 2

6 4 (0.6) (0.4)C , or

0.311 04.


a) Calculate the probabilities of a majority of people approving the transit system.

Sample Size Probability

7 P(x > 3) = 1 – P(0) + P(1) + P(2) + P(3)

= 1 – binomcdf(7, 0.35, 3)

≈ 0.1998

100 P(x > 50) = 1 – P(0) + P(1) + P(2) + + P(50)

= 1 – binomcdf(100, 0.35, 50)

≈ 7.3783 × 10–4

1000 P(x > 500) = 1 – P(0) + P(1) + P(2) + + P(500)

= 1 – binomcdf(1000, 0.35, 500)

= 0


b) The probability of the majority of people approving the transit system decreases with the

increase in sample size.


The expected payout per scratch card is 6 3 1

10 25 5010 10 10

, or $18.50.

Then, the expected cost to the store if it has 200 customers one particular day is 200(18.5), or

$3700.


Answers may vary.


Answers may vary. Example: 19% of the Canadian population live in rural areas. A group of 30

Canadians is selected at Pearson International Airport.

a) What is the probability that more than two of them live in a rural area?

b) What is the expected number of rural citizens?


a) outcome 1, outcome 2, outcome 3)!

(! ! !

a b cnP p q r

a b c where

P is the probability of outcome 1, outcome 2, and outcome 3

a is the number of times outcome 1 occurs

b is the number of times outcome 2 occurs

c is the number of times outcome 3 occurs

p is the probability of outcome 1

q is the probability of outcome 2

r is the probability of outcome 3

b) In this situation, n = 10, a = 3, b = 2, c = 5, p = q = r = 1

3.

10

3 2 510 1 1 1

3 2 53 2 5

!( , , )

! ! !

!( , , )

! ! !

1( 3, 2, 5) 2520

3

( 3, 2, 5) 0.042 676...

3 3 3

a b cnP X a Y b Z c p q r

a b c

P X Y Z

P X Y Z

P X Y Z

The probability of rolling three reds, two greens, and five yellows in 10 rolls of the die is about

0.0427.



1

0

0

1

1

1

1

1

( ) ( )

!(1 )

( )! !

!(1 )

( )! ( 1)!

!(1 )

( )!( 1)!

( 1)!(1 )

( )!( 1)!

( 1)!(1 )

(( 1) ( 1))!( 1)!

n

i i

i

nx n x

n x

x

nx n x

x

nx n x

x

nx n x

x

nx n x

x

x

E X x P x

x C p q

nx p p

n x x

nx p p

n x x x

np p

n x x

nnp p p

n x x

nnp p p

n x x

( 1) ( 1)

1

nn x

x

Let m = n – 1 and y = x – 1.

0

!(1 )

( )! !

my m y

y

mnp p p

m y y

This is a form of the binomial theorem (x + y)n with x = p, y = 1 – p, and n = m.

( (1 ))

(1)

m

m

np p p

np

np

Chapter 4 Section 4 Hypergeometric Distributions


There are 15 astronauts with 6 women and 9 men. Four are to be selected: n(S) = 15C4

For the probability of 2 men and 2 women, n(A) = 6C2 × 9C2.

6 2 9 2

15 4

( )

540

1365

0.3956

C CP A

C

The probability that 2 men and 2 women are selected is about 39.56%.


a) Use a graphing calculator to determine all ( ) a x n a r x

n r

C CP x

C

, where n = 10, r = 4, a = 6,

x = 0, 1, 2, 3, 4.



c) Compared to the shape of the histogram in Example 2, the probability is less bell-shaped with

the mode at x = 2. It is skewed to the left.

d) P(0) = 0.00476 or 0.476%, which means that getting no green jelly beans is very unlikely.

e) From the graphing calculator, E(X) = 2.4.

On average, there will be 2.4 green jelly beans in four selections.


a) In this situation, n = 24, r = 7, and a = 10. Use the indirect method to determine

P(x ≥ 3) = 1 – P(x < 3).

10 0 14 7 10 1 14 6 10 2 14 5

24 7 24 7 24 7

( 3) 1 (0) (1) (2)

1

0.6430

P x P P P

C C C C C C

C C C

The probability that at least three are male is about 0.6430.

b) For r = 7, a = 10, and n = 24,


7 10

( )

( )

24

2.9167

raE X

n

On average, about 2.9167 males will be chosen in a selection of seven students.


The 80 foxes caught were all different from each other. So, there were no repetitions and the trials

were dependent. If they were tagged, the trial was deemed a success; if not, the trial was deemed

a failure. This is represented by a hypergeometric distribution.

n = size of fox population

a = number originally tagged r = number later caught (sample size)

= 500 = 80

E(X) = number of foxes that had been tagged (expectation from the sample)

= 34

Use the expectation formula.

( )

( )

80(500)

34

1176.470

8 5004

6

03

raE X

n

n

n

n

The fox population is about 1176 foxes.


a) The random variable is the number of red marbles selected. The size of the sample space is 5

marbles. The size of the population is 10 marbles. The range of the random variable is 0 to 5.

b) The random variable is the number of hearts selected. The size of the sample space is 7 cards.

The size of the population is 52 cards. The range of the random variable is 0 to 7.


Since the trials are independent, this is not a hypergeometric probability situation.


From a bag containing five red and six blue blocks, the probability of getting three red blocks in

four selections is 5 3 6 1

11 4

C C

C

. Answer D.



Look for dependent trials whose outcomes are either success or failure. This is not the scenario in

the case of the probability of each sum when two dice are rolled, the probability of a given

number being randomly chosen from the numbers 1 to 10, or the probability of the number of

times a 3 occurs when rolling a die six times. An example of a hypergeometric distribution is the

probability of the number of aces in a seven-card hand. Answer A.


a) For 6 3 9 2

n r

C C

C

, n = 6 + 9 , or 15, and r = 3 + 2, or 5.

b) For 5 7

10 6

a bC C

C

, a = 10 – 7 , or 3, and b = 6 – 5, or 1.

c) For 6 3 2

25

c

d

C C

C

, c = 25 – 6 , or 19, and d = 3 + 2, or 5.



n r

C CP x

C

, where n = 15, r = 4, a = 7,

x = 0, 1, 2, 3, 4.

b) Use a graphing calculator to determine all ( ) a x n a r x

n r

C CP x

C

, where n = 8, r = 4, a = 4,

x = 0, 1, 2, 3, 4.




n r

C CP x

C

, where n = 20, r = 5, a = 5,

x = 0, 1, 2, 3, 4, 5.

b) Use the expectation formula with r = 5, a = 5, and n = 20.

( )

(5 5)

20

1.25

raE X

n

On average, 1.25 hearts will be in a five-card hand of honour cards.

Use a graphing calculator to calculate the sum of the x ∙ P(x) values.

On average, 1.25 hearts will be in a five-card hand of honour cards.


a) In this situation, n = 20, r = 3, and a = 5. Use the indirect method to determine

P(x ≥ 1) = 1 – P(x < 1).

5 0 15 3

20 3

( 1) 1 (0)

1

0.6009

P x P

C C

C

The probability that at least one light bulb is defective is about 0.6009.


b) For r = 3, a = 5, and n = 20,

( )

(3 5)

20

0.75

raE X

n

On average, 0.75 defective light bulbs will be chosen in a selection of 3 light bulbs.

c) 5 3 15 0

20 3

(3)

0.0088

C CP

C

The probability that all three light bulbs are defective is about 0.88%, which is very unlikely.


a) There are 4 aces in a deck of cards.

4 4 48 9

52 13

(4 aces)

0.0026

C CP

C

The probability of 4 aces in a 13-card hand is about 0.0026.

b) There are 4 kings in a deck of cards. Use the indirect method to determine

P(x ≥ 1) = 1 – P(x < 1).

4 0 48 13

52 13

( 1) 1 (0)

1

0.6962

P x P

C C

C

The probability of at least one king in a 13-card hand is about 0.6962.

c) There are 13 cards of each suit in a deck of cards.

13 5 13 8 26 0

52 13

6

(5 clubs, 8 diamonds)

2.6084 10

C C CP

C

The probability of 5 clubs and 8 diamonds in a 13-card hand is about 2.6084 × 10–6

.


There are 30 mice, 8 with a genetic mutation, 10 are selected.

The probability of fewer than three means P(0) + P(1) + P(2).

The probability of more than seven typically means P(8) + P(9) + P(10). However, there are only

8 mice with a defective mutation. So, the probability of more than seven means P(8). Knowing

the shape of the distribution, the ends are highly unlikely.

There is a greater probability of the mice having the genetic mutation in fewer than three.



n = size of fox population


= 200 = 100

E(X) = number of foxes that had been tagged (expectation from the sample)

= 14


( )

( )

100(20

100 20014

0)

14

1428.571 429

raE X

n

n

n

n

The fox population is about 1429 foxes.


a) n = size of fox population

= 120


= 80 = 25

E(X) = number of deer that had been tagged (expectation from the sample)


( )

( )25 80

12

16.666...

0

raE X

n

They would expect to find about 17 deer tagged.

b) Yes. The expected number of tagged deer is 17.


a) Answers may vary. I predict that a five-card hand will have a greater probability of no

spades.

b) For a seven-card hand, For a five-card hand,

13 0 39 7

52 7

(no spades)

0.114 967...

C CP

C

13 0 39 5

52 5

(no spades)

0.221533...

C CP

C

My prediction was correct. As the number of cards in a hand increases, so does the possibility

that it contains a spade.



a) There are 20 cars. It is not possible to send four cars from the 1920s or 1930s.

Four from 1940s, Four from 1950s, Four from 1960s,

4 4 16 0

20 4

(1940s)

0.000 206...

C CP

C

6 4 14 0

20 4

(1950s)

0.003 095...

C CP

C

7 4 13 0

20 4

(1960s)

0.007 223...

C CP

C

Sending four cars from the 1960s has the greatest probability.

b) There are 13 cars from the 1950s or 1960s. Use a graphing calculator to determine all

( ) a x n a r x

n r

C CP x

C

, where n = 20, r = 4, a = 13, x = 0, 1, 2, 3, 4.

c) Use the expectation formula with r = 4, a = 4, and n = 20.

4

( )

(4)

2

0.8

0

raE X

n

On average, 0.8 cars from the 1940s will be selected.

d) Increasing the number of cars from the 1920s changes the value of n to 23 and n – a becomes

10. With more choices from other decades, the mode should shift to the left.


a) There are 9 men and 11 women from which 12 are selected. Note: It is not possible to have a

jury with no men. Use a graphing calculator to determine all ( ) a x n a r x

n r

C CP x

C

, where n = 20,

r = 12, a = 9, x = 1, 2, 3, 4, 5, 6, 7, 8, 9.


b) Answers may vary.

c) Answers may vary.


There are 13 cards of each suit in a deck of cards. The bridge hand has 13 cards.

13 4 13 3 13 3 13 3

52 13

(4 one suit, 3 of each other suit)

0.0263

C C C CP

C

The probability of four cards of one suit and three of each other suit in a 13-card hand is about

0.0263.


There is only one way for the coins to total 75 cents, selecting three quarters.

4 3 8 0

12 3

(75 cents)

0.0182

C CP

C

The probability of that the value of three coins is 75 cents is about 0.0182.

Chapter 4 Section 5 Comparing and Selecting Discrete Probability Distributions

Chapter 4 Section 5 Example Your Turn Page 184

a)


b) There are 5 males and 6 female names from which 4 are selected.

Use a graphing calculator to determine all P(x) = nCxpxq

n−x , where n = 4, x = 0, 1, 2, 3, 4, p =

6

11,

and q = 5

11.

Use a graphing calculator to determine all ( ) a x n a r x

n r

C CP x

C

, where n = 11, r = 4, a = 6, x = 0,

1, 2, 3, 4.

c) The graphs have the same bell-like shape, with the x = 2 female names being the most likely

outcome. The hypergeometric graph is slightly taller than the binomial graph at x = 2 (0.455 vs

0.369) and x = 3 (0.303 vs. 0.295), and shorter at the other values of x. This occurs due to the

dependent nature of the hypergeometric distribution, causing probabilities to increase when fewer

choices are available.



Answers may vary.


No. For a binomial distribution, the probability of each success is the same, but with

hypergeometric the probability of success changes with each trial.


Look for independent versus dependent trials.

a) Hypergeometric: Trials are dependent.

b) Binomial: Trials are independent and given the probability of failure.

c) Hypergeometric: Trials are dependent.

d) Uniform: Trials are independent and all outcomes are equally likely.


If, in a probability distribution, the number of successes is counted, then the distribution may be

binomial or hypergeometric. Answer C.


The probability distribution for the number of squares chosen that contain money is

hypergeometric. Answer C.



Answers may vary.

a) Possible random variable is the number of cards of a particular suit or denomination, x = 0 to

5.

b) Possible random variable is the number of grade 11 students or grade 12 students, x = 0 to 4.

c) Possible random variable is the value of the outcome (a particular suit or denomination).

d) Possible random variable is the number of a particular digit, x = 0 to r.

e) Possible random variable is the number of defective bottles, x = 0 to r.

f) Possible random variable is the value of the outcome (winning square).


Answers may vary.


a) Since each outcome is equally likely, this is a uniform distribution.

b) Answers may vary. Example: Six people are asked to choose a number between 1 and 15.

What is the probability that two people choose the number 13?


a) Since these are independent success/failure trials, this is a binomial distribution.

b) Answers may vary. Example: At Bill’s Burger Barn, there is a hat with five free hamburger

tickets and ten free fries tickets. If three tickets are drawn, what is the probability of winning a

free hamburger?



a) Since these are dependent success/failure trials, this is a hypergeometric distribution.

b) Answers may vary. Example: For a random draw, 20 slips of paper containing people’s

names are placed into a bin, 20% of which are her friends. What is the probability that at least one

friend’s name will be drawn?

c) The probability of at least one friend winning a prize is P(x ≥ 1) = 1 – P(x < 1).

In the first scenario, In the second scenario,

4 0 16 5

20 5

( 1) 1 (0)

1

0.7183

P x P

C C

C

0 5

5 0

( 1) 1 (0)

1 (0.2) (0.8)

0.6723

P x P

C

d) Since there is a higher probability of a friend winning in the first scenario, Barb would be

happier with the hypergeometric distribution.


Answers may vary.


For the expectation for cutting a card from a deck four times, use np. For dealing four cards from

a deck, use ra

n. However, n = r and p =

a

n. So, the results are the same.

a) The expected number of aces in four cuts of the cards is 1

413

, or about 0.3077. The

expected number of aces in four dealt cards is 4(4)

52, or about 0.3077.

b) The expected number of red cards in four cuts of the cards is 1

42

, or 2. The expected

number of red cards in four dealt cards is 4(26)

52, or 2.

c) The expected number of hearts in four cuts of the cards is 1

44

, or 1. The expected number

of hearts in four dealt cards is 4(13)

52, or 1.


a) This can be modelled using a hypergeometric distribution because the trials are dependent.


b) Use a graphing calculator to determine all ( ) a x n a r x

n r

C CP x

C

, where n = 20, r = 6, a = 12,

x = 0, 1, 2, 3, 4, 5, 6.

c) Use the expectation formula with r = 6, a = 12, and n = 20.

( )

(6 12)

2

3.6

0

raE X

n

On average, 3.6 girls will win a prize.


On average, 3.6 girls will win a prize.

d) Answers may vary. A basket contains 20 slips of paper, each with a student’s name on it.

Eight of the names are males and 12 are females. Six names are selected at random, with

replacement, and the number of female names is counted.


Answers may vary.

a) Show the probability distribution for the number of successful ring-tosses in eight attempts.

b) There are 15 bottles, with five green and ten clear. Show the probability distribution for the

number of successful ring-tosses onto green bottles in four attempts.

c) There are eight bottles numbered 1 to 8, and the probability of a successful ring-toss on any

one is equally likely. Show the probability distribution for the bottle number.




n r

C CP x

C

, where n = 10, r = 6, a = 6,

x = 0, 1, 2, 3, 4, 5, 6.

Use a graphing calculator to determine all ( ) a x n a r x

n r

C CP x

C

, where n = 200, r = 6, a = 6,

x = 0, 1, 2, 3, 4, 5, 6.

b) Use a graphing calculator to determine all P(x) = nCxpxq

n−x , where n = 6, x = 0, 1, 2, 3, 4, 5, 6,

p = 0.6, and q = 0.4.

Let p = 0.6. Use a graphing calculator to determine all P(x) = nCxpxq

n−x , where n = 6, x = 0, 1, 2,

3, 4, 5, 6, p = 0.03, and q = 0.97.

c) The two distributions for n = 10 both have somewhat similar bell shapes, with a mode of

x = 4. The hypergeometric graph is slightly taller than the binomial graph at x = 3 and x = 4, and

shorter at the other values of x.


The two distributions for n = 200 both have almost identical half-bell shapes, with a mode of

x = 0. The hypergeometric graph is slightly taller than the binomial graph at x = 1, and slightly

shorter at the other values of x.

d) The binomial distribution gives a close approximation of a hypergeometric distribution when

r is small in relation to n.


Use a graphing calculator to determine the sum of all ( )

( )!

np xe npP x

x

, where n = 2000,

p = 0.015, x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Shown in L2.

Use a graphing calculator to determine the sum of all P(x) = nCxpxq

n−x , where n = 2000, x = 0, 1,

2, 3, 4, 5, 6, 7, 8, 9, p = 0.015, and q = 0.985. Shown in L3.

The Poisson distribution gives approximately 7.1 × 10

–6, where as the binomial distribution gives

approximately 6.4 × 10–6

.


a) i) 1

1( )

1 ( 1)

2

1

2

n

i

i

E X xn

n n

n

n

ii) ( )E X np

an

n

a

iii) ( )ra

E Xn

na

n

a

b) The expected value for the binomial distribution equals that of the hypergeometric

distribution.



a) i) The probability that your first red light will be on your first trip through the intersection is

0.4.

ii) The probability that your first red light will be on your second trip through the intersection is

0.6(0.4), or 0.24.

iii) The probability that your first red light will be on your third trip through the intersection is

0.6(0.6)(0.4), or 0.144.

iv) The probability that your first red light will be on your nth trip through the intersection is

(0.6)n−1

(0.4).

b) The probability of an event occurring for the first time after n initial failures, involving

independent trials is (probability of failure)n × (probability of success).

c) The probability of success after a waiting time of x failures is P(x) = qxp, where p is the

probability of success in each single trial and q is the probability of failure.

d) Use a graphing calculator to determine all P(x) = qxp, where x = 0, 1, 2, 3, 4, 5, 6, 7, p =

1

6,

and q = 5

6.

e) All three distributions have two possible outcomes, success or failure. The geometric and

binomial distributions involve independent trials where the probability of success is the same in

every trial. This is in contrast to the hypergeometric distribution, where trials are dependent and

the probability of success is not the same in every trial. Binomial and hypergeometric

distributions involve a fixed number of trials, whereas geometric distributions involve continued

trials until success.

Chapter 4 Review

Chapter 4 Review Question 1 Page 188

a) Continuous: The time spent playing a hockey game can be measured in fractions of a second.

b) Discrete: The number of times you make a basket in basketball must be a whole number.

c) Discrete: The number of candies in a bag must be a whole number.


d) Continuous: The mass of candies can be measured in fractions of a kilogram.


a) For ease of graphing, convert probabilities to decimal values.

b)


a) 3

0

1 5 1 1( ) 0 1 2 3

12 12 3 6

5 2 30

12 3 6

5 8 6

12

1.583...

x

x P x

The expected value of x is about 1.58.

b) ( ) 2(0.05) 4(0.13) 6(0.24) 8(0.38) 10(0.12) 12(0.05) 14(.03)

7.32

E x



A uniform distribution occurs when, in a single trial, all outcomes are equally likely.



a)


1 1

6

1

6

2 1

6

2

6

3 1

6

3

6

4 1

6

4

6

5 1

6

5

6

6 1

6

6

6

b) Use the sum of the values of x ∙ P(x) in the table in part a).

1 2 3 4 5 6( )

6 6 6 6 6 6

21

6

3.5

E x

The expectation is 3.5. The predicted average value of the random number will be 3.5.


a) Calculate the expectation. Amounts received indicate a positive value for the random

variable, x. Amounts lost indicate a negative value for x.

Number of

Green Balls

Amount

($), x P(x) x • P(x)

0 –360 0 3

3 0 (0.4) (0.6)C –77.76

1 –40 1 2

3 1(0.4) (0.6)C –17.28

2 280 2 1

3 2 (0.4) (0.6)C 80.64

3 600 3 0

3 3(0.4) (0.6)C 38.40

Sum 24


win $24 on each turn, on average.


b) Calculate the expectation. Amounts received indicate a positive value for the random

variable, x. Amounts lost indicate a negative value for x.

Number of

Green Balls

Amount

($), x P(x) x • P(x)

0 –360 3 0

15 14 13

25 24 23C

–71.22

1 –40 3 1

10 15 14

25 24 23C

–18.26

2 280 3 2

10 9 15

25 24 23C

82.17

3 600 3 3

10 9 8

25 24 23C

31.30

Sum 23.99


win $23.99 on each turn, on average.


Use a graphing calculator to determine all P(x) = nCxpxq

n−x , where n = 6, x = 0, 1, 2, 3, 4, 5, 6,

p = 1

6, and q =

5

6.


a) For type O, n = 120 and p = 0.46.

120 0.

( )

.

6

5

4

5 2

E X np

The expected number of people with type O blood is 55.2.

A binomial distribution has a specific number of identical independent trials in which the result is

success or failure. The binomial distribution is a suitable model in the situation because the

probability of success (type O) does not change with each trial.

b) For type A, n = 120 and p = 0.42. The expected number of people with type A blood is

120(0.42), or 50.4. For type B, n = 120 and p = 0.09. The expected number of people with type B

blood is 120(0.09), or 10.8. For type AB, n = 120 and p = 0.03. The expected number of people

with type AB blood is 120(0.03), or 3.6.



a) The probability of winning a prize once in 10 tries is 1 9

10 1(0.12) (0.88)C , or about 0.3798.

b) Use the indirect method to determine P(x ≥ 3) = 1 – P(x < 3).

0 10 1 9 2 8

10 0 10 1 10 2

( 3) 1 (0) (1) (2)

1 (0.12) (0.88) (0.12) (0.88) (0.12) (0.88)

0.1087

P x P P P

C C C

The probability of winning a prize at least three times in 10 tries is about 0.1087.

c) The expected number of winning cards in 10 tries is 10(0.12), or 1.2.



n r

C CP x

C

, where n = 25, r = 7, a = 10,

x = 0, 1, 2, 3, 4, 5, 6, 7.

b) Use the expectation formula with r = 7, a = 10, and n = 25.

( )

(7 10)

2

2.8

5

raE X

n






a) There are 56 coins, with 18 being rare. Select 10 coins.

18 10 38 0

56 10

6

(all rare)

1.228 10

C CP

C

The probability that all 10 coins are rare is about 1.228 × 10–6

.

b) There are 56 coins, with 18 being rare. Select 10 coins.

18 0 38 10

56 10

(0 rare)

0.0133

C CP

C

The probability that no coins are rare is about 0.0133.

c) Use the indirect method to determine P(x ≥ 2) = 1 – P(x < 2).

18 0 38 10 18 1 38 9

56 10 56 10

( 2) 1 (0) (1)

1

0.9043

P x P P

C C C C

C C

The probability that at least two coins are rare is about 0.9043.


n = size of seal; population


= 420 = 100

E(X) = number of seals that had been tagged (expectation from the sample)

= 42


100 420

( )

( )

100(420)

42

100

42

0

raE X

n

n

n

n

The seal population is 1000 seals.


Look for independent versus dependent trials.

a) Binomial: Trials are independent and given the probability of failure.

b) Hypergeometric: Trials are dependent.

c) None of these


d) Uniform: Trials are independent and all outcomes are equally likely.


a) There are 12 faces cards in a deck of 52 cards. The probability that five of seven cards dealt

are face card is 12 5 40 2

52 7

C C

C

, or about 0.0046.

b) There are 12 faces cards in a deck of 52 cards: p = 3

13 and q =

10

13. The probability that five

of seven cards chosen, with replacement are face card is

5 2

7 5

3 10

13 13C

, or about 0.0081.

c) There is a greater probability of five face cards with replacement. This is because the trials

are independent and the probability of a face card does not change.

Chapter 4 Test Yourself

Chapter 4 Test Yourself Question 1 Page 190

This is a uniform distribution. The expected outcome on a typical roll is

1(2 4 6 8 10 12 14 16)

8 , or 9. Answer C.


The binomial and hypergeometric distributions are similar in that they both involve counting

success. Answer D.


The expectation for a uniform distribution is calculated using 1

1 n

i

i

xn

. Answer A.


Counting the number of tails when a coin is flipped 20 times is an example of a binomial

distribution. Answer A.


The probability that exactly two students will be selected when five people are selected from four

students and three teachers is 4 2 3 3

7 5

C C

C

. Answer B.



This is a binomial distribution with p = 0.4 and n = 2(5), or 10. The expected number of red lights

is 10(0.4), or 4.


There are 12 face cards in an honour deck of cards.

12 2 8 1

20 3

(2 face)

0.4632

C CP

C

The probability that two are face cards is about 0.4632.


a) The situation in #7 is modelled by a hypergeometric distribution because the trials are

dependent (without replacement).

b) The situation can be changed to a binomial distribution by making it with replacement.


n = size of beaver population

= 452


= 200 = 65

E(X) = number of beaver that had been tagged (expectation from the sample)


( )

( )65 200

45

28.7610...

2

raE X

n

They would expect to find about 29 beaver tagged.



n−x , where n = 8, x = 0, 1, 2, 3, 4, 5, 6,

7, 8, p = 1

6, and q =

5

6.



c) Use a graphing calculator to calculate the sum of the x ∙ P(x) values.

The expected number of sums of 7 in eight rolls is about 1.3333.


a) The probability that the line is always busy is 15 0

15 15(0.95) (0.05)C , or about 0.4633.

Use the direct method to determine P(x ≥ 12).

12 3 13 2 14 1 15 0

15 12 15 13 15 14 15 15

( 12) (12) (13) (14) (15)

(0.95) (0.05) (0.95) (0.05) (0.95) (0.05) (0.95) (0.05)

0.9945

P x P P P P

C C C C

The probability that the line is busy at least 12 times is about 0.9945.

b) The expected number of times the line should be busy is 15(0.95), or 14.25.


There are 10 males and 5 females that have applied for 4 job promotions.

5 0 10 4

15 4

(0 women)

0.1538

C CP

C

5 0 10 4

15 4

(at least 1 women) 1 (0)

1

0.8462

P P

C C

C

The probability that no females were promoted is about 0.1538, while the probability that at least

one female was promoted is about 0.8462. There is a much greater chance that at least one female

was promoted, so the committee’s claim is unfounded.


a) There are 6 people and p = 0.12.

Use the indirect method to determine P(x ≥ 2).


0 6 1 5

6 0 6 1

( 2) 1 (0) (1)

1 (0.12) (0.88) (0.12) (0.88)

0.1556

P x P P

C C

The probability that at least two of the people will have the disease is about 0.1556.

b) The expected number of people with the disease is 6(0.12), or 0.72.


a) There are 30 players with 18 Canadians. Select 7 players.

18 1 12 6

30 7

(1 Canadian)

0.0082

C CP

C

The probability that only one draft pick is Canadian is about 0.0082.

b) There are 30 players with 18 Canadians. Select 7 players.

18 7 12 0

30 7

(all Canadian)

0.0156

C CP

C

The probability that all draft picks are Canadian is about 0.0156.

c) Use the direct method to determine P(x ≥ 4).

18 4 12 3 18 5 12 2 18 6 12 1 18 7 12 0

30 7 30 7 30 7 30 7

(most Canadian) (4) (5) (6) (7)

0.7335

P P P P P

C C C C C C C C

C C C C

The probability that most of the draft picks are Canadian is about 0.7335.


Answers may vary. In Section 4.5, question 13, you saw that the binomial distribution gives a

close approximation of a hypergeometric distribution when r is small in relation to n. If n is very

large, non-replacement of successes will not change the ratio of successes to the population that

much, and a binomial distribution will be a good approximation to the hypergeometric

distribution.


a) Four numbers are chosen from six positive and eight negative numbers. A product of four

numbers will be positive when 4 positive, 2 positive and 2 negative, or 4 negative.

6 4 8 0 6 2 8 2 6 0 8 4

14 4 14 4 14 4

(positive)

0.5045

C C C C C CP

C C C

The probability that the product of these four numbers will be positive, given that there is no

repetition of numbers is about 0.5045.

b) Four numbers are chosen from six positive and eight negative numbers. A product of four

numbers will be positive when 4 positive, 2 positive and 2 negative, or 4 negative.


4 0 2 2 0 43 4 3 4 3 4

(positive)7 7 7 7 7 7

0.2003

P

The probability that the product of these four numbers will be positive, given that repetition of

numbers is permitted is about 0.2003.

Documents

Chapter 4 Probability Distributions for Discrete Variables Chapter …bartlettstp.weebly.com/uploads/2/3/0/2/23023680/textbook... · 2018. 9. 9. · The probability that the marble