Probability and Statistics Solutions Chapter 3 - … · Probability and Statistics Solutions – Chapter 3 Problem Set 3.1 Exercises 1) EV 0 0.02 1 0.26 2 0.37 3 0.19 4 0.12 5 0.04

Probability and Statistics Solutions – Chapter 3

Problem Set 3.1 Exercises

1) 0 0.02 1 0.26 2 0.37 3 0.19 4 0.12 5 0.04 2.25EV

2) a) Cookie Dough =50

67%75

Butter Braids =15

20%75

Bread = 10

13%75

Product Cookie Dough

Butter Braids Bread Packs

Value $6 $10 $12

Probability 0.67 0.20 0.13

b) 6 0.67 10 0.20 12 0.13 $7.60EV In other words, the expected value of a sale (average sale) is $7.60.

Note: The table contains rounded values!

3) First build a probability model for this situation.

Value $0 $5 $10 $30

Probability 0.30 0.40 0.25 0.05

$0 0.30 $5 0.40 $10 0.25 $30 0.05 $6.00EV The owner should charge $6.


Value $2 $5 $10 $15

Probability 0.20 0.40 0.30 0.10

$2 0.20 $5 0.40 $10 0.30 $15 0.10 $6.90EV The owner should charge $6.90 + $1.50 = $8.40 for a player to play this game.

5) 1 0.29 2 0.38 3 0.17 4 0.11 5 0.05 2.25EV

The expected number of laps completed is 2.25.

6) a) Use ‘x’ to represent the missing amount (???).

$0 0.32 $1 0.47 $3 0.08 0.07 $21 0.06 $2.53x

$0 $0.47 $0.24 0.07 $1.26 $2.53x

0.07 $0.56x

$8.00x

The missing payout is $8.

b) They would pick $3 for the price instead of $2 because the expected payout is $2.53.

If they charged only $2, they would lose money on average.

c) They would pick $3 for the price instead of $6 because $6 is a very large amount to

charge when the average payout is $2.53. At $6, players would quickly recognize this as

a game in which you easily lose money and will be much less likely to play. Casinos

don’t make money unless people play.

7) 1 2 3 4 5 6

3.56

pips. Using the expected value calculation, we would have had

5.36

21

6

6

6

5

6

4

6

3

6

2

6

1

6

16

6

15

6

14

6

13

6

12

6

11 pips

8)

Result Heads Tails

Value $20 $30

Probability 0.5 0.5

$20 0.5 $30 0.5 $25EV

9) a) Use ‘x’ for the missing amount (???).

$3 0.25 $6 0.35 $10 $50 0.07 $6.95x

$0.75 $2.10 $10 $3.50 $6.95x

$10 $0.60x

0.06x This is impossible because the total of the probabilities must add up to 1 and the

probabilities would only add up to 0.73 if x=0.06.

b) We must make sure the probabilities add up to exactly 1.

0.25 0.35 0.07 1x

0.33x

c) $3 0.25 $6 0.35 $10 0.33 $50 0.07 $9.65EV

10) a) First notice that the total of the probabilities must add up to 1.

0.20 0.25 0.15 0.30 1x

0.10x The missing probability is 10%.

Value $1 $3 $4 $7 ??

Probability 0.20 0.25 0.15 0.30 x

b) Perform the expected value calculation. Let ‘y’ represent our missing prize amount.

$1 0.20 $3 0.25 $4 0.15 $7 0.30 0.10 4.75y

$0.20 $0.75 $0.60 $2.10 0.10 4.75y

0.10 $1.10y

$11y

11) a) 0 0.03 1 0.45 2 3 4 0.02 1.6x y

0 0.45 2 3 0.08 1.6x y

2 3 1.07x y

b) 0.03 0.45 0.02 1x y

0.50x y

c) We will now solve the system 2 3 1.07

0.50

x y

x y

using linear combinations (elimination).

2 3 1.07

2( 0.50)

x y

x y

which gives:

2 3 1.07

2 2 1

x y

x y

. When we add, the ‘x’ terms cancel.

2 3 1.07

2 2 1

0.07

x y

x y

y

Since 0.50x y we can also quickly determine that ‘x’ must equal 0.43.

x = 0.43 and y = 0.07

Problem Set 3.1 Review Exercises

12) a) 145 29

( ) 0.45325 65

P Male

b) 175 7

( ) 0.54325 13

P Cell

c) 55 11

( | ) 0.7375 15

P Male TV

d) 45 1

( | ) 0.25180 4

P Computer Female

13) There is no indication that order matters. The employee will pick out chairs and tables.

There are 10 3 4 2 120 6 720C C ways to do this.

14) a) 26 1

( e ) 0.552 2

P R d

b) 13 1

( ) 0.2552 4

P Spade

c) 12 3

( ) 0.2352 13

P Face

d) 13 1

( | e ) 0.526 2

P Heart R d


1) Consider the probability model shown below.

Color White Red Blue

Value $0 $10 $20

Probability 25

30

4

30

1

30

25 4 1$0 $10 $20 $2

30 30 30EV

This game should cost $2 if it is to be a fair game.


Roll 1 2 3 4 5 6

Value $1 $2 $3 $4 $5 $12

Probability 1

6

1

6

1

6

1

6

1

6

1

6

1 1 1 1 1 1$1 $2 $3 $4 $5 $12 $4.50

6 6 6 6 6 6EV

This game should cost $4.50 to be a fair game.

3)

Value 57 58 59 0 61 62 63 64 65 66 67

Prob. 1

36

2

36

3

36

4

36

5

36

6

36

5

36

4

36

3

36

2

36

1

36

1 2 3 4 5 6 5 4 3 2 157 58 59 0 61 62 63 64 65 66 67

36 36 36 36 36 36 36 36 36 36 36

1,992 16655.33

36 3EV

If you roll one more time, the average result is 55.33 which is better than stopping with

55 points. It is to your advantage to roll one more time.

4)

Value 62 63 64 0 66 67 68 69 70 71 72

Prob. 1

36

2

36

3

36

4

36

5

36

6

36

5

36

4

36

3

36

2

36

1

36

1 2 3 4 5 6 5 4 3 2 162 63 64 0 66 67 68 69 70 71 72

36 36 36 36 36 36 36 36 36 36 36

2,152 53859.78

36 9EV

If you roll one more time, the average result is 59.78 which is worse than stopping with

60 points. It is not to your advantage to roll one more time.

5) Using the fact that it is advisable to roll when you have 55 points but not advisable to

roll if you have accumulated 60 points, we know the ‘break-even’ point must be

somewhere between 55 and 60. Using guess and check we can find the correct

solution. Here are the results for a guess of 58 points:

Value 60 61 62 0 64 65 66 67 68 69 70

Prob. 1

36

2

36

3

36

4

36

5

36

6

36

5

36

4

36

3

36

2

36

1

36

1 2 3 4 5 6 5 4 3 2 160 61 62 0 64 65 66 67 68 69 70

36 36 36 36 36 36 36 36 36 36 36

2,08858

36EV

At 58 points, it makes no difference if you stop or roll once more.

6) In order to win, a player must get all three digits correct.

1 1 1 1( 3 ) 0.001

10 10 10 1000P All Correct

This implies that the chance of not winning is 0.999.

All three digits match At least one doesn’t match

Value $500 $0

Probability 0.001 0.999

1 999$500 $0 $0.50

1000 1000EV

The expected value is $0.50 and since they pay $1 to play, the average player loses

$0.50 every time they play.

7) In order to win, the player must pick blue and blue, red and red, or yellow and yellow.

P(match) = 12 11 10 9 8 7 278 139

0.3230 29 30 29 30 29 870 435

Marbles match Marbles do not match

Value $12 $0


The expected value is $3.83 as shown in the calculation below.

83.3$68.0$32.12$ Since the game costs $5 to play, the expected loss for a player is $5 - $3.83 = $1.17.

8) The only way the insurance company loses money is if the collection is lost, damaged, or

stolen. If this happens, they have to pay out $20,000. If this does not happen, they pay

nothing.

Result Lost, stolen, damaged etc.

No problem

Value (to insurance company) – $20,000 $0


𝐸𝑉 = −$20,000 ∙ 0.002 + $0 ∙ 0.998 = −$40

The insurance company expects to pay out $40 per year on this policy. Since they

collect $300, they would have an annual expected profit of $260 for this type of policy.

9) Build a probability model to organize the information. Note that the table takes into

consideration the original cost of the land.

Highly productive Moderately productive Unproductive

Value $130,000 $90,000 $0


$130,000 0.20 $90,000 0.70 $0 0.10 $89,000EV

The expected selling price for a parcel of land is $89,000 so the prospector made a good

investment since each parcel costs $50,000. The prospector will expect to make an

average profit of $39,000 per parcel of land like this.

10) Begin with by building a probability model that tracks what might happen.

Woman dies Woman survives

Value – $250,000 $0

Probability 0.000943 0.999057

𝐸𝑉 = −$250,000 ∙ 0.000943 + $0 ∙ 0.999057 = −$235. .75

Since the amount collected is $360 and the expected payout is $235.75, the insurance

company has an expected profit of $124.25 on this particular policy.

11) a)

b) P($100) =1 1 3 1 4 1

0.0120 20 20 20 400 100

P($20) =1 5 3 2 11

0.0320 20 20 20 400

P($5) =1 14 14 7

0.0420 20 400 200

P($2) =3 17 51

0.1320 20 400

Value $0 $2 $5 $20 $100

Probability 0.80 0.1275 0.035 0.0275 0.01

c) $0 0.80 $2 0.13 $5 0.04 $20 0.03 $100 0.01 $1.98EV This represents the average payout, however, the player had to pay $5 to play so there

is actually an average loss for the player of $1.98 – $5 = –$3.02 per play.

P($20)=20

2

P($2)=20

17

P($100)=1

20

P($5)=14

20

P($20)=5

20

P($100)=1

20

P(Silver)=3

20

P(Gold)=1

20

P(Red)= 16

20

12)

Color Red Blue Green Yellow

Value $2 $4 $5 $5

Probability 1/2 1/6 1/6 1/6

1 1 1 1$2 $4 $5 $5 $3.33

2 6 6 6EV

This game should cost $3.33 to be a fair game.

13) Start by building a tree diagram.

P($1)=0.6

P($20)=0.1

P($10)=0.3

P($1)=0.56

P($10)=0.33

P($20)=0.11

P($1)=0.67

P($10)=0.22

P($20)=0.11

P($1)=0.67

P($10)=0.33

P($1)=0.6

P($10)=0.3

P($20)=0.1

P(Silver)=0.3

P(Red)=0.6

P(Gold)=0.1

The amounts of money that may be won are $0, $1, $2, $10, $11, $20, $21, & $30.

Value $0 $1 $2 $10 $11 $20 $21 $30

Probability 0.6 0.18 0.03 0.09 0.04 0.04 0.01 0.01

The expected value of the payout is $2.80. Since the game costs $3 to play, you will lose

an average of $0.20 every time you play. You should not play.


14) a) Since the sum of the probabilities must equal 1, ‘X’ = 0.55.

b) $2 0.55 $4 0.25 $7 0.15 $11 0.05 $3.70EV

c) Add $2 to the average payout. Lulu should charge $5.70 per play.

15) The order people are placed on a committee does not matter so this is a combination

question. We must select 2 juniors and 4 seniors so we find that there are

8 2 8 4 28 70 1,960C C ways to select the committee.

16) The three cards must either be all red or all black. It does not matter what the color of

the first card is, we just must make sure cards 2 and 3 match it.

25 24 600 4( ) 1 0.24

51 50 2,550 17P All Same Color

17) a) 3*3 = 9 possible outcomes

b) S = {aa, ab, ac, ba, bb, bc, ca, cb, cc}


Note: Answers may vary because there are many ways that digits can be assigned.

1)

I will let the numbers 1-8 represent students who want to eliminate final exams.

I will let 9 and 0 represent students who do not want to eliminate final exams.

I will use line # 147 from the random digit table.

I will look at 1-digit at a time.

I will not need to ignore any numbers.

I will look at 20 “students” and record how many wish to eliminate final exams.

I will carry this out one time.

Results: By following the above method, I found that 19 of the 20 students selected (or 95%)

were in support of eliminating final exams.

2) a) Since the probabilities must add up to 1, the missing probability must be 0.1 or 10%.

b) I will assign 1 and 2 to represent the top 10%, 3 through 6 for the 10% to 25% group,

7 through 9 for the 25% to 50% group, and 0 for the bottom 50%.

c)


I will look at 1-digit at a time.


I will ask 20 “students from BEC” and record their class ranks.


Group Top 10% 10% - 25% 25% - 50% Bottom 50%

# of Results I IIII II IIII III IIII

% of Students 5% 35% 40% 20%

The results of this simulation are fairly close to the expected results for both middle

groups. The furthest from the expected is that of the top 10% category.

3) a)

I will let the numbers 01-74 represent Charlotte making a free-throw. I will let 75-99

and 00 represent Charlotte missing a free-throw.


I will look at 2-digits at a time.


I will select three numbers and record how many times Charlotte makes her free-

throws.

I will carry this out 20 times.

Number of free-throws Charlotte Makes

0 1 2 3

# of times none IIII IIII IIII IIII I

b) 16/20 or 80% of the time Charlotte made at least two of three free-throws.

c) 14/20 or 70% of the time Charlotte missed at least one of three free-throws.

4) a) Assign 01 to 20 for A’s, 21 to 49 for B’s, 50 to 84 for C’s, 85 to 99 & 00 to D’s and F’s.

b)


I will look at 2-digits at a time.


I will ask 30 “students” and record their grades in Probability & Statistics.


The table below summarizes the results for this simulation from line 106.

Grade A’s B’s C’s D’s or F’s

# of Results IIII I III IIII IIII IIII IIII I

% of Students 20% 10% 50% 20%

c) The frequency of A’s was exactly as expected. The frequency of B’s was much lower

than expected. The number of C’s was much higher than expected. And the number of

D’s or F’s was close to what was expected.

5) a) Assign the numbers 01 to 52 to represent the cards in the deck (as shown in the table

below).

b) Use line # 138. Look at 2-digits at a time. Ignore 53 to 99 and 00 because they do not

represent any cards. Also ignore any repeats because it is impossible to get the exact

same card two times in one hand. Choose five “cards” and see if you have any pairs.

Repeat this until you get a hand with a pair. Record how may hands it took to get a pair.

Hearts Digits

Diam. Digits

Clubs Digits

Spades Digits

A 01 A 14 A 27 A 40

2 02 2 15 2 28 2 41

3 03 3 16 3 29 3 42

4 04 4 17 4 30 4 43

5 05 5 18 5 31 5 44

6 06 6 19 6 32 6 45

7 07 7 20 7 33 7 46

8 08 8 21 8 34 8 47

9 09 9 22 9 35 9 48

10 10 10 23 10 36 10 49

J 11 J 24 J 37 J 50

Q 12 Q 25 Q 38 Q 51

K 13 K 26 K 39 K 52

Results:

Hand 1 = 16, 08, 40, 22, 15 3D, 8H, AS, 9D, 2D No Matches

Hand 2 = 47, 13, 35, 17, 28 8S, KH, 9C, 4D, 2C No Matches

Hand 3 = 04, 10, 43, 35, 34 4H, 10H, 4S, 9C, 8C This has both the 4 of Hearts & 4 of Spades.

In this simulation, it took 3 trials before there was a hand with two matching cards.

6) a) Even though it is unlikely to happen, four 6’s in a row are possible.

b) Roughly 10% of all digits should be a nine. However there is no guarantee that exactly

10% are nine’s. It is similar to flipping a coin. We might expect roughly 50% heads, but

it certainly could be slightly higher or lower than that as flips are completed.

c) Go to the next line in the table and continue your process.

7) a) Assign each day of the year a number. In this case, use 001 (January 1st) through 365

(December 31st). Starting on line #121, we will look at 3-digit numbers, ignoring any that

are greater than 365 and 000. We will record 30 “birthdates” and see if we have any

“students” who have the same birthdate.

b) The list below represents the 30 students selected using the method from part a).

077, 148, 168, 347, 052, 245, 102, 290, 136, 081, 271, 025, 330, 184, 281, 350, 143, 271,

276, 119, 159, 121, 184, 103, 360, 277, 099, 330, 008, 173

In this simulation, there was a match in the group of 30 students. In fact, there were

three pairs of students born on the same day of the year. These pairs of students were

born on the 184th, 330th, and 271st day of the year.


8) In one interpretation of this problem, there are 8 cards that would be good for the first

card. Once you have that card (a king or an ace), there would only be 4 cards that would

be acceptable for the second card. 8 4 32 8

( & ) 0.0152 51 2652 663

P K A

Another way to think about this problem is that you can get an Ace and then a King or a King

and then an Ace. So, two orders times 4/52 times 4/51.

𝑃(𝐾 𝑎𝑛𝑑 𝐴𝑐𝑒) = 2 ∙ (4

52∙

4

51) =

32

2652=

8

663≈ 0.0121

9) a) Remember, there are only 7 even pool balls in the bag. Of those, the 2, 4, 6, and 8

are solid while the other 3 are striped. Building a tree diagram will be helpful here.

Only the starred branches will matter. These give: 28 24 52 26

0.5398 98 98 49

.

b) We want the probability of the pool ball having an even number if we know it is striped. We

know there are seven striped pool balls in the bag. Of those, only 3 have even numbers. This

gives: 3

0.437 .

10) a) 36 + 11 + 8 = 55

b) 97 – 55 = 42 who play neither of these sports

𝑃(𝑛𝑒𝑖𝑡ℎ𝑒𝑟) = 42

97≈ 0.43

c) 11

( | ) 0.2347

P Hockey Football

P(Solid)= 7

14

P(Striped)= 7

14

P(Solid)= 8

14

P(Striped)= 6

14

P(Striped,Even)= 3

7

P(Solid,Even)= 4

7

***28

98

***24

98

Chapter 3 Review Exercises

1) a) The only way to win is to get a red and red or a blue and blue.

P(Red & Red) =10 9 90 3

0.1525 24 600 20

P(Blue & Blue) =15 14 210 7

0.3525 24 600 20

Therefore, there must be a 50% chance of losing if the marbles don’t match.

Result No Match Red & Red Blue & Blue

Value $0 $10 $5


b) $0 0.50 $10 0.15 $5 0.35 $3.25EV

c) The average payout is $3.25 but it costs $4 to play so you would expect to lose an

average of $0.75 every time you play. It is not to your advantage to play this game.

2) a)

Prize Value $500 $100 $50 $20 $0

Probability 1/1000 2/1000 5/1000 10/1000 982/1000

b) 𝐸𝑉 = 500 ∙1

1000+ 100 ∙

2

1000+ 50 ∙

5

1000+ 20 ∙

10

1000+ 0 ∙

982

1000= 1.15

However, it costs $5 per ticket, so the average prize is –$3.85 per ticket. In other words,

you expect to lose an average of $3.85 per ticket purchased.

c) The most likely prize is to win $0, or lose $5 after the cost is figured in.

3) a)

Assign each student a number from 01 to 38.

Use line #137.

Look at 2-digits at a time.

Ignore numbers 39-99 and 00. Also ignore any repeats on the same day,

because a student will not need to do two problems on the same day.

Select the 4 “students to do a problem on the board” and record the results.

Repeat this process five times to represent an entire week.

b) The four students that get selected for Monday are: numbers 12, 14, 36, and 11. (Note

that a result of 12 came up twice but the second 12 had to be ignored or we would have

picked the same student twice on the same day.)

The four students that get selected for Tuesday are: numbers 16, 08, 22, and 15.

The students for Wednesday are: numbers 13, 35, 17, and 28.

The students for Thursday are: numbers 04, 10, 35, and 34.

The students for Friday are: numbers 17, 19, 12, and 32.

4) a)

Assign the digits 1 to 6 to represent each side of a die.

Start on line # 119.

Look at 1-digit at a time.

Ignore 7, 8, 9, and 0.

Select two “dice” and record the total pips shown.

Run this simulation 36 times.

Note: Since we are rolling two dice, we must get two results for each trial. For example, the

first simulated roll results in a 5 and a 5, giving a total of ten.

The results are summarized in the table below (rounded to the nearest percent).

Total 2 3 4 5 6 7 8 9 10 11 12

# of Results 1 1 2 3 6 5 6 4 4 1 3

Percent 3% 3% 6% 8% 17% 14% 17% 11% 11% 3% 8%

≈Theoretical Percent

3% 6% 8% 11% 14% 17% 14% 11% 8% 6% 3%

b) The theoretical percentages are based upon the dice chart below. While the results

are not an exact match, a pattern emerges showing that our simulation has higher

percentages in the middle range of totals and lower percentages on the lower and

higher totals. This matches the theoretical percentage pattern quite well.

+ 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

5) a) Start by building the tree diagram.

The possible totals are $20, $40, $100, and $120. The probability model based upon the

tree diagram can now be constructed. All probability values shown are rounded.

b)

Amount Won $20 $40 $100 $120

≈ Probability 0.333 0.167 0.167 0.333

c) 𝐸𝑉 = 20 ∙ 0.333 + 40 ∙ 0.167 + 100 ∙ 0.167 + 120 ∙ 0.333 = 70

The expected value for this game is $70. In order for this to be a fair game it should cost

$70 per play.

P($20)= 0.6

0.16

0.16

0.16

P($100)= 0.3

P($20)= 0.6

P($20)=1

P($100)=0.5

P($20)=0.5

P(Tails)=0.5

P(Heads)=0.5

0.16

0.3

P($100)= 0.3

6) One way to solve this problem is by constructing a table that shows the Sample Space.

Bill Spin

$1 $5 $10

1 $1 $5 $10

2 $2 $10 $20

3 $3 $15 $30

Since each result is equally likely, they all have the same probability of 1/9. (1

3∙

1

3=

1

9)

1 1 1 1 1 1 1 1 1$1 $5 $10 $2 $10 $20 $3 $15 $30

9 9 9 9 9 9 9 9 9EV = 10.66667

This gives an expected value of approximately $10.67 per play. Therefore, in order to

make this a fair game, a player should be charged $10.67 per play.

7) a) Since the probability values must add up to 1, the “other” category must have a

probability of 0.16.

b) Assign digits as follows: ‘Bus’ will be 01 to 31, ‘Walk’ will be 32 to 45, ‘Car’ will be 46

to 84, and ‘Other’ will be 85 to 99 and 00.

c)

Start on line #104.

Look at 2-digit numbers.

Nothing needs to be ignored.

Select 10 “students” and record how they get to school in the morning.

Run three trials.

The results of this simulation are given in the chart below.

Method of getting to school Bus Walk Car Other

Probability 0.31 0.14 0.39 0.16

Trial #1 results 4 0 3 3



8) a) The most likely number of centers hit is 17.

b) 15 0.04 16 0.12 17 0.35 18 0.28 19 0.18 20 0.03 17.53EV

The expected number of centers hit is 17.53.

9) One way to organize the information is by using a Venn diagram. There are 4 circles,

reds, spades, faces, and aces. Note that the red circle and spade circle will not overlap

and the ace circle will not overlap the face circle.

Note that this represents 43 cards from a deck. The other 9 cards (2-10 of clubs) are not worth

anything.

Other Other Red

Other Spade

Other Face

Red Face

Spade Face

Other Ace

Red Ace

Spade Ace

Value $0 $2 $3 $5 $7 $8 $10 $12 $13

Prob. 9

52

18

52

9

52

3

52

6

52

3

52

1

52

2

52

1

52

9 18 9 3 6 3 1 2 1$0 $2 $3 $5 $7 $8 $10 $12 $13

52 52 52 52 52 52 52 52 52EV

The expected value for this game is approximately 3.673077.

Therefore, this game should cost $3.67 in order to be a fair game.

Face=3

($5)

Ace=1

($10)

Spade=9 ($3) Red=18 ($2)

1 ($13)

6 ($7) 3 ($8)

2 ($12)

Documents

Probability and Statistics Solutions Chapter 3 - … · Probability and Statistics Solutions – Chapter 3 Problem Set 3.1 Exercises 1) EV 0 0.02 1 0.26 2 0.37 3 0.19 4 0.12 5 0.04