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.
4) First build a probability model for this situation.
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
Problem Set 3.2 Exercises
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.
2) First build a probability model for this situation.
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
Probability 0.3195 0.6805
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
Probability 0.002 0.998
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
Probability 0.20 0.70 0.10
$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.
Problem Set 3.2 Review Exercises
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}
Problem Set 3.3 Exercises
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 use line # 103 from the random digit table.
I will look at 1-digit at a time.
I will not need to ignore any numbers.
I will ask 20 “students from BEC” and record their class ranks.
I will carry this out one time.
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 use line # 129 from the random digit table.
I will look at 2-digits at a time.
I will not need to ignore any numbers.
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 use line # 106 from the random digit table.
I will look at 2-digits at a time.
I will not need to ignore any numbers.
I will ask 30 “students” and record their grades in Probability & Statistics.
I will carry this out one time.
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.
Problem Set 3.3 Review Exercises
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
Probability 0.50 0.15 0.35
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
Trial #2 results 3 1 6 0
Trial #3 results 4 1 3 2
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)