Chapter 5
Section 5.1
Check Your Understanding, page 286:
1. (a) This probably means that if you asked a large sample of U.S. adults whether they usually eat
breakfast, about 61% of them will answer yes. (b) In a random sample of 100 adults, we would expect
that around 61 of them will usually eat breakfast. However, the exact number will vary from sample to
sample.
2. (a) This probability is 0. If an outcome can never occur then it will occur in 0% of cases. (b) This
probability is 1. If an outcome will occur on every trial, then it will occur in 100% of cases. (c) This
probability is 0.01. If an outcome is very rare, but will occur once in a while in a long sequence of trials,
this is consistent with a probability of 0.01 which means that the event occurs in about 1% of trials. (d)
This probability is 0.6. An outcome that will occur more often than not will occur in more than 50% of
trials which means a probability that is greater than 0.50. This leaves us with two choices: 0.60 and 0.99.
The wording suggests that the event occurs often but not nearly every time. This suggests something that
occurs 60% of the time.
Check Your Understanding, page 292:
1. Assign the members of the AP Statistics class the numbers 01-28 and the rest of the students numbers
29-95. Again ignore the numbers 96-99 and 00. In Table D read off 4 two-digit numbers, ignoring
duplicates. Record whether all four numbers are between 01-28 or not. Do this many times and compute
the percent of time that that all four students are part of the AP Statistics class.
2. Assign the numbers 1-10 to Jeff Gordon, 11-40 to Dale Earnhardt Jr., 41-60 to Tony Stewart, 61-85 to
Danica Patrick, and 86-99 and 00 to Jimmie Johnson. Do RandInt(1,100) until you have at least one
number (box) for each of the 5 drivers. Count how many numbers (boxes) you had to sample in order to
get at least one of each.
Exercises, page 293:
5.1 (a) If we use a polygraph machine on many, many people who are all telling the truth, about 8% of
the time, the machine will say that the people are lying. (b) Answers will vary. A false positive would
mean that a person telling the truth would be found to be lying. A false negative would mean that a
person lying would be found to be telling the truth. The U.S. judicial system is set up to think that a false
positive would be worse – that is, saying that someone is guilty (lying) who is not is worse than finding
someone to be truthful when they are, in fact guilty (lying).
5.2 (a) If we test many, many athletes who have not taken performance-enhancing drugs, about 3% of
the time the test will say that they have. (b) Answers will vary. One answer would be that the false
positive is worse because we will ruin the career of a good athlete who has, in fact, been following the
rules.
5.3 (a) If we look at many families where the husband and wife both carry this gene, in approximately
25% of them the first-born child will develop cystic fibrosis. (b) If the family has 4 children, this
constitutes a sample of size 4 which is very small. In order for the probability to be closely reflected in
the sample, the sample size must be very large.
5.4 (a) If we look at many, many hands of poker in which you hold a pair, the fraction of times in which
you can make four of a kind will be about 88/1000. (b) It does not mean that exactly 88 out of 1000 such
hands would yield four of a kind; that would mean , for example, that if you’ve been dealt 999 such hands
106
The Practice of Statistics for AP*, 4/e
and only had four of a kind 87 times, then you could count on getting four of a kind the next time you
held a pair.
5.5 (a) Answers will vary. For example, on one set of 25 spins, we obtained 16 tails and 9 heads. Based
9
on this set of 25 spins the probability of heads is
= 0.36. (b) You could get an even better estimate by
25
spinning the coin many more times.
5.6 (a) Answers will vary. For example, on one set of 25 observations, we obtained 11 tails and 14
14
= 0.56. (b) You could get an
heads. Based on this set of 25 observations the probability of heads is
25
even better estimate by knocking down the coin many more times.
5.7 In the short run there was quite a bit of variability to the percentage of free throws made. In fact, the
basketball player did not do as well early on, but his percentage both increased somewhat and became less
variable as time went on.
5.8 In the short term there was lots of variability to the proportion of heads. In the long term this
proportion settles down around 0.50.
5.9 No, the TV commentator is incorrectly applying the law of large numbers to a small number of at
bats for the player.
5.10 No, the weather in various years may not be independent. Plus the TV weather man is applying the
law of large numbers to a small number of years which is an incorrect application of it.
5.11 (a) There are 10,000 four-digit numbers (0000, 0001, 0002,…, 9999), and each is equally likely to
be chosen. One way to see this is to consider writing each of the numbers on a slip of paper and putting
all of the numbers in a big box. Then you randomly select any slip of paper. Each of the 10,000 slips in
the box, including the ones with 2873 and 9999 on them, is equally likely to be the one you select. (b)
Most people would say that 2873 is more likely than 9999 to be randomly chosen. To many it somehow
“looks” more random – we don’t “expect” to get the same number four times in a row. Choose a number
that most would avoid so that if you win, you don’t have to split the pot with as many other people.
5.12 (a) The wheel is not affected by its past outcomes—it has no memory; outcomes are independent.
So on any one spin, black and red remain equally likely. (b) The gambler is wrong again. Removing a
card changes the composition of the remaining deck, so successive draws are not independent. If you
hold 5 red cards, the deck now contains 5 fewer red cards, so your chance of another red decreases.
5.13 (a) Let 1, 2, and 3 represent the player making the free throw and 4 represent a miss. If 5 or 6
comes up, ignore it and roll again. (b) Let the two-digit numbers 01-75 represent making the free throw
and 76-99 and 00 represent missing. Read a two-digit number from Table D. (c) Let diamonds, spades,
and clubs represent making a free throw and hearts represent missing. Deal one cared from the deck.
5.14 (a) Let 1 and 2 represent a green light and 3-6 represent a red light. Roll the die once. (b) Let the
numbers 1, 2 and 3 represent a green light and 4-9 represent a red light. Ignore 0. Look up one number in
the table. (c) Let diamonds represent a green light and clubs and spades represent a red light. Deal one
card. If a heart is dealt, ignore that outcome and deal again.
Chapter 5: Probability: What Are the Chances?
107
5.15 (a) There are actually 19 numbers between 00 and 18, 19 numbers between 19 and 37, and 3
numbers between 38 and 40. This changes the proportions between the three different outcomes. (b)
There is no reason to skip numbers that have already been encountered in the table. These numbers just
represent the handedness, not a particular individual to select for the sample.
5.16 (a) There is no reason to skip numbers that have already been encountered in the table. These
numbers just represent the obesity, not a particular individual to select for the sample. (b) This will give
the numbers 0 through 9 an equal chance of occurring, but if boys and girls are equally likely, we would
expect it to be more likely to have 4, 5 or 6 boys than 0 or 9.
5.17 (a) This is a legitimate simulation. The chance of rolling a 1, 2 or 3 is 75% on a 4-sided die and the
rolls are independent of each other. (b) This is not a valid design because the chance of heads is 50%
(assuming the coin is fair) rather than the 60% that she hits the center of the target. This will
underestimate her percent of hitting the target.
5.18 (a) This is not a valid design because you are not putting the card back in the deck after dealing.
This means that on the second draw, the proportion of red and black cards has changed, depending on
what card was dealt first. (b) This is an appropriate design because there is a 95% chance of getting a
number between 00 and 94 and the selections are independent because Table D was used.
5.19 (a) What is the probability that, in a random selection of 10 passengers, none from first class are
chosen? (b) Number the first class passengers as 01-12 and the other passengers as 13-76. Ignore all
other numbers. Look up two-digit numbers in Table D until you have 10 unique numbers (no repetitions
because you do not want to select the same person twice). Count the number of two-digit numbers
between 01 and 12. (b) The numbers read in pairs are: 71 48 70 99 84 29 07 71 48 63 61 68 34 70 52.
The bold numbers indicate people who have been selected. The other numbers are either too large (over
76) or have already been selected. There is one person among the 10 selected who is in first class in this
sample. (d) Since in 15% of the samples no first class passenger was chosen, it seems plausible that the
actual selection was random.
5.20 (a) What is the probability that, in selecting 7 tiles from 100, all 7 are vowels? (b) Let the numbers
01-42 represent the vowels, 43-98 represent the consonants, and 99 and 00 represent the blank tiles. Look
up two-digit numbers in Table D until you have 7 unique numbers (no repetitions since once you pull one
tile from the bag you cannot pull it again). Record whether all 7 numbers are between 01 and 42 or not.
(c) The numbers read in pairs are: 00 69 40 59 77 19 66. The bold numbers indicate tiles that have been
selected. In this case only 2 are vowels, so the whole sample is not just vowels. (d) Since there were only
2 occasions among 1000 repetitions in which all tiles were vowels, that suggests that this would only
happen about 0.2% of the time. It is likely that the bag was not mixed properly.
5.21 (a) Read off 30 three-digit numbers from the table, ignoring numbers greater than 365 and 000.
Repeats are allowed because these numbers represent the actual birth date of an individual. Record
whether there were any repeats in the sample or not. (b) Answers will vary. One possible answer: We
used Minitab to select 5 samples. There were repeats in all 5 samples. (c) Answers will vary. One
possible answer: After the simulation we would not be surprised that the probability is 0.71 since we
found repeats in 100% of our samples. Before the simulation this may seem surprising since there are
365 different days that people could be born on.
5.22 (a) Answers will vary. One possible answer: In our sample we won 11 out of 25 times when we
stayed and we won 14 out of 25 times when we switched. (b) Answers will vary. Based on our
simulation we might suggest that the readers were correct. We won 56% of the time when we switched
108
The Practice of Statistics for AP*, 4/e
which is closer to 50%. However, Marilyn is, in fact, correct, so hopefully most students will find a
higher proportion of wins than we did.
5.23 (a) 43 of the 200 samples yielded at least 55% who say they recycle. This means that, according to
the simulation, if 50% recycle, we would see at least 55% of the sample saying the recycle in about
21.5%. This is not particularly unusual. (b) However, only 1 of the 200 samples yielded at least 63%
who said that they recycle. This means that it would happen in about 0.5% of samples. This seems rather
unusual. It would be much more likely that the actual percentage who recycle is larger than 50%.
5.24 (a) If 27 out of 60 say they leave the water running, this means that 45% leave the water running.
22% of the 200 samples in the simulation showed 45% or fewer leaving the water running. This is not
particularly unusual. (b) However, 18 out of 60 would represent a sample with 30% saying they leave
the water running. No samples in the simulation had a percentage this low. So finding such a sample
would be unusual if 50% of the population leave the water running. It would be much more likely that
the actual percentage in the population was smaller.
5.25 State: How many men would be expect to choose in order to find one who is red-green colorblind?
Plan: We’ll use technology to simulate choosing men. We’ll label the numbers 01-07 as colorblind men
and all other two-digit numbers as non-colorblind men. Use technology to produce two-digit numbers
until a number between 01 and 07 appear. Count how many two-digit numbers there are in the sample.
Do: We did 50 repetitions of the simulation using technology. The first repetition is given here: 17 33 49
41 02. The number in bold is the stopping point. For this repetition we chose 5 men in order to get the
one colorblind man. The dotplot below gives the number of men chosen to get a colorblind man in each
of 50 repetitions. The average of the number of men from these samples is 16.88. Conclude: Based on
our simulation, we would suggest that we would need to sample about 17 men, on average. That is, we
would have to choose about 16 men before getting a colorblind man.
5.26 State: How many games of Scrabble would you expect to play, on average, to get a “bingo”? Plan:
We’ll use technology to simulate games of Scrabble. We’ll label the numbers 01-03 as a bingo and 04-99
and 00 as not a bingo. Use technology to produce two-digit numbers until a number between 01 and 03
appear. Count how many two-digit numbers there are in the sample. Do: We did 50 repetitions of the
simulation using technology. The first repetition is given here: 14 33 12 10 05 42 95 81 72 21 31 55 74
79 70 23 63 17 65 62 81 11 92 97 07 38 08 33 29 05 06 85 28 63 44 46 99 18 47 52 48 06 36 88 83 01.
The number in bold is the stopping point. For this repetition we simulated playing 46 games with the first
bingo being on the last game. The dotplot below gives the number of games needed in each of 50
repetitions. The average number of games from these samples is 33.36. Conclude: Based on our
simulation, we would suggest that it would take playing about 33 games to get one bingo, on average.
Chapter 5: Probability: What Are the Chances?
109
5.27 State: What is the probability that, in a sample of 4 randomly selected U.S. adult males, at least one
of them is red-green colorblind? Plan: We’ll use Table D to simulate selecting samples of 4 men. Since
7% of men are red-green colorblind, let 01-07 denote a colorblind man and 08-99 and 00 denote a noncolorblind man. Read 4 two-digit numbers from the table for each sample and record whether the sample
had at least one red-green colorblind man in it. Do: We did 50 repetitions. We started on row 109 from
Table D. The first 5 samples are listed here: Sample 1: 36 00 91 93 (no colorblind); Sample 2: 65 15 41
23 (no colorblind); Sample 3: 96 38 85 45 (no colorblind); Sample 4: 34 68 16 83 (no colorblind); Sample
5: 48 54 19 79 (no colorblind). In our 50 samples, 15 had at least one colorblind man in them. Conclude:
Based on our 50 samples, we would suggest that about 30% of samples would have at least one colorblind
man in them.
5.28 State: What is the probability that none of the winning numbers will appear on any of 5 Lotto
tickets with numbers randomly selected. Plan: Pick the 6 numbers from 1-49 to be the “winning”
numbers. We’ll pick 01-06 but it could be any 6. Then, using technology, pick 6 two-digit numbers
(ignoring numbers from 50-99 and 00) for each of 5 tickets. Note that there cannot be repeats within a
ticket, but there can be repeats between tickets. Record whether the set of 5 tickets has any matches or
not. Do: We did 50 repetitions. The first sample was: ticket 1: 32 21 49 37 18 44; ticket 2: 23 19 10 33
44 31; ticket 3: 36 41 48 40 19 03; ticket 4: 43 25 33 06 40 08; ticket 5: 25 11 15 42 27 06. This did
include the picked numbers (in bold). In fact, in all 50 repetitions, at least one ticket shared a number
with those picked as “winning” numbers. Conclude: Based on our 50 samples, the player should have
been quite surprised not to see any winning numbers on the 5 tickets. This is very unlikely to occur.
5.29 State: What is the probability that a random assignment of 8 men and 12 women to two groups of
10 each will result in at least 6 men in the same group? Plan: Number the men 01-08 and the women 0920. Use technology to pick 10 numbers between 01 and 20 for one group, the remaining 10 numbers are
in the other group. Record whether at least 6 of the numbers between 01 and 08 are in one group. Do:
We did 50 repetitions. The first repetition resulted in Group 1: 17 05 04 14 11 10 03 15 08 01; Group 2:
02 19 12 06 18 09 16 13 07 20. The men are shown in bold. There were 5 men in the first group and
three men in the second group, so neither group had 6 or more men. In our 50 repetitions, 9 had one
group with 6 or more men in it. Conclude: Based on our 50 random assignments, the researchers should
not have been too surprised. 18% of our repetitions had one group with at least 6 men in it.
5.30 State: What is the probability that if the train arrives on time 90% of the time, the train will be late
at least two out of 6 randomly selected days? Plan: Let the digits 1-9 represent days that the train arrives
on time and the digit 0 represent days when the train does not arrive on time. Use Table D to pick 6 digits
and record whether there were at least two 0’s among the 6 digits. Do: We did 50 repetitions starting on
line 122. The first 5 repetitions are as follows: 138738 (no); 159895 (no); 052909 (yes); 087359 (no);
275186 (no). In our 50 repetitions we found 5 samples in which the train arrived late 2 or more of the 6
days. That is, in about 10% of our repetitions the train was late 2 or more times. Conclude: It seems that
it could be that the auditor had the experience just by chance.
5.31 c
5.32 a
5.33 d
5.34 c
5.35 c
110
The Practice of Statistics for AP*, 4/e
5.36 e
5.37 (a) The population is adult U.S. residents and the sample is the 178,545 adults who were
interviewed. (b) Since the interviews were conducted using the telephone, those people who do not have
a telephone were excluded. In general people who do not have phones tend to be poorer and therefore
may experience more stress in their lives than the population as a whole.
5.38 (a) Both distributions are skewed to the right with the drivers generally taking longer to leave when
someone is waiting for the space. The median time to leave was about 40 seconds for the drivers with no
one waiting and about 47 seconds for the drivers with someone waiting. There is more variability for the
drivers with someone waiting as well – they had an IQR of 23.03 seconds as opposed to an IQR of 12.87
seconds for those with no one waiting. Finally there were no outliers for those with someone waiting, but
there were two high outliers for those with no one waiting. (b) We cannot conclude that drivers take
longer when someone is waiting because this was an observational study. The researchers merely
observed what was happening, they did not randomly assign the treatments of either having a person
waiting or not to the drivers of the cars leaving the lot.
Section 5.2
Check Your Understanding, page 303:
1. A person cannot have a cholesterol level of both 240 or above and between 200 and 239 at the same
time.
2. A person has either a cholesterol level of 240 or above or they have a cholesterol level between 200
and 239. P (A or B) = P (A) + P (B) = 0.16 + 0.29 = 0.45.
3. P (C) =
1 − P (A or B) =
1 − 0.45 =
0.55.
Check Your Understanding, page 305:
1.
Heart
Non-heart
Face Card
3
9
Non-face Card
10
30
2. The probability of the event A and B means the probability of selecting a face card that is a heart. This
3
.
probability is
52
3. The probability of A or B means the probability of selecting a card that is either a heart, a face card, or
both. Since A and B are not disjoint, this is not equal to the sum of the probability of A and the
12 13 3 22 11
+
−
=
=
= 0.423.
probability of B. P (A or B) = P (A) + P (B) − P (A and B) =
52 52 52 52 26
Chapter 5: Probability: What Are the Chances?
111
Exercises, page 309:
5.39 (a) The table below illustrates the possible pair combinations in the sample space. (b) Each of the
1
16 outcomes has probability
.
16
First Roll
1
2
3
4
1
(1,1)
(2,1)
(3,1)
(4,1)
Second Roll
2
(1,2)
(2,2)
(3,2)
(4,2)
3
(1,3)
(2,3)
(3,3)
(4,3)
4
(1,4)
(2,4)
(3,4)
(4,4)
5.40 (a) The sample space is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} (b) Each of the 8
1
outcomes has probability .
8
5.41 There are four ways to get a sum of 5 from these two dice: (1,4), (2,3), (3,2), (4,1). Each of these
1
 1 1
has probability
so together the probability of getting a sum of 5 is 4   = .
16
 16  4
5.42 There are 4 ways to get more heads that tails: HHH, HHT, HTH, THH. Each of these has
1
1 1
probability so together the probability of getting more heads than tails is 4   = .
8
8 2
5.43 (a) Legitimate. (b) Not legitimate: the total is more than 1. (c) Legitimate (even if the deck of
cards is not!).
5.44 Model 1 is not legitimate because the probabilities have sum
6
≠ 1. Model 2 is legitimate. Model 3
7
7
≠ 1. Model 4 is not legitimate because
6
probabilities cannot be greater than 1 and the sum of the probabilities is more than 1.
is not legitimate because the probabilities have sum
5.45 (a) The given probabilities have sum 0.96, so P(type AB) = 1 − 0.96 = 0.04. The sum of all possible
outcomes is 1. (b) The probability that the chosen person does not have AB is the sum of all the other
probabilities which is 0.96. (c) P(type O or B) = 0.49 + 0.20 = 0.69.
5.46 (a) The given probabilities sum to 0.91 so
P (other) =
1 − 0.91 =
0.09. (b) P (non English) =
1 − 0.63 =
0.37. (c)
P (neither English nor French) =
1 − 0.63 − 0.22 =
0.15.
5.47 (a) The given probabilities have sum 0.72, so this probability must be 0.28. (b)
P (at least a high school education) =
1 − P(has not finished HS) =
1 − 0.13 =
0.87.
5.48 (a) 35% are currently undergraduates. This makes use of the addition rule of mutually exclusive
events because (assuming there are no double majors) “undergraduate students in business” and
112
The Practice of Statistics for AP*, 4/e
“undergraduate students in other fields” have no students in common. (b) 80% are not undergraduate
business students. This makes use of the complement rule.
5.49 (a) The individuals are the students in the urban school. The variables measured are the children’s
275
gender and whether or not they eat breakfast regularly. (b) P (female) =
.
595
300
110
P (Eats breakfast regularly) =
. P (Female and eats breakfast regularly) =
.
595
595
275 300 110 465
P (Female or eats breakfast regularly) =
+
−
=
.
595 595 595 595
5.50 (a) The individuals are the 100 U.S. senators. The variables being measured are the gender and the
(47 + 13) 60
political affiliation of the senators. (b) P (Democrat)
=
= = 0.6.
100
100
(13 + 4) 17
13
P(Female)
=
= = 0.17. P (Female democrat)
= = 0.13.
100
100
100
60 17 13
64
P(Female or Democrat) =
+
−
=
= 0.64.
100 100 100 100
5.51 (a)
Black
10
8
18
Not Black
10
10
20
Total
20
18
38
Even
Not Even
Total
18
20
(b) P (B) = ; P (E) = . (c) The event “B and E” would be that the ball lands in a black, even spot.
38
38
10
P (B and E) = . (d) The event “B or E” would be that the ball lands in either a black spot, or an even
38
spot, but it could be that it lands in both. If we added the separate probabilities for B and E we would
18 20 10 28
+
−
= .
double count the probability that it lands in a black, even spot. P (B or E) =
38 38 38 38
5.52 (a)
Jack Not a Jack
Total
Red Card
2
24
26
Black Card
2
24
26
Total
4
48
52
4
26
2
(b) P (J) = ; P (R) = . (c) The event “J and R” would be that we deal a red jack. P (J and R) = .
52
52
52
(d) The event “J or R” would be that the card was either a jack, a red card, or both. If we added the
separate probabilities for J and R we would double count the probability that the card is both.
4 26 2 28
P (J or R) =
+
−
= .
52 52 52 52
Chapter 5: Probability: What Are the Chances?
113
5.53 (a)
110 + 190 + 130 430
=
. The probability of being either a male or someone who eats
595
595
430
165
breakfast, or both is
. (c) P (BC ∩ M C ) =. The probability of being neither male, nor someone
595
595
165
who eats breakfast (therefore being a female who doesn’t eat breakfast) is
.
595
P (B ∪ M)
(b) =
5.54 (a)
36 + 4 + 13 53
=
. The probability of being either a Republican senator, a female senator
100
100
53
47
. (c) P (R C ∩ FC ) =. The probability of being neither a Republican senator or a
or both is
100
100
47
.
female senator (therefore being a male Democrat) is
100
(b) P (R
=
∪ F)
5.55 (a)
114
Facebook
MySpace 8,400,000
Not Myspace 8,600,000
Total 17,000,000
Not Facebook
Total
2,400,000 10,800,000
600,000 9,200,000
3,000,000 20,000,000
The Practice of Statistics for AP*, 4/e
(b)
(c) P (FB
∪ MS). (d) P (FB ∪ MS)
=
5.56 (a)
Undergraduate
Graduate
Total
Mac
3,000
1,000
4,000
8,600,000 + 8, 400,000 + 2, 400,000 19, 400,000
=
.
20,000,000
20,000,000
PC
3,700
2,300
6,000
Total
6,700
3,300
10,000
(b)
1,000
= 0.10. There are a total of 1,000 students who were both Mac
10,000
users and graduate students out of the population of 10,000 students.
(c) P (M ∩ G). (d) P (M ∩ =
G)
5.57 c
5.58 d
5.59 c
5.60 c
5.61 The scatterplot for the average crawling age and average temperature is given below.
Chapter 5: Probability: What Are the Chances?
115
In this scatterplot we see that there appears to be a moderately strong, negative linear relationship. In
fact, the correlation bears this out as it is -0.70. The equation for the least-squares regression line is
∧
age
= 35.7 − 0.077temp. We predict that babies will walk 0.077 weeks earlier for every degree warmer it
gets.
5.62 The simplest design would be a completely randomized design, assigning half of the
women to take strontium ranelate and half to take the placebo. A better design would block
according to the medical center (and country); that is, randomly assign the strontium
ranelate to half of the women from country A, half of those from country B, and so on.
This blocking would control for differences from one country to another.
Section 5.3
Check Your Understanding, page 314:
2, 268 + 800 + 588
3656
= = 0.3656.
10,000
10,000
800
800
800
| L) = 0.219. P (L=
| E)
= = 0.50. To answer the question about
2. P (E=
3656
368 + 432 + 800 1600
whether engineering students earn lower grades than students in the liberal arts, you would want to look at
P (L | E). This gives the probability of getting a lower grade if the student is an engineer. We would want
to compare that to the probability of getting a lower grade if the student is in liberal arts and social
sciences.
1. =
P (L)
Check Your Understanding, page 317:
1. A and B are independent. Since we are putting the first card back and then re-shuffling the cards
before drawing the second card, knowing what the first card was will not tell us anything about what the
second card will be.
2. A and B are not independent. Once we know the suit of the first card, then the probability of getting a
heart on the second card will change depending on what the first card was.
3. The two events, “female” and “right-handed” are independent. Once we know that the chosen person
is female, this does not tell us anything more about whether she is right-handed or not. Overall,
116
The Practice of Statistics for AP*, 4/e
24 6
18 6
= of the students are right-handed. And, among the women,
= are right-handed. So
28 7
21 7
P ( right-handed ) = P ( right-handed | female ) .
Check Your Understanding, page 321:
1.
2. P (laptop) =
P(laptop ∩ CA) + P(laptop ∩ TX) + P(laptop ∩ NY) =
0.30 + 0.175 + 0.175 =
0.65.
Check Your Understanding, page 323:
1. P (one returned safely) =
1 − P(one was lost) =
1 − 0.05 =
0.95. So, if there are 20 missions and whether
the bomber returned safely or not is independent for each mission, then
20
P (safe return on all 20 missions)
= P(1st safe) ∗ P (2nd safe) ∗…∗ P (20th =
safe) 0.95
=
0.3585.
2. No, we cannot conclude that 2.4% of adults 55 or older are college students. Whether one is a college
student and one’s age are not independent events. Far more younger people are college students than
older people.
Exercises, page 329:
5.63 (a) P (almost certain=
| M)
597
426
= 0.2428. (b) P (F | Some chance)
= = 0.5983.
2459
712
197
197
= = 0.6176. (b)
197 + 122 319
151
151
P (third class | survived)
=
= = 0.3416.
197 + 94 + 151 442
5.64 (a) P(survived | first class)
=
663
1421
= 0.2801. (b) P (good chance)
= = 0.2944. (c) The events “a
2367
4826
good chance” and “female” are not independent since the two probabilities in parts (a) and (b) are not the
same.
| F)
5.65 (a) P (good chance=
94
94
= = 0.3602. (b)
94 + 167 261
197 + 94 + 151
442
P (survived)
=
= = 0.3662. (c) The events “survived” and “second
197 + 122 + 94 + 167 + 151 + 476 1207
class” are not independent since the two probabilities in parts (a) and (b) are not the same.
5.66 (a) P (survived | second class)
=
Chapter 5: Probability: What Are the Chances?
117
13
13
= = 0.7647. This means that 76.47% of the females are democrats. (b)
13 + 4 17
13
13
P (F | D)
=
= = 0.2167. This means that 21.67% of the democrats are female.
47 + 13 60
5.67 (a) P(D | F)
=
5.68 (a) P (B | =
M)
P (M |=
B)
190
= 0.5938. This means that 59.38% of the males eat breakfast. (b)
320
190
= 0.6333. This means that 63.33% of those who eat breakfast are males.
300
60
= 0.60. From question 5.67 we saw that P (D | F) = 0.7647. Since these two
100
probabilities are not the same, D and F are not independent.
=
5.69 P (D)
300
= 0.504. From question 5.68 we saw that P (B | M) = 0.5938. Since these two
595
probabilities are not the same, B and M are not independent.
=
5.70 P (B)
5.71 (a) P (is studying other than English) =
1 − P ( none ) =
1 − 0.59 =
0.41. (b)
P ( Spanish | other than English
=
)
0.26
= 0.6341.
0.41
5.72 (a) P ( $50,000 or more ) = 0.215 + 0.100 + 0.006 = 0.321. (b)
P ( at least $100,000 | at least $50,000
=
)
0.106
= 0.3302.
0.321
5.73 P ( B ) < P ( B | T ) < P ( T ) < P ( T | B ) . There are very few professional basketball players, so
P ( B ) should be the smallest probability. It’s much more likely to be over 6 feet tall than it is to be a
professional basketball player if you’re over 6 feet tall. Lastly, if you are a professional basketball player,
it is quite likely that you are tall (larger than the probability that a randomly selected individual is over 6
feet tall).
5.74 Answers may vary. One possible answer is P ( T ) < P ( T | A ) < P ( A ) < P ( A | T ) . Nearly everyone
who’s career is teaching has a college degree so P ( A | T ) will have the largest probability (close to 1).
And more of the general population will have college degrees than will be teachers, so P ( T ) < P ( A ) .
The final question is where to put P ( T | A ) . Here we have assumed that the proportion of teachers
among the college educated is smaller than the proportion of college educated among all people.
5.75 There are 36 different possible outcomes of the two dice: (1,1), (1,2),…,(6,6). Let’s assume that the
second die is the green die. There are then six ways for the green die to show a 4: (1,4), (2,4), (3,4), (4,4),
1
(5,4), (6,4). Of those, there is only one way to get a sum of 7, so P ( sum of 7 | green is 4 =
) = 0.1667.
6
Overall, there are 6 ways to get a seven: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). So
118
The Practice of Statistics for AP*, 4/e
6
= 0.1667. Since these two probabilities are the same, the events “sum of 7” and “green
36
die shows a 4” are independent.
P ( sum of 7=
)
5.76 There are 36 different possible outcomes of the two dice: (1,1), (1,2),…,(6,6). Let’s assume that the
second die is the green die. There are then six ways for the green die to show a 4: (1,4), (2,4), (3,4), (4,4),
1
(5,4), (6,4). Of those, there is only one way to get a sum of 8, so P ( sum of 8 | green is 4 =
) = 0.1667.
6
5
Overall, there are 5 ways to get an eight: (2,6), (3,5), (4,4), (5,3), (6,2). So P ( sum of 8=
) = 0.1389.
36
Since these two probabilities are not the same, the events “sum of 7” and “green die shows a 4” are not
independent.
5.77 (a)
84 168
 14  6   6  14  84
+
=
= 0.4421.
(b) P ( one soft ∩ one hard ) =    +    =
 20  19   20  19  380 380 380
5.78 (a)
999,000
 1,000  999 
(b) P ( defective ∩ defective )= 
= 0.00999.

=
10,000
9,999
99,990,000



5.79 P ( download music ∩ don't care ) =
P ( don't care | download music ) P ( download music )
=
0.67 )( 0.29 )
(=
0.1943.
Chapter 5: Probability: What Are the Chances?
119
5.80 P ( belong to health club ∩ go twice a week ) =
P ( go twice/week | belong ) P ( belong )
=
0.40 )( 0.10 )
(=
5.81
0.04.
P ( college ∩ pros ∩ 3+years pro )
= P ( 3+years | college ∩ pros ) P ( college ∩ pros )
P ( 3+years | college ∩ pros ) P ( pros | college ) P ( college )
0.40 )( 0.017 )( 0.05 )
(=
0.0003.
5.82 P ( online ∩ profile ∩=
comment ) P ( comment | online ∩ profile ) P ( online ∩ profile )
P ( comment | online ∩ profile ) P ( profile | online ) P ( online )
0.76 )( 0.55 )( 0.93)
(=
0.3887.
5.83
State: What is the probability that a randomly chosen customer paid with a credit card? Plan: We need
to find the probability P ( credit card ) . Do: Look at the tree diagram. P ( credit card ) is the sum of the
three branches that involve people using their credit card. So
P ( credit card ) = 0.2464 + 0.0068 + 0.0420 = 0.2952. Conclude: About 29.5% of customers use their
credit card to pay for gasoline purchases.
120
The Practice of Statistics for AP*, 4/e
5.84
State: What percent of the overall vote does the candidate expect to get? Plan: We need to find the
probability P ( support ) . Do: Look at the tree diagram. P ( support ) is the sum of the three branches that
involve people supporting this candidate. So P ( support ) = 0.12 + 0.36 + 0.10 = 0.58. Conclude: The
candidate expects to get about 58% of the overall vote which means that he would win.
5.85 P ( premium gas | credit card ) =
P ( premium gas ∩ credit card )
P ( credit card )
card )
P ( credit card ) = 0.2952. So P ( premium gas | credit=
5.86 P ( black | support ) =
P ( black | support
=
)
P ( black ∩ support )
P ( support )
. From exercise 5.83,
0.0420
= 0.1423.
0.2952
. From exercise 5.84, P ( support ) = 0.58. So
0.36
= 0.6207.
0.58
Chapter 5: Probability: What Are the Chances?
121
5.87 (a)
(b) There are two branches in the tree that lead to the patient surviving 5 years. Add the probabilities of
these two branches. P ( survive ) = 0.378 + 0.180 = 0.558.
5.88 (a)
(b) There are two branches in the tree that lead to the server winning the point. Add the probabilities of
these two branches. P ( win point ) = 0.4307 + 0.2080 = 0.6387.
5.89 An individual light remains lit for 3 years with probability 1 − 0.02; the whole string remains lit
with probability (1 − 0.02 ) =
( 0.98)  0.6676 .
20
20
5.90 This would not be surprising: assuming that all the authors are independent (for example, none were
written by siblings or married couples), we can view the nine names as being a random sample. Then
P ( none common ) =
0.4032.
(1 − 0.096 ) =
9
5.91 State: If 10 people appear at random to give blood, what is the probability that at least one of them
is a universal donor (has blood type O-negative)? Plan: Since the people are appearing at random, it is
reasonable to assume that their blood types are independent of each other. Use the fact that
P ( at least one universal donor ) = 1 − P ( none are universal donors ) . Do: Start by finding
P ( none are universal donors ) . The probability that any one person is not a universal donor is
10
=
=
0.4737. This means that
1 − 0.072 =
0.928. So P ( none are universal donors
) 0.928
P ( at least one universal donor ) =
1 − 0.4737 =
0.5263. Conclude: There is about a 53% chance that at
least one of 10 random blood donors is a universal donor.
122
The Practice of Statistics for AP*, 4/e
5.92 State: What is the probability that at least one of seven internet references in a paper doesn’t work
two years later? Plan: We will assume that whether or not the internet references are there two years
later or not is independent. This is a reasonable assumption if they are from different domains. Use the
fact that P ( at least one site gone ) = 1 − P ( no sites are gone ) . Do: Start by finding
P ( no sites are gone ) = P ( all sites are still there ) . The probability that any one site is still there is
7
=
=
0.3773. This means that
1 − 0.13 =
0.87. So P ( all sites are still there
) 0.87
P ( at least one site gone ) =
1 − 0.3773 =
0.6227. Conclude: There is about a 62% chance that at least one
of the 7 internet reference sites is gone two years later.
5.93 We cannot simply multiply the probabilities together. If the first of the three consecutive shows
starts late, it is much less likely that the next show will start on time. These events are not independent of
each other.
5.94 We cannot simply multiply the probabilities together. It is likely that if one flight is late, that
whatever is causing it to be late (weather, backed-up airplanes at the airport, etc) will also be affecting the
other three flights. Therefore these four events are not independent of each other.
FB )
5.95 P ( MS =
P ( MS ∩ FB ) 0.42
= = 0.4941.
P ( FB )
0.85
5.96 We want to find P ( G | MAC ) . Putting the known probabilities into a table and solving for the
unknowns will aid us in finding this probability.
MAC
Graduate
10%
Undergraduate 30%
Total
40%
PC
23%
37%
60%
Total
33%
67%
100%
0.1
From the table, we find that P ( G | MAC
=
) = 0.25.
0.4
5.97 (a)
Chapter 5: Probability: What Are the Chances?
123
From the tree diagram, add the probabilities for all branches that end in lactose intolerant. This leads to
P ( lactose intolerant ) = 0.123 + 0.098 + 0.036 = 0.257. (b) We are looking for
P ( Asian | lactose intolerant
=
)
P ( Asian ∩ lactose intolerant ) 0.036
= = 0.1401.
P ( lactose intolerate )
0.257
5.98 (a) Add the probabilities for all branches that end in a contribution. This involves multiplying
along the branches.
P ( contribute ) = ( 0.5 )( 0.4 )( 0.8 ) + ( 0.3)( 0.3)( 0.6 ) + ( 0.2 )( 0.1)( 0.5 ) = 0.16 + 0.054 + 0.01 = 0.224.
=
(b) We are looking for P ( recent donor | contribute
)
P ( recent donor ∩ contribute ) 0.16
= = 0.7143.
P ( contribute )
0.224
5.99 (a)
(b) Add the probabilities for all branches that end in a positive test.
P ( positive ) = 0.009985 + 0.00594 = 0.015925. (c)
P ( antibody | =
positive )
P ( antibody ∩ positive ) 0.009985
= = 0.6270.
P ( positive )
0.015925
5.100 (a) A false-positive means that the technician identified someone as having the disease when, in
fact, they were healthy. This happened in 50 out of 750 healthy people so the false-positive rate was
50
= 0.0667. A false-negative means that the technician identified someone as being healthy when, in
750
fact, they had the disease. This happened in 10 out of 250 patients so the false-negative rate was
P ( had disease ∩ positive result )
10
.
= 0.04. (b) We are looking for P ( had disease | postive result ) =
P ( positive result )
250
Out of the 1000 people tested, there were 240 + 50 =
290 positive results, so
290
P ( positive result
=
) = 0.29. There were 240 people who had the disease and got a positive result so
1000
240
P ( had disease ∩ positive result ) =
=
0.24. This gives
1000
0.24
P ( had disease | postive result
=
) = 0.8276.
0.29
124
The Practice of Statistics for AP*, 4/e
5.101 (a) & (b) These probabilities are:
P(1st card ♠) =
13
= 0.25
52
12
 0.2353
51
11
P(3rd card ♠ | 2♠s picked) =
 0.22
50
10
P(4th card ♠ | 3♠s picked) =
 0.2041
49
9
P(5th card ♠ | 4♠s picked) =
= 0.1875
48
P(2nd card ♠ | ♠ picked) =
(c) The product of these conditional probabilities gives the probability of a flush in spades by the
extended multiplication rule: We must draw a spade, and then another, and then a third, a fourth, and a
fifth. The product of these probabilities is about 0.0004952. (d) Since there are four possible suits in
which to have a flush, the probability of a flush is four times the probability found in (c), or about
0.001981.
1
5.102 (a) There are 6 ways to get doubles out of 36 possibilities so P ( doubles ) = . (b) Since the rolls
6
are independent we can use the multiplication rule for independent events.
P ( no doubles first ∩ doubles second ) =
P ( no doubles first ) P ( doubles second )
=
30  6  5  1  5
= =
.
36  36  6  6  36
=
roll ) P ( no doubles ) P ( no doubles ) P (=
doubles )
(c) P ( first doubles on third
5  5  1  25
=
.

6  6 
 6  216
3
5 1
(d) For the first doubles on the fourth roll, the probability is     . For the first doubles on the fifth
6 6
4
5 1
roll, the probability is     . The general result is that the probability that the first doubles are rolled
6 6
5
on the kth roll is  
6
k −1
1
 .
6
5.103 (a)
=
P ( two boys | at least one boy )
=
0.25
= 0.333.
0.75
(b) P ( two boys | older is boy
=
)
P ( two boys ∩ at least one boy )
P ( two boys )
=
P ( at least one boy )
P ( at least one boy )
P ( two boys ∩ older is boy )
P ( two boys )
0.25
=
= = 0.5.
P ( older is boy )
P ( older is boy ) 0.50
5.104 c
5.105 e
Chapter 5: Probability: What Are the Chances?
125
5.106 d
5.107 The z-score corresponding to a BMI of 18.5 is
with z-score lower than -1.12 is 0.131.
18.5 − 26.8
= −1.12. Using Table A, the percent
7.4
5.108 Note that P ( one not underweight ) =
1 − P ( underweight ) =
1 − 0.131 =
0.869.
P ( at least one is underweight ) =
1 − P ( none are underweight ) =
1 − 0.8692 =
0.2448.
5.109 Generally, more education means more freedom. For example, 45% of those with bachelor’s
degrees felt free to organize their work, compared to 29.7% of those who finished high school and 24% of
those who did not finish high school. This can also be seen in the bar graph.
126
The Practice of Statistics for AP*, 4/e
Chapter Review Exercises (page 334)
R5.1 When the weather conditions are like those seen today, it has rained on the following day about
30% of the time.
R5.2 (a) Let the numbers 01-15 represent drivers who do not have their seat belts on and 16-99 and 00
represent drivers who do have their seat belts on. Using the table of random digits, read 10 sets of 2digits. We do not need to discard repeated numbers because they are not referring to specific individuals.
For each set of 10 2-digit numbers, record whether there are two consecutive numbers between 01-15 or
not. (b) The first sample is 29 07 71 48 63 61 68 34 70 52. Those not wearing seat belts are in bold. In
this sample there was only one not wearing a seat belt, so there were not two consecutive people not
wearing their seat belts. The second sample is 62 22 45 10 25 95 05 29 09 08. In this sample there were
two consecutive drivers not wearing their seatbelts (represented by 09 and 08). The third sample is 73 59
27 51 86 87 13 69 57 61. In this sample there were not two consecutive drivers not wearing their seat
belts. Based on these three simulations, we’d suggest that the probability of finding two consecutive
people out of a sample of 10 not wearing their seat belts to be about 0.33.
R5.3 Assign each of the eight teams a single-digit number. In particular, let the Sandblasters be the
number 1. Now use technology to pick 5 different random numbers between 1 and 8. These are the
teams that had their sculpture blown up last year. Pick 5 more numbers between 1 and 8 for those teams
to have their sculpture blown up this year. Record whether the number 1 was missing in both years or
not. Repeat this many times. The first sample in our simulation was: 1 2 4 3 8 for the first year and 6 8 3
2 5 for the second year. In this sample, team 1 was chosen in the first year. In our 50 simulations, team 1
was chosen in at least one year 47 times. Only three times out of 50, or 6% of the time, was team 1 not
chosen in either year. The other teams should probably be suspicious.
R5.4 (a) Die A can be either 6 or 2 and Die B can be either 5 or 1. So the difference (Die A – Die B) can
take on 3 different values: 6 − 5 =
1, 6 − 1 =
5, and 2 − 5 =−3. Since the rolling of the dice is
1 or 2 − 1 =
21 1
1 1  1
1 1
1
1)
=
, P ( 6 ∩=
1)
= , and
independent, P ( 6 ∩ 5=
) P ( 6 ) P ( 5=)  = , P ( 2 ∩=


3 2 3
3  2  6
3 2  6
21 1
P(2 ∩=
5)
= . Putting this all together we get
3  2  3
Difference
1
5
-3
Probability
1 1 3 1
1
1
+ = =
6 3 6 2
6
3
(b) Die A is more likely to be higher. Half of the time Die B rolls a 1 and Die A is automatically higher.
If Die B rolls a 5, there are still some times that Die A is higher – when it rolls a 6. So more than half of
the time Die A is higher.
R5.5 (a) It is legitimate because every person must fall into exactly one category, the probabilities are all
between 0 and 1, and they add up to 1. (b) The probability that a randomly chosen person is Hispanic is
0.001 + 0.006 + 0.139 + 0.003 =
0.149. (c) The probability that a randomly chosen person is not a NonHispanic white is 1 − 0.674 =
0.326. (d) Being white and being Hispanic are not mutually exclusive
P ( white or Hispanic ) = P ( white ) + P ( Hispanic ) − P ( white ∩ Hispanic )
events. So
= 0.813 + 0.149 − 0.139 = 0.823.
Chapter 5: Probability: What Are the Chances?
127
R5.6 (a)
(b) The probability that neither admits him is 0.3. (c) The probability that at least one admits him is 0.7.
We can see this from the general addition rule P ( P ∪ S) = P ( P ) + P ( S) − P ( P ∩ S) = 0.4 + 0.5 − 0.2 = 0.7.
R5.7 (a)
(b) Add the branches that end in a positive test. P (=
+) 0.095 + 0.027
= 0.122. (c)
P ( actually used steroids
=
| +)
P ( actually used steroids ∩ + ) 0.095
= = 0.7787.
P (+)
0.122
4
R5.8 (a) These events are not independent, because P ( pizza with mushrooms ) = , but
7
2
P ( pizza with mushrooms | thick crust ) = . If the events had been independent, these probabilities
3
4 1
would have been equal. (b) With the eighth pizza P ( mushrooms =
) = , and
8 2
2 1
P ( mushrooms | thick crust =
) = , so these events are independent.
4 2
R5.9 (a) There are a total of 871 pine seedlings, of which 209 had damage from deer, so the probability
209
= 0.24.
of a randomly selected pine seedling having damage by deer is
871
128
The Practice of Statistics for AP*, 4/e
60
= 0.2844,
211
1 2  44
1  76


P  damage | <=
 = 0.1991, and
 = 0.3248, P  damage | 3 to =
3  221
3  234


2  29

P  damage | >=
 = 0.1415. (c) Yes, knowing the amount of cover does change the probability of
3  205

deer damage. It appears that deer do much more damage when there is no cover or less than 1/3 cover
and much less damage when there is more cover than that.
(b) P ( damage | no cover
=
)
3
R5.10 (a) P ( up three consecutive =
years )  P ( up one year
=
)
0.65 ) 0.274625.
(=
3
3
P ( same direction for 3 years ) =
P ( up 3 years ) + P ( down 3 years ) =
( 0.65) + ( 0.35)
3
(b)
= 0.2746 + 0.0429 = 0.3175.
R5.11 (a)
The possible blood types and their probabilities are:
Blood type A
AB
B
Probability 0.25
0.5
0.25
(b)
The possible blood types and their probabilities are:
Blood type A
AB
B
Probability 0.5
0.25
0.25
So the probability that at least one of two children are type B is 1 − ( 0.75 ) =
0.4375.
2
Chapter 5: Probability: What Are the Chances?
129
AP Statistics Practice Test (page 336)
T5.1 c. Probability only tells us what happens approximately in the long run, not what will happen in the
short run.
T5.2 d. You need exactly 62 of the 100 2-digit numbers to represent the event “having heard of Cocacola.”
T5.3 c. Add the probabilities for 3, 4 and 5 cars.
T5.4 b. All 2-digit numbers among the first 10 are between 00 and 97 except 98.
T5.5 b. 255 of the 1000 students had a GPA below 2.
T5.6 c. There are 285 students who either have a GPA below 2, have skipped many classes or both.
T5.7 e. There are 110 students who have skipped many classes. 80 of them have a GPA below 2.
T5.8 e. If A and B are independent, then we don’t know whether B has occurred if A occurred. But if A
and B are mutually exclusive, then if B has occurred then we know that A couldn’t have occurred.
T5.9 b. P ( woman ∪ never married ) = P ( woman ) + P ( never married ) − P ( woman ∩ never married ) .
 12  11  10 
T5.10 c. We want P ( first is picture ∩ second is picture ∩ third is picture ) =     ≈ 0.01.
 52  51  50 
T5.11 (a) Since each outcome is equally likely and there are 48 outcomes, each outcome has probability
1
. There are 27 ways in which the teacher wins (all boxes above and to the right of the diagonal line
48
27
indicating ties). So the probability that the teacher wins is
. (b) We use the fact that
48
27
P ( A ∪ B=
) P ( A ) + P ( B ) − P ( A ∩ B ) . From part (a), P ( A ) = . There are 8 outcomes in which “you
48
8
get a 3” so P ( B ) =
and there are 5 outcomes in which “you roll a 3” and the teacher still wins so
48
5
27 8
5 30 5
+
−
=
= . (c) From part
P ( A ∩ B ) =. Putting all of this together gives P ( A ∪ B ) =
48 48 48 48 8
48
27
8
5
(b), we have P ( A ) = , P ( B ) =
and P ( A ∩ B ) =. Since
48
48
48
5  27  8 
P ( A ∩ B) =
≠
 = P ( A ) P ( B ) , A and B are not independent.
48  48 
 48 
T5.12 (a)
130
The Practice of Statistics for AP*, 4/e
(b) To get the probability that a part randomly chosen from all parts produced by this machine is
defective, add the probabilities from all branches in the tree that end in a defective part.
P ( defective ) = 0.06 + 0.09 + 0.04 = 0.19. (c) First compute the conditional probabilities that the part was
=
produced on a particular machine given that it is defective. P ( A | defective
)
0.06
= 0.3158.
0.19
0.09
0.04
= 0.4737. P ( C | defective
=
) = 0.2105. Since the largest of these three
0.19
0.19
conditional probabilities is for machine B, given that a part is defective, it is most likely to have come
from machine B.
P ( B | defective
=
)
P ( gets cancer ∩ smoker ) 0.08
= = 0.32. (b) The event that an
P ( smoker )
0.25
individual either smokes or gets cancer is the complement of the event that the individual neither smokes
nor gets cancer. So
P ( smokes ∪ gets cancer ) =
1 − P ( does not smoke ∩ does not get cancer ) =
1 − 0.71 =
0.29.
(c) First find the probability that one person gets cancer. From part (b) we know that
P ( smokes ∪ gets cancer ) =
0.29. But we also know that P ( smokes ) = 0.25 and
=
T5.13 (a) P ( gets cancer | smoker
)
P ( smokes ∩ gets cancer ) =
0.08. Now, using P ( S ∪ C )= P ( S) + P ( C ) − P ( S ∩ C ) we get
0.29 =
0.25 + P ( gets cancer ) − 0.08. Solving for P ( gets cancer ) we get 0.12. Finally
P ( at least one gets cancer ) =
1 − P ( neither gets cancer ) =
1 − 0.882 =
0.2256.
T5.14 (a) Assign the numbers 01-17 to represent cars with out-of-state plates and the remaining 2-digit
numbers between 00 and 99 to represent other cars. Start reading 2-digit numbers from a random number
table until you have found two numbers between 01 and 17 (repeats are allowed). Record how many 2digit numbers you had to read in order to get 2 numbers between 01 and 17. Repeat many times for the
simulation. (b) The first sample is 41 05 09. The numbers in bold represent cars with out-of-state
plates. In this sample it took three cars to find two with out-of-state plates. The second sample is 20 31
06 44 90 50 59 59 88 43 18 80 53 11. In this sample it took 14 cars to find two with out-of-state plates.
The third sample is 58 44 69 94 86 85 79 67 05 81 18 45 14. In this sample it took 13 cars to find two
with out-of-state plates.
Chapter 5: Probability: What Are the Chances?
131