CHAPTER 6 REVIEW
1) Given that P(A) = 0.6, P(B) = 0.3, and P(B|A) = 0.5.
a) P(A and B) = ?
b) P(A or B) = ?
c) Are events A and B independent?
ANS:
a) P(A and B) = (0.6)(0.5) = 0.3
b) P(A or B) = 0.6 + 0.3 – 0.3 = 0.6
c) Because P(B) ≠ P(B|A), events A and B are not independent.
2) Consider the experiment of drawing two cards from a standard deck of 52 cards. Let event A =
“draw a face card on the first draw,” B = “draw a face card on the second draw,” C = “the first card
drawn is a diamond.”
a) Are the events A and B independent?
b) Are the events A and C independent?
ANS:
a) NO, the events are not independent. The probability of B changes depending on what
happens with A. Because there are 12 face cards, if the first card drawn is a face card, then P(B) =
11/51. If the first card is not a face card, then P(B) = 12/51. Because the probability of B is affected
by the outcome of A, A and B are not independent.
b) P(A) = 12/52 = 3/13. P(A|C) = 3/13 (3 of the 13 diamonds are face cards). Because these
are the same, the events, “draw a face card on the first draw” and “the first card drawn is a diamond”
are independent.
3) Suppose 80% of the homes in Lakeville have a desktop computer and 30% have both desktop
computer and a laptop computer. What is the probability that a randomly selected home will have a
laptop computer given that they have a desktop computer?
ANS:
Let D = “a home has a desktop computer,” L = “a home has a laptop computer.” We are
P( D  L) 0.3
given that P(D) = 0.8 and P(D and L) = 0.3. Thus, P( L | D) 

 0.38
P ( D)
0.8
4) A contest is held to give away a free pizza. Contestants pick an integer at random from the
integers 1 through 100. If the picked number is divisible by 24 or by 36, the contestant wins the
pizza. What is the probability that a contestant wins a pizza?
ANS:
Let A = “the number is divisible by 24” = (24, 48, 72, 96}
Let B = “the number is divisible by 36” = {36, 72}
Note that P(A and B) = 1/100 (72 is the only number divisible by both 24 and 36)
P(A or B) = P(A) + P(B) – P(A and B) = 4/100 + 2/100 – 1/100 = 5/100 = 0.05
5) Based on concerns over the eating habits and fitness level of school-aged children, the School
Board of a large district decided to offer healthy choices in the school cafeterias. They randomly
selected students from all grade levels and provided them with proposed menus for the healthier
lunches. Students were asked if they would purchase these lunches. The results of the survey are
summarized in the table below by grade level.
K–5
6–8
9 – 12
Total
Yes
No
6231
5964
3493
15688
2016
1912
3939
7867
8247
7876
7432
23555
What is the probability that a high school student selected at random would not plan to purchase the
proposed healthier lunches?
ANS:
3939 / 7867 = 0.5300
6) According to a recent national survey of college students, 55% admitted to having cheated at
some time during the last year. What is the probability that for two randomly selected college
students, one or the other would have cheated during the past year?
ANS:
The question asks for the probability that at least one student chosen at random would have
cheated. That is the probability that the first student chosen or the second student chosen would
have cheated.
Let A = the event that the first student selected will have cheated
Let B = the event that the second student selected will have cheated
The question asks for P(A or B) = P(A) + P(B) – P(A and B)
So, the P(A and B) = P(A)P(B) and then, P(A or B) = 0.55 + 0.55 – (0.55)(0.55) = 0.7975
7) Given two events, A and B, if P(A) = 0.37, P(B) = 0.41, and the P(A or B) = 0.75, then the two
events are
a) independent but not mutually exclusive
b) mutually exclusive but not independent
c) mutually exclusive and independent
d) neither mutually exclusive nor independent
e) it cannot be determined from the given information if the two events are independent or mutually
exclusive.
ANS: D
Two events A and B are independent if P(A|B) = P(A)
P(A and B) = 0.37 + 0.41 – 0.75 = 0.03
P(A|B) = 0.03 / 0.41 = 0.07, while the P(A) = 0.37.
Since P(B|A) ≠ P(A), the two events are not independent.
8) Security procedures at the U.S. Capitol require that all bags--meaning briefcases, backpacks,
shopping bags, any carrying bag and purses--must be screened. Currently, it is reported that 95% of
all bags that contain illegal items trigger the alarm. 12% of the bags that do not contain illegal
trigger the alarm. If 3 out of every 1000 bags entering the Capitol contain an illegal item, what is the
probability that a bag that triggers the alarm will contain an illegal item?
ANS:
The question is, what is the probability that a bag contains an illegal item, given that it sets
off the alarm?
P(a bag contains an illegal item | triggers the alarm) = P(it triggers the alarm and contains an illegal
item) / P(it triggers the alarm)
P(it triggers the alarm and contains an illegal item) = (0.003)(0.95) = 0.00285
P(it triggers the alarm) = (0.003)(0.95) + (0.997)(0.12) = 0.12249
P(a bag contains an illegal item | triggers the alarm) = 0.00285 / 0.12249 = 0.0233
9) Suppose your teacher’s stash of calculators contains 3 defective calculators and 17 good
calculators. You select two calculators from the box for you and your friend to use on the AP
Statistics exam. What calculations would you use to determine the probability that one of the
calculators drawn will be defective?
ANS:
The question asks for P(first is good and second is defective or the first is defective and the
second is good).
Remember P(A or B) = P(A) + P(B).
P(A) = P(first is good and the second is defective) = (17/20)(3/19) = 0.1342
P(B) = P(second is good and the second is defective) = (3/20)(17/19) = 0.1342
P(A or B) = 0.1342 + 0.1342 = 0.2684
10) 75% of people who purchase hair dryers are women. Of these women purchasers of hair dryers,
30% are over 50 years old. What is the probability that a randomly selected hair dryer purchaser is a
woman over 50 years old?
ANS:
Let W = the purchaser of a hair dryer is a woman
Let F = the purchaser of a hair dryer is over 50 years old
It is known that P(W) = 0.75 and P(F|W) = 0.30
Thus, P(W and F) = (0.75)(0.30) = 0.0225
11) The local Chamber of Commerce conducted a survey of one thousand randomly selected
shoppers at a mall. For all shoppers, “gender of shopper,” and “items shopping for” was recorded.
The data collected is summarized in the following table:
Items shopping for
Clothing
Shoes
Other
Total
Male
75
25
150
250
Female
350
230
170
750
Total
425
255
320
1000
If a shopper is selected at random from this mall,
a) What is the probability that the shopper is a female?
b) What is the probability that the shopper is shopping for shoes?
c) What is the probability that the shopper is a female shopping for shoes?
d) What is the probability that the shopper is shopping for shoes given that the shopper is a female?
e) Are the events “female” and “shopping for shoes” disjoint?
f) Are the events “female” and “shopping for shoes” independent?
ANS:
a) P(female) = 750 / 1000 = 0.75
b) P(shopping for shoes) = 255 / 1000 = 0.255
c) P(female and shopping for shoes) = 230 / 1000 = 0.23
d) P(shopping for shoes | female) = 230 / 750 = 0.3067
e) There are 230 females shopping for shoes. Therefore, the events “female” and “shopping for
shoes” are not disjoint.
f) P(shopping for shoes | female) = 0.3067 and P(shopping for shoes) = 0.255. Since these two
probabilities are not ≠, events “female” and “shopping for shoes” are not independent.
12) Suppose that among the 6000 students at a high school, 1500 are taking honors courses and
1800 prefer watching basketball to watching football. If taking honors courses and preferring
basketball are independent, how many students are both taking honors courses and prefer basketball
to football?
ANS:
1500/6000 = 0.25 are honors students, and 1800/6000 = 0.30 prefer basketball. Because of
independence, their intersection is (0.25)(0.30) = 0.075 of the students, and (0.075)(6000) = 450.
13) As reported in The New York Times, the Russian Health Ministry announced that ¼ of the
country’s hospitals had no sewage system and 1/7 had not running water. What is the probability
that a Russian hospital will have at least one of these problems if the two problems are independent?
ANS:
P(at least no sewage system or no running water) = ¼ + 1/7 – (1/4)(1/7) = 5/14
14) Suppose that, in a certain part of the world, in any 50-year period the probability of major
plague is 0.39, the probability of a major famine is 0.52, and the probability of both a plague and a
famine is 0.15. What is the probability of a famine given that there is a plague?
ANS:
P(famine | plague) = P(famine and plague) / P(plague) = (0.15) / (0.39) = 0.385
17) Suppose that, for any given year, the probabilities that the stock market declines, that women’s
hemlines are lower, and that both events occur are, respectively, 0.4, 0.35, and 0.3. Are the two
events are independent?
ANS:
If E and F are independent, then P(A and B) = P(A) P(B); however, (0.4)(0.35) ≠ 0.3
18) If P(A) = 0.2 and P(B) = 0.1, what is P(AUB) if A and B are independent?
ANS:
P(A U B) = 0.2 + 0.1 – (0.2)(0.1) = 0.28
19) An insurance company charges $800 annually for car insurance. The policy specifies that the
company will pay $1000 for a minor accident and $5000 for a major accident. If the probability for
a motorist having a minor accident during the year is 0.2, and of having a major accident, 0.05, how
much can the insurance company expect to make on a policy?
ANS:
1000(0.2) + 5000(0.05) = 450
800 – 450 = 350
Questions 20 – 24 refer to the following study: 1000 students at city high school were classified both
according to GPA and whether or not they consistently skipped classes.
GPA
Many skipped classes
Few skipped classes
< 2.0
80
175
255
2.0 – 3.0
25
450
475
> 3.0
5
265
270
20) What is the probability that a student has a GPA between 2.0 and 3.0?
110
890
1000
ANS:
P(GPA between 2.0 and 3.0) = 475 / 1000 = 0.475
21) What is the probability that a student has a GPA under 2.0 and has skipped many classes?
ANS:
P(GPA under 2.0 and skipped many classes) = 80 / 1000 = 0.080
22) What is the probability that a student has a GPA under 2.0 or has skipped many classes?
ANS:
P(GPA under 2.0 or has skipped many classes) = (255/1000) + (110/1000) – (80/1000) =
0.285
23) What is the probability that a student has a GPA under 2.0 given that he has skipped many
classes?
ANS:
P(GPA under 2.0 | skipped many classes) = 80 / 110 = 0.727
24) Are “GPA between 2.0 and 3.0” and “skipped few classes” independent?
ANS:
P(GPA between 2.0 and 3.0 | skipped few classes) = P(GPA between 2.0 and 3.0)
P(GPA between 2.0 and 3.0 | skipped few classes) = 450/890 = 0.5056
P(GPA between 2.0 and 3.0) = 0.475
Since the two probabilities are not ≠, the two events are not independent.
25) Consider the following table of ages of U.S. senators:
Age (yr):
<40 40 – 49
50 – 59
60 – 69
70 – 79 >79
Number of senators:
5
30
36
22
5
2
What is the probability that a senator is under 70 years old given that he is at least 50 years old?
ANS:
P(under 70 years old | at least 50 years old) = 58/65 = 0.892
Questions 26 – 29 refer to the following study: 500 people used a home test for HIV, and then all
underwent more conclusive hospital testing. The accuracy of the home test was evidenced in the
following table.
Positive test
Negative test
HIV
35
5
40
Healthy
25
435
460
26) What is the probability that a person has HIV and tests positive?
ANS:
P(has HIV and test positive) = 35/500 = 0.070
60
440
500
27) What is the probability of testing positive given that the person does not have HIV?
ANS:
P(test positive | does not have HIV) = 25/460 = 0.054
28) What is the probability of testing positive given that the person has HIV?
ANS:
P(test positive | has HIV) = 35/40 = 0.875
29) What is the probability of testing negative given that the person does not have HIV?
ANS:
P(test negative | does not have HIV) = 435/460 = 0.946
30) Suppose that 2% of a clinic’s patients are known to have cancer. A blood test is developed that
is positive in 98% of patients with cancer but is also positive in 3% of patients who do not have
cancer. If a person who is chosen at random from the clinic’s patients is given the test and it comes
out positive, what is the probability that the person actually has cancer?
ANS:
P(cancer | positive test) = P(cancer and positive test) / P(positive test)
P(cancer and positive test) = (0.02)(0.98) = 0.0196
P(positive test) = P(cancer and positive test) or P(no cancer and positive test)
= (0.02)(0.98) + (0.03)(0.98) = 0.0490
P(cancer | positive test) = 0.0196/0.0490 = 0.4
31) A computer technician notes that 40% of computers fail because of the hard drive, 25% because
of the monitor, 20% because of a disk drive, and 15% because of the microprocessor. If the problem
is not in the monitor, what is the probability that it is in the hard drive?
ANS:
P(hard drive) = .4/(.4+.2+.15) = 0.533
32) Suppose that 60% of students who take the AP Statistics exam score 4 or 5, 25% score 3, and
the rest score 1 or 2. Suppose further that 95% of those scoring 4 or 5 receive college credit, 50% of
those scoring 3 receive such credit, and 4% of those scoring 1 or 2 receive credit. If a student who is
chosen at random from among those taking the exam receives college credit, what is the probability
that she received a 3 on the exam?
ANS:
P(3 | credit) = P(3 and credit) / P(credit)
P(3 | credit) = (0.25)(0.5) = 0.125
P(credit) = P(4 or 5 and credit) + P(3 and credit) + P(1 or 2 and credit)
= (0.6)(0.95) + (0.25)(0.5) + (0.15)(0.04)
= 0.57 + 0.125 + 0.006 = 0.701
P(3 | credit) = 0.125 / 0.701 = 0.178
33) Given the probabilities P(A) = 0.4 and P(A U B) = 0.6, what is the probability P(B) if A and B
are mutually exclusive? If A and B are independent?
ANS:
P(A U B) = P(A) + P(B)  P(B) = 0.2 if A and B are mutually exclusive.
If A and B are independent, then P(A ∩ B) = P(A)P(B)
P(A U B) = P(A) + P(B) – P(A ∩ B)
0.6 = 0.4 + P(B) – 0.4 P(B)  P(B) = 1/3
34) A sample of applicants for a management position yields the following numbers with regard to
age and experience:
Years of experience
Less than 50 years old
More than 50 years old
0–5
80
10
6 – 10
125
75
> 10
20
50
a) What is the probability that an applicant is less than 50 years old? Has more than 10 years’
experience? Is more than 50 years old and has 5 or fewer years’ experience?
b) What is the probability that an applicant is less than 50 years old given that she has between 6
and 10 years’ experience?
c) Are the two events “less than 50 years old” and “more than 10 years’ experience” independent
events? How about the two events “more than 50 years old” and “between 6 and 10 years’
experience”? Explain.
ANS:
a) P(age < 50) = 225 / 360 = 0.625
P(experience > 10) = 70 / 360 = 0.194
P(age > 50 ∩ experience ≤ 5) = 10 / 360 = 0.028
b) P(age < 50 | 6 < experience < 10) = 125 / 200 = 0.625
c) P(age < 50 | experience > 10) = P(age < 50)
20 / 70 = 225 / 360
0.286 ≠ 0.625 and so they are not independent.
P(age > 50 | 6 < experience < 10) = P(age > 50)
75 / 200 = 135 / 360
0.375 = 0.375 and so they are independent.
35) Assume there is no overlap between the 56% of the population who wear glasses and the 4%
who wear contacts. If 55% of those who wear glasses are women and 63% of those who wear
contacts are women, what is the probability that the next person you encounter on the street will be a
woman with glasses? A woman with contacts? A man with glasses? A man with contacts? A
person not wearing glasses or contacts? Explain your reasoning.
ANS:
P(a woman and glasses) = (0.56)(0.55) = 0.308
P(a woman with contacts) = (0.04)(0.63) = 0.0252
P(a man with glasses) = (0.56)(0.45) = 0.252
P(a man with contacts) = (0.04)(0.37) = 0.0148
P(not wearing glasses nor contacts) = 1 – (0.308 + 0.0252 + 0.252 + 0.0148) = 0.4
36) Explain what is wrong with each of the following statements:
a) The probability that a student will score high on the AP Statistics exam is 0.43, while the
probability that she will not score high is 0.47.
b) The probability that a student plays tennis is 0.18, while the probability that he plays basketball is
six times as great.
c) The probability that a student enjoys her English class is 0.64, while the probability that she
enjoys both her English and her social studies classes is 0.71.
d) The probability that a student will be accepted by his first choice for college is 0.38, while the
probability that he will be accepted by his first or second choice is 0.32.
e) The probability that a student fails AP Statistics and will still be accepted by an Ivy League
school is -0.17.
ANS:
a) The probability of the complement is 1 minus the probability of the event, but 1 – 0.43 ≠ 0.47.
b) Probabilities are never greater than 1, but 6(0.18) = 1.08.
c) The probability of an intersection cannot be greater than the probability of one of the separate
events.
d) The probability of a union cannot be less than the probability of one of the separate events.
e) Probabilities are never negative.
37) Two students are selected at random from the school without replacement. What is the
probability that they are boys?
Freshmen
Sophomores
Junior
Seniors
Total
Boys
9
7
5
7
28
Girls
7
8
7
10
32
Total
16
15
12
17
60
ANS:
P(first is boy ∩ second is boy) = (28/60) (27/59) = 0.192
38) At a particular school 47% of the students are boys and 53% are girls. Of the boys, 25% are
seniors. Of the girls, 31% are seniors. If a randomly selected student is a senior, what is the
probability that it is a girl?
ANS:
P(a girl | a senior) = P(a girl and a senior) / P(a senior)
P(a girl and a senior) = (0.53)(0.31) = 0.1643
P(a senior) = (0.47)(0.25) + (0.53)(0.31) = 0.2818
P(a girl | a senior) = 0.1643 / 0.2818 = 0.5830
39) Among a group of boys, 70% like chocolate ice cream, 40% like strawberry ice cream, and 30%
like both. If a boy is randomly selected from the group, what is the probability that he likes either
chocolate or strawberry ice cream, but NOT both?
ANS:
P(chocolate or strawberry ice cream) = 0.7 + 0.4 – 0.3 = 0.80
40) There are 13,000 students at a certain university, 55% of whom are women. Of the women,
58% live on campus while only 38% of the men do. If a random sample of 500 students is taken,
what is the expected number of students in the sample who live on campus?
ANS:
55% of students are women and 58% of them live on campus, so (0.55)(0.58) = 0.319 or
31.9% of the students are women who live on campus. Similarly, 45% of the students are men and
38% of them live on campus, so (0.45)(0.38) = 0.171 or 17.1% of the students are men who live on
campus. Thus, 31.9% + 17.1% = 49% of the students live on campus. In a sample of 500 students,
we would expect 49% of them, or 245, to live on campus.