AP Statistics
Unit 4 Test Review: Probability Theory and Simulation
1.
Which of the following are true regarding mutually exclusive events? (circle all that apply)
A.
Mutually exclusive events are also called disjoint.
B.
Two events are mutually exclusive if they cannot both occur at the same time.
C.
Two events that are mutually exclusive must also be independent.
D.
Two events that are mutually exclusive cannot be independent.
Use the following table for questions 2 – 6.
A researcher classified 1000 students according to both grade-point average (GPA) and the students’ participation in at least
one school-related club. A student is chosen at random from this group.
GPA
< 2.0
2.0 to 3.0
> 3.0
Total
At Least One Club
60
110
220
390
No Club
190
350
70
610
Total
250
460
290
1000
2.
What is the probability that a student chosen at random has a GPA greater than 3.0?
290/1000 = 0.290
3.
What is the probability that a student chosen at random has a GPA greater than 3.0 and joined at least one club?
220/1000 = 0.220
4.
What is the probability that a student chosen at random has a GPA greater than 3.0 or joined at least one club?
290/1000 + 390/1000 – 220/1000 = 460/1000 = 0.460
5.
What is the approximate probability that a student chosen at random has a GPA greater than 3.0 given that he or she
joined at least one club?
220/390 = 0.564
6.
Are the events GPA greater than 3.0 and participated in at least one club independent?
A.
No, because P(GPA greater than 3.0 | participated in at least one club)  P(participated in at least one club).
B.
No, because P(GPA greater than 3.0 | participated in at least one club)  P(GPA greater than 3.0).
C.
Yes, because P(GPA greater than 3.0 | participated in at least one club)  P(participated in at least one club).
D.
Yes, because P(GPA greater than 3.0 | participated in at least one club)  P(GPA greater than 3.0).
E.
No, because there were many students with GPAs greater than 3.0 who participated in at least one club.
7.
Suppose that among all U.S. students, 80% have gone to an amusement park and 45% have gone to a beach. Only 15%
have done neither. If you select a student at random, what is the probability that the student has gone to the beach but
not to an amusement park? Include a tree diagram or table that illustrates this situation.
Attended Amusement
Park?
Yes
Gone to
Beach?
No
Total
Yes
40
5
45
No
40
15
55
Total
80
20
100
(Fill in totals using given information. You know that
the No/No box is 15%. Use subtraction to fill in the
remaining 3 boxes.)
The probability of having gone to the beach but not
to an amusement park is 0.05.
8.
A recent study reported that math instructors estimate that 70% of their students do homework regularly and that if a
student does the homework regularly, she or he will have a 90% chance of passing the final exam. What is the
probability that a student chosen at random will have done homework regularly and passed the math final exam?
P(does homework  passed exam)  P(does homework) P(passed exam| does homework)
 (0.70)(0.90)
 0.63
9.
Suppose that 2% of a clinic’s patients are known to have Lyme disease. A test is developed that is positive in 98% of
patients with Lyme disease, but it is also positive in 3% of patients who do not have the disease. A patient is chosen at
random from the clinic.
a.
Construct a table or draw a tree diagram that reflects the situation.
Has Lyme
Disease
No Lyme
Disease
Tests
Positive
(0.98) · (.02)
= 0.0196
(0.03) · (0.98) =
0.0294
0.049
Tests
Negative
(0.02)(0.02)
= 0.0004
(0.97)(0.98) =
0.9506
0.951
0.0200
0.98
1
Total
13.
Total
b.
What is the probability that the patient’s test comes out positive for Lyme disease?
0.049
c.
What is the probability that the person actually has Lyme disease given that the test comes out positive?
0.0196/0.049 = 0.4
The proportion of voters who voted for George W. Bush in the 2000 presidential election was approximately 0.48.
a.
Describe how to use these lines from a random digit table to simulate taking a sample of 10 people who voted
in the 2000 presidential election and recording whether they voted for Bush.
11805
05431
39808 27732
50725
68248
83452
99634 06288 98083 13746
70078
88685 40200 86507 58401
36766
67951
Take 10 sets of two-digit numbers, letting 01-48 = voted for Bush and 49-00 = voted for another
candidate. Repeats are allowed, as these represent percentages and not individual voters. Determine
the proportion who voted for Bush.
b.
Start at the beginning of the first line, take three samples of size 10, and compute the sample proportion for
each sample. (Do not start a new line for each sample. Start where the previous sample finished.)
Sample 1:
11 80 50 54 31 39 80 82 77 32
4 votes for Bush => sample probability = 0.40
Sample 2:
50 72 56 82 48 83 45 29 96 34
4 votes for Bush => sample probability = 0.40
Sample 3:
06 28 89 80 83 13 74 67 00 78
3 votes for Bush => sample probability = 0.30
14.
If events A and B are independent and P(A) = 0.3 and P(B) = 0.5, then which of these is true?
A.
P(A and B) = 0.8
B.
P(A or B) = 0.15
C.
P(A or B) = 0.8
D.
P(A | B) = 0.3
E.
P(A | B) = 0.5
15.
For all events A and B, P(A and B) =
A.
P(A) · P(B)
B.
P(B | A)
C.
P(A | B)
D.
P(A) + P(B)
E.
P(B) · P(A | B)
16.
The number of team flags the booster club will sell at a football game has the probability distribution as shown in the
table below.
Note: “A” is the correct answer only if the events
are independent. “E” works for all events. (We
often use a version of “A” for dependent events
when we can predict P(B) based on the outcome
of event A.)
Number of team flags (X)
5
6
7
8
P(X)
0.20
0.15
0.10
0.25
9
0.18
10
0.12
If each team flag costs $8.00, what is the expected amount of money the booster club will take in at the football game?
A.
$7.42
5(0.20) + 6(0.15) + 7(0.10) + 8(0.25) + 9(0.18) + 10(0.12) = 7.42 expected # of flags sold
B.
$8.00
C.
$53.15
7.42($8.00) = $59.36
D.
$59.36
E.
$64.00
17.
Alex, Bryan and Charlie are all playing tennis matches in a tournament against different opponents. Based on previous
performances, there is a 0.4 probability that Alex will win his first match, a 0.3 probability that Bryan will win his first
match, and a 0.2 probability that Charlie will win his first match. If the chance that each wins his match is independent
of the other, what is the probability that none of them wins in their first matches?
A.
0.024
B.
0.304
P(none wins first match)  P A  P B  P C  (0.6)(0.7)(0.8)  0.336
C.
0.336
D.
0.700
E.
0.900
   
18.
Five homes from a subdivision will be randomly selected to receive 1 month of free cable TV. There are 80 homes in the
subdivision. The homes are assigned numbers 01 – 80 and the random number table below is used to select the five
homes. No home may receive more than one free month of service. Which of the following is a correct selection of the
five homes?
99154
92210
A.
B.
C.
D.
E.
9, 1, 5, 4, 7
15, 47, 03, 23, 23
15, 47, 03, 23, 35
99, 70, 23, 92, 08
99, 15, 47, 03, 92
70392
70439
23889
08629
92335
73299
19.
An inspection procedure at a manufacturing plant involves picking three items at random and then accepting the whole
lot if at least two of the three items are in perfect condition. If in reality 90% of the whole lot are perfect, what is
probability that the lot will be accepted?
(0.9)(0.9)(0.9) = 0.729
A.
0.600
P PP
B.
0.667
(0.9)(0.9)(0.1) = 0.081
PPN
C.
0.729
(0.9)(0.1)(0.9) = 0.081
PNP
D.
0.810
E.
0.972
(0.1)(0.9)(0.9) = 0.081
NPP
20.
Consider the following table of ages of U.S. senators:
Age (yr):
<40
40-49
Number of senators
5
30
0.729 + 0.081 + 0.081 + 0.081 = 0.972
50-59
36
60-69
22
70-79
5
>79
2
What is the probability that a senator is under 70 years old given that he is at least 50 years old?
A.
0.580
B.
0.624
P( 70   50) 58 / 100 58
P( 70|  50) 


 0.892
C.
0.643
P( 50)
65 / 100 65
D.
0.892
E.
0.969
Use the following for questions 21 – 24.
Five hundred 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.
HIV
Healthy
Positive Test
35
25
60
Negative Test
5
435
440
40
460
21.
What is the predictive value of the test? That is, what is the probability that a person has HIV and tests positive?
A.
0.070
B.
0.130
35
P(HIV  PT) 
 0.07
C.
0.538
500
D.
0.583
E.
0.875
22.
What is the false-positive rate? That is, what is the probability of testing positive given that the person does not have
HIV?
A.
0.054
B.
0.50
P(PT  Healthy) 25 / 500
25
P(PT |Healthy) 


 0.054
C.
0.130
P(Healthy)
460 / 500 460
D.
0.417
E.
0.875
23.
What is the sensitivity of the test? That is, what is the probability of testing positive given that the person has HIV?
A.
0.070
B.
0.130
P(PT  HIV) 35 / 500 35
P(PT | HIV) 


 0.875
C.
0.538
P(HIV)
40 / 500 40
D.
0.583
E.
0.875
24.
What is the specificity of the test? That is, what is the probability of testing negative given that the person does not
have HIV?
A.
0.125
P(NT  Healthy) 435 / 500 435
P(NT | Healthy) 


 0.946
B.
0.583
P(Healthy)
460 / 500 460
C.
0.870
D.
0.950
E.
0.946
25.
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.
The sum of complements must equal 1 (0.43 + 0.47 = 0.90)
26.
b.
The probability that a student plays tennis is 0.18, while the probability that he plays basketball is six times as
great.
Probability cannot exceed 1 (6*0.18 = 1.08)
c.
The probability that a student enjoys her English class is 0.64, while the probability that she enjoys both her
English and Social Studies classes is 0.71.
P(A  B) cannot be greater than either the P(A) or the P(B) alone.
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.
P(A  B) cannot be less than the P(A) or the P(B) alone.
e.
The probability that a student fails AP Statistics and will still be accepted by an Ivy League school is -0.17.
Probability cannot be negative.
Suppose the probability that someone will make a major mistake on an income tax return is 0.23. One day, and Internal
Revenue Service (IRS) agent plans to audit as many returns as necessary until she finds one with a major mistake.
Each box is a
group
needed to
get a major
mistake.
a.
b.
84177
59884
63395
54355
08989
29357
06757
31180
55041
52704
57024
23737
17613
53115
15866
90359
97284
67881
15582
84469
06589
02649
00637
03668
51506
94868
13119
74796
89283
38760
81435
57967
71020
71567
03514
35841
41050
05811
85940
94268
59195
52869
92031
84514
91932
08844
07635
23114
06449
75011
06488
26294
03309
15864
05059 continue to next row
13006
74987 continue to next row
64759 continue to next row
72605
38942
Use simulation with the random digit table above to estimate the probability that a major mistake will be found
before the fifth return.
Use 01-23 to be a major mistake, and 24-00 to be not a major mistake. Check groups of two digits until a major
mistake is found. Determine whether or not the mistake was made before or after the 5 th return.
major mistake found
22
22
before 5th return
P(mistake before 5th)   0.629
major mistake found
13
35
on/after 5th return
Use probability to calculate the probability that a major mistake will be found before the fifth return.
P(mistake found before 5th) = P(mistake found 1st) + P(found 2nd) + P(found 3rd) + P(found 4th)
=
0.23
+ (0.77)(0.23) + (0.77)2(0.23) + (0.77)3(0.23)
= 0.648
27.
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
a.
225
 0.625
360
>10
80
125
20
225
10
75
50
135
90
200
70
360
P( 10 years experience) 
P( 50   5 years' experience) 
70
 0.194
360
10
 0.028
360
What is the probability that an applicant is less than 50 years old given that she has between 6 and 10 years’
experience?
P( 50 | 6  experience  10) 
c.
6-10
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 five or fewer years’ experience?
P( 50) 
b.
0-5
P( 50  6  experience  10) 125 / 360 125


 0.625
P(6  experience  10)
200 / 360 200
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.
If the two events “less than 50 years old” and “more than 10 years’ experience” are independent, then:
P( 50 | experience  10)  P( 50)
20
225

70
360
0.286
 0.625
These probabilities are not equal, therefore the two events are not independent.
If the two events “more than 50 years old” and “between 6 and 10 years’ experience” are independent, then:
P(  50 | 6  experience  10)  P(  50)
75
135

200
360
0.375
 0.375
These probabilities are equal, therefore the two events are independent.