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Chapter 4
Introduction to Probability
Learning Objectives
1. Obtain an appreciation of the role probability information plays in the decision making process.
2. Understand probability as a numerical measure of the likelihood of occurrence.
3. Know the three methods commonly used for assigning probabilities and understand when they
should be used.
4. Know how to use the laws that are available for computing the probabilities of events.
5. Understand how new information can be used to revise initial (prior) probability estimates using
Bayes’ theorem.
Solutions:
1. Number of experimental Outcomes = (3)(2)(4) = 24
2.
66! 6 5 4 3 2 1 20
33!3! (3 2 1)(3 2 1)
= = =
ABC
ACE
BCD
BEF
ABD
ACF
BCE
CDE
ABE
ADE
BCF
CDF
ABF
ADF
BDE
CEF
ACD
AEF
BDF
DEF
4. a.
b. Let: H be head and T be tail
(H,H,H) (T,H,H)
c. The outcomes are equally likely, so the probability of each outcome is 1/8.
5. P(Ei) = 1/5 for i = 1, 2, 3, 4, 5
The classical method was used.
6. P(E1) = .40, P(E2) = .26, P(E3) = .34
8. a. There are four outcomes possible for this 2-step experiment; planning commission positive - council
approves; planning commission positive - council disapproves; planning commission negative -
council approves; planning commission negative - council disapproves.
H
T
H
T
H
T
H
T
H
T
H
T
H
T
(H,H,H)
(H,H,T)
(H,T,H)
(H,T,T)
(T,H,H)
(T,H,T)
(T,T,H)
(T,T,T)
1st Toss 2nd Toss 3rd Tos s
9.
50 50! 50 49 48 47 230,300
44!46! 4 3 2 1
= = =
10. a. Using the table provided, 86.5% of Delta flights arrive on time.
P(on-time arrival) = .865
11. a. Total motorcyclists = 350 + 170 = 520
158 =
.
Plannin g Commission Council
p
n
a
d
a
d
(p, a)
(p, d)
(n, a)
(n, d)
86 .6667
129 =
South:
92 .6525
141 =
West:
76 .8261
92 =
The West region shows the highest probability (.8261) of DOT-compliant helmet use.
12. a. Step 1: Use the counting rule for combinations:
59
5
æ
è
çö
ø
÷=59!
5!(59-5)! =(59)(58)(57)(56)(55)
(5)(4)(3)(2)(1) =5,006,386
Step 2: There are 35 ways to select the red Powerball from digits 1 to 35
13. Initially a probability of .20 would be assigned if selection is equally likely. Data does not appear to
confirm the belief of equal consumer preference. For example using the relative frequency method
we would assign a probability of 5/100 = .05 to the design 1 outcome, .15 to design 2, .30 to
14. a. P(E2) = 1/4
b. P(any 2 outcomes) = 1/4 + 1/4 = 1/2
b. S = {2 of clubs, 3 of clubs, . . . , 10 of clubs, J of clubs, Q of clubs, K of clubs, A of clubs}
c. There are 12; jack, queen, or king in each of the four suits.
16. a. (6)(6) = 36 sample points
b.
c. 6/36 = 1/6
d. 10/36 = 5/18
17. a. (4,6), (4,7), (4,8)
b. .05 + .10 + .15 = .30
18. a. Let C = corporate headquarters located in California
P(C)
= 53/500 = .106
b. Let N = corporate headquarters located in New York
T = corporate headquarters located in Texas
.
1
2
3
4
5
6
1 2 3 4 5 6
2
3
4
5
6
7
3
4
5
6
7
8
4
5
6
7
8
9
5
6
7
8
9
10 11
10
9
8
7
6 7
8
9
10
11
12
Die 1
Total for Both
Die 2
c. Let A =
corporate
headquarters
located in one of the
eight states
19. a. A summary of the data provided in the exercise follows:
Response
United States
Great Britain
Total
Yes
187
197
384
No
334
411
745
Unsure
256
213
469
Total
777
821
1598
Probability = 334/777 = .4299
d. The probability that an investor in the United States thinks the government is adequately protecting
investors is 187/777 =.2407; for investors in Great Britain the probability is 197/821 = .24. The two
probabilities are almost identical; thus, there does not appear to be a difference between the
20. a.
b.
P(Age <25) =P(E1)+P(E2)=.2023+.4947 =.6970
c.
P(Age >24) =P(E3)+P(E4)=.2585+.0445=.3030
Age
Experimental
Financially
Number of
Outcome
Independent
Responses
Probability
E1
16 to 20
191
191/944 =
.2023
E2
21 to 24
467
467/944 =
.4947
21. a.
PFall
( )
=645
4535 =.1422
b.
PTransportation Incident
( )
=1795
4535 =.3958
4535 =.0249
22. a. P(A) = .40, P(B) = .40, P(C) = .60
b. P(A B) = P(E1, E2, E3, E4) = .80. Yes P(A B) = P(A) + P(B).
23. a. P(A) = P(E1) + P(E4) + P(E6) = .05 + .25 + .10 = .40
P(B) = P(E2) + P(E4) + P(E7) = .20 + .25 + .05 = .50
e. Bc = {E1, E3, E5, E6}; P(Bc) = P(E1) + P(E3) + P(E5) + P(E6)
= .05 + .20 + .15 + .10 = .50
24. Let E = experience exceeded expectations
M = experience met expectations
a. Percentage of respondents that said their experience exceeded expectations
= 100 - (4 + 26 + 65) = 5%
25. Let W = person feels guilty about wasting food
L = person feels guilty about leaving lights on when not in a room
a.
P(WÈL)=P(W)+P(L)-P(WÇL)
