Math 141 Week-In-Review
7.5-7.6, 8.1-8.2
1. Past records in a certain city produced the following probability data on a driver being in an
accident on a particular holiday:
Rain (R)
No Rain (RC)
Totals
Accident (A)
.025
.015
.040
No Accident (AC)
.335
.625
.906
Totals
.360
.640
1.000
Find the probability of
a. An accident, rain or no rain. Solution: .040
b. Rain, accident or no accident. Solution: .360
c. An accident and rain Solution: .025
d. An accident, given rain. Solution: P(A|R) = P(AR)/P(R) = .025/.360 = .069
2. If 60% of a store’s customers are female and 75% of the female customers have charge
accounts at the store, what is the probability that a customer selected at random is a female
and has a charge account?
Solution: P(FC) = P(C|F)P(F) = .75(.60) = .45
3. A single card is drawn from a standard 52-card deck. If event E is the drawn card is a spade
and event F is the drawn card is a face card, are the two events independent?
Solution: P(E) P(F) = (1/4)(12/52) = 3/52
P(EF) = 3/52
Yes
4. A single card is drawn from a standard 52-card deck. If event G is the drawn card is a club
and event H is the drawn card is a heart, are the two events independent?
Solution: P(G) P(H) = (1/4)(1/4) = 1/16
P(GH) = 0
No
5. A space shuttle has four independent computer control systems. If the probability of failure
(during flight) of any one system is .001, what is the probability of failure of all four systems?
Solution: (.001)^4 = 1 x 10^(-12)
Since independent, P(ABCD) = P(A)P(B)P(C)P(D)
6. A fair die is rolled 5 times. What is the probability of getting a 6 on the 5 th roll, given that a 6
turned up on the preceding 4 rolls?
Solution: Since independent, P(A|B) = P(A) = 1/6
7. A fair die is rolled 5 times. What is the probability that the same number turns up every time?
Solution: 6/(6^5) = 1/(6^4)
8. Find the probabilities by referring to the following tree diagram:
C
.4
U
.6
.2
CC
C
.5
.2
V
.8
.3
CC
C
.6
W
.4
CC
a. P(U|C) Solution: (.2*.4)/(.2*.4+.5*.2+.3*.6) = 2/9
b. P(V/ C C ) Solution: (.5*.8)/(.5*.8+.2*.6+.3*.4) = 5/8
c. P(W|C) Solution: (.3*.6)/(.3*.6+.5*.2+.2*.4) = 1/2
d. P(U\ C C ) Solution: (.2*.6)/(.2*.6+.5*.8+.3*.4) = 3/16
e. P(V|C) Solution: (.5*.2)/(.5*.2+.2*.4+.3*.6) = 5/18
f.
P(W| C C ) Solution: (.3*.4)/(.3*.4+.5*.8+.2*.6) = 3/16
9. In a study to determine frequency and dependency of color-blindness relative to females and
males, 1000 people were chosen at random and the following results were recorded:
Female (F)
Male ( F C )
Totals
Color-Blind, C
2
24
26
C
Normal, C
518
456
974
Totals
520
480
1000
a. What is the probability that a person is a woman, given that the person is color-blind?
Solution: P(F|C) = 2/26 = 1/13
b. What is the probability that a person is color-blind, given that the person is a male?
Solution: P(C|FC) = 24/480 = 1/20
c. Are the events color-blindness and male independent?
Solution: P(CFC) = 24/1000 P(C)P(FC) = (26/1000)(480/1000) = 39/3125 No
d. Are the events color-blindness and female independent?
Solution: P(CF) = 2/1000 P(C)P(F) = (26/1000)(520/1000) = 169/12500 No
10. A new, inexpensive skin test is devised for detecting tuberculosis. To evaluate the test before it
is put into use, a medical researcher randomly selects 1000 people. Using precise but more
expensive methods already available, it is found that 8% of the 1000 people tested have
tuberculosis. Now each of the 1000 subject is given the new skin test and the following
results are recorded: The test indicates tuberculosis in 96% of those who have it and in 2%
of those who do not. Based on these results, what is the probability of a randomly chosen
person having tuberculosis given that the skin test indicates the disease? What is the
probability of a person not having tuberculosis given that the skin test indicates the disease?
Solution: P(T|+) = (.08*.96)/(.08*.96+.92*.02) = 96/119  .807
P(TC|+) = (.92*.02)/(.08*.96+.92*.02) = 23/119  .193
OR 1 – P(T|+) = 1-96/119 = 23/119
11. One of two urns is chosen at random with one as likely to be chosen as the other. Then a ball
is withdrawn from the chosen urn. Urn 1 contains 1 white and 4 red balls, and urn 2 has 3
white and 2 red balls.
a. If a white ball is drawn, what is the probability that it came from urn 1?
Solution: (.5*.2)/(.5*.2+.5*.6) = 1/4
b. If a red ball is drawn, what is the probability that it came from urn 2?
Solution: (.5*.4)/(.5*.4+.5*.8) = 1/3
12. An urn contains 5 white and 4 red balls. Two balls are drawn in succession without
replacement. If the second ball is white, what is the probability that the first ball was white?
Solution: [(5/9)(4/8)]/[(5/9)(4/8)+(4/9)(5/8)] = 1/2
13. Urn 1 contains 3 white and 7 red balls, and urn 2 has 5 white and 4 red balls. A ball is drawn
from urn 1 and placed in urn 2. Then a ball is drawn from urn 2. If the ball drawn from urn 2 is
red, what is the probability that the ball drawn from urn 1 was red?
Solution: (.7*.5)/(.7*.5+.3*.4) = 35/47
14. If two cards are drawn in succession from a standard 52-card deck without replacement and
the second card is a heart, what is the probability that the first card is a heart?
Solution: [(1/4)(12/51)]/[(1/4)(12/51)+(3/4)(13/51)] = 4/17
15. A manufacturer obtains clock-radios from three different subcontractors: 20% from A, 40%
from B, and 40% from C. The defective rates for these subcontractors are 1%, 3%, and 2%,
respectively. If a defective clock-radio is returned by a customer, what is the probability that it
came from subcontractor A? From B? From C?
Solution:
From A: (.2*.01)/(.2*.01+4*.03+.4*.02) = 1/11
From B: (.4*.03)/(.2*.01+.4*.03+.4*.02) = 6/11
From C: (.4*.02)/(.2*.01+4*.03+.4*.02) = 4/11
16. A carton of 20 calculator batteries contains 2 dead ones. A random sample of 3 is selected
from the 20 and tested. Let X be the random variable associated with the number of dead
batteries found in a sample.
a. Find the probability distribution of X.
Solution: P(0) = C(18,3)/C(20,3) = 68/95
P(1) = C(18,2)C(2,1)/C(20,3) = 51/190
P(2) = C(18,1)C(2,2)/C(20,3) = 3/190
X
0
1
2
3
P(X)
68/95
51/190
3/190
0
b. Find the expected number of dead batteries found in a sample.
Solution: E(X) = 0(68/95) + 1(51/190) + 2(3/190) + 3(0) = 57/190 = 3/10 or .3
17. What is the expected value (long-run average) of the number of dots facing up for the roll of a
single die?
Solution: E(X) = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 3.5
18. A spinner device is numbered from 0 to 5, and each of the 6 numbers is as likely to come up
as any other. A player who bets $1 on any given number wins $4 (and gets the bet back) if
the pointer comes to rest on the chosen number; otherwise, the $1 bet is lost. What is the
expected value of the game (long-run average gain or loss per game)?
Solution: E(x) = 4(1/6) – 1(5/6) = -1/6  - 17 cents
19. An outdoor concert featuring a very popular musical group is scheduled for a Sunday afternoon
in a large open stadium. The promoter, worrying about being rained out, contacts a long-range
weather forecaster who predicts the chance of rain on that Sunday to be .24. If it does not
rain, the promoter is certain to net $100,000; if it does rain, the promoter estimates that the net
will be only $10,000. An insurance company agrees to insure the concert for $100,000 against
rain at a premium of $20,000. Should the promoter buy the insurance?
Solution: If they insure & it rains, they get 100,000 – 20,000 premium + 10,000 gate = 90,000
.24 (rain)
.76 (no rain)
Insure
90,000
80,000
Don’t Insure
10,000
100,000
Expected value if insure:
E(X) = 90,000(.24) + 80,000(.76) = $82,400
Expected value if don’t insure:
E(X) = 10,000(.24) + 100,000(.76) = $78,400