Math 166 Week-in-Review - © S. Nite 9/26/2012
WIR #3
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Week in Review #3 (L.1-L.2, 1.1-1.7)
1. Construct a truth table for (p ∨ q) ∧ ∼ (p ∧ ∼ r)
p
Q
R
T
T
T
∼r
T
p∨q
T
F
p∧∼r
F
∼( p ∧ ∼ r)
T
(p ∨ q) ∧ ∼ (p ∧ ∼ r)
T
T
F
T
T
T
F
F
T
F
T
T
F
F
T
T
T
F
F
T
T
T
F
F
F
T
T
T
F
F
T
T
F
T
F
T
T
F
T
T
F
F
T
F
F
F
T
F
F
F
F
F
T
F
T
F
2. Shade Venn diagrams to represent each of the following:
a.
(A ∪ BC) ∪ C
U
b. (AC ∪ CC) ∩ B
U
A
B
C
A
B
C
3. Consider the experiment, “A wheel with 17 numerals on the perimeter is spun and allowed to come to a
rest so that a pointer points within a numbered sector.”
a. Write the sample space.
Solution: S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}
P(S) = 1
b. What is the event that the outcome is a prime number?
Solution: E = {2, 3, 5, 7, 11, 13, 17}
7
P(E) =
17
c. What is the event that the outcome is the square of 4?
Solution: F = {16}
1
P(F) =
17
d. What is the event that the outcome is a number divisible by 6?
Solution: G = {6, 12}
2
P(G) =
17
Math 166 Week-in-Review - © S. Nite 9/26/2012
WIR #3
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3. Consider an experiment of rolling a pair of dice. What is the event that
e. A sum of eight turns up?
Solution: E = {(2,6), (3,5), (4,4), (5,3), (6,2)}
5
P(E) =
36
f. A sum less than five turns up?
Solution: F = {(1,1), (1,2), (1,3), (2,1), (2,2), (3,1)}
6
1
=
P(F) =
36 6
g. A sum of eight turns up or a two is showing?
Solution: G = {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (2,1), (2,3), (2,4), (2,5), (2,6), (3,5),
(4,4), (5,3)}
14
7
P(G) =
=
36 18
h. A sum of 4 or a sum of 10 turns up?
Solution: H = {(1,3), (2,2), (3,1), (4,6), (5,5), (6,4)}
6
1
P(H) =
=
36 6
2. Consider an experiment of rolling a pair of dice.
a. Give an example of two mutually exclusive events.
Solution: The event that a sum of 4 turns up and the event that a 5 is showing.
b. Give an example of two events that are not mutually exclusive.
Solution: The event that a sum of 7 turns up and the event that a 5 is showing.
3. What is the probability of each of the events in # 1-2?
4. Give the probability distribution for the following situation: A die was cast 200 times with the
following results:
Probability
Outcome
Number of Times
1
35
.175
.14
2
28
3
29
.145
.18
4
36
.165
5
33
.195
6
39
5. From a survey involving 1,000 people in Houston, it was found that 500 people had tried Diet
Pepsi, 600 had tried regular Pepsi, and 200 had tried both. If a resident of the city is selected
at random, what is the probability that
a. The resident has tried diet or regular Pepsi?
Solution:
Math 166 Week-in-Review - © S. Nite 9/26/2012
WIR #3
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U
R
400
200
300
D
100
P(D∪R) = P(D) + P(R) - P(D∩R)
OR
P(D∪R) = 1 – P[(D∪R)C]
=.5 + .6 - .2
= 1 - .1
= .9
= .9
b. The resident has tried either diet or regular Pepsi, but not both?
Solution: P[(D∩RC) ∪ (DC∩R)] = .3 + .4 = .7
6. If 65% of a store’s customers are female and 80% 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) = .80(.65) = .52
7. 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 8 or 9, are the two events independent?
Solution: P(E) P(F) = (1/4)(8/52) = 1/26
P(E∩F) = 2/52 = 1/26
Yes, they are independent.
8. If the odds in favor of Jamie attending class on Friday is 7/11, what is the probability that Jamie will skip
class on Friday? Express your answer to the nearest tenth of a percent.
Solution: 11/18 ≈ 61.1%
9. A survey indicated that 68% of college students like a certain product. Based on the survey, find the
probability that at least 25 college students in a sample of 30 like the product.
Solution: Binomial Experiment---.0488… Sorry, this is in a later chapter.
10.
Find P(A∪C) by referring to the following tree diagram:
C
.7
A
.3
.62
D
.45
C
.38
B
.55
D
Solution: P(A∪C) = P(A)+P(C)-P(A∩C) = .791
Math 166 Week-in-Review - © S. Nite 9/26/2012
WIR #3
11.
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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
Solution: .08 + .1 +.18 = .36
a.
P(C)
Solution: .4
b.
P(C|U)
c.
P(V∪C)
Solution: P(V) + P(C) – P(V∩C) = .5 + .36 - .1 = .76
Solution: .08/.36 = 2/9
d.
P(U|C)
C
e.
P(V/ C ) Solution: .4/(.12+.4+.12) = 5/8
Solution: .1
f.
P(V∩C)
Solution: .6
g.
P(C|W)
12.
The probability of a pencil lead breaking during next test is .085. What are the odds in favor of a
pencil lead breaking during the next test?
Solution: .085/.915 = 17/183
14. Suppose P(E∪F) = .68, P(E) = .38 and P(F) = .3. Are the events E and F independent? Why or why
not?
Solution: P(E) P(F) = .38*.3 = .114
P(E∩F) = P(E∪F) – P(E) – P(F) = .68 - .38 - .3 = 0
No, because P(E) P(F) ≠ P(E∩F)
15. A pair of fair dice is cast. What is the probability that the sum of the numbers falling uppermost is even
given that one of the numbers is a 4?
Solution: 5/11
An experiment consists of selected a letter at random from the words SAMPLE SPACE.
16. Write the sample space for the experiment.
Solution: {A, C, E, L, M, P, S}
17. How many events are possible for the experiment?
Solution: 2^7 = 128
18. What is the event that a letter chosen in the situation is a vowel?
Solution: {A, E}
Math 166 Week-in-Review - © S. Nite 9/26/2012
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19. Past records in a certain city produced the following probability data on a driver being in an accident on
a particular holiday:
Accident (A)
No Accident (AC)
Totals
Rain (R)
.025
.335
.360
C
No Rain (R )
.015
.625
.640
Totals
.040
.906
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 = 5/72
20. 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
21. 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)
22. A fair die is rolled 5 times. What is the probability of getting a 6 on the 5th roll, given that a 6 turned up
on the preceding 4 rolls?
Solution: Since independent, P(A|B) = P(A) = 1/6
23. 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)
24. 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
Math 166 Week-in-Review - © S. Nite 9/26/2012
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24. 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
518
456
974
Normal, C
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
25. 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
26. 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
27. 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
28. 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
29. 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