Download Ch. 6 Review Questions

Practice questions for Chapter 6
Questions on the Range of Probabilities
Questions 1-5 refer to the following data.
The table below shows the results of a hypothetical survey on participation in sports activities by men and women.
Those surveyed answered yes to activities they had done at least twice in the previous 12 months. Assume that the
sample is representative of the overall population:
Total
Aerobic Exercising
Baseball
Hunting
Softball
Exercise Walking
Men
109,059
3,717
12,603
18,512
11,535
25,146
Women
115,588
19,535
2,974
2,343
8,541
46,286
1. If you randomly select one man, what's the probability that he hunts, expressed as a percentage?
A. 16.02%
B. 16.97%
C. 589.13%
D. 8.24 %
2. If you randomly select one woman, what's the probability that she does aerobics?
A. .034
B. .169
C. .201
D. .087
E. 5.917
3. If you randomly select one man, what's the probability that he doesn't do exercise walking?
A. .231
B. 1.300
C. .400
D. .660
E. .769
4. If you randomly select one person from the study, what would be the probability that he or she plays baseball for
exercise, expressed as a percentage?
A. 11.56%
B. 2.57%
C. 6.93%
D. 7.07%
E. 3.47%
5. True or False: In this study, there are 18,512 male hunters. And, there are 12,603 male baseball players. If you
add the male hunters and male baseball players together, you get 18,512 + 12,603 = 31,115. The total number of
men surveyed was 109,059. Therefore, assuming the sample is random and representative of the U.S. population,
the probability that a randomly selected man in the U.S. either hunts or plays baseball is 31,115/109,059 = 28.5%.
(Hint: Look carefully at how the survey was answered.)
Questions on Calculating Probabilities
1. You have a bag of red, white, and blue marbles. You know there are 11 red marbles, 8 blue marbles, and 3 white
marbles in the bag. What's the probability of randomly selecting a blue marble from the bag?
A. .275
B. .364
C. .375
D. .500
E. .727
2. In Sausha's pocket, she has: 7 pennies, 3 nickels, 3 dimes, 5 quarters, 2 half-dollars, and 1 Susan B. Anthony
dollar coin. If Sausha selects one coin from her pocket, what's the probability that it's divisible by $0.10? (Note:
Ignore the fact that it's possible to distinguish the coins by the size and shape of the coin)
A. 1/3
B. 1/2
C. 0.300
D. 0.286
E. 1/7
Questions on Concepts of Probability
1. The set of all possible outcomes of a (probability) experiment is called the:
A. experiment
B. complement
C. Venn Diagram
D. sample space
E. probability of the event
2. A jar contains five coins: 2 pennies, 1 nickel, and 2 dimes. You draw one coin at random, what's the probability
of selecting the nickel? What's the probability of selecting a dime?
A. 1/5, 2/5
B. 1/3, 1/3
C. 2/5, 1/5
D. 1/5, 2/4
E. 1/3, 2/4
Questions 3-4 refer to the following information:
You have a bag of different flavored candies. The probabilities of randomly selecting a particular flavored candy are
shown below:
Flavor
Cherry
Lime
Lemon
Orange
Grapefruit
Probability
.245
.325
.180
.215
.035
3. The probability of selecting an orange candy OR a cherry candy is an example of:
A. complementary events
B. disjoint events
C. independent events
D. conjunctive events
E. ancillary events
4. Your buddy Ear1likes the lime candies. He eats a bunch of them and hands the bag back to you. The complement
of the event P (selecting a lime candy from the bag) would:
A. increase.
B. decrease.
C. stay the same.
D. become O.
E. There isn't enough information to answer this question. .
5. You have six different cards with colors on both sides of the card:
Card 1: Green and Green
Card 3: Green and Red Card 5: Green and Green
Card 2: Green and Blue Card 4: Red and Red
Card 6: Green and Blue
You're shown a card with a green face showing. What is the probability that the other side of the card is not green?
A.
B.
C.
D.
E.
20%
50%
60%
66.7%
85%
Questions on Rules of Probability and Conditional Probability
1. One card is randomly selected from a standard 52-card deck. Find the probability that the card is an ace or a black
card.
A.
B.
C.
D.
E.
1/26
1/13
7/13
15/26
1/2
For questions 3-5, refer to the following table of outcomes for the roll of a black and a white die. Ordered pairs
represent (black, white).
(1,1)
(1,2) (1,3) (2,1) (2,2) (2,3) (3,1)
(3,2) (3,3) (4,1) (4,2) (4,3)
(5,1) (5,2)
(5,3) (6,1) (6,2)
(6,3) (1,4)
(2,4) (3,4) (4,4)
(5,4) (6,4)
(1,5) (2,5) (3,5) (4,5) (5,5)
(6,5) (1,6)
(2,6) (3,6) (4,6) (5,6) (6,6)
3. Which entry (entries) in the table represent(s) the event (black die shows a 3 | white die shows a 4)?
A. (4,3)
B. {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)}
C. {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)}
D. {(3,4), (4,3)}
E. (3,4)
4. P(black die shows a 2 and sum of the faces is 5) =
A. 1/6 + 4/36 ( 1/6)(1/6)
B. (1/6)(1/6)
C. 1/6
D. 1/6 + 4/36 - 1/36 4/36
E. 4/36
5. P(black die shows 6 or white die shows 2) =
A. (1/6)(1/6)
B. 1/3
C. 1/6 + 1/6
D. 1/6 + 1/6 - 1/36
E. 1/6 + 1/6 - 2/36
Questions on Conditional Probabilities and Tree Diagrams
Use the tree diagram below to answer questions 3-6:
3. P(Yes) =
A. .16
B. .9
C. .12
D. .70
E. .012
4. P(B|Yes) =
A. .018
B. .2
C. .375
D. .3
E. .06
5. Which statement about the information in the tree diagram is incorrect?
A. The probability of A is .60.
B. The probability of C given the occurrence of Yes is .40.
C. The probability .54 at the end of path from A and No is a compound probability.
D. The probability of (C and Yes) is .04.
E. All these statements are incorrect.
6. If the probabilities in this diagram changed, and the likelihood of A" B, and C were now equivalent, the
probability of getting a No would be:
A. 0.33
B. 0.50
C. 0.76
D. 0.84
E. Can't be determined
Questions on Calculating Conditional Probabilities Graphically
Ouestions 1-4 refer to this diagram:
The tree diagram below shows the probabilities for the simple events combining weather patterns and days of the
week.
1. What is the probability of having a clear day, given that it is a weekend?
A. .215
B. .536
C. .75
D. .751
E. .714
2. What is the probability of a weekday that is overcast?
A. .10
B. .071
C. .029
D. .286
E. 1 - .071
3. True or False: Days of the week and weather patterns are independent.
4. Find the probability of a day being a weekday, conditional on it raining (to the nearest thousandth).
A. .150
B. .107
C. .286
D. .714
E. Not enough information to answer the question
ANSWERS:
Questions on the Range of Probabilities
1. Answer B. 16.97%."'It's the number of men who hunt, divided by the population at risk, all men in the
sample. Then, multiply this number by 100 to get a percentage. So, the solution is: (18,512/109,059) * 100
2. Answer B. .169. It's the number of events, which is women who do aerobics, divided by the population
at risk, which is the total number of women in the sample. So, the solution is: 19,535/115,588 .
3. Answer E. .769. You could calculate this by subtracting the number of men who walk from the total
number of men to get the number who don't walk, then divide this by the total number of men: (109,059 25,146)/(109,059). Or, you could calculate the probability that a man walks for exercise and then subtract
it from 1 to get the probability that he does not walk for exercise: 1 - (25,146/109,059).
4. Answer C. 6.93%. You're looking at all people in the study. The answer is the sum of the men and
women who play baseball, or 12,603 + 2,974 = 15,577, divided by the total number of people in the study,
or 109,059 + 115,588 = 224,647, and multiplied by 100 to get a percentage. So, the equation is:
(15,577/224,647) * 100.
5. Answer False. Those surveyed were allowed to say yes to more than one activity. Some of the baseball
players may also be hunters.
Questions on Calculating Probabilities
1. Answer B. .364. In total there are 22 marbles in the bag. Since 8 marbles are blue, the probability of
selecting a blue marble from the bag is 8/22, or .364. This means there's a 36.4% chance of selecting a
blue marble.
2. Answer D. .286. Three different coins are divisible by $0.10: the dime, half-dollar and dollar coin. Of
the 21 coins in her pocket, 6 of them are a dime, half-dollar, or dollar coin. Therefore, the probability is
6/21 or 0.286.
Questions on Concepts of Probability
1. Answer D. sample space. The sample space shows all the possible outcomes of a probability
experiment.
2. Answer A.1/5, 2/5. Since there five coins and only one is a nickel, the probability of selecting that
nickel is 1/5. And, 2 of the 5 possible outcomes are dimes, giving a probability of 2/5.
3. Answer B. disjoint events. Disjoint, or mutually exclusive, events are events that have no outcomes in
common. It isn't possible for a candy to be both orange and cherry, so there's no intersection of these
events.
4. Answer A. increase. Removing lime candies would reduce the probability of drawing a lime candy from
the bag. In turn, the complement of the event (the probability of selecting a non-lime candy) will increase
since the complement equals one minus the probability of the original event.
5. Answer C. 60%. There are five possible cards with a green face. You know that of these five, three have
a different color on the other side. Therefore, the probability of not finding green or the other side is 3/5,
or 60%.
Questions on Rules of Probability and Conditional Probability
1. Answer C. 7/13. If you apply the general rule P(Ace or Black Card) = P(Ace) + P(Black Card) - P(Ace
and Black Card), you find that P(Ace) = 4/52 = 1/13 P(Black Card) = 1/2, and P(Ace and Black Card) =
2/52 (which accounts for the two black aces in the deck). So, P(Ace or Black Card) = 1/13 + 1/2 - 2/52 =
7/13.
3. Answer E. (3,4). This is the only entry in the table for which the value on the black die is 3, given that
you know that the value on the white die is 4.
4. Answer B. (1/6)(1/6). This solution for the intersection of these two independent events is
P(A and B) = P(A) * P(B | A) = P(B) * P(B | A). You solved for P(A) * P(B | A). So, (1/6) * (1/6) =
(1/6)(1/6).
5. Answer D. 1/6 + 1/6 - 1/36. You know that the probability of each event is 1/6. (There's a 1/6 chance of
getting a 6 on a die, and there's a 1/6 chance of getting of 2 on a die). Since . there's one entry where these
events intersect, (6,2), you don't want to double count this entry. So, you have to subtract the probability
of this intersection from the union. Therefore, 1/6 + 1/6 - 1/36 = 11/36.
Questions on Conditional Probabilities and Tree Diagrams
3. Answer A. .16. In this case you want to sum all the paths where a Yes occurred, or in other words (P(A
and Yes) = .06) + (P(B and Yes) = .06) + (P(C and Yes) = .04) = .16.
4. Answer is C. .375 in other words: P(B I Yes) = P(B and Yes)/ P(Yes), or .06/.16 = .375.
5. Answer is B. The probability of C given the occurrence of Yes is .40. This conditional probability is
equivalent to P( C and Yes)/ P(Yes), or .04/.16 or .25.
6. Answer is C.0.76. The probability of A, B, and C now all equal .33. If you compute the paths for each
of these possibilities with No, you get P(A and No) = .297, P(B and No) = .264, and P(C and No) = .198.
The sum i-s .76.
Questions on Calculating Conditional Probabilities Graphically
1. Answer C. .75. The path leading from weekends to clear gives the probability of a clear day conditional
on that day being a weekend.
2. Answer B. .071. This is simply the probability of the simple event, weekday and overcast.
3. Answer True. If you look at the conditional probabilities of weather given days of the week, they're
the same regardless of whether the path comes off of weekdays or weekends. If the probability of a day
being rainy, for example, given that it's a weekday, is the same given it's a weekend, weather and days of
the week are independent. Knowing the day of the week does not give any additional information about
the behavior of weather.
4. Answer D. .714. The answer is just the probability of it being a weekday, or .714. Since days of the
week and weather patterns are independent, the probability of a weekday given rain is the same as the
probability of a weekday.