BUAD310 HW#3

Subjective

The host of a daily financial radio program stated that in his opinion, "There is a 25 percent chance that AT&T Wireless and Verizon will merge." What kind of probability is this?

(a) Empirical (b) Results from a study

A researcher found that out of 500 searches on major search engines for the keyword phrase "ring tone," more than 40 percent were fake pages created by spammers. (a) What kind of probability is this? (b) How would it have been derived?

(a) S = {(S,L), (S,T), (S,B), (P,L), (P,T), (P,B), (C,L), (C,T), (C,B)} (b)No, different likelihood

A survey asked tax accounting firms their business form (S = sole proprietorship, P = partnership, C = corporation) and type of risk insurance they carry (L = liability only, T = property loss only, B = both liability and property). (a) Choose the correct sample space from the given elementary events. (b) Would each elementary event be equally likely?

P(A∩B) = P(A) × P(B) = 0.40 × 0.50 = 0.2000.

Given P(A) = 0.40, P(B) = 0.50. If A and B are independent, find P(A ∩ B).

P(A ∩ B) = .25 *General Rule of Addition: P(A or B) = P(A) + P(B) − P(A ∩ B)*

If P(A) = .35, P(B) = .60, and P(A or B) = .70, then

P(A ∩ B) = .10 *General Rule of Addition: P(A or B) = P(A) + P(B) − P(A ∩ B*

The following probabilities are given about events A and B in a sample space: P(A) = .30, P(B) = .40, P(A or B) = .60. We can say that

(a) P(A∪B)=P(A)+P(B)−P(A∩B)=.7+.3−0=1.0. (b) P(A|B)=P(A∩B)/P(B)=.00 / .30=.00. (c) P(B|A)=P(A∩B)/P(A)=.00 / .7=.00.

Given P(A) = .70, P(B) = .30, and P(A ∩ B) = .00: (a) Find P(A U B). (b) Find P(A | B). (c) Find P(B | A).

(a) S = {(1,H), (2,H), (3,H), (4,H), (5,H), (6,H), (1,T), (2,T), (3,T), (4,T), (5,T), (6,T)} (b) Yes

A die is thrown (1, 2, 3, 4, 5, 6) and a coin is tossed (H, T). (a) Choose the sample space for the die/coin combination for the elementary events given. (b) Are the elementary events equally likely?

a. P(V) = 116/158 = 0.7342. b. P(A) = 47/158 = 0.2975. c. P(A ∩ V) = 32/158 = 0.2025. d. P(A U V)= P(A) + P(V) - P(A ∩ V) = 47/158 + 116/158 - 32/158 = 0.8291. e. P(A | V) = P(V ∩ A)/P(V) = 32/116 = 0.2759. f. P(V | A) = P(V ∩ A)/P(A) = 32/47 = 0.6809.

A survey of 158 introductory statistics students showed the following contingency table. Find each event probability. a. P(V) b. P(A) c. P(A ∩ V) d. P(A U V) e. P(A | V) f. P(V | A)

Empirical Subjective

An entrepreneur who plans to open a Cuban restaurant in Nashville has a 20 percent chance of success. Which kind of probability is it?

(a) There is 25% chance that a clock will not wake Bob (a failure, F). Both clocks would have to fail in order for him to oversleep. Assuming independence: P(F1 ∩ F2) = P(F1) × P(F2) = .25 × .25 = .0625. There is a 6.25% chance that Bob will oversleep. (b) No The probability that at least one of the clocks wakes Bob is 1 - (P(F1) × P(F2) × P(F3)) = 1 - (.25 × .25 × .25) = .9844, which is less than 99%.

Bob sets two alarm clocks (battery-powered) to be sure he arises for his Monday 8:00 a.m. accounting exam. There is a 75 percent chance that either clock will wake Bob. (a) What is the probability that Bob will oversleep? (Round your answer to 4 decimal places.) (b) If Bob had three clocks, would he have at least a 99 percent chance of waking up?

(a) P(Success | At least 60) = P(Success ∩ At least 60) = 0.005 = 0.125. P(At least 60)*0.04 There is a 12.50% chance of success, given that he or she is at least 60 years old. (b) No P(Success) × P(At least 60) = (0.31)(0.04) = 0.0124 which does not equal P(Success ∩ At least 60) = 0.005 therefore the two events are not independent.

Over 1,000 people try to climb Mt. Everest every year. Of those who try to climb Everest, 31 percent succeed. The probability that a climber is at least 60 years old is 0.04. The probability that a climber is at least 60 years old and succeeds in climbing Everest is 0.005. (a) Find the probability of success, given that a climber is at least 60 years old. (b) Is success in climbing Everest independent of age?

When two events A and B are independent, the joint probability of the events can be found by multiplying the probabilities of the individual events

Regarding probability, which of the following is correct?

(a) P(A) × P(B) = 0.40 × 0.60 = 0.24 and P(A ∩ B) = 0.24; therefore, A and B are independent. (b) P(A) × P(B) = 0.90 × 0.20 = 0.18 and P(A ∩ B) = 0.18; therefore, A and B are independent. (c) P(A) × P(B) = 0.50 × 0.70 = 0.35 and P(A ∩ B) = 0.25; therefore, A and B are not independent.

Which pairs of events are independent? (a) P(A) = 0.60, P(B) = 0.40, P(A ∩ B) = 0.24. (b) P(A) = 0.90, P(B) = 0.20, P(A ∩ B) = 0.18. (c) P(A) = 0.50, P(B) = 0.70, P(A ∩ B) = 0.25.

(a) P(a viewer is aged 18-34) = 69/100 = 0.69. (b) P(a viewer prefers watching TV videos) = 48/100 = 0.48. (c) P(a viewer is aged 18-34 and prefers watching videos on mobile/laptop device) = 39/100 = 0.39 or 39%. (d) P(a viewer prefers videos on mobile/laptop device | they are 18-34) = P(prefers mobile or laptop device ∩ Age 18-34)/P(Age 18-34) = 0.39/0.69 = 56.52% (e) P(aged 35-54 ∪ mobile/laptop device) = P(aged 35-54) + P(mobile/laptop device) - P(aged 35-54 ∩ mobile/laptop device) = (0.20 + 0.52) - 0.10 = 0.62 or 62%.

The contingency table below shows the results of a survey of video viewing habits by age. Find the following probabilities or percentages: (a) Probability that a viewer is aged 18-34. (Round your answer to 2 decimal places.) (b) Probability that a viewer prefers watching videos on a TV screen. (Round your answer to 2 decimal places.) (c) Percentage of viewers who are 18-34 and prefer videos on a mobile or laptop device. (d) Percentage of viewers aged 18-34 who prefer videos on a mobile or laptop device. (Round your answer to 2 decimal places.) (e) Percentage of viewers who are 35-54 or prefer videos on a mobile or laptop device.

P(A ∩ B) = P(A | B) P(B) *The definition of conditional probability: P(A | B) = P(A ∩ B) / P(B).*

The following relationship always holds true for events A and B in a sample space.

.769 Given that 775 have survived to 75, the probability is 596 divided by 775.

The following table shows the survival experience of 1,000 males who retire at age 65: Based on these data, the probability that a 75-year-old male will survive to age 80 is

.72 Use the General Rule of Addition: P(A or B) = P(A) + P(B) − P(A ∩ B) = .25 + .65 − .18.

The manager of Ardmore Pharmacy knows that 25 percent of the customers entering the store buy prescription drugs, 65 percent buy over-the-counter drugs, and 18 percent buy both types of drugs. What is the probability that a randomly selected customer will buy at least one of these two types of drugs?

a. P(D3) = 15/38 = 0.3947. b. P (Y3) = 15/38 = 0.3947. c. P(Y3 | D1) = P(Y3 ∩ D1) / (D1) = (2/38)/(11/38) = 0.1818. Another way is 2/11 = 0.1818 because the two 38s cancel each other out. d. P(D1 | Y3) = P(Y3 ∩ D1) / P(Y3) = (2/38)/(15/38) = 0.1333. Another way is 2/15 = 0.1333 because the two 38s cancel each other out.

This contingency table shows average yield (rows) and average duration (columns) for 38 bond funds. For a randomly chosen bond fund, find the probability of the following: (Round your answers to 4 decimal places.) a. The bond fund is of long duration. b. The bond fund has high yield. c. The bond fund has high yield given that it is of short duration. d. The bond fund is of short duration given that it has high yield.