STATS 350 HW 4

Records show that 9% of all college students are foreign students who also smoke. It is also known that 60% of all foreign college students smoke. What percent of the students at this university are foreign?

Answer Let F correspond to "Foreign student" and S to "Smoke". We have P(F ∩ S) = 0.09 and P(S | F) = 0.60.From P(S | F) = P(F ∩ S)/P(F), we get P(F) = P(F ∩ S)/P (S | F) = 0.09/0.60 = 0.15 Therefore, 15% of the student body at the university is foreign.

An analyst estimates that the probability of default on a seven-year AA-rated bond is 0.50, while that on a seven-year A-rated bond is 0.50. The probability that they will both default is 0.27. a. What is the probability that at least one of the bonds defaults? (Round 2 decimal places.) b. An analyst estimates that the probability of default on a seven-year AA-rated bond is 0.50, while that on a seven-year A-rated bond is 0.50. The probability that they will both default is 0.27.a. What is the probability that at least one of the bonds defaults? (Round 2 decimal places.) c. Given that the seven-year AA-rated bond defaults, what is the probability that the seven-year A-rated bond also defaults? (Round 2 decimal places.)

Answers Let event A correspond to "Default on a seven-year AA bond", and B to "Default on a seven-year A bond".We have P(A) = 0.50, P(B) = 0.50, and P(A ∩ B) = 0.27. a. P(A U B) = P(A) + P(B) − P(A ∩ B) = 0.50 + 0.50 − 0.27 = 0.73 b. P((A U B)^c) = 1 − P(A U B) = 1 − 0.73 = 0.27 c. P(B | A) = P(A ∩ B)/P(A) = 0.27/0.50 = 0.54

Dr. Miriam Johnson has been teaching accounting for over 26 years. From her experience, she knows that 60% of her students do homework regularly. Moreover, 93% of the students who do their homework regularly pass the course. She also knows that 83% of her students pass the course. Let event A be "Do homework regularly" and B be "Pass the course". a. What is the probability that a student will do homework regularly and also pass the course? (Round your answer to 2 decimal places.) b. What is the probability that a student will NEITHER do homework regularly nor will pass the course? (Round your answer to 2 decimal places.) c. Are the events "pass the course" and "do homework regularly" mutually exclusive? d. Are the events "pass the course" and "do homework regularly" independent?

Answers Let event A correspond to "Do homework regularly" and B to "Pass the course". P(A) = 0.60, P(B | A) = 0.93, P(B) = 0.83 a. P(A ∩ B) = P(B | A)P(A) = 0.93(0.60) = 0.56 b. P((A ∪ B)c) = 1 − P(A U B);P(A U B) = P(A) + P(B) − P(A ∩ B) = 0.60 + 0.83 − 0.56 = 0.87; Therefore, P((A ∪ B)c) = 1 − 0.87 = 0.13 c. NO , because P(A ∩ B) = 0.56 ≠ 0 d. NO, because P(B | A) = 0.93 ≠ 0.83 = P(B)

Apple products have become a household name in America. Suppose that the likelihood of owning an Apple product is 61% for households with kids and 48% for households without kids. Suppose there are 1,200 households in a representative community, of which 820 are with kids and the rest are without kids. Let event A correspond to "Household with kids", and B to "Household owns an Apple product". a. Are the events "household with kids" and "household without kids" mutually exclusive and exhaustive? b. What is the probability that a household is without kids? (Round 4 decimal places.) c. What is the probability that a household is with kids and owns an Apple product? (Do not round intermediate calculations. Round 4 decimal places.) d. What is the probability that a household is without kids and does not own an Apple product? (Do not round intermediate calculations. Round 4 decimal places.)

Answers Let event A correspond to "Household with kids", and B to "Household owns an Apple product".We have P(A) = 820/1,200 = 0.6833, P(B | A) = 0.61, and P(B | Ac) = 0.48. a. YES , P(A ∩ Ac) = 0. Events corresponding to "household with kids" and "household without kids" are mutually exclusive. They are also exhaustive because P(A) + P(Ac) = 1. b. P(Ac) = (1,200 − 820)/1,200 = 380/1,200 = 0.3167 c. P(A ∩ B) = P(B | A) P(A) = (0.61)(0.6833) = 0.4168. d. P(Ac ∩ Bc) = P(Ac) − P(Ac ∩ B) = P(Ac) − P(B | Ac)P(Ac) = 0.3167 − 0.48(0.3167) =0.1647

The probabilities that stock A will rise in price is 0.43 and that stock B will rise in price is 0.57. Further, if stock B rises in price, the probability that stock A will also rise in price is 0.53. a. What is the probability that at least one of the stocks will rise in price? (Round 2 decimal places.) b. Are events A and B mutually exclusive? c. Are events A and B independent?

Answers P(A) = 0.43, P(B) = 0.57, P(A | B) = 0.53. a. P(A U B) = P(A) + P(B) − P(A ∩ B) = P(A) + P(B) − P(A | B)P(B) = 0.43 + 0.57 − 0.53(0.57) = 0.70 b. NO because P(A ∩ B) ≠ 0. -Events A and B are NOT mutually exclusive since P(A ∩ B) = 0.30 ≠ 0. c. NO because P(A | B) ≠ P(A). - Events A and B are NOT independent since P(A | B) = 0.53 ≠ 0.43 = P(A).

Let P(A) = 0.55, P(B) = 0.10, and P(A ∩ B) = 0.09. a. Are A and B independent events? b. Are A and B mutually exclusive events? c. What is the probability that neither A nor B takes place? (Round your answer to 2 decimal places.)

Answers P(A) = 0.55, P(B) = 0.10, and P(A ∩ B) = 0.09. a. NO because P(A | B) ≠ P(A). - P(A | B) = P(A ∩ B)/P(B) = 0.09/0.10 = 0.90 ≠ 0.55 = P(A);A and B are NOT independent events. b. NO because P(A ∩ B) ≠ 0. - P(A ∩ B) = 0.09 ≠ 0;A and B are NOT mutually exclusive events. c. P((A U B)c) = 1 − P(A U B);P(A U B) = P(A) + P(B) − P(A ∩ B) = 0.55 + 0.10 − 0.09 = 0.56;Therefore, P((A U B)c) = 1 − 0.56 = 0.44

Let P(A) = 0.58, P(B) = 0.23, and P(A ∩ B) = 0.13. a. Calculate P(A | B). (Round 2 decimal places.) b. Calculate P(A U B). (Round 2 decimal places.) c. Calculate P((A U B)^c). (Round 2 decimal places.)

Answers P(A) = 0.58; P(B) = 0.23; P(A ∩ B) = 0.13 a. P(A | B) = P(A ∩ B)/P(B) = 0.13/0.23 = 0.57 b. P(A U B) = P(A) + P(B) − P(A ∩ B) = 0.58 + 0.23 − 0.13 = 0.68 c. P((A U B)^c) = 1 − P(A U B) = 1 − 0.68 = 0.32

Let P(A) = 0.67, P(B) = 0.32, and P(A | B) = 0.47. a. Calculate P(A ∩ B). (Round 3 decimal places.) b. Calculate P(A U B). (Round 3 decimal places.) c. Calculate P(B | A). (Round 3 decimal places.)

Answers P(A) = 0.67; P(B) = 0.32; P(A | B) = 0.47. a. P(A ∩ B) = P(A | B)P(B) = (0.47)(0.32) = 0.150 b. P(A U B) = P(A) + P(B) − P(A ∩ B) = 0.67 + 0.32 − 0.150 = 0.840 c. P(B | A) = P(A ∩ B)/P(A) = 0.150/0.67 = 0.224

Consider the following probabilities: P(Ac) = 0.69, P(B) = 0.48, and P(A ∩ Bc) = 0.12. a. Find P(A | B^c). (Do not round intermediate calculations. Round your answer to 2 decimal places.) b. Find P(B^c | A). (Do not round intermediate calculations. Round your answer to 3 decimal places.) c. Are A and B independent events?

Answers P(Ac) = 0.69, P(B) = 0.48, and P(A ∩ Bc) = 0.12. a. P(A | Bc) = P(A ∩ Bc)/P(Bc) = 0.12/(1 − 0.48) = 0.23 b. P(Bc | A) = P(A ∩ Bc)/P(A) = 0.12/(1 − 0.69) = 0.387 c. NO because P(A | Bc) ≠ P(A) - P(A) = 1 − P(Ac)= 1 − 0.69 = 0.31P(A | Bc) = 0.23 ≠ 0.31 = P(A)Events A and Bc are not independent. Therefore, A and B are NOT independent either.

You roll a die with the sample space S = {1, 2, 3, 4, 5, 6}. You define A as {1, 4, 6}, B as {1, 3, 4, 5, 6}, Cas {1, 5}, and D as {2, 3, 5}. Determine which of the following events are exhaustive and/or mutually exclusive. a. A and B b. A and C c. A and D d. B and C

Answers a. A U B = {1, 3, 4, 5, 6} ≠ {1, 2, 3, 4, 5, 6} = S; the events A and B are NOT exhaustive.A ∩ B = {1, 4, 6}; the events A and B are NOT mutually exclusive. b. A U C = {1, 4, 5, 6} ≠ {1, 2, 3, 4, 5, 6} = S; the events A and C are NOT exhaustive.A ∩ C = {1}; the events A and C are NOT mutually exclusive. c. A U D = {1, 2, 3, 4, 5, 6} = S; the events A and D ARE exhaustive.A ∩ D = {∅}; the events A and D ARE mutually exclusive. d. B U C = {1, 3, 4, 5, 6} ≠ S; the events B and C are NOT exhaustive .B ∩ C = {1, 5}; the events B and C are NOT mutually exclusive.

Determine whether the following probabilities are best categorized as subjective, empirical, or classical probabilities. a. Before flipping a fair coin, Sunil assesses that he has a 50% chance of obtaining tails. b. At the beginning of the semester, John believes he has a 90% chance of receiving straight A's. c. A political reporter announces that there is a 48% chance that the next person to come out of the conference room will be a Republican, since there are 88 Republicans and 97 Democrats in the room.

Answers a. A classical probability, because it is based on logical analysis rather than on observation or personal judgment b. A subjective probability, because it is drawn on personal judgment. c. An empirical probability, as 0.48 represents the relative frequency of Republicans in the room, that is, 88/(88 + 97) = 0.48.

Consider the following scenarios to determine if the mentioned combination of attributes represents a union or an intersection. a. A marketing firm is looking for a candidate with a business degree and at least five years of work experience. b. A family has decided to purchase a Toyota minivan or a Honda minivan.

Answers a. An Intersection b. A Union

A sample space, S, yields four simple events, A, B, C, and D, such that P(A) = 0.35, P(B) = 0.10, and P(C) = 0.25. a. FindP(D). (Do not round intermediate calculations. Round your answer to 2 decimal places.) b. Find P(Cc). (Round your answer to 2 decimal places.) c. Find P(A U B). (Round your answer to 2 decimal places.)

Answers a. P(D) = 1 − [P(A) + P(B) + P(C)] = 1 − [0.35 + 0.10 + 0.25] = 1 − 0.70 = 0.30 b. P(Cc) = 1 − P(C) = 1 − 0.25 = 0.75 c. P(A U B) = P(A) + P(B) = 0.35 + 0.10 = 0.45

A sample space S yields eight equally likely events, I, J, K, L, M, N, O, and P. a. Find P(L). (Round your answer to 2 decimal places.) b. Find P(Jc). (Round your answer to 2 decimal places.) c. Find P(I U K U M). (Do not round intermediate calculations. Round your answer to 2 decimal places.)

Answers a. P(L) = 1/8 = 0.13 (The events are equally likely) b. First define Jc = {I, K, L, M, N, O, P}Then P(Jc) = 7/8 = 0.88. c. P(I U K U M) = P(I) + P(K) + P(M) = (1/8) + (1/8) + (1/8) = 3/8 = 0.38

At four community health centers on Cape Cod, Massachusetts, 15,164 patients were asked to respond to questions designed to detect depression. The survey produced the following results. Diagnosis Number Mild 3,257 Moderate. 1,546 Mod Severe 975 Severe 773 No Depression 8,613 a. What is the probability that a randomly selected patient suffered from mild depression? (Round 3 decimal places.) b. What is the probability that a randomly selected patient did not suffer from depression? (Round 3 decimal places.) c. What is the probability that a randomly selected patient suffered from moderately severe to severe depression? (Round 3 decimal places.) d. Given that the national figure for moderately severe to severe depression is approximately 6.7%, does it appear that there is a higher rate of depression in this summer resort community?

Answers a. P({Mild}) = 3,257/15,164 = 0.215 b. P({No Depression}) = 8,613/15,164 = 0.568 c. P({Moderately Severe} U {Severe}) = (975/15,164) + (773/15,164) = 0.064 + 0.051 = 0.115 d. YES , According to the data, Cape Cod appears to have a higher level of "moderately severe to severe depression" with 11.5% of its residents compared to 6.7% at the national level.

An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the basis of body mass index (BMI), expressed as weight/height2. An adult is considered overweight if the BMI is 25 or more but less than 30. An obese adult will have a BMI of 30 or greater. A recent study suggests that 33.1% of the adult population in the United States is overweight and 35.7% is obese. Use this information to answer the following questions. a. What is the probability that a randomly selected adult is either overweight or obese? (Round your answer to 3 decimal places.) b. What is the probability that a randomly selected adult is neither overweight nor obese? (Round your answer to 3 decimal places.) c. Are the events "overweight" and "obese" exhaustive? d. Are the events "overweight" and "obese" mutually exclusive?

Answers a. P({overweight} U {obese}) = P({overweight}) + P({obese}) = 0.331 + 0.357 = 0.688 b. Using the complement rule,P(({overweight} U {obese})c) = 1 − P({overweight} U {obese}) = 1 − 0.688 = 0.312 c. NO because you may be neither overweight nor obese. -The events {overweight} and {obese} are NOT exhaustive. The U.S. adult population also comprises individuals that are neither overweight nor obese. Furthermore,P({overweight}) + P({obese}) = 0.331 + 0.357 = 0.688 ≠ 1 d. YES because you cannot be both overweight and obese. - According to the BMI classification, an individual is either "overweight" or "obese" (but not both). As a result, the two corresponding events ARE mutually exclusive.

Jane Peterson has taken Amtrak to travel from New York to Washington, DC, on six occasions, of which three times the train was late. Therefore, Jane tells her friends that the probability that this train will arrive on time is 0.50. a. Would you label this probability as empirical or classical? b. Why would this probability not be accurate?

Answers a. This is an EMPIRICAL probability, since it is based on observed outcomes of an experiment. b. The experiment must be repeated a large number of times for empirical probabilities to be accurate. - This probability is UNLIKELY to be accurate due to the small sample.