STATS 300 - Section 5.3 Homework

5.3.4 - When two events are​ disjoint, they are also independent A. True B. False

B. False

5.3.31 - According to a​ poll, about 19​% of adults in a country bet on professional sports. Data indicates that 47.1​% of the adult population in this country is male. a. Are the events​ "male" and​ "bet on professional​ sports" mutually​ exclusive? Explain A. No. A person can be both male and bet on professional sports at the same time. B. Yes. A person cannot be both male and bet on professional sports at the same time. C. No. A person cannot be male and bet on professional sports at the same time. D. Yes. A person can be both male and bet on professional sports at the same time

A. No. A person can be both male and bet on professional sports at the same time.

5.3.7 - Determine whether the events E and F are independent or dependent. Justify your answer b. - E: A randomly selected person planting tulip bulbs in October. ​F: A different randomly selected person planting tulip bulbs in April A. E cannot affect F because​ "person 1 planting tulip bulbs in October​" could never​ occur, so the events are neither dependent nor independent. B. E can affect the probability of​ F, even if the two people are randomly​ selected, so the events are dependent. C. E cannot affect F and vice versa because the people were randomly​ selected, so the events are independent. D. E can affect the probability of F because the people were randomly​ selected, so the events are dependent.

C. E cannot affect F and vice versa because the people were randomly​ selected, so the events are independent.

5.3.7 - Determine whether the events E and F are independent or dependent. Justify your answer a. - E: A person living at least 70 years. ​F: The same person regularly handling venomous snakes. A. E and F are independent because regularly handling venomous snakes has no effect on the probability of a person living at least 70 years. B. E and F are dependent because regularly handling venomous snakes can affect the probability of a person living at least 70 years. C. E and F are independent because living at least 70 years has no effect on the probability of a person regularly handling venomous snakes. D. E and F are dependent because living at least 70 years has no effect on the probability of a person regularly handling venomous snakes.

B. E and F are dependent because regularly handling venomous snakes can affect the probability of a person living at least 70 years.

5.3.31 - According to a​ poll, about 19​% of adults in a country bet on professional sports. Data indicates that 47.1​% of the adult population in this country is male. The poll data indicated that 11.5​% of adults in this country are males and bet on professional sports. What does this indicate about the assumption in part​ (b)? A. The assumption was correct and the events are independent. B. The assumption was incorrect and the events are not independent.

B. The assumption was incorrect and the events are not independent

5.3.31 - According to a​ poll, about 19​% of adults in a country bet on professional sports. Data indicates that 47.1​% of the adult population in this country is male. e. How will the information in part​ (d) affect the probability you computed in part​ (c)?

Find P(male or bets on professional sports). Use P(male and bets on professional sports) = 0.115 instead of the value you found in part​ (b). P(E or ​F) = ​P(E) + ​P(F) − ​P(E and​ F) P(male or bets on professional sports) = 0.471 + 0.19 - 0.115 = 0.546

5.3.21 - In airline​ applications, failure of a component can result in catastrophe. As a​ result, many airline components utilize something called triple modular redundancy. This means that a critical component has two backup components that may be utilized should the initial component fail. Suppose a certain critical airline component has a probability of failure of 0.0074 and the system that utilizes the component is part of a triple modular redundancy. ​ a. Assuming each​ component's failure/success is independent of the​ others, what is the probability all three components​ fail, resulting in disaster for the​ flight? 0.00000041 b. What is the probability at least one of the components does not​ fail? 0.99999959

Use the fact that P(E1) = P(E2) = P(E3) = 0.0074 to fill in the correct numbers. P(E1 and E2 and E3) = P(E1)•P(E2)•P(E3) =​ (0.0074)(0.0074​)(0.0074​) = 0.00000041 b. 1 - 0.00000041 = 0.99999959

For a parallel structure of identical​ components, the system can succeed if at least one of the components succeeds. Assume that components fail independently of each other and that each component has a 0.07 probability of failure. a. Would it be unusual to observe one component​ fail? Two​ components? would not, 0.07, greater, would, 0.0049, less b. What is the probability that a parallel structure with 2 identical components will​ succeed? 0.9951 c. How many components would be needed in the structure so that the probability the system will succeed is greater than 0.9998​? 4

a. It would not be unusual to observe one component fail, since the probability that one component​ fails 0.07 is greater than 0.05. It would be unusual to observe two components fails because (0.07)(0.07) = 0.0049 is less than 0.05. b. P(at least one component succeeds) = 1 − P(2 components fail) = 1 − 0.0049 = 0.9951 c. Trial and error P(system succeeds) = 1 − (0.07)^n For this problem, the probability that the system will succeed is greater than 0.9998 is when n = 4

5.3.25 - Suppose Dan wins 44​% of all staring contests. ​a. What is the probability that Dan wins two staring contests in a​ row? 0.1936 b. What is the probability that Dan wins six staring contests in a​ row? 0.0073 ​ c. When events are​ independent, their complements are independent as well. Use this result to determine the probability that Dan wins six staring contests in a​ row, but does not win seven in a row. 0.0041

a. P(E and F) = (P(E))(P(F)) = (0.44)(0.44) = 0.1936 b. (0.44)(0.44)(0.44)(0.44)(0.44)(0.44) = 0.0073 c. PE^c = 1 −​ P(E) 1 - 0.44 = 0.56 (0.0073)(0.56) = 0.0041

5.3.13 - About 20​% of the population of a large country is nervous around strangers. If two people are randomly​ selected, what is the probability both are nervous around strangers​? What is the probability at least one is nervous around strangers​? Assume the events are independent. a. The probability that both will be nervous around strangers is 0.04 b. The probability that at least one person is nervous around strangers is 0.64

a. P(E) = 0.20 P(F) = 0.20 P(E and F) = (P(E))(P(F)) = (0.20)(0.20) = 0.04 b. Find P(at least 1 nervous) = 1 −​ P(neither nervous) PE^c = 1 − P(E) PE^c = 1 - 0.20 = 0.80 P(neither E) = (P(person 1 not E))(P(person 2 not E) = (0.80)(0.80) = 0.64 P(at least 1 nervous) = 1 −​ P(neither nervous) P(at least 1 nervous) = 1 −​ 0.64 = 0.36

5.3.11 - What is the probability of obtaining three tails in a row when flipping a​ coin? Interpret this probability a. The probability of obtaining three tails in a row when flipping a coin is 0.125. b. Consider the event of a coin being flipped three times. If that event is repeated ten thousand different​ times, it is expected that the event would result in three tails about 1,250

a. P(tail) = 1/2 P(tail on all three flips) = (1/2)(1/2)(1/2) = 0.125 b. (0.125)(10,000) = 1,250

5.3.31 - According to a​ poll, about 19​% of adults in a country bet on professional sports. Data indicates that 47.1​% of the adult population in this country is male. b. Assuming that betting is independent of​ gender, compute the probability that an adult from this country selected at random is a male and bets on professional sports. 0.0895

b. P(male and bets on professional sports) = (P(male))(P(bets on professional sports) = (0.471)(0.19) = 0.0895

5.3.31 - According to a​ poll, about 19​% of adults in a country bet on professional sports. Data indicates that 47.1​% of the adult population in this country is male. c. Using the result in part​ (b), compute the probability that an adult from this country selected at random is male or bets on professional sports. 0.5715

c. P(E or ​F) = ​P(E) + ​P(F) − ​P(E and​ F) P(male or bets on professional sports) = 0.471 + 0.19 - 0.0895 = 0.5715