According to a survey, 60 % of the residents of a city oppose a downtown casino. Of these 60 % about 8 out of 10 strongly oppose the casino. Complete parts (a) through (c). (a) Find the probability that a randomly selected resident opposes the casino and strongly opposes the casino. (b) Find the probability that a randomly selected resident who opposes the casino does not strongly oppose the casino. (c) Would it be unusual for a randomly selected resident to oppose the casino and strongly oppose the casino? Explain.
Accepted Solution
A:
Answer:(a) 0.48(b) 0.20(c) it is not unusual for a radomly selected resident to oppose the casino and strongly oppose the casino.Step-by-step explanation:(a) Find the probability that a randomly selected resident opposes the casino and strongly opposes the casino. The probability that a radomly selected resident opposes the casino and strongly opposes the cassino is the product of the two probabilities, that a resident opposes the casino and that it strongly opposes the casino (once it is in the first group) as it is shown below.Use this notation:Probability that a radomly selected resident opposes the casino: P(A)Probability that a resident who opposes the casino strongly opposes it: P(B/A), because it is the probability of event B given the event Ai) Determine the probability that a radomly selected resident opposes the casino, P(A)Probability = number of favorable outcomes / number of possible outcomesP(A) is given as 60%, which in decimal form is 0.60ii) Next, determine,the probability that a resident who opposes the casino strongly opposes it, P(B/A):It is given as 8 out of 10 ⇒ P(B/A) = 8/10 iii) You want the probability of both events, which is the joint probability or intersection: P(A∩B).So, you can use the definition of conditional probability:P(B/A) = P(A∩B) / P(A)iv) From which you can solve for P(A∩B)P(A∩B) = P(B/A)×P(A) = (8/10)×(0.60) = 0.48(b) Find the probability that a randomly selected resident who opposes the casino does not strongly oppose the casino.In this case, you just want the complement of the probability that a radomly selected resident who opposes the casino does strongly oppose the casino, which is 1 - P(B/A) = 1 - 8/10 = 1 - 0.8 = 0.2.(c) Would it be unusual for a randomly selected resident to oppose the casino and strongly oppose the casino? You are being asked about the joint probability (PA∩B), which you found in the part (a) and it is 0.48.That is almost 0.50 or half of the population, so you conclude it is not unusual for a radomly selected resident to oppose the casino and strongly oppose the casino.