Consider purchasing a system of audio components consisting of a receiver, a pair of speakers, and a CD player. Let A1 be the event that the receiver functions properly throughout the warranty period, A2 be the event that the speakers function properly throughout the warranty period, and A3 be the event that the CD player functions properly throughout the warranty period. Suppose that these events are (mutually) independent with P(A1) 5 .95, P(A2) 5 .98, and P(A3) 5 .80.a. What is the probability that all three components function properly throughout the warranty period?b. What is the probability that at least one component needs service during the warranty period?c. What is the probability that all three components need service during the warranty period?d. What is the probability that only the receiver needs service during the warranty period?e. What is the probability that exactly one of the three components needs service during the warranty period?f. What is the probability that all three components function properly throughout the warranty period but that at least one fails within a month after the warranty expires?

Answer:a) 0.6698b) 0.3302c) 0.0014d) 0.0662e) 0.2918f) 0.5861Step-by-step explanation:Part a) Probability that all three components function properlyThe probability that each of the component functions properly within the warranty period is given as:P(A1) = 0.91P(A2) = 0.92P(A3) = 0.80We have to find the probability that all 3 function properly. Since, working of one component is independent of the other, the probability that all 3 function properly will be:Probability that all three components function properly = P(A1) x P(A2) x P(A3)Probability that all three components function properly = 0.91 x 0.92 x 0.80Probability that all three components function properly = 0.6698Part b) Atleast one component needs serviceThe event "atleast one" is complement of the event "None"This means, probability that atleast one component needs service is complement of the event that none of the component needs service.None of the component needs service means that all 3 function properly. The probability that all 3 function properly is calculated in previous part which is 0.6698.So,Probability that atleast one component service = 1 - Probability that none of the component needs serviceProbability that atleast one component service = 1 - 0.6698Probability that atleast one component service = 0.3302Part c) Probability that all three components need service.Since,P(A1) = Probability that Receiver functions proper = 0.91Probability that it does not function properly and needs service = P(A1)' = 1 - P(A1) = 1 - 0.91 = 0.09Similarly,P(A2)' = 1 - P(A2) = 1 - 0.92 = 0.08P(A3)' = 1 - P(A3) = 1 - 0.80 = 0.20These are the individual probabilities that the components will need the service during the warranty period.So,The probability that all 3 will need the service = P(A1)'  x P(A2)'  x P(A3)' The probability that all 3 will need the service = 0.09 x 0.08 x 0.20The probability that all 3 will need the service =  0.0014Part d) Probability that only the receiver needs serviceSince only the receiver needs the service, the rest two components will function properly.So, we have to multiply the probability of receiver not functioning properly with probabilities that other two components will function properly. i.e.Probability that only the receiver needs service = P(A1)' x P(A2) x P(A3)Probability that only the receiver needs service = 0.09 x 0.92 x 0.80Probability that only the receiver needs service = 0.0662Part e) Probability that exactly one of the three components needs serviceWe have to find the probability that only one of the 3 components needs service. This component can be any of the 3 components, so there will be 3 cases:i) Only Receiver needs service:Probability of this event = P(A1)' x P(A2) x P(A3) = 0.09 x 0.92 x 0.80 = 0.0662ii) Only Speaker needs service:Probability of this event = P(A1) x P(A2)' x P(A3) = 0.91 x 0.08 x 0.80 = 0.0582iii) Only CD player needs service:Probability of this event = P(A1) x P(A2) x P(A3)' = 0.91 x 0.92 x 0.20 = 0.1674The probability that exactly one component needs service will be the summation of these probabilities.So,The probability that exactly one of the three components needs service during the warranty period = 0.0662 + 0.0582 + 0.1674 = 0.2918Part f) Probability that all three components function properly throughout the warranty period but that at least one fails within a month after the warranty expires.The probability that all three components function properly throughout the warranty period is calculated in part a.Now we need to find the probability that atleast one fails within one month after the warranty expires.When the warranty period is over, there is an equal chance of working properly and failing to function properly. So there is a 50% chance if the component will function properly after the warranty is over.Since, "atleast one" is complement of "none" first we find that none of the component fails:Probability that none of the component fails = 0.5 x 0.5 x 0.5 = 0.125So,The probability that atleast one component fails = 1 - 0.125 = 0.875Now, the probability that all three components function properly throughout the warranty period but that at least one fails within a month after the warranty expires = Probability that all 3 function properly during warranty x Probability that atleast one fails within one month after warranty= 0.6698 x 0.875= 0.5861