Consider the system of components connected as in the accompanying picture. Comp
ID: 3071624 • Letter: C
Question
Consider the system of components connected as in the accompanying picture. Components 1 and 2 are connected in parallel, so that subsystem works if and only if either 1 or 2 works; since 3 and are connected in series, that subsystem works if and only if both 3 and 4 work. If components work independently of one another and P(component i works) = 0.86 for i = 1, 2 and = 0.77 for i = 3, 4 calculate P(system works). (Round your answer to four decimal places.) 4. Need Help? ReadtTalk to a Tuter Talk to a Tutor + -11 points DevoreStat9 2.E.084 My Notes Ask Your Tea r 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 Ag be the event that the CD player functions properly throughout the warranty period. Suppose that these events are (mutually) independent with P(A1) 0.92, P(A2)-0.97, and P(A3)-0.80. (Round your answers to four decimal places.) (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?Explanation / Answer
1)
P(system works)=P(at least one of 1,2 or 3-4 works) =1-(1-0.86)*(1-0.86)*(1-0.77*0.77)=0.9920
2)
a)P(all 3 works)=0.92*0.97*0.8=0.7139
b)
P(at least one component need service)=1-P(all 3 works) =1-0.7139=0.2861
c)
P(all 3 need service)=(1-0.92)*(1-0.97)*(1-0.80)=0.0005
d)
P(only receiver needs service =(1-0.92)*0.97*0.80 =0.0621
e)
P(exactly one)==(1-0.92)*0.97*0.80+0.92*(1-0.97)*0.80+0.92*0.97*(1-0.80)=0.2626