Please show work for this so I can understand. What I have so far is that I coul
ID: 3041076 • Letter: P
Question
Please show work for this so I can understand. What I have so far is that I could either do this by p(works) or P(1-doesn't work). For P(works), I did P(works)=P(A) or P(B) OR (P(C) and P(D)). Please help me. Thank you.
Consider the system of components connected as depicted below. The system can be thought of as being comprised of two subsystems: one with components A and B, and the other with components C and D. Components A and B are connected in parallel, therefore that subsystem works iff either A or B works. Since C and D are connected in series, that subsystem works iff both C and D work. Components work independent of each other (that is, operation/failure of one component doesn't influence the operation/failure of another). If event A is defined as "component A works" and so on, then P(A) = P(B) = P(C) = P(D) = 0.90. a. Calculate P(system works) as a function of the P() notation. b. Calculate the quantity for P(system works).Explanation / Answer
a)here P(subystem AB works) =1-P(none of A and B owrks) =1-(1-P(A))*(1-P(B))
=(P(A)+P(B)-P(A)*P(B)
P(subystem CD works) =P(both C and D works) =P(C)*P(D)
hence probability that system works) =P(system works)
=1-P(none of system AB and CD works)
=1-(1-(P(A)+P(B)-P(A)*P(B)))*(1-P(C)*P(D))
b) P(system works) =1-(1-(0.9+0.9-0.9*0.9))*(1-0.9*0.9)=0.9981