A fluid flow system connecting points X and Y is illustrated
Solution
a)
P(flow) = P[(A or B) and C and (D or E)]
As all events are independent, then intersections are just the products of the probabilities,
= P(A or B) P(C) P(D or E)
= [P(A) + P(B) - P(A)P(B)] P(C) [P(D) + P(E) - P(D)P(E)]
As all probabilities for components are 0.80, evaluating,
P(flow) = 0.73728 [ANSWER]
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b)
If switch D is open, it is like P(D) = 0,
P(flow) = P[(A or B) and C and (D or E)]
P(flow) = P[(A or B) and C and E]
= [P(A) + P(B) - P(A)P(B)] P(C) P(E)
= 0.6144 [answer]
************
c)
There are two cases:
1. Both A and E are closed.
2 Both A and E are open. [Like P(A) and P(E) = 0]
Each has probability 1/2.
Thus,
P(flow) = 1/2 P(flow|both closed) + 1/2 P(flow|both open)
= 1/2 {[P(A) + P(B) - P(A)P(B)] P(C) [P(D) + P(E) - P(D)P(E)]} + 1/2 {P(B)] P(C) [P(D)]}
= 0.62464 [ANSWER]
