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A fluid flow system connecting points X and Y is illustrated schematically below

ID: 3440564 • Letter: A

Question

A fluid flow system connecting points X and Y is illustrated schematically below. Fluid can only flow through each labeled junction when that "switch" is closed. Each switch is independent of the others, and the probability that any given switch is open is 80%. What is the probability that fluid flows from X to Y? What is the probability that fluid flows from X to Y. given that switch D is open? Suppose switch E always copies the status of switch A. What is the probability that fluid flows from X to Y?

Explanation / Answer

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]

************

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]

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