The shown below is a segment of a series electrical circuit which has three rela

Question

The shown below is a segment of a series electrical circuit which has three relays. The current will flow from point (a) to point (b), that is system is working, if there is a closed path when the relays are switched to "on". However, the relays may malfunction. Suppose for each relay when it is twitched to "on" the probability that each relay does not close and fail to work properly is 0.02, where relays are working independent of each other's. Find the probability that system is not working. Work: Find the probability that the component 1 is working given that the system is not working. (That is, find P(A_1 | W): where A_1 denote the event that the component 1 is working properly and W denotes the event that system is not working.) Work:

Explanation / Answer

Solution

Back-up Theory

Let Ai represent the event that Relay i is ON and working, i = 1, 2, 3

Let W represent the event that System is ON and working.

Let Wc and Ric represent the respective complementary events.

Since it is a series connection. W => A1A2A3

So, P(W) = P(A1A2A3) = P(A1) x P(A2) x P(A3) …………………………… (1)

[given the relays work independently]

Also, A1, A2 and A3 are independent => A1c, A2c and A3c are also independent …(2)

P(A/B) = P(AB)/P(B) …………………………………………………………….(3)

P(at least one of A and B) = P(A) + P(B) - P(AB) = P(A) + P(B) - P(A)P(B) if A and B are independent ………….(4)

Now, to work out the solution,

Given P(A1c) = P(A2c) = P(A3c) = 0.02

Part (a)

Probability System is not working = P(Wc) = 1 - P(W) = 1 - P(A1) x P(A2) x P(A3) [by (1)]

= 1 – [{1 - P(A1c)}x{1 - P(A2c)}x{1 - P(A3c)}] = 1 – 0.983 = 1 – 0.941192

= 0.058808 ANSWER

Part (b)

Probability of Component 1 working given that the System is not working = P(A1/Wc)

= P(A1Wc)/P(Wc) [from (3)] = P(A1 and at least one of A2c and A3c)/0.058808 [from Part (a) answer]

= P(A1) x P(at least one of A2c and A3c)/0.058808

[given the relays work independent and by(2)]

= {1 – 0.02}x{P(A2c) + P(A3c) - P(A2c)xP(A3c)}/ 0.058808 [by (4)]

= (0.98)(0.02 + 0.02 – 0.0004)/0.058808 = 0.6599 ANSWER

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