A relay has a series of 5 circuits in a line. The even-numbe

Here is the original question.

varun289 wrote:A relay has a series of 5 circuits in a line. The even-numbered circuits are control circuits; the odd are buffer circuits. If both a control circuit and the buffer circuit immediately following it both fail in that order, then the relay fails. The probability of circuit one failing is 3/8; circuit two, 3/8; circuit three, 3/10; circuit four, 3/4; and circuit five, 2/5 .What is the probability that the relay fails?

Let the circuit series : BCBCB
Probability of the relay failing = Probability that the first combo of CB fails + Probability that second combo of CB fails - Probability that both the combos fails = (3/8)*(3/10) + (3/4)*(2/5) - (3/8)*(3/10)*(3/4)*(2/5) = 9/80 + 6/20 - 27/800 = 90/800 + 240/800 - 27/800 = 303/800

tarunjohri wrote:If we go from either direction, once one pair of CB has failed
the question of second pair would not matter. SO why are you including the
intersection part?

Let us indicate a failure by red and success by blue.

So, if the first pair of CB fails, the following scenarios are possible :

• 1A. BCBCB
  1B. BCBCB

And, if the second pair of CB fails, the following scenarios are possible :

• 2A. BCBCB
  2B. BCBCB

Now, 1B and 2B are same scenario.
Hence, if we just add the individual probability of 1st pair of CB failing and 2nd pair of CB failing, we'll end up in considering both failing scenario twice.

That's why we need to subtract the intersection once from the sum of probability.

Hope that helps.

A relay has a series of 5 circuits in a line. The even-numbered circuits are control circuits; the odd are buffer circuits. If both a control circuit and the buffer circuit immediately following it both fail in that order, then the relay fails. The probability of circuit one failing is 3/8; circuit two, 3/8; circuit three, 3/10; circuit four, 3/4; and circuit five, 2/5 .What is the probability that the relay fails?

Failure occurs if both circuit2 and circuit3 fail OR if both circuit4 and circuit5 fail

This is an OR probability

P(Event A OR Event B occurs) = P(Event A occurs) + P(Event B occurs) - P(Events A and B both occur)

So, P(complete failure) = P(circuit2 and circuit3 fail OR circuit4 and circuit5 fail)
= P(c2 and c3 fail) + P(c4 and c5 fail) - P(c2 and c3 fail AND c4 and c5 fail)
= (3/8)x(3/10) + (3/4)x(2/5) - (3/8)x(3/10)x(3/4)x(2/5)
= 9/80 + 6/20 - 27/800
= [spoiler]303/800[/spoiler]

Cheers,
Brent