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?	Choose the correct choice from the following
A) 9/80  B) 3/10  C) 303/800  D) 35/80  E) 497/800
The relay fails when "2 and 3 fail" or "4 and 5 fail".   P(2 and 3 fail) = (3/8)*(3/10) = 9/80   P(4 and 5 fail) = (3/4)*(2/5) = 3/10   Both these cases include the cases in which all 4 circuits fail. So we need to subtract that probability once out of the sum.   P(2, 3, 4 and 5 fail) = (9/80)*(3/10) = 27/800   P(Relay fails) = 9/80 + 3/10 - 27/800 = 303/800   Answer (303/800)
<result>C</result>
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?