I get 357/800..

the only way the relay fails is if (2&3 fail) or (4&5 fail) or (2,3,4&5 all fail).

plug in the appropriate proabilities and signs...
(3/8*3/10) + (3/4*2/5) + (3/8*3/10*3/4*2/5)

Am i missing something ?

The first circuit doesn't matter.

Prob(relay fails) = 1 - Prob(relay succeeds)

Prob(2+3 work) = 1 - 9/80 = 71/80
Prob(4+5 work) = 1 - 3/10 = 7/10

Prob(relay fails) = 1 - Prob(2+3 work AND 4+5 work) = 1 - (71/80)(7/10) = 1 - 497/800 = 303/800
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One small thing, yes. When you work out the probability that 2+3 fail, you're also including the possibility that 2+3+4+5 all fail. When you work out the probability that 4+5 fail, you're also including the possibility that 2+3+4+5 fail. So when you write the first two terms in your sum above: (3/8*3/10) + (3/4*2/5)
you've actually counted twice the possibility that 2+3+4+5 all fail. You actually want to subtract the last term, not add it, to correct for the double-counting:

(3/8*3/10) + (3/4*2/5) - (3/8*3/10*3/4*2/5)

It's really a Venn diagram situation, where you need to take out the overlap between two groups.
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I believe the question is poorly written.
does the failure count if F -> F-> F? or
W->F->F?

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 (C)
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