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[crazy7437]

There are two boxes, each containing two components. Each component is defective with probability 1/4, independent of all other components. The probability that exactly one box contains exactly one defective component equals
a) 3/8
b) 5/8
c) 15/32
d) 17/32
i am getting 27/32. can anybody help please ?

[Amit Goyal]

Suppose there are two boxes A and B, with components A1 & A2, and B1 & B2, respectively.
Let D(x) denote the event that component x is defective and N(x) be the event that component x is not defective.
Probability that exactly one box contains exactly one defective component
= (Probability that box A has exactly one defective component - Probability that both boxes A and B have exactly one defective component)
+
(Probability that box B has exactly one defective component - Probability that both boxes A and B have exactly one defective component)
= 2 (Probability that box A has exactly one defective component - Probability that both boxes A and B have exactly one defective component)    -- By symmetry
= 2 (Probability that box A has exactly one defective component (1 - Probability that box B has exactly one defective component)) -- By independence of the defectiveness of components
= 2((Pr(D(A1), N(A2)) + Pr(N(A1), D(A2)))  (1-(Pr(D(B1), N(B2)) + Pr(N(B1), D(B2))))
= 2((3/8)(5/8)) = 15/32

[crazy7437]

Thank you sir!
I was getting 27/64 sorry , I was not considering the case where the other box can contain both defective components.