Probability question on finding a defective ball in a specific box There are two boxes, each containing two balls. Each ball is defective with probability 1/4, independent of other balls. The probability that exactly one box contains exactly one defective ball is 
(A) 3/8  (B) 5/8  (C) 15/32  (D) 17/32
One box can be picked in 2 ways and 1 ball in that box can be picked in 2 ways. So, 4 ways. The probability of 1 defective and 3 normal balls is 27/256.
The total number of possibilities is 4C0(81/256)+4C1(27/256)+4C2(9/256)+4C3(3/256)+4C4(1/256)=256/256=1.
So, I thought the answer should be 27/64. This is different from the given choices. Is my method wrong ? 
 A: The probability that Box A  has exactly $1$ defective is $2(1/4)(3/4))$, that is, $3/8$. The same is true for Box B. 
So the probability Box A has exactly $1$ defective and Box B doesn't (because it has $0$ or $2$) is $\frac{15}{64}$. Double this.
Remark: The event "exactly $1$ box has exactly $1$ defective" is not the same as the event "$1$ defective and $3$ good." For we could have $3$ defective and $1$ good.  That would add $\frac{3}{64}$ to your $\frac{27}{64}$, giving $\frac{30}{64}$, which simplifies to the answer C).
A: You have to read the question very carefully, and in fact, I'm not entirely sure that it is stated unambiguously.  I would interpret it as asking for the probability that one box contains one defective ball, and the other does not (that is, the other contains $0$ or $2$ defective balls).
Now the probability that one box contains exactly one defective ball is


*

*$\frac14$ (the first ball defective). . .

*times $\frac34$ (the second ball OK). . .

*times $2$ (the balls could be the other way around)


which is $\frac38$.  The probability that the other box is not in this situation is $\frac58$.  Multiply these, and multiply by $2$ as the boxes could be the other way round, to give the final probability
$$\frac{15}{32}\ .$$
Check: the probability of a box not having one defective is the probability that it has $0$ or $2$ defectives, which is
$$\Bigl(\frac34\Bigr)^2+\Bigl(\frac14\Bigr)^2=\frac{10}{16}=\frac58$$
as found above.
