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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 ?

Day Dreamer
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  • You were on the right track. You missed out one part - the part where you allow both balls in the other box to be defective. This part corresponds to the 4th term in the sum that gives the total number of possibilities. Hence, required probability $= {4 \choose 1}(27/256)+{4 \choose 3}(3/256)=15/32$ – Deepak Gupta Sep 10 '15 at 17:29

2 Answers2

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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).

André Nicolas
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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.

David
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  • Awesome .. This was a very good question .. I was thinking this could be misprinted answer options .. Thanks a lot :) – Day Dreamer Mar 27 '15 at 06:44