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In this question the general formula P(A|B) = P(A ∩ B)/P(B) is used. I understand through intuition why the answer should be 1/3. What I don't understand is why P(both girls, at least one girl) is 1/4. I know that P(A ∩ B) = P(A)P(B) if A and B are independent. Here P(Both girls) = 1/4 and P(at least one girl) = 3/4 so why isn't P(Both girls, at least one girl) = 3/16?

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Tharuka Devendra
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  • Please add the `[self-study]` tag & read its [wiki](https://stats.stackexchange.com/tags/self-study/info). – gung - Reinstate Monica Aug 18 '17 at 15:28
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    This problem has been discussed a few times on math.stackexchange.com. I have often found these questions confusing. I found this answer explains why I have had trouble with it: https://math.stackexchange.com/a/15344/5844 – John Aug 18 '17 at 15:54

2 Answers2

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Another way to approach this problem is Bayes' theorem.

$$P(\text{both girls}\mid\text{at least one girl}) = \frac{P(\text{at least one girl}\mid \text{both girls})P(\text{both girls})}{P(\text{at least one girl})}$$

$$P(\text{both girls}\mid\text{at least one girl}) = \frac{1\cdot\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$$

Sven Hohenstein
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Like you said, $P(A \cap B) = P(A) P(B)$ only when $A$ and $B$ are independent. In this instance if $A = $ Both girls, and $B$ = at least one girl, these events are not independent. More specifically, if $A$ holds, then both are girls, which automatically implies that there is at least one girl, so $B$ holds. Thus, you don't have independence.

From the above logic it is thus clear that, if $A$ holds, then $B$ holds, which means that the event $A \subset B$. Thus $A \cap B = A$. So $P(A \cap B) = P(A) = 1/4$ in this case.

Greenparker
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