# Elementary Problem on Conditional Probability

Let there be a family consisting of 2 children such that :

B : Event in which both children in a family are girls.

L : Event in which at least one child is a girl
\begin{align} P(B\mid L)&=?\\ \text{I found it by enumeration as:}& \{(g,g),(g,b),(b,g),(b,b)\}\\ P(B|L)=1/3 \end{align} Is there a way of arriving at this answer without enumerating? \begin{align} P(B|L)&=P(BL)/P(L)\\ P(L)&=1-P(\text{No Girls})\\&=1-1/4=3/4\\P(BL)&=? \end{align} Obviously, the answer has to be $1/4$ and I can see that from the enumeration but I can't deduce why.

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## 2 Answers

You’re almost there: the event $BL$ is the same as the event $B$, since $B\subseteq L$. Thus, $$P(BL)=P(B)=\frac14\;,$$ and your second calculation yields the result

$$P(B\mid L)=\frac{P(BL)}{P(L)}=\frac{1/4}{3/4}=\frac13\;,$$

just as it should.

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In this case the event $B$ is a subset of the event $L$. Another way of saying this is: if both are girls, then at least one is a girl. This means that $P(BL) = P(B)$. From here you can see the probability is $\frac{1}{4}$.

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