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here's a simple question

There are ten families in a village. Two without kids, one with one kids, five with two kids and two with 3 kids. I randomly knock on a door. A kid opens the door and says he lives there. What is the probability that he's the only one?

So, there are 8 families with kids, one of them only has one kid. Its easy to see that the answer is $\frac{1}{8}$. How do I define two events $A,B$ such that $P(A|B)=\frac{1}{8}$?

At first I thought about $A_i=\{I\, picked\, a\, house\, with\, i\, kids\}$ $B=\{I\, picked\, a\, house\, with\, at\, least\, one\, kid\}$ but then what is $P(A_1|B)$? To me it seems that $P(A_1\cap B)=\frac{1}{8}$ and $P(B)=\frac{4}{5}$ which means that $P(A_1|B)=\frac{1}{8}\cdot \frac{5}{4}$

Where is my mistake?

Thanks

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    These are all important questions :) but I've got it. Thanks2012-11-11

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Since $A_1\subset B$, $P(A_1\cap B)=P(A_1)=\frac1{10}$. The rest is correct.