# Conditional probability thoughts

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

• If you pick a house where a kid lives, what is the probability that the kid (rather than one of his parents) opens the door? Without some assumption about that I don't think you can reach a definite answer. – hmakholm left over Monica Nov 11 '12 at 18:55
• And are kids who have siblings more likely to be at home when you come knocking, because they don't need to leave the house in order to find someone to play with? – hmakholm left over Monica Nov 11 '12 at 18:56
• These are all important questions :) but I've got it. Thanks – Yotam Nov 11 '12 at 19:40

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