# Conditional probability Bayes Theorem

I am trying to solve this problem but I am not sure how to obtain the formula given below. Any help would be appreciated.

A boy is selected at random from among the children belonging to families with $n$ children. It is known that the boy has at least $2$ sisters. Show that the probability that he has $k-1$ brothers is:

$$\frac {(n-1)!}{(2^{n-1}-n)(k-1)!(n-k)!}$$

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Can you identify at least part of the formula? For example, what is $\frac{(n-1)!}{(k-1)!(n-k)!}$? Perhaps that might give you an idea as to how to get to the answer. – Dilip Sarwate May 7 '13 at 23:17

Let's let $B$ be the number of brothers and $S$ be the number of sisters. Then we want to find $P(B = k-1 \mid S \geq 2)$.

$$P(B = k-1 \mid S \geq 2) = \frac{P(B = k-1 \cap S \geq 2)}{P(S \geq 2)} = \frac{P(S \geq 2 \mid B = k-1)P(B = k-1)}{P(S \geq 2)},$$

which is from Bayes.

Now, since there are $n-1$ children besides the boy in question, for the probability that $B = k-1$ we have the binomial probability $$P(B = k-1) = \binom{n-1}{k-1}\cdot \frac{1}{2^{n-1}} = \frac{(n-1)!}{(k-1)!(n-k)!\left(2^{n-1}\right)},$$ assuming genders are equally likely among remaining children.

For the denominator, the probability of at least two sisters we have

$$P(S \geq 2) = 1 - P(S = 0) - P(S = 1) = 1 - \frac{1}{2^{n-1}} - \frac{n-1}{2^{n-1}} = \frac{2^{n-1}-n}{2^{n-1}}.$$

So now we're already getting pretty close to the expression in your question. Can you take it from here?

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Yes, I understood now. Thank you very much. – Whats My Name May 8 '13 at 8:37
@Cubic, my pleasure. glad it helped. – sol May 9 '13 at 17:05