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A couple gives birth to a girl with probability $p$, a boy with probability $q=1-p$.
Let $N$ be the number of children needed so that at least both a girl and a boy are born.
What is the distribution of $N$?

For $ P(N=k), k \ge 2, $ two children should be picked such that one is girl and one is boy, so $ P(N=k)={k \choose 2} pq. $ I feel this is not correct.

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    @mjqxxxx can you turn, as is, your two comment into an answer. I will accept it.2013-01-15

2 Answers 2

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To expand on the question a bit, let $N$ be a random variable equal to the number of children at which you first have children of both genders. This takes on a particular value $k\ge 2$ when the first $k-1$ children are girls and $k$-th is a boy, or vice versa. The distribution is therefore $ P[N=k]=p^{k-1}q+pq^{k-1} $ for $k\ge 2$. The probability that you have children of both genders by the time you have $k$ children is $ P[N\le k]=\sum_{i=2}^{k}\left(p^{i-1}q+pq^{i-1}\right)=q\frac{p-p^k}{1-p}+p\frac{q-q^{k}}{1-q}=1-p^k-q^k, $ using the fact that $p+q=1$. This approaches $1$ as $k\rightarrow\infty$, as it should, since you will eventually have both boys and girls, a.s. (that is, with probability $1$).

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Assuming you have $k>0$ children, the probability that all the babies are girls is $p^k$. The probability that all the babies are boys is always $(1-p)^k$. Otherwise, you have both genders.

Therefore, the probability to have both genders with $k$ children is $1-p^k-(1-p)^k$

Specifically, for any number of chilren, there's always a finite probability that they'll all be of the same gender, and therefore there's no such $N$.

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    @yohBS I'm also puzzled by the wording of the question, and I can't explain how to define $N$. However, mjqxxxx pointed that your formula is $P(N \le k)$, which seems true as $1-p^k-q^k \to 1$ as $k\to \infty$.2012-12-28