# Distribution of the number of children needed so that at least both a girl and a boy are born

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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This is not correct. The situation you want is for the first $k-1 \ge 1$ children to be one gender, and the $k$-th child to be the other. This is just $P[N=k]=p^{k-1}q + q^{k-1}p$ for $k\ge 2$. You should verify that $\sum_{k=2}^{\infty} P[N=k]=1$. – mjqxxxx Dec 27 '12 at 22:16
And take care, $P(N=1)=P(N=0)=0$. Don't skip that the condition that it takes a minimum of two kids. @mjqxxxx – Thomas Andrews Dec 27 '12 at 22:17
In particular, the formula $\binom k 2 pq$ is unbounded - as $k$ gets larger, so does the probability. In particular, you can get $$\binom k 2 pq>1$$ – Thomas Andrews Dec 27 '12 at 22:19
The probability that there are k girls is $p^k$ isn’t the possibility for any N that there is at least one boy and one girl equal to $1-(p^N+q^N)$?? – Carry on Smiling Dec 27 '12 at 22:21
What does P(N=k) mean? – Carry on Smiling Dec 27 '12 at 22:32

## 2 Answers

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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(1) Take care when $k=0$. (2) What question are you answering? The current one asks for a "distribution" of $N$. (I believe this question needs further clarification in order for the meaning of this request to be properly understood.) – whuber Dec 27 '12 at 22:26
if $k=0$ then clearly yuo don't have both genders... – yohBS Dec 27 '12 at 22:28
Right, but your formula gives a "probability" of $-1$ in the case $k=0$, whereas it should be $0$. – whuber Dec 27 '12 at 22:29
OK. fixed. Thanks. – yohBS Dec 27 '12 at 22:29
@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 $k\to \infty$. – Nicolas Essis-Breton Dec 28 '12 at 13:02