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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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    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 $\sum_{k=2}^{\infty} P[N=k]=1$.2012-12-27
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    And take care, $P(N=1)=P(N=0)=0$. Don't skip that the condition that it takes a minimum of two kids. @mjqxxxx2012-12-27
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    In particular, the formula $\binom k 2 pq$ is unbounded - as $k$ gets larger, so does the probability. In particular, you can get $$\binom k 2 pq>1$$2012-12-27
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    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 $1-(p^N+q^N)$??2012-12-27
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    What does P(N=k) mean?2012-12-27
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    $N$ is a random variable; it is the number of children at which you first have children of both genders. $P[N=k]$ is the probability that the random variable has value $k$; i.e., it is the probability distribution of $N$.2012-12-27
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    @mjqxxxx I verified your expression $P(N=k)=p^{k-1}q+q^{k-1}p,k \ge 2$, and it sums to one. But, comparing to Khromonkey formula, $P(N=k)=1-p^k-q^k$, your formula seems to be missing some case. What am I missing?2012-12-27
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    Khromonkey's formula is $P[N\le k]$.2012-12-27
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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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    (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.)2012-12-27
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    if $k=0$ then clearly yuo don't have both genders...2012-12-27
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    Right, but your formula gives a "probability" of $-1$ in the case $k=0$, whereas it should be $0$.2012-12-27
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    OK. fixed. Thanks.2012-12-27
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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