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The problem is: In a sequence of independent bernoulli trials let X be the number of successes in the first m trials, and Y be the number of successes in the first n trials, $m<n$. Show that the conditional distribution of X, given $Y = y$ is hypergeometric. I thought that $$p(x, y) = \frac{\binom{m}{x}\binom{n-m}{y-x}}{\binom{n}{y}},$$ and that $p_y$ is equal to the summation of the above over all appropriate x. Then for the conditional distribution, I wanted to divide the former by the latter, but I am not sure that's right. Can someone help me solve this thanks!

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up vote 0 down vote accepted

You need to find: $P_{X|Y}(x,y)={P_{X,Y}(x,y)\over P_Y(y)}$.

Computing the required probabilities:

$\eqalign{P_{X,Y}(x,y) &= P[ x\ {\rm succ.\ in\ first\ }m {\rm\ trials\ and\ } y-x{\rm\ succ.\ in\ last\ } n-m{\rm\ trials}]\cr &= {m\choose x} p^x q^{m-x} \cdot {n-m\choose y-x} p^{y-x} q^{n-m-(y-x)}. }$

$P_Y(y) = {n\choose y} p^y q^{n-y}$.

Taking the quotient gives what you have in your post; which is the correct solution.

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