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
Bernoulli trials hypergeometric relation
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probability
1 Answers
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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.