The question is from a typical example for E-M algorithm.
Let's say $(y_1,y_2,y_3)$ $\sim$ $\text{multinomial}(n;p_1,p_2,p_3)$, where $p_1+p_2+p_3=1$.
How can we derive the conditional distribution of $y_2$ given $y_2+y_3=n$?
The answer is $y_2|y_2+y_3 \sim \text{binomial}(n, p_2/(p_2+p_3)$).
Any idea on how to derive this rigorously?
Here's a more general result; yours is the special case $m=0$:
If $Y_1,Y_2,Y_3$ $\sim$ $\text{multinomial}(n;p_1,p_2,p_3)$ then the conditional
distribution of $Y_2$ given
$Y_1=m$ $(m\leq n)$
is $\text{binomial}( n-m, p_2/(p_2+p_3) )$:
\begin{eqnarray} p( Y_2=y_2| Y_1=m) &=& p(Y_2=y_2, Y_1=m)/ P(Y_1=m)\\ &=&\frac{\binom{n}{m}\binom{n-m}{y_2} p_2^{y_2} p_3^{n-m-y_2}p_1^m } {\binom{n}{m} p_1^m (p_2+p_3)^{n-m}}\\ &=&\binom{n-m}{y_2} p_2^{y_2} p_3^{n-m-y_2} /(p_2 + p_3)^{n - m} \end{eqnarray}
