Let X have binomial distribution with parameters $n = 4$ and $p = 0.75$, and Y have have binomial distribution with parameters $n = 5$ and $p = 0.5$. Suppose $X, Y$ are independent. Let $Z = X + Y$
For $2 ≤ k ≤ 7$, how would I obtain $P(X = 2 | Z = k)$. How would I proceed with this?
I am aware that this is
$$\frac{P(X=2, Z=k)}{P(Z=k)}$$ $$\frac{P(X=2)P(Y=k-2)}{P(X+Y=k)}$$
If this is right, how would I proceed with this?
More generally assume that $X\sim\mathsf{Bin}(n,p)$ and $Y\sim\mathsf{Bin}(m,q)$ are independent, and $Z=X+Y$. For $0\leqslant i\leqslant k\leqslant n+m$ we would have by definition of conditional probability \begin{align} \mathbb P(X=i\mid Z=j) &= \frac{\mathbb P(X=i,Z=k)}{\mathbb P(Z=k)}\\ &= \frac{\mathbb P(X=i,Y=k-i)}{\mathbb P(Z=k)}\\ &= \frac{\mathbb P(X=i)\mathbb P(Y=k-i)}{\mathbb P(Z=k)}\\ \end{align} Assuming there are no $j$ such that $\mathbb P(X=j)=1$ or $\mathbb P(Y=j)=1$ then $$ \mathbb P(Z=k) = \begin{cases} (1-p)(1-q),\ k=0\\ \frac1k\sum_{i=1}^k(-1)^{i-1}\mathbb P(Z=k-i)\left(\frac p{1-p}\right)^i\left(\frac q{1-q}\right)^i,\ k\ne 0 \end{cases} $$ Here $Z$ has Poisson binomial distribution. The computation would be quite tedious to do by hand.