I knew there are lots of similar probability problems. However when the questions vary a bit, I got confused about the concepts. For example, I understand the concept of conditional probability between two dependent variables but I don't know how to interpret the following situation:
Given: $$X = binomial(n,p)$$ $$Y = binomial(m,p)$$ $$Z = X+Y$$ $$p_X(X=i) = {n \choose i} \cdot p^i \cdot (1-p)^{n-i}$$ $$p_Y(X=j) = {n \choose j} \cdot p^j \cdot (1-p)^{n-j}$$
I found that: $$\because Z = X+Y$$ $$\therefore Z=z = i+j$$ $$p_Z(X,Y) = {n+m \choose z} \cdot p^z \cdot (1-p)^{n+m-z}$$
How do I find: $$Pr(Z=12|X=10)$$
My approach:
$$Pr(Z|X)=\dfrac{Pr(Z(X=10,Y=2))Pr(X=10))}{Pr(X=10)}$$
in which the Pr(X=10) can be dropped but I don't know how to proceed to next step.
Also, is the following approach correct?
$$\because Pr(B|A)=\dfrac{Pr(A \cap B)}{Pr(A)}$$
$$\therefore Pr(Z|X)=\dfrac{Pr(Z=12,X=10)}{Pr(X=10)}$$
by substituting the given value into the pmf of $p_z$ and $p_x$
$\begin{align}\mathsf P(X+Y=z\mid X=x)~&=~\dfrac{\mathsf P(Y=z-X, X=x)}{\mathsf P(X=x)} \\[2ex]&=~\dfrac{\mathsf P(Y=z-x, X=x)}{\mathsf P(X=x)} \\[2ex]&=~\mathsf P(Y=z-x\mid X=x) \\[4ex]\therefore\qquad\mathsf P(Z=12\mid X=10)~&=~\mathsf P(Y=2\mid X=10)\end{align}$
Now consider that $X,Y$ are independent, so...