Let $X$ and $Y$ be independent random variables and let $Z = X + Y$. To find $P(Z = z | X = x)$ this is the same as finding $P(Y = z - x)$ (correct me if I'm wrong).
However, if we wish to find $P(X = x | Z = z)$, we cannot simply take $P(Y = z - x)$. Why is that? I know that we can solve for $P(X | Z)$ using Bayes' theorem, but is there an intuitive explanation why $P(Y = z - x)$ won't work?
$P(Z=z|X=x)=P(Y=z-x|X=x)$ and independnec allows us to drop the conditionin on the right side. However, $P(X=x|Z=z)=P(Y=z-x|Z=z)$ and we cannot drop the conidtioning since $Y$ and $Z$ are not independent.
$P(X=x|Z=z)=P\frac {(Y=z-x)} {P(Z=z)}$ so you have to compute $P(Z=z)$ to find this conditinal probability.