Conditional PMF of two independent Poisson r.v.s

Question

Problem:

My solution: $$P(X=k\mid X+Y=n)=P(Y=n-k)=e^{-\lambda}\frac{\lambda^{n-k}}{(n-k)!}$$ Official Solution:

My question: How does the first equality even work. I suppose they are using the independence of $X$ and $Y$ to say something like:

$$P(X=k \cap X+Y=n)=P(X=k)P(Y=n-k)$$

But I don't think this is right because in this specific probability you do not know that $X=k$.

My question(2): Why is my answer wrong. These two expressions are definitely not equivalent, but I cannot see where I went wrong.

Answer 1

1. The event $\{X=k\} \cap \{X+Y = n\}$ is equal to $\{X = k\} \cap \{Y = n-k\}$. Then you can use independence.
2. You can see that the official solution is your answer multiplied by $\frac{P(X=k)}{P(T=n)}$. The error in your solution attempt is the first equality: you haven't justified why $P(X=k \mid X+Y=n) = P(Y=n-k)$. Your intuition is correct that somehow the event $Y=n-k$ plays a role, but you need to go back to the definition of the conditional probability $P(X=k \mid X+Y=n) = \frac{P(X=k, X+Y=n)}{P(X+Y=n)}$ to see how the event $Y=n-k$ enters the calculation.
3. Side note: what is true is $P(X=k \mid X+Y = n) = P(Y = n-k \mid X+Y = n)$. But $P(Y = n-k \mid X+Y = n) \ne P(Y = n-k)$.