Probability of at most 2 men entering a drugstore

Question

I am trying to solve the following probability problem:

The number of men that enter a drugstore in a given hour is a Poisson random variable with parameter $ \lambda = 5 $. The number of women that enter the drugstore in a given hour is identically and independently distributed. Please compute the conditional probability that at most 2 men entered the drugstore, given that 5 people entered in that hour.

My attempt:

Let number of men who entered the store be $ X $ and the number of women be $ Y $.

Since $ X $ and $ Y $ are identically distributed, it is fair to say that:

$$ f_X(x) = f_Y(y) = \frac{5^{k}e^{-5}}{k!} $$

So we must compute $p(X \le 2 | X + Y = 5)$. But since $X$ and $Y$ are independent, we can say that $p(X \le 2 | X + Y = 5) = p(X \le 2)$. Hence we can do the following: $$ p(X \le 2) = e^{-5}\left[1 + 5 + \frac{5^2}{2!}\right] = 0.124652019 $$

However, the suggested answer is $0.5$

I cannot understand why this is so. Could someone please explain what is wrong with my approach?

Answer 1

We can say that $p(X \le 2 | X + Y = 5) = p(X \le 2)$.

Just because $X$ and $Y$ are independent doesn't mean $X$ and $X+Y$ are. $X+Y\sim Po(10)$. So $$P(X\leq2|X+Y=5)=\frac{P(X\leq2\text{ and }X+Y=5)}{P(X+Y=5)}=\frac{P(X=0)P(Y=5)+P(X=1)P(Y=4)+P(X=2)P(Y=3)}{P(X+Y=5)}=\frac{0.00118+0.00591+0.01182}{0.03783}=0.49987$$