Q:

Compute $\ P\{X_1 = 3 | X = 10 \} $, $\ X \sim Pois(10) $

Suppose $\ X$ is the number of people entering a store in an hour and $\ X \sim Pois(10) $.  Compute the probability that at most 3 men enter the store if it is known that 10 women already entered.  What assumptions do you make? 

I guess the assumption need to make is that the chance of a female \ male entering the store is the same $\ (0.5) $ and therefore the number of men entering the store is also poisson distribution with $\ \lambda = 5 $ ? 
Not sure how to proceed from here (if it is true)?  Lets say  $\ X_1 $ is number of men entering the store (with the values $\ 0,1,2,3 $ ) and $\ X_2 $ number of women. 
$\ P\{X_1 = 3 | X_2 = 10 \} = \frac{P\{X_1 = 3, X_2 = 10 \}}{P\{X_2 = 10 \}} $

A:

Since further info lacks you are entitled to go for the following:
$X=X_1+X_2$ where $X_1,X_2$ are iid and have Poisson distribution with rate $\lambda=5$.
The independence tells us that: $$P(X_1\leq3\mid X_2=10)=P(X_1\leq3)$$

