How to show that the conditional law of Y knowing that ${X=n}$ is a binomial law without showing it litteraly? 
In a bank, the number of cheques emitted everyday by clients is a random variable $X\sim P(\lambda), \lambda>0$. We assume that cheques are independently emitted. For a random cheque, the probability for the bank having a no founding cheque is $p, p\in ]0,1[$. We call $Y$ the random variable corresponding to the number of cheques emitted without provision,

Let be $n \in N_+$. Can I show that the conditional law of Y knowing that ${X=n}$ is a binomial law without showing it literally? (We have n cheques emitted everyday, from wich ...)
I tried $P(Y|X=n)=$ but ... I'm not even sure it's what we are searching for...
 A: If I understand your question correctly, the PDF of $Y$ will be a Poisson process with density $(1-p)\lambda$. The process is called thinning. To see how it works, assume that you drop each point of your Poisson process with probability $p$. Then the number of remaining points $Y$ is:
$$
\Pr(Y=m)=\sum_{i=m}\Pr(X=i)\binom{i}{m}p^{i-m}(1-p)^m\\
=\sum_{i=m}\frac{\lambda^ie^{-\lambda}}{i!}\binom{i}{m}p^{i-m}(1-p)^m\\
=\frac{(\lambda(1-p))^me^{-\lambda}}{m!}\sum_{i=m}\frac{\lambda^{i-m}}{(i-m)!}p^{i-m}\\
=\frac{(\lambda(1-p))^me^{-\lambda}}{m!}.e^{\lambda p}=\frac{(\lambda(1-p))^me^{-\lambda(1-p)}}{m!}.
$$
So the PDF of $Y$ is also Poisson with modified density.
A: You can write $$Y = \sum_{i=1}^{X} Z_i$$ where $Z_i=1$ if the $i-$th cheque is emitted without provision, $Z_i=0$ otherwise. If $X=n$, then the sum has a fixed number ($n$) of variables $Z_i$, each of which is Bernoulli random variable with parameter $p$. 
Now, you are supposed to know that the sum of such $n$ Bernoulli vars is a Binomial ($n,p$) - because this is the same as counting the number of "successes" in $n$ trials with probability of success $p$.
