The daily number $\displaystyle X$ of cheques presented at a local branch of a certain bank to be cashed has the following probability mass function (a Poisson distribution with parameter $\displaystyle \lambda$):

$\displaystyle p_X(k)=e^{-\lambda}\frac{\lambda^k}{k!}$ for $\displaystyle k=0,1,2,...$

If each cheque presented has a probability $\displaystyle p$ to be refused (due to the lack of sufficient funds), compute the pmf, $\displaystyle p_S(s)$ for $\displaystyle s=0,1,2,...$, of the daily number $\displaystyle S$ of refused cheques.
A hint was provided as follows: You may first compute the conditional probability $\displaystyle P(S=s|X=k)$ and then use the total probability formula by summing over all possible values of $\displaystyle k$.

I'm still grasping the concepts behind Poisson distributions, so any help would be appreciated.