The daily number 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 \lambda):

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

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

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