Thanks so much for your reply

The question is:

P(Y=k)=$\displaystyle \sum$ $\displaystyle \binom{n}{k}$ $\displaystyle p^k$(1-p)^(n-k)$\displaystyle e^{-x}$ $\displaystyle \frac{x^n}{n!}$

where the sum runs from n=k to infinity (sorry i'm rubbish at using latex)

is this right?

P(Y=k)=$\displaystyle \sum$ x^n/(n-k)!k! $\displaystyle p^k$ (1-p)^(n-k) $\displaystyle e^{-x}$

P(Y=k)=$\displaystyle p^k$ $\displaystyle e^{-x}$ $\displaystyle \sum$ x^n/(n-k)!k! . (1-p)^(n-k)

then change the sum so that it runs from n=0 to infinity

P(Y=k)=(p^k)/k! $\displaystyle e^{-x}$ $\displaystyle \sum$

then I'm not sure what I'll have left inside? x^n...

any hints appreciated