I get using linearity. Also, . But this is no longer a Poisson since the mean and variance are different?? If that's true then . Is this right?
Indeed is no longer a Poisson r.v., but you could have seen that directly from the question about the support of . While are integers, their mean may not be an integer. The support of the distribution of is "the set of possible values" for Y (You can remove the " " in the present situation). Therefore, it is...
And in order to determine the probability mass function of Y, you should first find that (or recall that) of . It is Poisson with parameter . Then giving the probability mass function of Y should be simple. Tell us if you still have problems.
(and yes of course )