If Y has a geometric distribution with success probability p, show that
P(Y= an odd integer)= p/(1-q^2)
By the way, if you make what would seem the obvious assumption that success is getting an odd number, then the probability of getting an odd number eventually is going to be 1, I'd have thought (unless p = 0) ......
Are you sure there's not some lead in, some introduction, some broader context ...... to the question. Because as is, the question is