The probability that a bird deposits $r$ eggs in its nest is given by the Poisson distribution with parameter $\lambda$. Assume that the probability of a egg to survive in nature is $p$ and that the survival of each egg is independent.
Knowing that a bird deposited at most $3$ eggs, what is the probability that there is exactly one survival egg?
My attempt to solve this was as follows:
P(exactly one survives| at most $3$ deposited) = P(exactly one survives and at most $3$ deposited)/P(at most $3$ deposited)
P(at most $3$ deposited) = $\sum_{r=0}^{3}\frac{e^{-\lambda}{\lambda}^r}{r!}$
P(exactly one survives and at most $3$ deposited) = $\sum_{r=1}^{3}\frac{e^{-\lambda}{\lambda}^r \cdot {r \choose 1} \cdot p \cdot (1-p)^{r-1}}{r!}$ where the $r=0$ term vanishes because we can't have a egg surviving if there is none in the nest.
Finally the result follows dividing 3./2. is that correct, could someone help me?