# Probability involving two Poisson constants.

A discrete stochastic variable $X$ is defined as the number of eggs laid by a bird at a certain time. Depending on the specie of the bird, the Poisson constant for this distribution is $\lambda = \lambda_1$ or $\lambda_2$. What is $\text {Pr}(X = k)$?

Am I crazy to think that it is the sum of the Poisson functions on both constants?

The question requires interpretation. One reasonable thing is to suppose that the bird laying the eggs is of Species $1$ with probability $p$, and of Species $2$ with probability $1-p$. Perhaps, quite unreasonably, you are expected to assume that $p=\frac{1}{2}$.
Then if $X$ is the number of eggs, $$\Pr(X=k)=pe^{-\lambda_1}\frac{\lambda_1^k}{k!}+(1-p)e^{-\lambda_2}\frac{\lambda_2^k}{k!}.$$ If $\lambda_1\ne \lambda_2$, this is not Poisson for any "mixed" $\lambda$, except in the extreme cases $p=0$ or $p=1$.
Remark: I am not sure what you meant by the sum of the Poisson functions. Maybe you meant $\Pr(X=k)=e^{-\lambda_1}\frac{\lambda_1^k}{k!}+e^{-\lambda_2}$. If so, that is not quite right, though if we take $p=\frac{1}{2}$, as you may be intended to do, it is very close. But we need to divide by $2$. In general $\Pr(X=k)$ is a weighted average of the two mass functions.
• True, I changed the wording. The bird that laid the eggs might be of Species $1$ or it might be of Species $2$. – André Nicolas Oct 27 '12 at 4:08
In a general case, you can always write the interested probability as follows $${\rm{Pr}}(X=k) = {\rm{Pr}}(X=k|\lambda_1){\rm{Pr}}(\lambda=\lambda_1)+{\rm{Pr}}(X=k|\lambda_2){\rm{Pr}}(\lambda=\lambda_2),$$ which uses the total probability formula.