# poisson random variable + conditional probability

Let Y be the number of calls to a particular hotline within 10 min. Suppose Y is a Poisson random variable with mean of 3. Find the probability that there are at most 4 calls given that there are already 2 calls within the 10 min.

I'm not really sure how to go about solving this. Am I supposed to be looking at this in terms of a binomial random variable? If not, then how would I set up the conditional probability?

By definition of conditional probability, \begin{align} \mathbb P(Y\leqslant 4\mid Y\geqslant 2) &= \mathbb P(Y=2\mid Y\geqslant 2) + \mathbb P(Y=3\mid Y\geqslant 2) + \mathbb P(Y=4\mid Y\geqslant 2)\\ &= \frac{\mathbb P(Y=2,Y\geqslant 2)}{\mathbb P(Y\geqslant 2)} +\frac{\mathbb P(Y=3,Y\geqslant 2)}{\mathbb P(Y\geqslant 2)} +\frac{\mathbb P(Y=4,Y\geqslant 2)}{\mathbb P(Y\geqslant 2)}\\ &= \frac{\mathbb P(Y=2)+\mathbb P(Y=3)+\mathbb P(Y=4)}{1 - (\mathbb P(Y=0) + \mathbb P(Y=1))}\\ &= \frac{e^{-3}(3^2/2! + 3^3/3! + 3^4/4!)}{1 - e^{-3}(1 + 3)}\\ &\approx 0.7693246. \end{align}
• From the wording of the question, it's given that there were at least two calls in the first $10$ minutes. – Math1000 Feb 18 at 3:12