# Probability of probability of Poisson Distribution

Jimmy Butler (shooting guard for the Minnesota Twolves) makes about p =30% of his three-point shot attempts. In any game, the number of three-point shots he attempts is a random variable Z whose distribution is approximately Poisson with mean = 5. What is the probability that in a random game he makes exactly 2 three-point shots?

I can easily find the probability of taking n shots given the distribution, but I have trouble finding how to find the probability of making n shots, since p = .3, and 6.6666 attempted shots is not a valid input for the Poisson distribution formula.

Any help is appreciated!

This is a distribution mixture problem. Given the number of shots $n$ he attempts, the number of shots he makes $m\sim B(n,p)$ is governed by a binomial distribution with $p=0.3$. Then the number of shots $n\sim\mbox{Poisson}(\lambda)$ he attempts is governed by a Poisson distribution with $\lambda=5$. The goal is to find the mixed distribution for $m$, and in particular, the probability of $P(m=2)$. Give it a try!

Hint: show that $\,m\sim\mbox{Poisson}(p\lambda)\,$ after the mixture.

Let $N$ be the number of shots he takes and $M$ be the number of shots he makes.

The thing you find it easy to compute is $P(N=n).$

You should also be able to write down the probability he makes $m$ shots given that he takes $n$ since you know that the $n$ shots are independent trials with probability $p=0.3.$ This is the conditional distribution of $M$ given $N$, denoted as $P(M=m\mid N=n).$

You want $P(M=m),$ the distribution of the number of shots he makes, which can be obtained from the above quantities via the law of total probability as $$P(M=m) = \sum_{n=0}^\infty P(M=m\mid N=n)P(N=n).$$

• Thanks! That was very useful. – user493197 Oct 19 '17 at 17:29