A salesperson has a 20% probability of selling a product during each call. The average rate of calls made per hour is 20 and follows a Poisson distribution.
a) Find the probability that there are 3 sales made in a period of 15 minutes.
b) Find the probability that it takes more than 20 minutes to make 3 sales.
For the first one, do I simply say the average # of sales per hour is 20/100 * 20 = 4, and then just assume that it follows a Poisson distribution?
For the 2nd part, I have to use the exponential distribution somehow but I'm not sure precisely what to do.
Best Answer
For part $(b),$ you do not need to use the exponential distribution if you know that the average number of sales in $20$ minutes is $4/3$ and the number is Poisson-distributed. Let $X$ be the number of sales in $20$ minutes. You have $$ \Pr(X\le 2) = \Pr(X=0)+\Pr(X=1) + \Pr(X=2). $$ That's the probability that it takes more than $20$ minutes to make three sales.
If $Y\mid X \sim\operatorname{Binomial}(X,0.2)$ and $X\sim\operatorname{Poisson}(20),$ then $Y\sim\operatorname{Poisson}(20(0.2)) = \operatorname{Poisson}(4),$ as you guessed. If you need to prove that, then there's some algebra to do. If instead you can rely on that, then just go on from there.