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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.

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    $\begingroup$ (a) You don't assume things, you prove them. So this might be true, but you need to prove it. And for this, condition on the amount of calls and use the total probability formula. $\endgroup$ – Arnaud Mortier Jul 28 '18 at 21:28
  • $\begingroup$ (b) In the process of doing (a) you can prove that the amount of sales does follow a Poisson then you will have your answer. Otherwise you can again go old school with conditioning. $\endgroup$ – Arnaud Mortier Jul 28 '18 at 21:30
  • $\begingroup$ I don't see how b) follows from a).. if the time between two sales is exponentially distributed, do I just declare one variable $X$ and another $Y$ corresponding to the time between 2 sales, and then find $P(Z > 20)$ where $Z = X+Y$? Is the sum of two exponential variables also exponential? $\endgroup$ – Saad Jul 28 '18 at 21:45
  • $\begingroup$ Also, the way to go about proving what you said would be: P(sales = x) = P(sales = x | calls = x) * P(calls = x) + P(sales = x | calls = x+1| * P(calls = x+1) ... etc, correct? $\endgroup$ – Saad Jul 28 '18 at 21:46
  • $\begingroup$ @Saad Your last comment is correct. The infinite sum can be evaluated nicely. $\endgroup$ – Mike Earnest Jul 28 '18 at 21:48
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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.

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  • $\begingroup$ Thanks, Michael. Is the final answer to the second part, 0.8478? I simply did poisson of x=0, x=1, x=2 and added them all. $\endgroup$ – user585380 Sep 8 '18 at 15:07
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For the (a) part, you can see that the mean sales in 15 minutes is only 1. Therefore, it will (e^-1 1^3)/6

I'm extremely sorry for the bad syntax. I will learn using MathJax in a few days.

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