Prime customers

Probability Level 2

Eddie L'Amore gets an average of 3 customers a day.

On any given day what is the probability that he has a prime number of customers?

Assumptions :

The occurrence of customers is uncorrelated, and the customer probability distribution doesn't vary from day to day, so you can assume the number of daily customers follows a Poisson distribution .

The answer is 0.57.

1 solution

Geoff Pilling
Dec 15, 2018

You can assume a Poisson distribution. So

$P(n) = \dfrac {k^ n e^{-k}}{n!}$ . where $k=3$ .

So $P$ (prime) $= P(2) + P(3) + P(5) + P(7) + .... = 0.57$

I think it would be $P(n) = \dfrac{k^{n} e^{-k}}{n!}$ where in this case $k = 3$ .

Anyway, it's fortunate that the series converges pretty quickly, (only need to add up to $P(11)$ to get the correct value to 4 decimal places). Any idea of a closed form for this series? $0.5707...$ looks familiar .....

Brian Charlesworth - 2 years, 5 months ago

Ah... Correct you are!

I've updated the solution... Somehow I had transposed the n and k... Thanks for pointing it out!

Geoff Pilling - 2 years, 5 months ago

Prime customers

The answer is 0.57.

1 solution

1 pending report

Vote up reports you agree with