Prime customers

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.

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1 solution

Geoff Pilling
Dec 15, 2018

You can assume a Poisson distribution. So

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

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

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

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

Brian Charlesworth - 2 years, 5 months ago

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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

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