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Customers enter a store according to a Poisson Process of rate = 6 per hour.

Individuals who enter the shop have (independently of each other) probability $\theta$ of buying something.

  • If exactly n people enter the shop during a certain period, write down the probability that precisely k of them will buy something.

  • Hence show that the probability that precisely k purchase something in time t hours is

$(60\theta t)^k exp(-6\theta t)\over k!$ $k=0,1,2,...$

For the first part, I think that the probability of $k$ people buying something would be $\theta^k(1-\theta)^{n-k}$ but I'm not sure how that helps me do the 2nd part.

user123965
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1 Answers1

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Suppose that $B(T)$ is the number of users buying during the interval $T$, and $N(T)$ the number of people arriving during $T$.

For the first part the probability of $k$ people buying something is: $$ \Pr(B(T)=k\mid N(T)=n)=\binom{n}{k}\theta^k(1-\theta)^{n-k}. $$ Note that you need $\binom{n}{k}$ to choose $k$ people out of $n$ people who are going to buy something.

For the second part you can use Bayesian decomposition of the probability and use the fact that $N(T)$ is a Poisson RV: $$ \Pr(B(T)=k)=\sum_{n=1}^\infty \Pr(B(T)=k|N(T)=n)\Pr(N(T)=n)\\ =\sum_{n=k}^\infty \binom{n}{k}\theta^k(1-\theta)^{n-k}\frac{e^{-\lambda T}(\lambda T)^n}{n!}\\ =\sum_{n=k}^\infty \theta^k(1-\theta)^{n-k}\frac{e^{-\lambda T}(\lambda T)^n}{k!(n-k)!}\\ =\frac{(\lambda\theta T)^k}{k!}\sum_{n=k}^\infty \frac{e^{-\lambda T}(\lambda T(1-\theta))^{n-k}}{(n-k)!}\\ =\frac{(\lambda\theta T)^ke^{-\lambda T\theta}}{k!} $$ where $\lambda$ is the intensity of PP.

Arash
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