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Background

A typical Poisson process denoted by $N_t$ refers to the number of arrivals (discrete) at time $t$, with rate $\lambda$, where

$$P(N_t = k) = \frac{e^{-\lambda t} (\lambda t)^k}{k!}$$

Question

A department store has three doors. Arrivals at each door form Poisson processes with rates $\lambda_1 = 110, \lambda_2 = 90, \lambda_3 = 160$ customers per hour. 30% of all customers are male. The probability that a male customer buys something is .8, and the probability of a female customer buying something is .1.

What is the probability that the third female customer to purchase anything arrives during the first 15 minutes? What is the expected time of her arrival?

I think I know how to approach one door but not three. Consider the first door. Then $\lambda = 110, t = .25$, and $k = 3$. Then

$$P = \frac{e^{-27.5}27.5^3}{3!}(.7)(.1)$$

How do you approach the other doors? It is not additive is it?

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1 Answer 1

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Q1 Hint. Let $X_t$ be the number of women that have entered and made a purchase up to time $t$. Then, $X_t\sim\operatorname{Poisson}(\lambda t p q)$ where $\lambda \equiv \lambda_1 + \lambda_2 + \lambda_3 = 360$, $p = 0.7$ is the probability that a customer is female, and $q = 0.1$ is the probability that a female customer purchases an item.

Q2 Hint. Let $\tau$ be the first time at which the third woman who purchases something enters the store. Let $\Phi(t)\equiv\mathbb{P}(\tau\leq t)=\mathbb{P}(X_{t}\geq3)$. Then, $\mathbb{E}\tau=\int_{0}^{\infty}t\Phi^\prime(t)dt$.

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  • $\begingroup$ For Q2 would you take $1 - P(X = 0) - P(X = 1) - P(X = 2)$ and find the derivative with respect to $t$? $\endgroup$
    – Vahan
    Jul 23, 2019 at 1:04
  • $\begingroup$ That works. If you already know the CDF of the Poisson process you can also just use the fact that $1 - \mathbb{P}(X_t = 0) - \mathbb{P}(X_t = 1) - \mathbb{P}(X_t = 2) = 1 - \mathbb{P}(X_t \leq 2) = 1 - \Phi_{X_t}(2)$. $\endgroup$
    – parsiad
    Jul 23, 2019 at 1:17

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