Poisson Customers

Probability Level 2

Damon is working the evening shift at the register of his retail job. There are currently two registers open, but his coworker is about to go home for the day and close her register.

The number of customers approaching each register is an independent Poisson random variable. If each register was getting an average of 2 customers per minute, what is the probability that Damon will have more than 4 customers approaching his register in minute after his coworker goes home?

Round your answer to 3 decimal places.

The answer is 0.371.

1 solution

Andy Hayes
Nov 15, 2016

Since the numbers of customers are independent Poisson random variables, the sum of these variables is also a Poisson random variable with $\lambda=2+2=4.$

The probability of having up to 4 customers in the next minute is calculated below:

$\begin{aligned} P(X=0) &= \frac{4^0 e^{-4}}{0!}\approx 0.0183 \\ \\ P(X=1) &= \frac{4^1 e^{-4}}{1!}\approx 0.0733 \\ \\ P(X=2) &= \frac{4^2 e^{-4}}{2!}\approx 0.1465 \\ \\ P(X=3) &= \frac{4^3 e^{-4}}{3!}\approx 0.1954 \\ \\ P(X=4) &= \frac{4^4 e^{-4}}{4!}\approx 0.1954 \\ \\ P(X\le 4) &= \sum\limits_{k=0}^{4}P(X=k) \approx 0.629 \end{aligned}$

Then, the probability that he will have more than 4 customers in the next minute is:

$P(X>4)=1-P(X \le 4)\approx \boxed{0.371}$

Poisson Customers

The answer is 0.371.

1 solution

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