How to Calculate Crime Probability Using Poisson and Binomial Distributions?

[zzod]

Hey guys, I'm kind of stuck on this question.

In a certain town, crimes occur at a Poisson rate of 2.4 per month (i.e. according to a Poisson process with a rate of 2.4 per month). What is the probability of having exactly 2 months (not necessarily consecutive) with exactly 4 crimes during the next year? Assume that every month has the same length.

I know that first you have to find the probability of 4 crimes in 2 months using Poisson distribution then use binomial distribution to answer the question. But I'm not sure how to do the first part! >.<

 

[moonman239]

Hey guys, I'm kind of stuck on this question.

In a certain town, crimes occur at a Poisson rate of 2.4 per month (i.e. according to a Poisson process with a rate of 2.4 per month). What is the probability of having exactly 2 months (not necessarily consecutive) with exactly 4 crimes during the next year? Assume that every month has the same length.

I know that first you have to find the probability of 4 crimes in 2 months using Poisson distribution then use binomial distribution to answer the question. But I'm not sure how to do the first part! >.<

If the rate is 2.4 per month, then the probability of a crime happening in any given month = 1 - (1/2.4) (or so I think) = 58.33% approximately. Then the probability of four crimes in two months = the probability of exactly one crime in two months^4.

 

[CRGreathouse]

Poisson chance of 4 crimes in a month is
$$\frac{2.4^4e^{-2.4}}{4!}$$. Use binomial from here.

 

[zzod]

Poisson chance of 4 crimes in a month is
$$\frac{2.4^4e^{-2.4}}{4!}$$. Use binomial from here.

I don't think this would work as the question is asking for the probability of exactly 4 crimes occurring in 2 months.

 

[The Investor]

Poisson chance of 4 crimes in a month is
$$\frac{2.4^4e^{-2.4}}{4!}$$. Use binomial from here.

This is correct. You do mean 4 crimes occurring in each of 2 months right? If not you need to word the question more clearly.