Probability of Poisson event happening twice, consecutively

[cmkluza]

Homework Statement

The number of telephone calls, T, received each minute can be modeled by a Poisson distribution with a mean of 3.5.

Find the probability that at least three telephone calls are received in each of two successive one-minute intervals.

Homework Equations

P = \frac{e^{-μ}μ^x}{x}

The Attempt at a Solution

I realize that I can calculate the probability of getting three phone calls during one minute using \frac{e^{-μ}μ^x}{x} = \frac{e^{-3.5}3.5^3}{3} (or, more simply, by using a calculator), but I don't currently have any intuition on how to find the probability of finding the probability of this happening twice, consecutively. Can anyone give me any suggestions on how to think about this in order to arrive at an answer? Thanks!

 

[Ray Vickson]

Homework Statement

The number of telephone calls, T, received each minute can be modeled by a Poisson distribution with a mean of 3.5.

Find the probability that at least three telephone calls are received in each of two successive one-minute intervals.

Homework Equations

P = \frac{e^{-μ}μ^x}{x}

The Attempt at a Solution

I realize that I can calculate the probability of getting three phone calls during one minute using \frac{e^{-μ}μ^x}{x} = \frac{e^{-3.5}3.5^3}{3} (or, more simply, by using a calculator), but I don't currently have any intuition on how to find the probability of finding the probability of this happening twice, consecutively. Can anyone give me any suggestions on how to think about this in order to arrive at an answer? Thanks!

You are not asked about 3 calls in each minute; you are asked about at least 3 calls in each minute; that is, in each minute the number of calls is 3 or 4 or 5 or 6 or ... .

 

[haruspex]

After you have addressed Ray's point, consider whether the two events (calls in first minute, calls in second minute) are independent or correlated.

 

[cmkluza]

You are not asked about 3 calls in each minute; you are asked about at least 3 calls in each minute; that is, in each minute the number of calls is 3 or 4 or 5 or 6 or ... .

Thanks, looks like I read the question a little too quickly. Anyhow, at least 3 just changes it to a cumulative distribution function, right?

After you have addressed Ray's point, consider whether the two events (calls in first minute, calls in second minute) are independent or correlated.

It would appear that the events are independent. Based on the old example of tossing a coin, I'd guess that I find the probability of getting at least 3 calls, and square it?

 

[haruspex]

Thanks, looks like I read the question a little too quickly. Anyhow, at least 3 just changes it to a cumulative distribution function, right?
It would appear that the events are independent. Based on the old example of tossing a coin, I'd guess that I find the probability of getting at least 3 calls, and square it?

Yes.

 

[cmkluza]

Yes.

Thanks for your help!