Thread: Poisson random variable

Suppose that the number of accidents occurring on a highway each day is a Poisson random variable with the expected number of accidents a day equaling 3. Find the probability that 3 or more accidents occur today. Then, find the probability that 3 or more accidents occur today under the assumption that at least 1 accident occurs today.

Originally Posted by holly123

Suppose that the number of accidents occurring on a highway each day is a Poisson random variable with the expected number of accidents a day equaling 3. Find the probability that 3 or more accidents occur today. Then, find the probability that 3 or more accidents occur today under the assumption that at least 1 accident occurs today.

The first part is simple substitution into the pmf: .

The second part is conditional probability: . Again, all that's required now is simple substitution into the pmf. Note that .

If you need more help, please show your work and say where you get stuck.

The whole "Poisson" thing is throwing me off. How do I find probabilities such as X less than or equal to 2?

Originally Posted by holly123

The whole "Poisson" thing is throwing me off. How do I find probabilities such as X less than or equal to 2?

Have you studied the Poisson distribution? It is odd that you would be given a question about it if you have not studied it.

Have you looked up the pmf in your textbook or used Google?

We just started it today so I'm very new to it. I'm trying looking at my book and google but I think it's only confusing me more. Google says probability is P(X = k) = e ^(- µ) * µ^k / k!

Originally Posted by holly123

[snip]
Google says probability is P(X = k) = e ^(- µ) * µ^k / k!

Yes. Now apply it.

So 1- Pr(0) + Pr(1) + Pr(2)= 1-(e^-3 +3e^-3 + 9/2 e^-3)
Is that right so far?

Originally Posted by holly123

So 1- Pr(0) + Pr(1) + Pr(2)= 1-(e^-3 +3e^-3 + 9/2 e^-3)
Is that right so far?

Yes. (Why would it not be?)

Originally Posted by mr fantastic

Yes. (Why would it not be?)

Hi

I'm having the same problem trying to grasp Poisson.
I don't see how you got from

P(X = k) = e ^(- µ) * µ^k / k!

to

1- Pr(0) + Pr(1) + Pr(2)= 1-(e^-3 +3e^-3 + 9/2 e^-3)

could you please explain

Originally Posted by unterglayben

Hi

I'm having the same problem trying to grasp Poisson.
I don't see how you got from

P(X = k) = e ^(- µ) * µ^k / k!

to

1- Pr(0) + Pr(1) + Pr(2)= 1-(e^-3 +3e^-3 + 9/2 e^-3)

could you please explain

Did you try substituting the given mean and the values of k into the pmf?

Hi

Thanks for getting back to me.

I got this online for the pmf of a poisson.



I still don't see how you get

1- Pr(0) + Pr(1) + Pr(2)= 1-(e^-3 +3e^-3 + 9/2 e^-3)

could you provide me with the formula into which I sub the values please. I see that on the left it's each value for the probability up to and including 3 and then subtract from 1 to get probability of greater than or equal to 4, but the 9/2 on the right hand side of the equation? No idea where that came from.

Sorry, I was using the values from my problem. I meant, I see the left side as being the values to figure out less than or equal to 2.

Originally Posted by unterglayben

Hi

Thanks for getting back to me.

I got this online for the pmf of a poisson.



I still don't see how you get

1- Pr(0) + Pr(1) + Pr(2)= 1-(e^-3 +3e^-3 + 9/2 e^-3)

could you provide me with the formula into which I sub the values please. I see that on the left it's each value for the probability up to and including 3 and then subtract from 1 to get probability of greater than or equal to 4, but the 9/2 on the right hand side of the equation? No idea where that came from.

Substitute and k = 2 into the pmf.