# Poisson distribution probability problem

Seneka
Homework Statement:
The number of mistakes a teacher makes while marking homework has a poisson distribution with mean of 1.6 errors per piece of homework. Find the probability that in a class of 12 pupils fewer than half of them have errors in their marking.
Relevant Equations:
P(X=r)= (e to the power - lambda)(lamba to the power r) all divided by r factorials
where lamba is the average rate of the event.
the expected and variance are both equal to the average rate of the event.
Mentor edit: ##P(X = r) = e^{-\lambda}\frac{\lambda^r}{r!}##
LaTeX script: # #P(X = r) = e^{-\lambda}\frac{\lambda^r}{r!}# #
So I thought you would find the probability of having 0 errors when the mean rate is 1.6. Square that by 5 and multiply that by one minus the probability of having 0 errors to the power of 7. So that is basically the probability of having 0 errors to the power of 5 multiplied by the probability of having one or more errors to the power of 7 for the 7 students that do get errors in their marking. This gave me 6.919...x10 to the power -5 which is not the answer.

Homework Helper
First, you are calculating the probability that exactly 5 have no errors. The question asks for the probability that fewer than half have errors, that is the total probability that the number with no errors is from 7 to 12.

Homework Helper
Sorry, some problem posting reply. Your calculation for 5 with no errors is wrong - which 5? You need to multiply by the number of combinations of 5 from 12.

Seneka
Sorry, some problem posting reply. Your calculation for 5 with no errors is wrong - which 5? You need to multiply by the number of combinations of 5 from 12.

Are you suggesting doing a cumulative binomial distribution up to 5 where the probability of no error is e to the power -1.6?

Homework Helper
No - from 7 to 12. Read the question - fewer than half have errors. I mentioned 5 because that was the calculation you did, but you need 7 to 12.

Seneka
No - from 7 to 12. Read the question - fewer than half have errors. I mentioned 5 because that was the calculation you did, but you need 7 to 12.
Yeah I realized but now I am getting 2.03..x10 to the power -34
My calculation as 12choose7(e to the power -1.6) to the power 7 multiplied by (1-(e to the power -1.6)) to the power of 5 and carried this pattern until I came to 12...is this wrong?

Seneka
Yeah I realized but now I am getting 2.03..x10 to the power -34
My calculation as 12choose7(e to the power -1.6) to the power 7 multiplied by (1-(e to the power -1.6)) to the power of 5 and carried this pattern until I came to 12...is this wrong?
I think it was an error. I got the right answer when I used my calculator to find the probability of less than and equal to 6 no errors and then subtracting from 1.

Seneka
No - from 7 to 12. Read the question - fewer than half have errors. I mentioned 5 because that was the calculation you did, but you need 7 to 12.
Thanks