Hello. I recently had this question on my Midterm "Suppose the number of students coming to
Thomas’s Monday office hours follows a Poisson
distribution with mean 5. Further suppose that 30%
of these students wear glasses. What is the probability
that for the next Monday office hours, 6 students
will come and at most 3 of them wear glasses?"
How would I go about solving this? The answer is supposed to be around 0.136. I don't want to miss this again on my final :(
Poisson distribution (m = mean):
P(x) = e^(-m) m^x / x!
Find P(0), P(1), P(2) and P(3). Add together for the probability.
Here's P(0):
P(0) = (e^-5) (5^0) / 0!
= .0067
Here's P(1):
P(1) = (e^-5) (5^1) / (1!)
= (.0067) (5) / (1)
= .0336
Here's P(2):
P(2) = (e^-5) (5^2) / (2!)
= (.0067) (25) / (2)
= .0837
Here's P(3):
P(3) = (e^-5) (5^3) / (3!)
= (.0067) (125) / (6)
= .1395
Adding: .0067 + .0336 + .0837 + .1395 = .2635 -->this is the probability.
I get a different answer, so double check my calculations.