Poisson distribution, no call on 3 days out of 5 between 10.00am and 10.10am

Hi, im having some issues with the following question.

Question) A sales manager receives 6 phone calls on average between 9.30am and 10.30am on a weekday. Find the probability that during a 5 day working week, there will be exactly 3 days on which she receives no calls between 10am and 10.10am.

X ~ Po(6) for a single day. Am i correct in assuming i should be using nCr to determine the possible combination of 3 days in 5. I tried to multiple this by the probability of no calls for a single day ie X ~ Po(1) due to the fact that the time period is 1/6 that originally stated but didnt get the correct answer.

correct answer is 0.199 btw

Any help would be appreciated.

Last edited:


Ambassador to the humans
This is sort of a multi-part problem it sounds like.

You need to first just figure out the probability of getting no call during that time frame (you'll need to use a poisson with a different lambda).

Once you know that you need to use the binomial distribution to actually answer the question.
Your a legend, thanks very much!

For anyone else who is curious below is the working.

Lambda = 6 for 60min therefore lambda for 10mins = (10/60)*6 = 1

P(No calls between 10:00 to 10:10) = (e^-1)((1^0)/(0!)) => e^-1 = 0.368

As Dason has pointed out you then need to use the Binomial distribution. To use this we need p,q,r and n

p = 0.368
q=1-p = 0.632
n = 5 (5 days)
r = 3 (3 days when no calls occur)

P(no calls) = (nCr)(p^r)(q^(n-r)) => 10 x (0.368^3)(0.632^2) = 0.199