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Patients arrive randomly and independently at a doctor's clinic from 8·0 A.M. at an average rate of one in Five minutes.The waiting room holds 12 persons. What is the probability that the room will be full when the doctor arrives at 9·0 A.M.?

I know how to solve some Poisson distribution based problem. But for this one I am not sure what am I missing. Can someone please help to exactly where am I wrong. According to me: Lambda = 12/hour. Number of events (desired)= 12 so Probability(12 arrivals in an hour) = e^(-12)*(12^12)/(12!). (not confident in this method though) Please help.

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  • $\begingroup$ Why are you computing the probability of 12 arrivals in an hour? $\endgroup$
    – Juho Kokkala
    Commented Sep 30, 2017 at 19:26
  • $\begingroup$ Because, this is how the queue would be full, right. $\endgroup$
    – user179025
    Commented Sep 30, 2017 at 19:27
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    $\begingroup$ This question requires you to apply common sense to characterize the event "the room is full." What happens if a 13th person arrives before 9:00 am? A 14th? What if a crowd of 30 arrives (it's possible, although unlikely)? $\endgroup$
    – whuber ♦
    Commented Sep 30, 2017 at 19:41

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@whuber is right, the probability of the event "the room is full" is the probability of 12 persons or more entering the waiting room. This is the same as calculating the complement of the event of 11 persons. Below I illustrate with some R code. 

# rate is 1/5 = 12/60
d <- 12

# set lower.tail = FALSE to calculate the complement event
p <- ppois(q = 11, lambda = d, lower.tail = FALSE)
format(p, scientific = FALSE)
# [1] "0.5384027"
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