Trying to figure out what the probability is that in a room of 200 people what the probability that at least one will get a phone call during a certain time window... In this case 2 hours


  • Average person gets 5 calls a day (distributed randomly over 16 hours).
  • Those calls happen during a 16 hour time window.
  • Movie length = 2 hours (120 minutes)

If there was only one person in the room, it's pretty easy to calculate...

If caller only got 1 call a day, the chances of it happening while in the movie would be 1/(16/2) = 1/8.

= 5 calls/day * 1/8 = 5/8

But, now how do I then apply this if there is 200 people in the room. I think I need to do some type of binomial coefficient?

Real Life Application My actual application is to determine what the likelyhood of someone getting a call is during a 5 minute presentation at a conference. But, thought the movie example was more universal.

  • $\begingroup$ Your 5/8 argument is not quite correct. $\endgroup$ Apr 5 '13 at 16:52
  • $\begingroup$ This is similar to the "Birthday Problem," about which so much has been asked. math.stackexchange.com/search?q=birthday+problem $\endgroup$ Apr 5 '13 at 16:53
  • $\begingroup$ Whenver you see "at least", think "Unity - none" $\endgroup$
    – DJohnM
    Apr 5 '13 at 19:42
  • $\begingroup$ @HagenvonEitzen How so? $\endgroup$
    – blak3r
    Apr 7 '13 at 20:18
  • $\begingroup$ @blak3r With 9 calls per day, you would obtain $\frac 98>1$. $\endgroup$ Apr 7 '13 at 20:45

Think about the complementary event $$ E^c = \{ \text{nobody gets a phone call during the movie} \} $$ Assuming that the events of each person getting a call are independent, then the joint probability of everybody not getting a call is the product of the individual probabilities. $$ P(E^c) = \left( \frac{3}{8} \right)^{200} \approx 6.40108 \cdot 10^{-86} \approx 0. $$ In other words, there's essentially a $100\%$ probability that somebody gets a call.


We look at the problem for numbers in a more reasonable range, where the answer is less obvious.

Using information about the number of people, the length of the presentation, and the mean number of calls a person receives per day, you can calculate the expected number $\lambda$ of calls in your time interval. In the movie case, your estimate of $\lambda$ is $\dfrac{5}{8}\cdot 200=125$.

A common model would be that the total number $X$ of calls is a random variable that has Poisson distribution with parameter $\lambda$. The probability $\Pr(X=k)$ of exactly $k$ calls is then given by $$\Pr(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}.$$ In particular, the probability of no calls is $e^{-\lambda}$. With your numbers, for the $5$ minute length and $200$ people, the probability of no calls should be around $e^{-125/24}$, about $0.0055$.

  • $\begingroup$ Explanation of why I accepted this answer: I thought all answers had great points. I thought Andre's was the most clear. Each generated a slightly different outcome and i'm not a mathematician so keep that in mind. Thanks everyone for the responses. $\endgroup$
    – blak3r
    Apr 8 '13 at 4:24
  • $\begingroup$ We need to remember that these are probabilistic models of a complex situation. Such a model should not be identified with "the truth," even in a high accuracy field such as Physics. And certainly not in your phone call situation. As an example, the people at your conference are presumably colleagues of sorts, and tend to call each other. But they certainly will not during the talk! $\endgroup$ Apr 8 '13 at 4:42

The porbability that John Doe receives all his five calls in the remaining 14 hours, is $(\frac{14}{16})^5$. The probability, that all 1000 calls of all people occur then, is $(\frac{14}{16})^{1000}\approx10^{-58}$, hence it is nearly certain that at least one phone will disturb your cineastic pleasure.

It would already be quite unusual to have no phone call for 50 people during any 15 minutes (happens only with $2\,\%$).

With a 5 minute presentation for 15 people, you already have a 2/3 chance of silence. :)


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