Suppose there are three persons: A, B and C (randomly picked from a cohort which has the same birth year). Now three mutually exclusive events can happen: $X$: all of them born on different dates, the probability of which is: $$P(X)=\frac{365\cdot364\cdot363}{365^3}.$$

The second event $Y$ is that all of them have same birthday. If I fix the birthday of A to January 1, this can happen in: $\frac{1}{365^3}$ ways. But A can be born on any of the $365$ days, so the probability is: $$P(Y)=\frac{365}{365^3}.$$

The third event $Z$ is when exactly two of them have the same birthday. If we club two persons together, say AB, then AB is born on any day and C on another. This can happen in $365\cdot364$ ways. But there can be three such pairs, so the required probability seems to be: $$P(Z)=\frac{3\cdot365\cdot364}{365^3}.$$

Problem is, these three probabilities are not summing up to 1. Kindly point out if there is a mistake in logic/calculation. If this is a duplicate question (I am not sure of this), I shall be glad if you point that out as well.


closed as off-topic by Peter Flom Sep 3 '18 at 12:32

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  • $\begingroup$ I think these probabilities do sum up to 1, because 365*364*363 + 365 + 3*365*364 = 48627125 = 365^3. Do you agree or did I misunderstand something? $\endgroup$ – fabiob Sep 3 '18 at 12:08
  • $\begingroup$ ah, and I think your solution is correct. $\endgroup$ – fabiob Sep 3 '18 at 12:09
  • $\begingroup$ I'm voting to close this question as off-topic because a comment pointed out that the problem is not a problem. $\endgroup$ – Peter Flom Sep 3 '18 at 12:32

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