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I've the following problem.

In a class, there are $x$ female and $y$ male students. Their birthdays are uniformly distributed over $N = 365$.

  1. What's the probability that there are at least two different female students with same birthday ?
  2. What's the probability that there is at least one female-male pair of students who share the same birthday given that no female students share a birthday ?
  3. What's the probability that at least one female student shares same birthday with another student ?

My attempt

  1. $Pr(\text{At least 2 fem. share a bday}) = 1 - Pr(\text{No fem. share a bday})$ where $$ Pr(\text{No fem. share a bday}) = \frac{N\times(N-1)\times \dots (N-x+1)}{N^x} $$

  2. $Pr(\text{At least one pair has the same bday}) = 1 - Pr(\text{no pair has the same bday})$ where

$$ Pr(\text{no pair has the same bday}) = \prod_{i = 1}^y\prod_{j = 1}^xPr(\text{Male i do not share a bday with fem. j}) $$ and

$$ \prod_{i = 1}^y\prod_{j = 1}^xPr(\text{Male i do not share a bday with fem. j}) = (\frac{N-1}{N})^{xy}$$ Of course I use the fact that there are $x$ distinct female birthdays under the given condition.

  1. $Pr(\text{A female shares a bday with another student}) = Pr(\text{At least 2 fem. share a bday}) + Pr(\text{A female shares a bday with a male|No 2 fem. share a bday)} = Answer1 + Answer 2 $

I've basically 2 questions regarding this:

  • Is it correct ?
  • How can I simplify it ? I feel much more comfortable when I think in terms of negation but I feel there's a simpler approach to this problem.
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1 Answer 1

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  1. is straightforward textbook stuff

  2. it is given that no two females have the same birthday. So the females leave $(365-x)$ birthdays open. The chance that no male birthday 'hits' this set is $1-({\frac{365-x}{365}})^y$

  3. 1 - probability that no two females share the same birthday times the probability that all males have birthdays in the set of birthdays that are left open.

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  • $\begingroup$ I see. This is indeed a better approach. However, is mine correct as well ? If not, what is flawed in my logic ? $\endgroup$
    – SpiderRico
    Dec 4, 2016 at 21:42
  • $\begingroup$ I think the flaw is that for male $i$ the probability of sharing birthday with $j_1$ and $j_2$ is conditional. E.g. if it is $j_1$ it can't also be $j_2$, and if it is not $j_1$, $j_2$'s chance is increased. $\endgroup$
    – Pieter21
    Dec 4, 2016 at 21:54

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