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If there are 12 strangers in a room, whats the probability that no two of them celebrate their birthday in the same month??

Try

Im trying to solve this by using the binomial distribution. Let $X$ be the number of couples that celebrate their birthday in the same month. we want to calculate $P(X=0)$.

Notice that this is binomial because each couple is a trial and we get either they have same birthday or no. The number of couples is ${12 \choose 2 } = 66$ and the probability of success is

$$ P = 12 \frac{1}{12} \cdot \frac{1}{12} = \frac{1}{12} $$

Threfore, according to the binomial, we have

$$ P(X=0) = {66 \choose 0} \frac{1}{12}^0 (\frac{11}{12} )^{66} = \boxed{0.003} $$

which differs from the book answer which is $0.00005$. What is the mistake here? To me it seems correct my answer to this problem.

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  • $\begingroup$ The error is that the events you are considering (person $a$ and $b$ don't share a birthday, person $a$ and $c$ don't share a birthday, etc...) are not independent events which is a requirement to use the binomial distribution in the first place. This should be obvious considering that if $a$ and $b$ share a birthday and $b$ and $c$ share, so too must $a$ and $c$. $\endgroup$
    – JMoravitz
    Oct 17, 2018 at 4:22

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The binomial distribution fails to apply here because there is no event with only two outcomes that occurs repeatedly and independently, with the same probability of success each time. Instead we put the strangers into the room sequentially:

  • The first one can have any birth month out of 12. The probability that no two in the room share a birth month is $\frac{12}{12}$.
  • The second can have any birth month but that of the first stranger. The probability of this is $\frac{11}{12}$.
  • The third can have any birth month but that of the first two. The probability is $\frac{10}{12}$, etc.

We continue until the last stranger, where the associated probability is $\frac1{12}$, then multiply to get the correct result of $\frac{12!}{12^{12}}=0.000537$.

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  • $\begingroup$ got it! thank you! $\endgroup$ Oct 17, 2018 at 4:38
  • $\begingroup$ There is an event that occurs repeatedly with same probability each time. The problem was not that. The problem was that the events were not independent. $\endgroup$
    – JMoravitz
    Oct 17, 2018 at 4:40

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