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Poisson provides an estimate to binomial problems. To understand Poisson distribution better, would it be possible to solve this poisson problem using the binomial approach?

Approximately 80,000 marriages took place in the state of New York last year. Estimate the probability that for at least one of these couples (a) both partners were born on April 30;

Solution. (a) The probability that an arbitrary couple were both born on April 30 is, assuming independence and an equal chance of having being born on any given date, $(\frac{1}{365})^2$. Hence, the number X of such couples is approximately Poisson with mean $\frac{80000}{365^2} =0.6$. Therefore, the probability that at least one pair were born on this date is approximately $1-e^{-.6}$

(b) both partners celebrated their birthday on the same day of the year. $1 -e^{-219.18} =1$

Now, is it possible to arrive the same conclusion solving with the Binomial Approach?

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up vote 1 down vote accepted

Yes: the probability is $1$ minus the probability of no couple sharing that birthday.

$$1-\left(1-\frac{1}{365^2}\right)^{80000} \approx 0.451457$$

$1-e^{-0.6}$ gives a similar value. $1-e^{-80000/365^2}$ is even closer.

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Hi Henry, question (b) asks (b) both partners celebrated their birthday on the same day of the year. the solution says The probability that an arbitrary couple were born on the same day of the year is $\frac{1}{365}$. why is that? why isn't it the same $\frac{1}{365^2}$ – user133466 Oct 6 '12 at 23:14
@user133466: Because there are 365 different days they can share a birthday so you need to look at $365\times \frac{1}{365^2}$ – Henry Oct 6 '12 at 23:16
i got it! going back to question (a) i tried to find the prob $(\frac{364}{365})^{80000}$ why is that wrong? – user133466 Oct 6 '12 at 23:19
ive added the second part of the problem, can you help convert that into binomial? Thank you very much!! – user133466 Oct 6 '12 at 23:39
For the second part you would want $1-\left(1-\frac{1}{365}\right)^{80000} = 1-\left(\frac{364}{365}\right)^{80000}\approx 1- 4.8\times 10^{-96}$ which is very close to $1$. – Henry Oct 7 '12 at 8:52

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