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So for a group a 7 people, find the probability that all of their birthdays do not occur in the winter. That is, all of their birthdays occur either in the spring, summer or fall. Assume that the probability of being born in each season is equally likely.

So the answer to this one is pretty simple as it is just $\frac{3}{4}^7=0.133$. However, I thought I would try it with a different method. I did it by using the stars and bars counting method.

I counted how many ways there are to arrange 7 people into 3 seasons and also how many ways there are to arrange 7 people into 4 seasons.

Ie. $$P(no\; birthdays\; in\; the\; winter) = \frac{\binom{7+3-1}{3}}{\binom{7+4-1}{4}}=0.4$$

Why is this not getting the same answer?

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    $\begingroup$ Not every arrangement has the same probability to occur. More simple think of a couple that gets two children. Splitting up in boys and girls there are $3$ arrangements: BB,BG,GG. But BG has probability $\frac12$ to occur and the others have probability $\frac14$ to occur. $\endgroup$ – drhab May 21 '16 at 7:23
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As I understand the question is about the number of bdays rather than distinct people. In such case we select for example 2 distinct days for 2 birthdays but do not need to multiply by 2 (people) because birthdays are identical (balls of the same color).

If im right the solution would be $\frac{\binom{281}{7}}{\binom{371}{7}}$

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  • $\begingroup$ Hmmm, that's a very interesting way to think of it and the answer is very close. Though one of the issues is that you can't split 365 days evenly. I think if the question didn't have the assumption that the 4 seasons are equally as likely, your solution would be the one to use. That is, there are about 90 days in winter each year, so we get $\frac{\binom{275}{7}}{\binom{365}{7}}=0.135$. This is really close. $\endgroup$ – jlcv May 21 '16 at 16:42
  • $\begingroup$ Nevermind, if there are 90 days of winter each year it would just be $\frac{275}{365}^7=0.1378$. $\endgroup$ – jlcv May 21 '16 at 16:58
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There are $3^7$ ways to assign $3$ seasons to $7$ people. $\binom{7+3-1}{3}$ counts multisubsets, which assumes that the people can't be distinguished, and isn't appropriate for this problem.

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