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[SOLVED] Another Birthday Probability Problem
Given 20 people, what is the probability that among the 12 months in the year there are 4 months containing exactly 2 birthdays and 4 containing exactly three birthdays.
Axioms and basic theorems of probability.
The sample space of this problem is a set of 20-tuples (n_1, ..., n_20) where n_i is an integer from 1 to 12 representing the month person i was born in. The sample space has u = 12^20 elements.
Divide the people into 8 distinct groups, 4 of 2 and 4 of 3. This can be done in
[tex]v = \frac{20!}{2!^4 3!^4}[/tex]
ways. Now each of these 8 groups needs to be assigned to a distinct month. Pick any ordering of the groups. The first group has 12 months to choose from, the second has 11 months to choose from, etc. There are thus w = 12 * 11 * ... * 5 = 12!/4! assignments.
The probability sought is vw/u. Unfortunately, when I calculate this I get a different answer than the book's 1.0604 x 10^-3. Where could I have gone wrong?
Homework Statement
Given 20 people, what is the probability that among the 12 months in the year there are 4 months containing exactly 2 birthdays and 4 containing exactly three birthdays.
Homework Equations
Axioms and basic theorems of probability.
The Attempt at a Solution
The sample space of this problem is a set of 20-tuples (n_1, ..., n_20) where n_i is an integer from 1 to 12 representing the month person i was born in. The sample space has u = 12^20 elements.
Divide the people into 8 distinct groups, 4 of 2 and 4 of 3. This can be done in
[tex]v = \frac{20!}{2!^4 3!^4}[/tex]
ways. Now each of these 8 groups needs to be assigned to a distinct month. Pick any ordering of the groups. The first group has 12 months to choose from, the second has 11 months to choose from, etc. There are thus w = 12 * 11 * ... * 5 = 12!/4! assignments.
The probability sought is vw/u. Unfortunately, when I calculate this I get a different answer than the book's 1.0604 x 10^-3. Where could I have gone wrong?