Given $20$ people, what is the probability that, among the $12$ months
in the year, there are $4$ months containing exactly $2$ birthdays and
$4$ months containing exactly $3$ birthdays?
I am less concerned about knowing the answer than making sure I have thought about the problem correctly.
The general problem seems to be finding the distinct integer-valued vectors that satisfy
$$x_1 + x_2 + … x_{12} = 20\ \ \ \ \ \ \ \ \ x_i \ge 0, i = 1, …, 12$$
It can be proven that there are $\binom{20+12-1}{12-1}$ such vectors. I am presuming this to be the sample space. (Please correct me if I'm wrong)
$4\cdot 2 + 4\cdot 3 + 4 \cdot 0= 20$, so there are $3$ distinct groups of $4$ months. Therefore we have $\binom{12}{4\ 4\ 4}$ possible ways to get the desired distribution of months.
The probability I get is $$\frac{\binom{12}{4\ 4\ 4}}{\binom{20+12-1}{12-1}} $$. The answer in the back of the textbook is $1.0604 \times 10^{-3}$ so my answer is off. I just want to know where I went wrong.
Best Answer
Your count of the sample space treats the $20$ people as indistinguishable, when in fact they are not. The actual size of the sample space is $12^{20}$: each of the $20$ people can have a birthday in any of $12$ months. (We are making the simplifying assumption that the probability of being born in any given month is $\frac1{12}$; this is in fact false.)
To get the correct numerator, observe (as in effect you did) that there are $\binom{12}4$ ways to choose the $4$ months containing exactly $2$ birthdays and then $\binom84$ ways to choose the $4$ months containing exactly $3$ birthdays, so there are $\binom{12}4\binom84=\binom{12}{4\;4\;4}$ ways to choose the months. Then, however, the $20$ people have to be assigned to these months, and there are $$\binom{20}{3\;3\;3\;3\;2\;2\;2\;2}=\frac{20!}{12^4}$$ ways to do that. The desired probability is therefore
$$\frac{\binom{12}8\binom8420!}{12^{24}}\approx0.00106042\;.$$