Calculating bounds with multiple random variables. I have this problem: 
Suppose there are 4 students (who we'll refer to as A, B, C, and D) in a class and each student is equally likely to have been born in any of the twelve months of the year. For any subset T of the students, RT be the event that that all the students in T are born in January. (Note that the event RT doesn't imply that students that are not in T are not born in January.) For example R{A,B} is the event that students A and B were both born in January. Let X be the random variable corresponding to the number of students born in January.
(RT here is supposed to be with T as a subscript but I do not know how to do that on stack)
I am trying to figure out this following problem: 
Let Xi be the number of students born on the ith month of the year, e.g., X1 = X, 
and let Y = max(X1, X2, X3, ... , X12). What's the exact value of P(Y >= 3)?
I am really stuck on how to do begin this process. I was thinking that I should find the expected value of each X1,X2,X3...X12. But I am not sure exactly how to do this/if I have to do that. 
Any help is appreciated thank you. 
 A: There are many symbols, but there are only $4$ students. We want $\Pr(Y\ge 4)$, which is $\Pr(Y=3)+\Pr(Y=4)$.
We go after $\Pr(Y=4)$ first, since it is easier. We want the probability the students were all born in the same month. The month can be chosen in $\binom{12}{1}$ ways. If all months are equally likely (unlikely!), and we have independence, the the probability the students were all born in a given month is $\left(\frac{1}{12}\right)^4$. Thus $\Pr(Y=4)=\binom{12}{1}\left(\frac{1}{12}\right)^4$. 
For $\Pr(Y=3)$, the lucky month with $3$ births can be chosen in $\binom{12}{1}$ ways. The lucky $3$ students who were born that month can be chosen in $\binom{4}{3}$ ways. The probability these students were born in that month is $\left(\frac{1}{12}\right)^3$. The remaining month can be chosen in $\binom{11}{1}$ ways, and the probability the remaining student was born that month is $\left(\frac{1}{12}\right)$. Thus $\Pr(Y=3)=\binom{12}{1}\binom{4}{3}\left(\frac{1}{12}\right)^4\binom{11}{1}\left(\frac{1}{12}\right)$.
Finally, calculate $\Pr(Y=4)+\Pr(Y=3)$.
Remark: We can do the same thing in what looks superficially another way. Have the students write down their birth month, in the order A then B then C then D. Under our assumptions, all $12^4$ possible strings are equally likely.
Now we need to count the favourables, the number of ways that the sequence written down has all months the same, plus the number of ways that the sequence can have $3$ months the same and one different. 
By the same argument as above, the number of strings with all months equal is $\binom{12}{1}$.
The number of strings with $3$ identical months is $\binom{12}{1}\binom{4}{3}\binom{11}{1}$.
Add, and for the probability divide by $12^4$. 
