Probability of someone being born the same day as you in a group of N people. I got into a discussion with a friend about this simple question:
Provided that I went to a class with $N$ people, where everyone was born the same year in the same city where there are $K$ hospitals, what's the probability that someone was born in the same hospital and in the same day as me?
My approach would be as follows:
$$\left( 1- \left(\frac{364}{365} \right)^N \right) \cdot \left(1-\left(\frac{K-1}{K} \right)^N \right)$$
But I'm not really sure that this is the correct answer :)
 A: Under appropriate independence assumptions, and assumptions about equality of hospitals, the probability that someone chosen at random was born on the same day as you and in the same hospital is
$$\frac{1}{365}\cdot\frac{1}{K}=\frac{1}{365K}.$$
So the probability the person was born on a different day or a different hospital than you is 
$$1-\frac{1}{365K}.$$
The probability that all $N$ were born on a different day or different hospital than you is 
$$\left(1-\frac{1}{365K}\right)^N.$$
So the probability that at least one person "double matches" with you is
$$1-\left(1-\frac{1}{365K}\right)^N.$$
Remark: This is exactly the same reasoning as the pure days problem, except that we now have $365K$ abstract "days."
A: REVISED: Your answer is correct
If the event "birth in a specific hospital" is independent of "birth in a specific day":A: Sameness of hospitals
B: Sameness of birthdays
P(A and B)=?
P(A)= 1- ((k-1)/k)^N
p(B)= 1- (364/365)^N
Independence: P(A and B)=P(A) * P(B) =[1- ((k-1)/k)^N  ][1- (364/365)^N ]
