Expectation of the sum - birthday anniversary problem. For a group of $100$ random people, find:


*

*The expected number of days that are the birthday of at least 3 people, birthyear not being significant.

*The expected number of distinct birthday days.
(Consider the year as having $365$ days so assume nobody has a birthday on Feb 29th)
My attempt: 
For the first question, my try consists of writing a random variable that counts the number of days that are the birthday of at least $3$ people as a sum of r.v
$$X=X_1+...+X_{365}$$
Such that
$$X_i=1 \text{ if at least $3$ people have birthday on the ith day}$$
$$X_i=0,\text{ otherwise}$$
My problem is to find the probability of $X_i=1$. I've tried to find the probability of nobody having a birthday on day i and exactly $1$ and $2$ having a birthday on day i and grab the complement, but it didn't work. I thought, for example, that the probability of exactly $2$ people having a birthday on a  pre-established day i is$$\frac{\binom{100}{2} (364)^{98}}{(365)^{100}}$$is that correct?
The answer is $0.9301$
For the second, I've tried something similary, but in this case
$$Y_i=1 \text{ if exactly 1 people have birthday on day i}$$
And
$$P(Y_i=1) = 100 \frac{1}{365} (\frac{364}{365})^{99}$$
Didnt work.
Thanks in advance.
 A: For (1) you could try a Poisson approximation: Number of birthdays that fall on a given day is a random variable that is approximately Poisson$\left(\lambda=\frac{100}{365}\right)$.  
For (2) you could use the same approach and take $365-$ (expected number of days with no birthdays).
I think these would give you good approximations to the exact values.
A: Two hints:
For 1, the number of people who have a birthday on January 1 has a Binomial distribution.
For 2, try defining $$
X_i = \begin{cases}
  1      &\text{if at least one person has a birthday on day i}\\
  0      &\text{otherwise}
\end{cases}$$
then consider $\sum_{i=1}^{365} X_i$.
A: *

*You have $\frac{\binom{100}{2} 364^{98}}{365^{100}} \approx 0.02839582
$ as the probability exactly $2$ out of $100$ share day $i$

*Similarly $\frac{\binom{100}{1} 364^{99}}{365^{100}} \approx 0.20880964$ is the probability exactly $1$ out of $100$ has birthday $i$ 

*and $\frac{\binom{100}{0} 364^{100}}{365^{100}} \approx 0.76006707$ is the probability that $0$ out of $100$ has birthday $i$ 

*Adding these up and subtracting from $1$ gives about $0.00272747$ for the probability at least $3$ out of $100$ share day $i$, as Henno Brandsma commented 

*Multiplying this by $365$ gives about $0.9955$ expected days that at least $3$ out of $100$ share as birthdays 

*Similarly the expected number of distinct birthdays (i.e the days that at least $1$ out of $100$ has as a birthday is about $365(1-0.76006707) \approx 87.5755$
A Poisson approximation on the probabilities might suggest these numbers could be close to about $0.02853642$, $0.20831586$, $0.76035291$, $0.00279481$, $1.0201$ and $87.4712$ respectively, which are not that far away 
