Take a look at a randomly chosen group of $n \in \Bbb N$ people. Let $X$ be the number of days (out of $365$ days), where at least one person celebrates his birthday. Think of $X$ as a random variable and let $n = 50$.
My attempt:
I don't think it's possible to give a definite answer here since we don't know how many people there are who celebrate their birthday on the same day. I would argue like this:
First, let us define
$X := \sum_{k=1}^{365} X_k \{0, 1\}$
with
$X_k := 1$, when a person celebrates his birthday and $0$ otherwise.
This means that $X_k \{0, 1\} \le 50$ since $n = 50$.
The probability for one person celebrating his birthday on day $j \in \{1, ..., 365\}$ is $1 \over 365$, hence, we get
$E(X)$ = $\sum_{k=1}^{365} X_k \{0, 1\}$ $*$ $50 \over 365$ $\le$ $2500 \over 365$ $= 6,85$.
Additionally, I would like to understand what this specific value is actually giving me here.
Edit:
After a discussion in the comments, I came to the conclusion that it would be better to calculate $\sum_{k=1}^{365} X_k \{0, 1\} \left(364 \over 365\right)^n$. Do I still have to write this as an inequality?
There seems to be some confusion here, so I will write up my calculation in greater detail. It is possible that I am answering the wrong question so, to be clear, I am answering this question:
Question: Let $X$ be the random variable defined to be the number of distinct birthdays in sample of $n$ randomly chosen individuals (assuming, as usual, that only $365$ birthdays are possible and that each is equally probable). Compute $E[X]$, the expected value of $X$. What is this value when $n=50$?
To do it, we introduce the indicator variables $Y_i$ for each date $i$. Thus $Y_i=1$ if $i$ is a birthday for at least one person in the sample, and $Y_i=0$ otherwise. (Note: I believe that$Y_i=X_i$ in the OP's notation, but I am not sure).
We remark that $$X=\sum_{i=1}^{365}Y_i\implies E[X]=\sum_{i=1}^{365}E[Y_i]$$
Here, of course, we have used the Linearity of Expectation.
We need to compute $E[Y_i]$. Of course this is just $p_i$ where $p_i$ denotes the probability that at least one person in the sample was born on day $i$. Since the probability that an individual was born on a day different than $i$ is $\frac {364}{365}$, we see that $$p_i=1-\left( \frac {364}{365} \right)^n$$
We deduce that $$E[X]=365\times \left(1-\left( \frac {364}{365} \right)^n\right)$$
Sanity check: For $n=1$ this is $1$, as it should be. For $n=2$ we get $1.997260274$ which makes sense (it should be slightly less than $2$ as there is a small chance that your two random people share a birthday). Indeed, with $n=2$ a simple computation shows that $X=1$ with probability $\frac 1{365}$ and $X=2$ with probability $\frac {364}{365}$ so, in this case, $E[X]=1\times \frac 1{365}+2 \times \frac {364}{365}=\frac {729}{365}=1.997260\dots$ as predicted. And, as $n\to \infty$ this expression approaches $365$, as of course it ought to.
For $n=50$ we get $46.78633607$ which seems sensible enough. It is less than $50$ as it must be, but it is still fairly close to $50$ reflecting the plausible fact that we expect a small number of redundancies.