Finding expected value of random variable You and a friend go to a wine bar. The wine bar offers the following wine styles for ordering:


*

*2 of the wines offered are a Rosé style wine

*3 of the wines offered are a White style wine

*and 5 of the wines offered are a Red style wine


All in all there are 10 unique wines to choose from.
You and your friend order a uniformly random subset of seven wines (no duplicates).
Y = the number of white wines in this order
What is the expected value of Y?
Here is what I think:
So for this I first break down the types of events and what the values are


*

*Y = 0 => { $XXXXXXX$ } (X meaning not a rosé style wine)

*Y = 1 => { $XXXXXXR, XXXXXRX, XXXXRXX, XXXRXXX, XXRXXXX, XRXXXXX, RXXXXXX$ } => $1 \choose 7$

*Y = 2 => { $XXXXXRR, 
XXXXRXR,
...
RXRXXXX,
RRXXXXX$ } => $2 \choose 7$

*Y = 3 => { $XXXXRRR,
XXXRXRR,
XXRXXRR,
..
RRRXXXX$ } => $3 \choose 7$


So now to get the expected value do I add up all the probabilities of each ie. Pr(Y=1) = $|Y=1|/|S|$ = $1 \choose 7$$/$$ 3 \choose 7$, so I would then do something like $Pr(Y = 0) + Pr(Y = 1) + Pr(Y = 2) + Pr(Y = 3)$
Thanks 
 A: Close, but a few details need attention.
Assuming the selection is unbiased and without repetition, then the probability of choosing $y$ of $3$ white wines, and $7-y$ of $7$ not white wines, out of all the ways to choose $7$ of $10$ wines is:
$$\Pr(Y=y) \;=\; \mathbf 1_{y\in\{0,1,2,3\}}\cdot{\dbinom{3}{y}\dbinom{7}{7-y}}\Big/{\dbinom{10}{7}}$$
So the expectation is:$$\mathsf E(Y) = \sum\limits_{y=0}^3\,y\;\mathsf P(Y=y) \\ = \sum\limits_{y=0}^3\,y{\dbinom{3}{y}\dbinom{7}{7-y}}\Big/{\dbinom{10}{7}}$$

 This is a Hypergeometric Distribution, so I just look that up and find that the mean is $7\cdot 3/10$.   That is $2.1$

A: We first do it more or less in the way you attempted. There are $\binom{10}{7}$ equally likely ways to choose $7$ wines. 
First we find the number of ways to choose $0$ Pinks. There is only one way, so $\Pr(Y=0)=\frac{1}{\binom{10}{7}}$. For the future, note that this is $\frac{\binom{3}{0}\binom{7}{7}}{\binom{10}{7}}$.  So now we know $\Pr(Y=0)$.
Next we find the number of ways to choose $1$ Pink, and therefore $6$ non-Pinks. There are $\binom{3}{1}$ ways to choose the Pink, and for each of these there are $\binom{7}{6}$ ways to choose the non-Pinks. It follows that $\Pr(Y=1)=\frac{\binom{3}{1}\binom{7}{6}}{\binom{10}{7}}$.
A similar analysis shows that $\Pr(Y=2)=\frac{\binom{3}{2}\binom{7}{5}}{\binom{10}{7}}$, and $\Pr(Y=3)=\frac{\binom{3}{3}\binom{7}{4}}{\binom{10}{7}}$.
Finally, $E(Y)=0\cdot\Pr(Y=0)+1\cdot \Pr(Y=1)+2\cdot \Pr(Y=2)+3\cdot \Pr(Y=3)$. Calculate.
Another way: Imagine we pick the wines one at a time. Let random variable $X_i$ be equal to $1$ if on the $i$-th pick we choose Pink, and let $X_i=0$ otherwise. Then $Y=X_1+\cdots +X_7$, and by the linearity of expectation $E(Y)=E(X_1)+\cdots+E(X_7)$. For any $i$, $\Pr(X_i=1)=\frac{3}{10}$, so $E(X_i)=\frac{3}{10}$. It follows that $E(X_i)=7\cdot \frac{3}{10}$. Easier!
