Expectation Value Question If I have 50 pieces of papers of which $x$ are red and select randomly 15 how many, out of these fifteen pieces will be red on average? 
Is this a binomial problem where success*$=$*select a red piece of paper paper and then can I find the expectation value?
An answer with explanation would be highly appreciated.
 A: The expectation for the average number of red pieces of paper if you draw one is $\frac x{50}$.  By the linearity of expectation, the expectation for $15$ pieces is $\frac {15x}{50}=\frac {3x}{10}$
A: It's not a binomial problem since the probability of getting a red piece of paper is not fixed for each draw (for instance the probability of getting a red paper in the first draw is $\frac{x}{50}$, while in the second draw it can either be $\frac{x}{49}$ or $\frac{x-1}{49}$ depending on whether you get a red paper in the first).
However, take a look at the hypergeometric distribution.
A: It is not a binomial distribution, since we are choosing without replacement. 
There are $\binom{50}{15}$ equally likely ways to choose $15$ pieces of paper. And the number of ways of choosing $k$ red (and therefore $15-k$ non-red) is 
$$\binom{x}{k}\binom{50-x}{15-k}.$$
Divide by $\binom{50}{15}$ for the probability that the number of reds is $k$.  The distribution of the number of red pieces is an instance of a  hypergeometric distribution.
We can write down the usual expectation formula, and then try to simplify it. The details are rather messy, so we describe a simple way to calculate the expectation.  
Imagine picking the pieces of paper one at a time. Let $X_1=1$ if the first piece picked is red, and let $X_1=0$ otherwise. Similarly, let $X_2=1$ if the second piece picked is red, and let $X_2=0$ otherwise. Define $X_3,X_4,\dots, X_{15}$ analogously. Let
$$X=X_1+X_2+\cdots+X_{15}.$$
Then $X$ is the number of red pieces picked. We want $E(X)$. 
By the linearity of expectation, we have
$$E(X)=E(X_1+\cdots+X_{15})=E(X_1)+\cdots +E(X_{15}).$$
For any $X_i$, the probability that $X_i=1$ is $\frac{x}{50}$. Thus $E(X_i)=\frac{x}{50}$, and therefore
$$E(X)=15\frac{x}{50}.$$
