Find expected red drawn balls when draw $n$ balls? I have $k_1$ red balls and $k_2$ blue balls in a box. Randomly select $n$ balls in the box without replacement. How to find expected number of red (or blue) balls. 
This is my solution
The probability to get r red balls from $k_1+k_2$ balls in the box (event X):
$$P_1(X)=\frac{\binom{k_1}{r}\binom{k_2}{n-r}}{\binom{k_1+k_2}{r}}$$
The $r$ can be from $0 $ to $k_1$ balls, so the prob. of the event is
$$P_2(X)=\sum_{r=0}^{k_1}\frac{\binom{k_1}{r}\binom{k_2}{n-r}}{\binom{k_1+k_2}{n}}$$
Finally, we got the expected value as
$$E(X)= \sum_{r=0}^{k_1} r\times \frac{\binom{k_1}{r}\binom{k_2}{n-r}}{\binom{k_1+k_2}{n}}$$
Is it right?
 A: The probability to draw $\ell$ red balls out of $r$ total draws is
$$
\frac{\binom{k_1}{\ell}\binom{k_2}{r-\ell}}{\binom{k_1+k_2}{r}}.
$$
This follows a hypergeometric distibution, and thus has expectation 
$$
r\frac{k_1}{k_1+k_2}.
$$
Note that, similarly, the expected value of drawn blue balls equals
$$
r\frac{k_2}{k_1+k_2}
$$
and thus that the sum equals $r$.
A: For $i=1,\dots,n$ let $R_i$ take value $1$ if trial $i$ results in a red ball, and let it take value $0$ otherwise. 
Then $$R:=R_1+\cdots+ R_n$$ stands for the number of red balls. 
With linearity of expectation and symmetry we find:$$\mathbb ER=n\mathbb ER_1=\frac{nk_1}{k_1+k_2}$$

edit for explanation:
As described in your question we will randomly select $n$ of the $k_1+k_2$ balls, but in order to create clarity before doing so we first reserve $n$ spots that are labeled by the numbers $1,2,\dots,n$. The first ball selected is placed on spot $1$, the second on spot $2$, et cetera.
Then event $\{R_i=1\}$ can be recognized as the event that a red ball will be placed on spot $i$. 
Essential is here that at the stage of the beginning of the selection all balls have equal chance to become the one that will be placed on spot $i$.
There are $k_1+k_2$ balls in total, so that chance must be $\frac{1}{k_1+k_2}$.
Further $k_1$ of the balls are red, so the probability that a red ball will be placed on spot $i$ equals: $$\frac{k_1}{k_1+k_2}$$
If that is not clear enough yet then number the red balls with $1,2,\dots,k_1$ and let $A_j$ denote the event that ball $j$ will be placed on spot $i$. 
Then the event that a red ball is placed on spot $i$ is $$A_1\cup\cdots\cup A_{k_1}$$ and its probability is: $$P(A_1\cup\cdots\cup A_{k_1})=\sum_{j=1}^{k_1}P(A_j)=\sum_{j=1}^{k_1}\frac1{k_1+k_2}=\frac{k_1}{k_1+k_2}$$
