There are 10 boxes, 15 balls; 10 red, 5 blue. Each is randomly placed in a box in an independent manner. What's E[X=the number of empty boxes?] There are 10 boxes, 15 balls; 10 red, 5 blue. 
Each is randomly placed in a box in an independent manner. The red balls are placed in boxes 1-10, blue balls are placed in 1-6.  What is the expected value of the number of empty boxes?
I have no real ideas as to how to approach this problem as it seems very "layered." A hint in the right direction would be great.
 A: Following lulu's suggestion, let $X_i=1$ if box $i$ is empty, and 0 if it is not; and let $\displaystyle X=\sum_{i=1}^{10}X_i$.
If $7\le i\le 10$, then $E(X_i)=P(\text{box i is empty})=\left(\frac{9}{10}\right)^{10},\;\;$ and
if $1\le i\le 6$, then $E(X_i)=P(\text{box i is empty})=\left(\frac{9}{10}\right)^{10}\left(\frac{5}{6}\right)^5$;
so $\displaystyle E(X)=\sum_{i=1}^{10}E(X_i)=4\left(\frac{9}{10}\right)^{10}+6\left(\frac{9}{10}\right)^{10}\left(\frac{5}{6}\right)^5\approx 2.2$
A: Let $A_i=1$ if box $i$ is empty, 0 if not. Consider the four boxes numbered 7-10. They can only get red balls. For each one we have $p(A_i=1)=\left(\frac{9}{10}\right)^{10}$. So $E(A_i)=\left(\frac{9}{10}\right)^{10}$. But expectation is linear, so $E(\sum_7^{10}A_i)=4\left(\frac{9}{10}\right)^{10}\approx1.395 $.
The remaining six boxes numbered 1-6 can get both red and blue balls. We have $p(A_i=1)=\left(\frac{5}{6}\right)^{5}\left(\frac{9}{10}\right)^{10}$ and $E(A_i)=\left(\frac{5}{6}\right)^{5}\left(\frac{9}{10}\right)^{10}$, and hence $E(\sum_1^6A_i)=6\left(\frac{5}{6}\right)^{5}\left(\frac{9}{10}\right)^{10}\approx0.841$.
So the expected number of empty boxes is $4\left(\frac{9}{10}\right)^{10}+6\left(\frac{5}{6}\right)^{5}\left(\frac{9}{10}\right)^{10}\approx2.235$
