Finding the average sum of 15 numbers out of 100 numbers I've got a question.
For the numbers 1,2,...,100 we choose with the same probability 15 numbers.
1) What is the average sum of those numbers?
2) What is the average sum of those numbers when the probability to choose an even number is 3 times greater than choosing an odd number?
 A: There are two reasonable interpretations of the problem: (i) the sampling is done without replacement and (ii) the sampling is done with replacement. However, the answers turn out to be the same under both interpretations!
Imagine the sampling is done one at a time. Let random variables $X_i$ be the number pulled on the $i$-th trial. We want $E(X_1+\cdots+X_{15})$, which is $E(X_1)+\cdots+E(X_{15})$.  (Note that the linearity of expectation does not require independence.) 
The expected value of $X_i$ is $\frac{1+\cdots+100}{100}$, which is $\frac{101}{2}$. Thus the expected value of the sum is $15\cdot\frac{101}{2}$.
We now look at the evens-odds problem. We interpret "three times greater" as meaning that the probability of pulling an even is three times the probability of pulling an odd. Then the probability of an even is $\frac{3}{4}$ and the probability of an odd is $\frac{1}{4}$.
Again we find $E(X_i)$ and multiply by $15$. Given that  we pull an even, the expected value is $\frac{2+4+\cdots+50}{50}$, which is $51$. Given we pull an odd, the expected value is $50$. So $E(X_i)=\frac{3}{4}\cdot 51+\frac{1}{4}\cdot 50$. Multiply by $15$. 
