There are two yellow and three blue balls in a bag. Three balls are randomly drawn
a) without replacement;
b) with replacement.
It is asked to find the expected number of drawn blue balls.
I have computed the value in both cases to be $\frac{9}{5}$. Surprisingly the "official" answer states that in the second case the expected value is larger (without specifying the value).
Can anybody check my result? If it is correct, is it just a coincidence or there is a deep reason for expected value in both cases to be equal?
Your result is correct. There is a good reason why the expected value is the same in each case: the linearity of expectation. Let the random variable $X_j$ be the number of blue balls drawn on the $j^{th}$ pick. You're asked to compute $E[X_1 + X_2 + X_3]$. When drawing with replacement, the $X_j$ are independent random variables, so it's clear that $E[X_1 + X_2 + X_3] = E[X_1] + E[X_2] + E[X_3] = 9/5$. However, because expectation is, by definition, a linear operator, $E[X_1 + X_2 + X_3] = E[X_1] + E[X_2] + E[X_3]$ even when the $X_j$ are not independent, as in the case of drawing without replacement.