Expected values of Binomial? 
A group of $d$ students tries to solve $k$ problems in the following way. Each student picks only one problem at random and tries to solve it. Assume that all the attempts have probability $p$ of success and are independent of each other and of the students choice. Let $X$ denote the number of solved problems. Find $\Bbb E(X)$.

I am not sure if I just overestimating this, but would the answer simply be $E(X) = \dbinom{d}{k}p$
Because there are exactly $\dbinom{d}{k}$ independent trials and p is the success rate
Thank you

The probability that $\ge j$ problems are solved is the probability that:


*

*At least one student solves one of the problems

*At least one of the remaining students solves a second problem


$\qquad \qquad \vdots$


*

*At least one of the remaining students solves a $j^{\text{th}}$ problem


And these events are all independent. And the probability that any given student chooses and solves any given problem is $\dfrac{p}{k}$, so we obtain:
$$\mathbb{P}(X \ge j) = dk\dfrac{p}{k} \cdot (d-1)(k-1)\dfrac{p}{k} \cdot (d-j+1)(k-j+1) \dfrac{p}{k}$$
which is precisely $\displaystyle \dfrac{p^j}{k^j} \binom{d}{j} \binom{k}{j}$.
Hence we have
$$\displaystyle \mathbb{E}(X) = \sum_{j=0}^k \mathbb{P}(X \ge k) = \sum_{j=0}^k \dfrac{p^j}{k^j} \binom{d}{j} \binom{k}{j}$$
I'm not sure how you could simplify this, but no doubt there is a way.

Another way of visualising it is this: you have a $d \times k$ grid, where the $d$ rows represent students and the $k$ columns represent the problems. You select one square from each row, and you need to work out the expected number of columns which have at least one square chosen from them.
