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

Fix one of the problems. The probability that a particular student solves that problem is $\frac{p}{k}$ (the probability that the student selected the fixed problem, multiplied by the probability that the student solved the problem correctly).  Since there are $d$ students, the probability that NOBODY solved the fixed problem correctly is $\left(1-\frac{p}{k}\right)^d$.  It follows that the probability the fixed problem is solved by at least one person is $1-\left(1-\frac{p}{k}\right)^d$.  The expected number of problems that are solved is then $k\left(1-\left(1-\frac{p}{k}\right)^d\right)$
