Finding expected value of a random variable X I've been given the following task:
There are 10 question in a test.
7 of them are "easy questions", and 3 are "hard questions".
a student randomly picks 6 questions.
What is the expected value for the hard questions that the student will pick?
This is my work so far:
I denote X to be the random variable wich counts the number of "hard questions" that the student picked, out of 6 questions that he choose.
The expected value then will be:
$$E(X)=\sum_{k\in Range(X)}k\cdot P_X(k)=\sum_{k=0}^{6}k\cdot P_X(k)=\sum_{k=1}^{6}k\cdot P_X(k)=\sum_{k=1}^{3}k\cdot P_X(k)=\sum_{k=1}^{3}k\binom{6}{k}p^kq^{6-k}=\sum_{k=1}^{3}6\binom{5}{k-1}p^kq^{6-k}=\sum_{k=1}^{3}6\cdot 5\cdot 4\cdot 3\binom{2}{k-1}p^kq^{6-k}=360\sum_{k=1}^{3}\binom{2}{k-1}p^kq^{6-k}=360pq^3\sum_{k=1}^{3}\binom{2}{k-1}p^{k-1}q^{3-k}=360pq^3\sum_{k=1}^{3}\binom{2}{k-1}p^{k-1}q^{2-(k-1)}=360pq^3\sum_{j=0}^{2}\binom{2}{j}p^{j}q^{2-j}=360pq^3$$
Now, since p=0.3, we'll get that:
$$360pq^3=37.044$$
Shouldn't I expect a much smaller expected value?
What did I do wrong here?
Is it because I didn't choose the right random variable, or is it some calculation error?
Thank you in advance.
 A: It is probably not worthwhile to try to locate a computational error, since the analysis is incorrect from the beginning.
Let $X$ be the number of hard questions chosen. We are taking $6$ questions from the $10$, without replacement. There are $\binom{10}{6}$ equally likely ways to do this. 
The random variable $X$ can take on values $0$, $1$, $2$, or $3$. The number of ways to choose $6$ questions, of which $k$ are hard and $6-k$ are easy is $\binom{3}{k}\binom{7}{6-k}$. It follows that 
$$E(X)=\frac{1}{\binom{10}{6}}\left(0\cdot \binom{3}{0}\binom{7}{6}+1\cdot \binom{3}{1}\binom{7}{5}+2\cdot \binom{3}{2}\binom{7}{4}+3\cdot\binom{3}{3}\binom{7}{3}                        \right).$$
Now calculate.
Another way: Imagine picking the questions one at a time. For $i=1$ to $6$, let $X_i=1$ if the $i$-th question chosen is hard, and let $X_i=0$ otherwise. Then
$$X=X_1+X_2+\cdots +X_6.$$
By the linearity of expectation, we have 
$$E(X)=E(X_1)+E(X_2)+\cdots+E(X_6).$$
The probability that $X_i=1$ is $\frac{3}{10}$. It follows that $E(X_i)=\frac{3}{10}$, and therefore
$$E(X)=6\cdot \frac{3}{10}.$$
Remark: The argument above is quite general, and can be used to compute quickly the mean of any random variable that has hypergeometric distribution.
Alternately, one can find the mean of a general hypergeometric by a binomial coefficient manipulation. We will not write out that calculation, since the procedure we used with indicator random variables $X_i$ is more efficient. 
If we assume (unreasonably) that the questions are chosen with replacement, the number of hard questions will have binomial distribution. The same indicator random variable argument will show that in this case the mean is again $1.8$.
