Expected number of questions when the student knows 50 out of 250 questions and the teacher selects 25 A teacher has 250 sample test questions and he selects 25 of which to put on the test. Johnny only practices 50 out of the 250 questions.
a) What's the expected number of questions that appear on the exam that Johnny have practiced on?
My answer:
I'm quite confused so I don't think it's the right answer. I know that the expected value is just
$$\sum_{k=1}^{25} \left[kp(k)\right]$$
However, I'm not sure what $p(k)$ is. I think it's $$ \frac{25}{250} \frac{50}{250}  = 0.02 $$ which would make $E(X) = 6.5$
b) What's the probability that Johnny hasn't practiced any of the exam questions?
My answer:
Here I tried using the hypergeometric distribution and set N = 250, r = 25, n = 50, and x = 0. Obviously assuming here that if Johnny sees a question he's practiced he'll be able to solve it.
This means the final answer will be
$$\frac {{50 \choose 0}{225 \choose 50}} {{250 \choose 50}}$$
which is
$$\frac {{225 \choose 50}} {{250 \choose 50}}$$
Are those answers correct? Any help would be appreciated! Thank you.
 A: 
What's the expected number of questions that appear on the exam that Johnny have practiced on?

Johnny practices on $50$ out of $250$ questions.  This implies that at random, any question has a $(1/5)$th probability of being a question that Johnny has practiced on.
Linearity of Expectation does not require that the events be independent.
Consider the following $2$ events:

*

*$E_1$ : Johnny practiced on the 1st question chosen by the teacher.

*$E_2$ : Johnny practiced on the 2nd question chosen by the teacher.

When computing the expected number of the questions, out of the $(25)$ chosen by the teacher, that Johnny will have practiced on, it is irrelevant (for example) that events $E_1$ and $E_2$ above are not independent events.
So, the expected number of questions, out of the $(25)$ questions chosen by the teacher, that Johnny will have practiced on is
$$(1/5) \times 25 = 5.$$


What's the probability that Johnny hasn't practiced any of the exam questions?

The probability may be expressed as
$$\frac{N\text{(umerator)}}{D\text{(enominator)}}.$$
$N$ will denote the number of ways of selecting $25$ questions from the $200$ that Johnny did not practice on.
$D$ will denote the number of ways of selecting $25$ questions from any of the $250$ questions.
So, the probability is
$$\frac{\binom{200}{25}}{\binom{250}{25}}.$$
Edit
See the comment of JMoravitz, following my answer.  In fact, I did think that the OP's (i.e. original poster's) answer of
$$\frac{\binom{225}{50}}{\binom{250}{50}}$$
was automatically wrong because it involved Combinations different from mine.  It never occurred to me that the $2$ answers are equivalent.

For what it's worth, my approach was to focus on the $25$ questions selected, reasoning that they had to be part of the $200$ questions that Johnny did not practice on.
The OP took the valid (but opposite) viewpoint.  He reasoned that the $50$ questions that Johnny did practice on had to be contained in the $225$ questions that were not selected.
