Precal Probability Expected Number 
A committee of four students will be selected from a list that contains six Grade 9 students and eight Grade 10 students. What is the expected number of Grade 10 students on the committee?

Since there can be $0$, $1$, $2$, $3$, or $4$ grade 10 students selected, where there will be $4$, $3$, $2$, $1$, $0$ grade 9 students selected respectively, I calculated the probability for each case in the form $$\binom{8}{n}\binom{6}{4 - n}\left(\frac{4}{7}\right)^4\left(\frac{3}{7}\right)^{4-n}$$ where $n$ is an integer between $0$ and $4$ inclusive.
However, the probability soon becomes more than $1$ (which shouldn't be), and I don't know how to approach this problem. 
 A: Alternative route (and not really an answer (for that see @Henno), but I cannot withhold myself ) :
Give the Grade 10 students the numbers $1,2,\dots,8$ and for $i\in\{1,\dots,8\}$ let $X_i$ take value $1$ if the Grade 10 student with number $i$ is chosen, and let $X_i$ take value $0$ otherwise. Then: $$X:=X_1+\cdots+X_{8}$$ is the number of Grade 10 students that are chosen. 
With linearity of expectation and symmetry we find:$$\mathbb EX=8\mathbb EX_1=8\Pr(X_1=1)$$
In total there are $14$ students and $4$ of them will be chosen, so $\Pr(X_1=1)=\frac4{14}=\frac27$. So we end up with: $$\mathbb EX=\frac{16}7$$

If you are asked to calculate an expectation, then do not start with calculation of distribution. First examine whether you can do it by applying e.g. linearity of expectation. In many case expectations are easyer to find than probabilities/distributions.
A: For $X$ is the number of grade 10 students to pick $x\in \{0,1,2,3,4\}$ many grade 10 students, we get:
$$P(X=x)= \frac{\binom{8}{x}\binom{6}{4-x}}{\binom{14}{4}}$$
because we pick 4 students out of 14 total, of which x from the 8 grade 10 ones and the rest from the grade 9 group. 
You seem to be using a sort of hybrid of this distribution and the binomial one, where the latter does not apply as we don't do independent trials.
This is the so-called hypergeometric distribution which has mean/expected value equal to $4 \frac{8}{14}=  \frac{16}{7}$ according to this link or a computation like @drhab's.
