Solving for an expected value from discrete random variables I'm having trouble seeing where I'm going wrong with a problem.  The is the question:
An urn contains 30 marbles of which 8 are black, 12 are red, and 10 are blue.  Randomly, select four marbles without replacement.  Let X be the number of black marbles in the sample of four.


*

*a. What is the probability that no black marble was selected?

*b. What is the probability that exactly one black marble was selected?

*c. Compute E(X).


I've solved a and b as follows:


*

*a. 
$\displaystyle {{8!\over 0!(8-0)!}\cdot{22!\over 4!(22-4)!} \over {30!\over 4!(30-4)!}}$
so basically, ${22*21*20*19 \over 30*29*28*27}$
->
approximately $0.267$

*b.
similarly,
$\displaystyle{{8!\over 1!(8-1)!}{22!\over 3!(22-3)!}\over {30!\over 4!(30-4)!}}$
->
approximately $0.449$
Right now, I'm trying to solve for E(X), which from what I think I understand is the total expected outcome of every way a black marble would be drawn...
so it would be E(X)=(0 drawn)+(1 drawn)+(2 drawn)+(3 drawn)+(4 drawn)
I've continued to solve in the same format as the previous questions for the last three possibilities of a black marble being drawn, and I get an answer of 
1.058.
The correct answer should be 1.067.
Am I making a simple calculation error, or is my understanding of how to solve for the expected value completely wrong from the beginning?
 A: The correct answer is indeed $1.067$.  I can't say how you are wrong without seeing what you've done.  But the easy way is to utilise Linearity of Expectation and indicator random variables.
$\mathsf E(X)$ is the expected number of black marbles in a sample of 4 marbles, which is drawn without replacement from a population of 30 containing 8 black marbles.
We consider that the drawn marbles can be placed in a line, and on doing so let $X_i$ be the indicator that the $i$-th marble in this line is black.   The probability that a particular marble in the line will be black is: $8/30$.   Thus $\mathsf E(X_i)=8/30$ for all four marbles.
By the Linearity of Expectation then: $\mathsf E(X) $ $= \sum\limits_{i=1}^4 \mathsf E(X_i) \\[2ex] = 16/15$

Alternatively we note that $\mathsf P(X=x) = \dfrac{\dbinom{8}{x}\dbinom{22}{4-x}}{\dbinom{30}{4}} = \cfrac{\cfrac{8!}{x!(8-x)!}\cfrac{22!}{(4-x)!(18+x)!}}{\cfrac{30!}{4!26!}}$ .
$$\begin{align}\mathsf E(X) & = \sum_{x=0}^4 x\;\mathsf P(X{=}x) \\[2ex] & = 0+\dfrac{\dbinom{8}{1}\dbinom{22}{3}}{\dbinom{30}{4}} + \dfrac{2\dbinom{8}{2}\dbinom{22}{2}}{\dbinom{30}{4}} + \dfrac{3\dbinom{8}{3}\dbinom{22}{1}}{\dbinom{30}{4}}+ \dfrac{4\dbinom{8}{4}\dbinom{22}{0}}{\dbinom{30}{4}}\end{align}$$
Crunching the numbers should give the same result, and if it doesn't you have indeed made a calculation error.  (Or a typo, and we're all capable of making won.)
A: Your reasoning for parts (a) and (b) looks correct.
The formula for expected value should be:
$E(X)=0*($probability 0 drawn$) + 1*($probability 1 drawn$) + 2*($probability 2 drawn$) + 3*($probability 3 drawn$) + 4*($probability 4 drawn$)$
Note that, of course, you can drop the first term as it is a product of zero, but I thought including it makes the formula more intuitive. It looks like to forgot to multiply each of the probabilities in the formula by 0, 1, 2, 3, and 4, respectively.
Since you already calculated the probabilities, why don't you plug in the numbers and see what you get?
