Basic Probability Question (Expected Value) We are given a bag of marbles containing 6 blue marbles, 7 red marbles, and 5 yellow marbles. We select 4 marbles without replacement. How can we calculate the expected value of the number of blue, red and yellow marbles taken?
If we select 1 marble, then it becomes much easier: 6/18 blue marbles, 7/18 red marbles, 5/18 yellow marbles. But I'm not sure how to extend this concept to selecting more marbles.
Edit: Can I just multiply each of those values by 4-- as in we'll get a distribution of 4 marbles that is the same as the distribution of the 18?
 A: Hint: expectation is linear, i.e., $E(X_1 + X_2 + \ldots + X_n) = E(X_1) + E(X_2) + \ldots + E(X_n)$. Think about how this applies here.
A: The number of blue marbles taken from the bag has the hypergeometric distribution and the expected value of the hypergeometric distribution is equal to
$$
n\cdot\frac KN,
$$
where $n$ is the number of draws, $K$ is the number of blue marbles and $N$ is the number of all marbles in the bag.
So the expected value of blue marbles taken from the bag is equal to
$$
4\cdot\frac 6{18}.
$$
Similarly, the expected value of red marbles is equal to $4\cdot\frac7{18}$ and the expected value of yellow marbles is equal to $4\cdot\frac5{18}$.
A: just go step by step.
Expected value for the blue marbles:
four times blue marbles are selected: BBBB
calculate for this event the probability. Now multiply it with 4 (blue marbles).
Next possible selections: BBBR (variations)
Calculate for this event the probability. Now multiply it with 3 (blue marbles). And multiply the result with 4, because you have here four possible variations: BBBR, BBRB, BRBB, RBBB
You go on like this. 
A hint: The amount of variations of BBRY is: $\frac{4!}{2!\cdot 1! \cdot 1!}=12$
Mybe it is not the easiest way, but you will come to a result.
greetings,
calculus.
A: Hint:
Be $X_1$ the result of the first selection which can be $B,R,Y$. Then:
$$P(X_1=B)=\frac{6}{18}$$
$$P(X_1=R)=\frac{7}{18}$$
$$P(X_1=Y)=\frac{5}{18}$$
Then :
$$E(Number of Blue Balls In The First Select)=P(X_1=B)\cdot 1+P(X_1=R)\cdot 0 + P(X_1=Y)\cdot 0=P(X_1=B)$$
Be $X_2$ the result of the second selection which can be $B,R,Y$, but the result is conditional on the number of marbles, so in the result of the first selection:
$$P(X_2=B)=P(X_2=B|X_1=B)P(X_1=B)+P(X_2=B|X_1=R)P(X_1=R)+P(X_2=B|X_1=Y)P(X_1=Y)$$
$$=\frac{5}{17}\frac{6}{18}+\frac{6}{17}\frac{7}{18}+\frac{6}{17}\frac{5}{18}$$
With that you can obtain the expectation of number of blue balls in the second selection:
$$E(Number of Blue Balls In The Second Select)=P(X_2=B)\cdot 1+P(X_2=R)\cdot 0 + P(X_2=Y)\cdot 0=P(X_2=B)$$
I think at that point you can continue an obtain a general expression.
