There is a jar with 4 marbles numbered 2, 4, 5, 6 and I pull out two without replacement. Given that the random variable X denote the larger of the two numbers on the chips drawn, what is the E(X) and Var(X)?
The way I approached this problem was trying to find the probabilities each outcome was greater than the next. So for P(2), since nothing is greater than it, it has 0 probability. And for P(4), it only has one number it can be greater than so the probability is 1/4. And so on for P(5) = 2/4 and P(6) = 3/4.
So for E(X) I calculated = 2(0) + 4(1/4) + 5(/4) + 6(3/4) = 8
$Var(X) = E(X^2) - (E(X))^2 $
$E(X^2) = 2^2(0) + 4^2(1/4) + 5^2(/4) + 6^2(3/4) = 43.5 $
$Var(X) = 43.5 - (8)^2 = - 25.5 $
As you can see, my variance turns out to be negative which is not possible. So I am thinking my probabilities for X are wrong. I'm not sure how Sampling without Replacement should be taken accounted for when calculating X probabilities.
There are $6$ possible outcomes (i.e. ways of choosing $2$ marbles from $4$), so $P(4)=\frac{1}{6}$, $P(5)=\frac{2}{6}$ and $P(6)=\frac{3}{6}$. You will find that the variance is positive.