Consider an urn with four balls: one ball is worth 3, two balls are worth 8 each, and one ball is worth 11. Suppose you randomly draw two balls from the urn at the same time. Let random variable $X$ denote the sum of the values of these two balls. What is the variance of $X$?
I made a value set for the sum of balls drawn, $X= \{11,14,16,19\}$ I made a sample space for possible combinations of balls drawn $S= \{(3,8)(8,3)(3,11)(11,3)(8,8)(11,8)(8,11)\}$
Probability of scoring 11 = probability of selecting $(3,8)$ or $(8,3)$: $P(X=11) = 2/7$
Probability of scoring 14 = probability of selecting $(3,11)$ or $(11,3)$: $P(X=14) = 2/7$
Probability of scoring 16 = probability of selecting $(8,8)$: $P(X=16) = 1/7$
Probability of scoring 19 = probability of selecting $(11,8)$ or $(8,11)$: $P(X=19) = 2/7$
$E(X) = (2/7 \cdot 11 + 2/7 \cdot 14 + 1/7 \cdot 16 + 2/7 \cdot 19 ) = 14.857$
Required to find Variance \begin{align*} V(X) &= (2/7 * (11^2) + 2/7 * (14^2) + 1/7 * (16^2) + 2/7 * (19^2) ) - (E(X))^2 \\ & = 230.286 - 220.730 \\ & = 9.556 \end{align*}
Is this approach and my answer correct?
To use a counting argument, you need to check that all draws are equally likely. Since you are choosing two from four possible balls without replacement, there should be ${4 \choose 2}=6$ equally likely possibilities (or $4\times 3=12$ if order matters).
These $6$ are
$$\{3,8_a\}, \{3,8_b\}, \{3,11\}, \{8_a,8_b\}, \{8_a,11\}, \{8_b,11\}$$
while the $12$ are
$$(3,8_a), (3,8_b), (3,11), (8_a,3), (8_a,8_b), (8_a,11), \\(8_b,3), (8_b,8_a), (8_b,11), (11,3), (11, 8_a), (11, 8_b)$$
Either will do: you can now add up the pairs and count the sums, and get the probabilities of $\frac13,\frac16,\frac16,\frac13$