Draw two numbers, A and B, from a set {1, 2, 3, 4, 5, 6}. A and B are drawn sequentially without replacement. Find the variance Var(3A+B). You draw two numbers, $A$ and $B$ from a set of integers $\{1,2,3,4,5,6\}$. The numbers are drawn sequentially from the set, without replacement. Find the variance $Var(3A+B)$.
I have tried working on this question for a bit but cannot seem to find a quick way to compute it - perhaps there is some insight that would allow us to skip a lot of the computation?. I do not have a lot of experience with similar questions and so I have just been trying a brute force approach so far. This quickly gets quite convoluted for me, as expanding to $Var(3A+B) = Var(3A) + Var(B) + Cov(3A,B)$ all require further sub steps. Any help greatly appreciated!
My approach so far

Variance and covariance can have scaling factors taken out:
$$\newcommand{Var}{\operatorname{Var}}\newcommand{Cov}{\operatorname{Cov}}\Var(3A)=9\Var(A)$$
$$\Cov(3A,B)=3\Cov(A,B)$$
Thus
$$\Var(3A+B)=9\Var A+\Var B+6\Cov(A,B)$$
where $6$ and not $3$ is the correct multiplier for $\Cov$ because the variance of a sum involves an $(a+b)^2$-type expansion. $A$ and $B$ are identically distributed because $(A,B)$ is a uniform random $2$-sample from the set, so the expression becomes
$$\Var(3A+B)=10\Var(A)+6\Cov(A,B)$$
We now derive the explicit numbers:
$$E(A)=E(B)=\frac{1+\cdots+6}6=\frac72$$
$$E(A^2)=\frac{1^2+\cdots+6^2}6=\frac{91}6$$
$$\Var(A)=\frac{91}6-\frac{49}4=\frac{35}{12}$$
$$E(AB)=\frac{(1+\cdots+6)^2-1^2-\cdots-6^2}{30}=\frac{35}3$$
$$\Cov(A,B)=\frac{35}3-\frac{49}4=-\frac7{12}$$
Finally
$$\Var(3A+B)=10\cdot\frac{35}{12}-6\cdot\frac7{12}=\frac{308}{12}=\frac{77}3$$
