Two dice $A$ and $B$ are rolled. Let the numbers obtained on $A$ and $B$ be $\alpha$ and $\beta$ respectively. If the variance of $\alpha -\beta$ is $\frac{p}{q},$ where $p$ and $q$ are co-prime, then the sum of the positive divisors of $p$ is equal to

Given that variance of $\alpha -\beta$ is $\frac{p}{q}$

$\alpha \in \left\{1, 2, 3, 4, 5, 6\right\}$

$\beta \in \left\{1, 2, 3, 4, 5, 6\right\}$

$\left(\alpha -\beta \right)=0$ ($6$ case)

$\left(\alpha -\beta \right)=-1$ ($5$ case)

$\left(\alpha -\beta \right)=-2$ ($4$ case)

$\left(\alpha -\beta \right)=-3$ ($3$ case)

$\left(\alpha -\beta \right)=-4$ ($2$ case)

$\left(\alpha -\beta \right)=-5$ ($1$ case)

$\left(\alpha -\beta \right)=1$ ($5$ case)

$\left(\alpha -\beta \right)=2$ ($4$ case)

$\left(\alpha -\beta \right)=3$ ($3$ case)

$\left(\alpha -\beta \right)=4$ ($2$ case)

$\left(\alpha -\beta \right)=5$ ($1$ case)

Mean $=0$

We know that variance $=\frac{{\displaystyle \sum _{i=1}^n}{f}_i{\left(x_i-\overline{x}\right)}^2}{{\displaystyle \sum _{i=1}^n}{f}_i}$

Variance$={\sigma }^2=\frac{0^2\times 6+2\times 1^2\times 5+2\times 2^2\times 4+2\times 3^2\times 3+2\times 4^2\times 2\times 5^2\times 1}{36}$

$=\frac{2}{36}\times \left(5+16+27+32+25\right)=\frac{105}{18}=\frac{35}{6}$

$\therefore p=35$

Sum of divisors of $p=1+5+7+35=48$

Option $\left(3\right)$ is correct.