Two dice A and B are rolled. Let the numbers obtained on A and B be...

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

$\alpha \in \left\{1, 2, 3, 4, 5, 6\right\}</p><p>$\beta \in \left\{1, 2, 3, 4, 5, 6\right\}$</p><p>\left(\alpha - \beta\right) = 0$ ($6$ case)

\left(\alpha - \beta\right) = - 1$($5$case)</p><p>\left(\alpha - \beta\right) = - 2$ ($4$ case)

\left(\alpha - \beta\right) = - 3$($3$case)</p><p>\left(\alpha - \beta\right) = - 4$ ($2$ case)

\left(\alpha - \beta\right) = - 5$($1$case)</p><p>\left(\alpha - \beta\right) = 1$ ($5$ case)

\left(\alpha - \beta\right) = 2$($4$case)</p><p>\left(\alpha - \beta\right) = 3$ ($3$ case)

\left(\alpha - \beta\right) = 4$($2$case)</p><p>\left(\alpha - \beta\right) = 5$ ($1$ case)

Mean $= 0$

We know that variance $= \dfrac{\mathop{∑}_{i = 1}^{n}{f}_{i}{\left({x}_{i} - x\right)}^{2}}{\mathop{∑}_{i = 1}^{n}{f}_{i}}$

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

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

$∴ p = 35$

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

Option \left(3\right) is correct.