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Two dice\mathrm{A} \text{ and } \mathrm{B} are rolled. Let numbers obtained on\mathrm{A} \text{ and } \mathrm{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 divisior of p is equal to

Option: 1

36


Option: 2

31


Option: 3

48


Option: 4

72


Answers (1)

best_answer

\begin{tabular}{|c|c|c|} \hline$\alpha-\beta$ & Case & $P$ \\ \hline 5 & $(6,1)$ & $1 / 36$ \\ \hline 4 & $(6,2)(5,1)$ & $2 / 36$ \\ \hline 3 & $(6,3)(5,2)(4,1)$ & $3 / 36$ \\ \hline 2 & $(6,4)(5,3)(4,3)(3,1)$ & $4 / 36$ \\ \hline 1 & $(6,5)(5,4)(4,3)(3,2)(2,1)$ & $6 / 36$ \\ \hline 0 & $(6,6)(5,5) \ldots \ldots(1,1)$ & $5 / 36$ \\ \hline-1 & $\ldots$ & $4 / 36$ \\ \hline-2 & $\ldots-\ldots / 36$ \\ \hline-3 & $(2,6)(1,5)$ & $3 / 36$ \\ \hline-5 & $(1,6)$ & $2 / 36$ \\ \hline \end{tabular}

\begin{aligned} & \sum\left(x^2\right)=\sum x^2 \mathrm{P}(\mathrm{x})=2\left[\frac{25}{36}+\frac{32}{36}+\frac{27}{36}+\frac{16}{36}+\frac{5}{36}\right] \\ & =\frac{105}{18}=\frac{35}{6} \\ & \mu=\sum(x)=0 \text { as data is symmetric } \\ & \sigma^2=\sum\left(\mathrm{x}^2\right)=\sum \mathrm{x}^2 \mathrm{P}(\mathrm{x})=\frac{35}{6} \mathrm{P}=35=5 \times 7 \end{aligned}
Sum of divisors =\left(5^{\circ}+5^1\right)

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HARSH KANKARIA

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