The variance of the number obtained in a throw of an unbiased die is :

$\frac{35}{12}$

The correct answer is Option (2) → $\frac{35}{12}$

The possible outcomes when rolling an unbiased die 1, 2, 3, 4, 5, 6.

Mean, $μ=E(X)=∑XP(X)$

$=\left(1×\frac{1}{6}+2×\frac{1}{6}+3×\frac{1}{6}+4×\frac{1}{6}+5×\frac{1}{6}+6×\frac{1}{6}\right)=\frac{21}{6}$

Variance, $σ=E(X^2)-[E(X)]^2$

$[E(X)]^2=\left(1^2×\frac{1}{6}+2^2×\frac{1}{6}+3^2×\frac{1}{6}+4^2×\frac{1}{6}+5^2×\frac{1}{6}+6^2×\frac{1}{6}\right)=\frac{91}{6}$

$σ^2=\frac{91}{6}-(\frac{21}{6})^2=\frac{35}{12}$