The expected value of the number obtained, on throwing a die having 2 written on three faces, 4 on two faces and 6 on one face, is :

$\frac{10}{3}$

The correct answer is Option (2) → $\frac{10}{3}$

The expected value (Mean) of a discrete random variable X is,

$E(X)=∑X.P(X)$

2 appears on 3 face → Probability = $\frac{3}{6}=\frac{1}{2}$

4 appears on 2 face → Probability = $\frac{2}{6}=\frac{1}{3}$

6 appears on 1 face → Probability = $\frac{1}{6}=\frac{1}{6}$

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

$=2+\frac{4}{3}=\frac{10}{3}$