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 :
Answer & explanation
Correct answer: option 2
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 = $\frac{1}{6}=\frac{1}{6}$
$E(X)=\left(2×\frac{1}{2}\right)+\left(4×\frac{1}{3}\right)+\left(6×\frac{1}{6}\right)$
$=2+\frac{4}{3}=\frac{10}{3}$