The mean of the numbers obtained on throwing a dice having written 1 on three faces, 2 on two faces and 5 on one face

(A) 1

(B) 2

(C) 5

(D) $\cfrac{8}{3}$

Option B is correct

$X$ denotes the random variable which represents the number obtained on throwing a dice

$\therefore$ $X$ can assume values 1, 2 and 5
$P(X = 1) = P\,({\rm{getting }}1{\rm{ on three faces }}) = \cfrac{3}{6} = \cfrac{1}{2}$

$P(X = 2) = P\,{\rm{\;}}\left( {{\rm{getting 2 on three faces}}} \right)$ $= \cfrac{2}{6} = \cfrac{1}{3}$

$P(X = 5) = P{\rm{ (getting }}5{\rm{ on three faces }}) = 1/6$

Hence, the probability distribution is :