A die is thrown three times. Let X be the 'number of twos seen', find the expectation of X.

We have, $X =$ number of twos seen
So, on throwing a die three times, we will have $X = 0,1,2,3$.

$\therefore$ $P(X = 0) = {P_{({\rm{not}}2)}} \cdot {P_{({\rm{not}}2)}} \cdot {P_{({\rm{not}}2)}} = \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} = \frac{{125}}{{216}}$

$P(X = 1) = {P_{({\rm{(not 2) }}}} \cdot {P_{{\rm{(not 2) }}}} \cdot {P_{(2)}} + {P_{{\rm{(not 2) }}}} \cdot {P_{(2)}} \cdot {P_{{\rm{(not 2) }}}} + {P_{(2)}} \cdot {P_{{\rm{(not 2) }}}} \cdot {P_{{\rm{(not 2) }}}}$

$= \frac{5}{6} \cdot \frac{5}{6}\frac{1}{6} + \frac{5}{6} \cdot \frac{1}{6} \cdot \frac{5}{6} + \frac{1}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} = \frac{{25}}{{36}} \cdot \frac{3}{6} = \frac{{25}}{{72}}$

$P(X = 2) = {P_{({\rm{not }}2)}} \cdot {P_{(2)}} \cdot {P_{(2)}} + {P_{(2)}} \cdot {P_{(2)}} \cdot {P_{({\rm{not 2) }}}} + {P_{(2)}} \cdot {P_{({\rm{not}}2)}} + {P_{(2)}}$

$= \frac{5}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{5}{6} + \frac{1}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}$

$= \frac{1}{{36}} \cdot \left[ {\frac{{15}}{6}} \right] = \frac{{15}}{{216}}$

$P(X = 3) = {P_{(2)}} \cdot {P_{(2)}} \cdot {P_{(2)}} = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{{216}}$

We know that, $E(X) = \Sigma XP(X) = 0 \cdot \frac{{125}}{{216}} + 1 \cdot \frac{{25}}{{72}} + 2 \cdot \frac{{15}}{{216}} + 3 \cdot \frac{1}{{216}}$

$= \frac{{75 + 30 + 3}}{{216}} = \frac{{108}}{{216}} = \frac{1}{2}$