Explanation

(d)
Let $A =$Event of getting $i$ on first dice
and $B =$Event of getting more than $i$ on second dice
$\therefore$ Required probability $= \sum_{i = 1}^{5}{P(A_{i}} \cap B_{1})$
$= \frac{1}{6}\lbrack P\left( B_{1} \right) + P\left( B_{2} \right) + P\left( B_{3} \right) + P\left( B_{4} \right) + P\left( B_{5} \right)\rbrack$
$= \frac{1}{6}\left\lbrack \frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6} \right\rbrack$
$= \frac{15}{36} = \frac{5}{12}$