Explanation

(d)
Total number of cases$= 6^{3} = 216$
Let $A$ denote the event of getting at least 6, $\overline{A}$ will denote the event of getting less than 6 for three dice.
$\therefore\overline{A}$={(1,1,1),(1,12),(1,1,3,),(1,2,1),(1,3,1),(2,1,2),(1,2,2),(2,2,1),(2,1,1),(3,1,1)}
$\Rightarrow n(\overline{A}$)=10
Probability of getting less than $6 = \frac{10}{216} = \frac{5}{108}$
Now, $P(A) = 1 - P\left( \overline{A} \right) = 1 - \frac{5}{108} = \frac{103}{108}$