Explanation

(a)
The total number of possible ways
$n(S) = 6 \times 6 \times 6 = 216$
Now, we find out how many ways are favourable to the total of 16 points in one throw. Let $A,\ B,\ C$ be three dice, then favourable ways happen only as follows (6,6,4),(6,4,6),(6,5,5),(5,5,6),(5,6,5),(4,6,6).
Hence, the total number of favourable ways
$n(A) = 6$
$\therefore$ Required probability$= \frac{n(A)}{n(S)} = \frac{6}{216} = \frac{1}{36}$