\(C^3 - C\) is divisible by \(12\)
i.e. \(C(C^2 - 1) = (C - 1)(C)(C + 1)\) is divisible by \(12\)

Lets check some numbers;

C = 20
\((19)(20)(21)\) is divisible by \(12\) - YES

C = 21
\((20)(21)(22)\) is divisible by \(12\) - YES

C = 22
\((21)(22)(23)\) is divisible by \(12\) - NO

C = 23
\((22)(23)(24)\) is divisible by \(12\) - YES

C = 24
\((23)(24)(25)\) is divisible by \(12\) - YES

C = 25
\((24)(25)(26)\) is divisible by \(12\) - YES

C = 26
\((25)(26)(27)\) is divisible by \(12\) - NO

C = 27
\((26)(27)(28)\) is divisible by \(12\) - YES

We can clearly see a pattern here - out of every \(4\) values, \(3\) values are divisible and \(1\) value is not

Let's see how many terms we have, \(n = 99 - 20 + 1 = 80\)
Since \(80\) fits in our pattern of \(4\), Therefore \(60\) terms will be divisible by \(12\)

Required Probability \(= \frac{3}{4}\)

Hence, option C