Note that n, (n + 1), and (n + 2) are 3 consecutive integers.
If the product of three consecutive integers are divisible by 8, then the following two scenarios are possible...

• 1. n is even --> (n + 1) is odd and (n + 2) is even --> Either n or (n + 2) will be divisible by 4
  2. (n + 1) is divisible by 8 --> n and (n + 2) are odd

Now, the above two scenarios are mutually exclusive, i.e. no three consecutive integers can be part of both the scenarios. So, there are no repetitions and we can easily calculate number of triplets in each case and add them to get the total number of triplets.

Case 1: n is even
Number of even integers between 1 ans 96, inclusive = {(96 - 2)/2} + 1 = 48

Case 2: (n + 1) is divisible by 8
n + 1 = 8a, where a ≥ 1
n = 8a - 1
8a - 1 ≤ 96
8a ≤ 97
a ≤ 12.1 implies 12 integers.

Total number of n = (48 + 12) = 60
Therefore, probability that n(n + 1)(n + 2) will be divisible by 8 = 60/96 = 5/8

The correct answer is D.