Raman has three unbiased dice, and he rolls them simultaneously. Find the probability that the sum of numbers obtained on the three unbiased dice is not more than 14.

The sum of numbers obtained should not be more than 14.

We will calculate all the possibilities to get a sum of 15,16,17 and 18 to find the probability of getting the sum of more than 14, and subtract it from 1 to get the probability that the sum of numbers obtained on the three unbiased
dice is not more than 14.

When three dice are rolled, the maximum sum that can be obtained is 18.
So, in the given question, the sum can be 15 or 16 or 17 or 18
For the sum to be 15, there are possibilities:
1. (6,6,3) - total cases = $$\dfrac{3!}{2!}$$ = 3
2. (6,5,4) - total cases = 3! = 6
3. (5,5,5) - total cases = 1

For the sum to be 16, there can be two possibilities:
1. (6 , 6 , 4) - total cases = $$\dfrac{3!}{2!}$$ = 3
2. (6, 5, 5) - total cases = $$\dfrac{3!}{2!}$$ = 3

For the sum to be 17, there can be only one possibility:
1. (6, 6, 5) - total cases = $$\dfrac{3!}{2!}$$ = 3
For the sum to be 18, there is only one case, which is (6, 6, 6)

So, the no. of favourable outcomes such that sum is greater than 14 = (10+ 3 + 3 + 3 + 1) = 20
Total no. of outcomes = 6 * 6 * 6 = 216

Required probability =1-  $$\dfrac{20}{216}$$ = $$\dfrac{196}{216}$$ = $$\dfrac{49}{54}$$

Hence, option A is the correct answer.