If 3 dice are rolled, what is the probability that the sum is 16 ?
 A). \[\dfrac{7}{12}\]
B). \[\dfrac{5}{36}\]
C). \[\dfrac{2}{9}\]
D). \[\dfrac{1}{36}\]

Hint: In this question, we need to find the probability that the sum is \[16\] . Totally \[3\] dice are rolled. The probability of an event is a number between \[0\] and \[1\] . It cannot be greater than \[1\] . First we need to find \[n(S)\ \] i. e). Total number of outcomes.
Formula used :
The formula for the probability of an event is:
\[P\left( A \right) = \dfrac{n\left( A \right)}{n\left( S \right)}\]
Where,
 \[P(A)\] is the probability of the event ``\[A\] ’’
\[n\left( A \right)\] is the number of favourable outcomes
\[n\left( S \right)\] is the total number of the events in the sample space.

Complete step-by-step solution:
Given that, 3 dice are rolled.
Let us consider the probability that the sum is 16 be \[{\ A}\]
First, We need to find the total outcomes,
\[S\ = \{\ (1,1,1)\ldots\ \ldots\ \ldots\ \]
              \[\ldots\ \ldots\ (6,6,6)\ \}\]
\[\ n\left( S \right) = 6^{3}\]
 \[n\left( S \right) = \ 216\]
We need to find the probability that the sum is \[16\]
For the sum to be \[16\] , the favourable outcomes are
\[A = \{\ (4,6,6)\ ,(6,4,6)\ ,(6,6,4)\ ,(5,6,5)\ ,(6,5,5)\ ,(5,5,6)\ \}\]
\[{\ n}\left( A \right) = \ 6\]
Now ,
Probability,
\[P\left( A \right) = \dfrac{n\left( A \right)}{n\left( S \right)}\]
=\[\dfrac{6}{216}\]
By simplifying,
We get,
\[P\left( A \right) = \dfrac{1}{36}\]
ie) . The probability of getting the sum \[16\] is \[\dfrac{1}{36}\]
Final answer:
 Option D. \[\dfrac{1}{36}\]

Note: The simple rule for the probability is the number of desired outcomes divided by the number of possible outcomes . The probability of dice for a particular number is one–sixth. The probability of an event lies between 0 and 1 . The probability of an event never occurs greater than 1 .