In a simultaneous throw of a pair of dice, find the probability of getting neither 9 nor 11 as the sum of the numbers on the faces.

Given:

Two dice are thrown simultaneously. 

To do:

We have to find the probability of getting neither 9 nor 11 as the sum of the numbers on the faces.

Solution:

When two dice are thrown, the total possible outcomes are $6\times6=36$.

This implies,

The total number of possible outcomes $n=36$

Outcomes, where we get 9 or 11 as the sum of the numbers on the faces, are $[(3, 6), (4, 5), (5, 4), (6, 3), (5, 6), (6, 5)]$

Number of outcomes where we get 9 or 11 as the sum of the numbers on the faces $=6$

Number of outcomes where we get neither 9 nor 11 as the sum of the numbers on the faces $=36-6=30$

Total number of favourable outcomes $=30$

Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

Therefore,

Probability of getting neither 9 nor 11 as the sum of the numbers on the faces $=\frac{30}{36}$

$=\frac{5}{6}$

The probability of getting neither 9 nor 11 as the sum of the numbers on the faces is $\frac{5}{6}$.