Question

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

Answer 1

We know that in a single throw of two dices, the total number of possible outcomes is (6 × 6) = 36.
Let S be the sample space.
Then n(S) = 36

 Let E₈ = event of getting neither 9 nor 11 as the sum of the numbers on the faces
Then 

\[\bar{{E_8}}\] = event of getting either 9 or 11 as the sum         
Thus,

\[E_8\]   = {(3, 6), (4, 5), (5, 4) , (5, 6), (6, 3), (6, 5) }

\[i . e . n\left( \bar{{E_8}} \right) = 6\]

\[\therefore P\left( \bar{{E_8}} \right) = \frac{n\left( \bar{{E_8}} \right)}{n\left( S \right)} = \frac{6}{36} = \frac{1}{6}\]

\[\text{ Hence } , P\left( E_8 \right) = 1 - P\left( \bar{{E_8}} \right)\]

\[= 1 - \frac{1}{6} = \frac{5}{6}\]