Question

In a simultaneous throw of a pair of dice, find the probability of getting neither a doublet nor a total of 10

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 E₁₂ = event of getting neither a doublet nor a total of 10
 Thus

\[\bar{{E_{12}}}\] = event of getting either a doublet or a total of 10
Then

\[\bar{{E_{12}}}\] = {(1, 1), (2, 2), (3, 3), (4, 4), (4, 6), (5, 5), (6, 4), (6, 6)}

\[i . e . n\left( \bar{{E_{12}}} \right) = 8\]

\[\therefore P \bar{\left( E_{12} \right)} = \frac{n \bar{\left( E_{12} \right)}}{n\left( S \right)} = \frac{8}{36} = \frac{2}{9}\]

\[\text{ Hence } , P\left( E_{12} \right) = 1 - P\left( \bar{{E_{12}}} \right)\]

\[= 1 - \frac{2}{9} = \frac{7}{9}\]