Question

In a simultaneous throw of a  pair of dice, find the probability of getting:

a number other than 5 on any dice.

Answer 1

\[\text{ When a pair of dice is thrown simultaneously, the sample space will be as follow }: \]
\[S = \left\{ \left( 1, 1 \right), \left( 1, 2 \right), \left( 1, 3 \right), \left( 1, 4 \right), \cdots\left( 6, 5 \right), \left( 6, 6 \right) \right\}\]
\[\text{ Hence, the total number of outcomes is 36 } . \]

\[\text{ Let A be the event of getting pairs that has the number 5 } . \]
\[\text{ The pairs that has the number 5 are } \left( 1, 5 \right), \left( 2, 5 \right), \left( 3, 5 \right), \left( 4, 5 \right), \left( 5, 1 \right), \left( 5, 2 \right), \left( 5, 3 \right), \left( 5, 4 \right), \left( 5, 5 \right), \left( 5, 6 \right), \left( 6, 1 \right), \left( 6, 2 \right), \left( 6, 3 \right), \left( 6, 4 \right) and \left( 6, 6 \right) . \]
\[\text{ Hence, the number of favourable outcomes is 11 } . \]
\[ \therefore P\left( A \right) = \frac{\text{ Number of favourable outcomes }}{\text{ Total number of outcomes }} = \frac{11}{36}\]
\[ \therefore P\left( A \right) = 1 - P\left( A \right) = 1 - \frac{11}{36} = \frac{25}{36}\]