Question

Find the chance of throwing at least one ace in a single throw with two dice.

Answer 1

The possible number of cases is $6\times 6$, or $36$.
An ace on one die may be associated with any of the $6$ numbers on the other die, and the remaining $5$ numbers on the first die may each be associated with the ace on the second die$;$ thus the number of favourable cases is $11$.
Therefore the required chance is $\frac{11}{36}$.
Or we may reason as follows:
There are $5$ ways in which each die can be thrown so as not to give an ace$;$ hence $25$ throws of the two dice will exclude aces. That is$,$ the chance of not throwing one or more aces is $\frac{25}{36};$ so that the chance of throwing one ace at least is $1-\frac{25}{36},$ or $\frac{11}{36}$.