Question

a pair of fair dice is rolled. what is the probability that the second die lands on a higher value than the first?

Answer 1

The required probability is
`(5)/(12)`.

Explanation:

If a fair dice is rolled then total outcomes are

{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

We need to find the probability that the second die lands on a higher value than the first.

So, total favorable outcomes are

{(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)}

Formula for probability:

`Probability=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}`

`Probability=(15)/(36)`

`Probability=(5)/(12)`

Therefore, the required probability is
`(5)/(12)`.