In a single throw of two dice, detemine the probability of not getting the same number on the two dice.

Answer

`5/10`

Explanation

`n(S) = (6 xx 6) = 36`.
Let E = event of getting the same number on two dice
or `E = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}`.
`therefore P(E) = (n(E))/(n(S)) = 6/36 = 1/6`
`rArr` P (not getting the same number on the two dice) `= (1 - 1/6) = 5/6.`