Three unbiased dice are thrown. Find the probability that, i) they all show different numbers, ii) at least two show the same number.

An unbiased dice has 6 faces with a different number each
of which has an equal probability of showing when the die is tossed. In the problem we
have 3 unbiased dice.

1) First we need to find the
probability that all show different numbers. This can be done by considering that when
the 1st die is tossed it can show any of the six numbers. For the next die there are
only 5 options and for the 3rd there are only 4 options. This gives the total number of
viable options as 6*5*4 = 120. The total number of possible options when 3 dice are
thrown is 6*6*6 = 216. This gives the probability of the given condition taking place as
120/216 = 5/9

2) At least two dice have to show the same
number. The probability of this happening can be written as 1 - probability that no dice
have the same number = 1 - 5/9 = 4/9

The
probability that all 3 dice show a different number is 5/9 and the probability that at
least 2 of them show the same number is 4/9.