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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.
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