prop the ability

Probability Level pending

Let w be a complex cube root of unity with w≠1. A fair dice is thrown three times if a,b,and c are the number obtained on the dice , then the propability that w^{a}+w^{b}+w^{c}=0 is??


The answer is 0.2222222222.

Saurabh Gupta by Saurabh Gupta , 24, India

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1 solution

Saurabh Gupta
Jun 4, 2014

In complex numbers
In Cube root of unity w¹+w²+w³=0 so it s easy that a=1,b=2,andc=3 in this problem In short u have to find the propability that on throwing the dice 3 times 1,2,3 comes in any pattern which is simple and after long calculation it will be= 2÷9=0.222222…………………

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If {a, b, c} = {1, 2, 3}, or any permutation of it, then it adds up to 0. But {4, 5, 6} leaves the powers the same as {1, 2, 3}, so that the 1st die can be any number {1, 2, 3, 4, 5, 6}. The 2nd can only be one of 4 numbers not the same as the first, viz, e.g., if a = 1, then b = {2, 3, 5, 6}. The 3rd can only be one of 2 numbers not the same as the 1st and 2nd. Hence, the total number possible is 6 x 4 x 2 = 48, which, divided by 6 x 6 x 6, give us 2/9 = 0.222222....

Michael Mendrin - 6 years, 11 months ago

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