Find a solution to the third problem.

Assuming that 1 is the same as the second dice and the third dice is different, then:

The number of the first dice is arbitrary and the probability is1;

The second dice has the same number of points as the first, and the probability is1/6;

The third dice is different from the first two with a probability of 5/6;

So the total probability is1× (1/6) × (5/6) = 5/36.

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Assuming that 1 is the same as the third dice, it also gets 5/36;

Assuming that the second and third dice are the same, 5/36 is also obtained;

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So the final probability is 3×(5/36)=5/ 12, so choose B.

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