ExplanationThere are a total of 63 = 216 total possible rolls for the three dice.
First figure out the probability of getting exactly two 1’s. There are 5 × 3 = 15 ways this could happen: 112, 113, 114, 115, 116; 121, 131, 141, 151, 161; or 211, 311, 411, 511, 611.
You could repeat this list of 15 possibilities in the obvious way for exactly two 2’s, exactly two 3’s, and so on. Thus, the total number of favorable rolls is 6 × 15 = 90. Because there are 216 possible rolls, 90 of which are favorable, the probability of getting exactly two of the three dice to show the same number is \(\frac{90}{216}=\frac{5}{12}\),
choice A.
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