chaitanya.mehrotra wrote:A gambler rolls three fair six-sided dice. What is the probability that two of
the dice show the same number, but the third shows a different number?
P(exactly n times) = P(one way) * total possible ways.
P(one way):
One way to get exactly 2 of the same number is to get the same number on the first two rolls but not on the third roll.
P(second number is the same as the first) = 1/6.
P(third number is different) = 5/6.
Since we want both of these events to happen, we multiply the fractions:
P(first two rolls are the same but the third is different) = 1/6 * 5/6 = 5/36.
Total possible ways:
The number that is different could appear on the first, second or third roll.
Thus, the result above must be multiplied by 3.
P(exactly 2 of the same number) = 5/36 * 3 = 5/12.
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