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?	Choose one
A) 1/36
B) 5/36
C) 10/36
D) 15/36
E) 20/36
First off, we want to find the probability of rolling a pair of any number. There are six   pairs and a total of 36 different ways to roll a pair of dices: 6/36 = 1/6.   Next, we want to account for the third die - it cannot be the same as the first two,   so it can be any of five numbers out of 6: 5/6.   Therefore probability is 5/6 x 1/6 = 5/36.   It is tempting here to think that we have solved the problem. But if we read carefully, the   question does not say that the first two dice have to be the same and the third die has   to be different.   the question says a gambler rolls 3 dice. So the first two do not have to be pairs.   As long as 2 of the 3 dice are pairs and the last die is a different number. To find the   number of way this can occur, we use combinations formula: 3!/2!1! = 3.   We multiply this to the numerator to give us 3 x 5/36 = 15/36.   ANSWER:15/36
<result>D</result>
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?