Given that A gambler rolls three fair six-sided dice and We need to find What is the probability that two of the dice show the same number, but the third shows a different number?

As we are rolling three dice => Number of cases = \(6^3\) = 216

Now, out of the three rolls lets pick the two rolls which show the same number. We can do that in 3C2 ways
= \(\frac{3!}{2!*(3-2)!}\) = \(\frac{3*2!}{2!*1!}\) = 3 ways

Now, out of the 6 numbers the two dice which show the same number can show any of these 6 numbers in 6 ways

The third die can show any number apart from the number which these two dice are showing in 5 ways. (5 numbers out of 6 except the number which the two dice are showing)

=> Total number of ways = 3 * 6 * 5

=> Probability that two of the dice show the same number, but the third shows a different number = \(\frac{3*6*5}{216}\) = \(\frac{5}{12}\)

So, Answer will be D
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems