2014 AMC 10B Problem 16

Question

Four fair six-sided dice are rolled. What is the probability that at least three of the four dice show the same value?

\dfrac{1}{36}

\dfrac{7}{72}

\dfrac{1}{9}

\dfrac{5}{36}

\dfrac{1}{6}

Count exactly three equal dice and four equal dice separately

For exactly three equal dice, choose the odd die position

There are

6^4

equally likely ordered outcomes.

If exactly three dice show the same value, choose the repeated value in

6

ways, the different value in

5

ways, and the position of the different die in

4

ways. This gives

6\cdot5\cdot4=120

outcomes.

If all four dice match, there are

6

outcomes.

The probability is

\frac{120+6}{6^4}=\frac{126}{1296}=\frac7{72}

.

Thus, the correct answer is B .

Problem 16 in other years