2014 AMC 10B Problem 16

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Concepts:dice (probability)casework

Difficulty rating: 1420

16.

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

136 \dfrac{1}{36}

772 \dfrac{7}{72}

19 \dfrac{1}{9}

536 \dfrac{5}{36}

16 \dfrac{1}{6}

Solution:

There are 646^4 equally likely ordered outcomes.

If exactly three dice show the same value, choose the repeated value in 66 ways, the different value in 55 ways, and the position of the different die in 44 ways. This gives 654=1206\cdot5\cdot4=120 outcomes.

If all four dice match, there are 66 outcomes.

The probability is 120+664=1261296=772\frac{120+6}{6^4}=\frac{126}{1296}=\frac7{72}.

Thus, the correct answer is B .

Problem 16 in Other Years