2014 AMC 10B 第 16 题

Question

掷四枚公平的六面骰子。至少三枚骰子显示相同点数的概率是多少？

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}

分别计数恰好三枚相同和四枚全相同。

Count exactly three equal dice and four equal dice separately

恰好三枚相同时，选择不同骰子的位置。

For exactly three equal dice, choose the odd die position

6^4

个等可能有序结果。

恰好三枚相同时，重复点数有

6

种，不同点数有

5

种，不同骰子的位置有

4

种，共

6\cdot5\cdot4=120

种。

四枚全相同有

6

种。

所求概率为

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

。

所以正确答案是 B。

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 .

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