AMC10 2014 B

AMC10 2014 B · Q16

AMC10 2014 B · Q16. It mainly tests Basic counting (rules of product/sum), Probability (basic).

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

掷四个公平的六面骰子。四骰子中至少有三骰子显示相同数值的概率是多少？

(A) $\frac{1}{36}$ $\frac{1}{36}$

(B) $\frac{7}{72}$ $\frac{7}{72}$

(D) $\frac{5}{36}$ $\frac{5}{36}$

(E) $\frac{1}{6}$ $\frac{1}{6}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

If exactly three of the four dice show the same number, then there are 6 possible choices for the repeated value and 5 possible choices for the non-repeated value. The non-repeated value may appear on any one of the 4 dice, so there are 6·5·4=120 possible ways for such a result to occur. There are 6 ways for all four dice to show the same value. There are 6^4 total possible outcomes for the four dice. The probability of the desired result is $\frac{120+6}{6^4} = \frac{7}{72}$.

如果恰好三骰显示相同数字，则重复数值有6种选择，非重复数值有5种选择。非重复数值可出现在4个骰子中的任意一个，因此有$6\cdot5\cdot4=120$种可能方式发生这种情况。四骰全相同有6种方式。四骰总可能结果为$6^4$。所需结果的概率为$\frac{120+6}{6^4}=\frac{7}{72}$。

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