AMC8 2011

AMC8 2011 · Q18

AMC8 2011 · Q18. It mainly tests Basic counting (rules of product/sum), Probability (basic).

A fair six-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number?

一个公平的六面骰子掷两次。第一骰子显示的数字大于等于第二骰子显示的数字的概率是多少？

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

(B) $\frac{5}{12}$ $\frac{5}{12}$

(D) $\frac{7}{12}$ $\frac{7}{12}$

(E) $\frac{5}{6}$ $\frac{5}{6}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

There are $6\cdot6=36$ ways to roll the two dice, and 6 of them result in two of the same number. Out of the remaining $36-6=30$ ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are $30/2=15$ ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is \[\frac{15+6}{36}=\frac{21}{36}=\boxed{\textbf{(D)}\ \frac{7}{12}}\]

掷两个骰子共有$6\cdot6=36$种方式，其中6种是两个数字相同。其余$36-6=30$种方式中，第一骰子大于第二骰子的次数应该等于第二骰子大于第一骰子的次数。也就是说，有$30/2=15$种方式第一骰子大于第二骰子。第一骰子大于等于第二骰子的概率为 \[\frac{15+6}{36}=\frac{21}{36}=\boxed{\textbf{(D)}\ \frac{7}{12}}\]

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