AMC10 2011 A

AMC10 2011 A · Q14

AMC10 2011 A · Q14. It mainly tests Linear equations, Probability (basic).

A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle’s circumference?

掷一对标准的6面骰子一次。所掷数字之和决定圆的直径。圆的面积的数值小于圆的周长的数值的概率是多少？

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

(B) $\frac{1}{12}$ $\frac{1}{12}$

(D) $\frac{1}{4}$ $\frac{1}{4}$

(E) $\frac{5}{18}$ $\frac{5}{18}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let $d$ be the sum of the numbers rolled. The conditions are satisfied if and only if $\pi (\frac{d}{2})^2 < \pi d$, that is, $d < 4$. Of the 36 equally likely outcomes for the roll of the two dice, one has a sum of 2 and two have sums of 3. Thus the desired probability is $\frac{1+2}{36} = \frac{1}{12}$.

设 $d$ 为掷得数字之和。条件等价于 $\pi (\frac{d}{2})^2 < \pi d$，即 $d < 4$。两骰子36种等可能结果中，和为2的一种，和为3的有两种。故概率为 $\frac{1+2}{36} = \frac{1}{12}$。

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