AMC10 2011 B

AMC10 2011 B · Q13

AMC10 2011 B · Q13. It mainly tests Probability (basic), Coordinate geometry.

Two real numbers are selected independently at random from the interval [−20, 10]. What is the probability that the product of those numbers is greater than zero?

从区间 [-20, 10] 中独立随机选取两个实数。它们的乘积大于 0 的概率是多少？

(A) $\frac{1}{9}$ $\frac{1}{9}$

(B) $\frac{1}{3}$ $\frac{1}{3}$

(D) $\frac{5}{9}$ $\frac{5}{9}$

(E) $\frac{2}{3}$ $\frac{2}{3}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Consider all ordered pairs $(a,b)$ with each of the numbers $a$ and $b$ in the closed interval $[-20,10]$. These pairs fill a $30\times 30$ square in the coordinate plane, with an area of $900$ square units. Ordered pairs in the first and third quadrants have the desired property, namely $a\cdot b>0$. The areas of the portions of the $30\times 30$ square in the first and third quadrants are $10^2=100$ and $20^2=400$, respectively. Therefore the probability of a positive product is $\frac{100+400}{900}=\frac{5}{9}$.

答案（D）：考虑所有有序对$(a,b)$，其中$a$和$b$都在闭区间$[-20,10]$内。这些点对在坐标平面上构成一个$30\times 30$的正方形，面积为$900$平方单位。第一象限和第三象限中的有序对满足所需性质，即$a\cdot b>0$。该$30\times 30$正方形位于第一象限和第三象限的部分面积分别为$10^2=100$和$20^2=400$。因此，乘积为正的概率为$\frac{100+400}{900}=\frac{5}{9}$。

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