AMC8 2007

AMC8 2007 · Q25

AMC8 2007 · Q25. It mainly tests Probability (basic), Conditional probability (basic).

On the dart board shown in the figure, the outer circle has radius 6 and the inner circle has radius 3. Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd?

图示飞镖盘外圆半径6，内圆半径3。三条半径将每个圆分成三个相等区域，显示分数。飞镖击中某区域的概率与该区域面积成正比。两支飞镖击中该盘，分数为两个区域分数的和。分数为奇数的概率是多少？

(A) $\frac{17}{36}$ $\frac{17}{36}$

(B) $\frac{35}{72}$ $\frac{35}{72}$

(D) $\frac{37}{72}$ $\frac{37}{72}$

(E) $\frac{19}{36}$ $\frac{19}{36}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) The outer circle has area $36\pi$ and the inner circle has area $9\pi$, making the area of the outer ring $36\pi-9\pi=27\pi$. So each region in the outer ring has area $\frac{27\pi}{3}=9\pi$, and each region in the inner circle has area $\frac{9\pi}{3}=3\pi$. The probability of hitting a given region in the inner circle is $\frac{3\pi}{36\pi}=\frac{1}{12}$, and the probability of hitting a given region in the outer ring is $\frac{9\pi}{36\pi}=\frac{1}{4}$. For the score to be odd, one of the numbers must be 1 and the other number must be 2. The probability of hitting a 1 is \[ \frac{1}{4}+\frac{1}{4}+\frac{1}{12}=\frac{7}{12}, \] and the probability of hitting a 2 is \[ 1-\frac{7}{12}=\frac{5}{12}. \] Therefore, the probability of hitting a 1 and a 2 in either order is \[ \frac{7}{12}\cdot\frac{5}{12}+\frac{5}{12}\cdot\frac{7}{12}=\frac{70}{144}=\frac{35}{72}. \]

（B）外圆面积为 $36\pi$，内圆面积为 $9\pi$，因此外环面积为 $36\pi-9\pi=27\pi$。所以外环中每个区域的面积为 $\frac{27\pi}{3}=9\pi$，内圆中每个区域的面积为 $\frac{9\pi}{3}=3\pi$。击中内圆某一给定区域的概率为 $\frac{3\pi}{36\pi}=\frac{1}{12}$，击中外环某一给定区域的概率为 $\frac{9\pi}{36\pi}=\frac{1}{4}$。要使得得分为奇数，其中一个数必须是 1，另一个数必须是 2。击中 1 的概率是 \[ \frac{1}{4}+\frac{1}{4}+\frac{1}{12}=\frac{7}{12}, \] 击中 2 的概率是 \[ 1-\frac{7}{12}=\frac{5}{12}. \] 因此，以任意顺序击中一个 1 和一个 2 的概率为 \[ \frac{7}{12}\cdot\frac{5}{12}+\frac{5}{12}\cdot\frac{7}{12}=\frac{70}{144}=\frac{35}{72}. \]

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