2007 AMC 8 第 25 题

Question

下图的飞镖盘中，外圆半径为

6

，内圆半径为

3

。三条半径把每个圆分成三个全等区域，并标出分值。飞镖击中某区域的概率与该区域面积成正比。当两支飞镖击中这个靶时，得分为击中区域分值之和。得分为奇数的概率是多少？

On the dart board shown in the figure below, 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 of the regions hit. What is the probability that the score is odd?

\dfrac{17}{36}

\dfrac{35}{72}

\dfrac{1}{2}

\dfrac{37}{72}

\dfrac{19}{36}

答案：B

知识点：几何概率奇偶性

难度评级：1680

小提示：

奇数得分来自一支飞镖落在

1

分区域，另一支落在

2

分区域

An odd score comes from one dart landing on a

1

-region and the other on a

2

-region.

大提示：

先按面积权重求落在

1

分区域的总概率，再求落在

2

分区域的总概率

Compute the total area-weighted probability of landing on

1,

then of landing on

2.

解答：

外圆面积为

6^2\pi = 36\pi

，内圆面积为

3^2\pi = 9\pi

。因此外环面积为

36\pi - 9\pi = 27\pi

。

所以击中一个外环扇区的概率为

\dfrac{9\pi}{36\pi} = \dfrac{1}{4}

。类似地，击中一个内圆扇区的概率为

\dfrac{3\pi}{36\pi} = \dfrac{1}{12}

。

汇总所有可能，击中

1

分区域的概率为

\dfrac{1}{12} + 2 \cdot \dfrac{1}{4} = \dfrac{7}{12}\text{。}

击中

2

分区域的概率为

2 \cdot \dfrac{1}{12} + \dfrac{1}{4} = \dfrac{5}{12}\text{。}

得到奇数得分的唯一方式是一支击中

1

分，另一支击中

2

分。

两种顺序的总概率为

2 \cdot \dfrac{5}{12} \cdot \dfrac{7}{12} = \dfrac{35}{72}\text{。}

所以正确答案是 B。

The area of the outer circle is

6^2\pi = 36\pi,

and the area of the inner circle is

3^2\pi = 9\pi.

Therefore, the area of the outer ring is

36\pi - 9\pi = 27\pi.

This means that the probability of hitting an outer segment is

\dfrac{9\pi}{36\pi} = \dfrac{1}{4}.

Similarly, the probability of hitting an inner segment is

\dfrac{3\pi}{36\pi} = \dfrac{1}{12}.

Summing over all possibilities, the probability of hitting a

1

is

\dfrac{1}{12} + 2 \cdot \dfrac{1}{4} = \dfrac{7}{12}.

Similarly, the probability of hitting a

2

is

2 \cdot \dfrac{1}{12} + \dfrac{1}{4} = \dfrac{5}{12}.

The only way to get an odd score is to hit one

1

and one

2.

The probability of hitting these numbers in either order is

2 \cdot \dfrac{5}{12} \cdot \dfrac{7}{12} = \dfrac{35}{72}.

Thus, B is the correct answer.

2007 AMC 8 第 25 题

答案：B

难度评级：1680

小提示：

大提示：

解答：

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