1991 AMC 12 第 13 题

Question

马

X

、

Y

和

Z

参加一场不会出现平局的三马赛跑。若

X

获胜的反向赔率为

3

比

1

，而

Y

获胜的反向赔率为

2

比

3

，那么

Z

获胜的反向赔率是多少？（“

H

获胜的反向赔率为

p

比

q

”是指

H

获胜的概率为

\frac{q}{p+q}

。）

Horses

X,

Y,

and

Z

are entered in a three-horse race in which ties are not possible. If the odds against

X

winning are

3

-to-

1

and the odds against

Y

winning are

2

-to-

3,

what are the odds against

Z

winning? (By “odds against

H

winning are

p

-to-

q

” we mean that the probability of

H

winning the race is

\frac{q}{p+q}.

)

3

比

20

3

-to-

20

5

比

6

5

-to-

6

8

比

5

8

-to-

5

17

比

3

17

-to-

3

20

比

3

20

-to-

3

答案：D

知识点：probability odds complementary events

难度评级：1520

小提示：

把每组赔率换算为概率

Convert each set of odds to a probability

大提示：

三匹马的获胜概率之和为

1

The three winning probabilities sum to

1

解答：

已知概率为

P(X)=\frac14

和

P(Y)=\frac35

。因此

P(Z)=1-\frac14-\frac35=\frac3{20}\text{。}

Z

失败的概率为

\frac{17}{20}

，所以

Z

获胜的反向赔率为

17

比

3

。

因此正确答案为 D。

The given probabilities are

P(X)=\frac14

and

P(Y)=\frac35.

Therefore

P(Z)=1-\frac14-\frac35=\frac3{20}.

The probability that

Z

loses is

\frac{17}{20},

so the odds against

Z

are

17

-to-

3.

Thus the correct answer is D.

1991 AMC 12 第 13 题

答案：D

难度评级：1520

小提示：

大提示：

解答：

其他年份的第 13 题