1972 AMC 12 第 17 题

Question

在一根绳子上随机选一点将其剪成两段。较长的一段至少是较短一段的

x

倍的概率为：

A piece of string is cut in two at a point selected at random. The probability that the longer piece is at least

x

times as large as the shorter piece is:

\dfrac12

\dfrac2x

\dfrac1{x+1}

\dfrac1x

\dfrac2{x+1}

将绳长归一化为

1

，并设剪切点距某一端为

t

Normalize the string length to

1

and let the cut be

t

units from one end

对靠近任一端的区间，求解

1-t\ge xt

Near either endpoint, solve

1-t\ge xt

令绳长为

1

。在靠近一端的区间内，条件为

1-t\ge xt\text{，}

所以

t\le\frac{1}{x+1}

。另一端也有同样长度的区间。两区间总长度即所求概率，为

\frac2{x+1}\text{。}

因此，正确答案是 E。

Take the string to have length

1.

Near one endpoint, the condition is

1-t\ge xt,

so

t\le\frac{1}{x+1}.

The same interval occurs at the other endpoint. Their total length, hence the probability, is

\frac2{x+1}.

Therefore, the correct answer is E.

其他年份的第 17 题