1989 AMC 12 第 20 题

Question

设

x

是从

100

到

200

之间均匀随机选取的实数。已知

\lfloor\sqrt{x}\rfloor=12

，求

\lfloor\sqrt{100x}\rfloor=120

的概率。（

\lfloor v\rfloor

表示不大于

v

的最大整数。）

Let

x

be a real number selected uniformly at random between

100

and

200.

If

\lfloor\sqrt{x}\rfloor=12,

find the probability that

\lfloor\sqrt{100x}\rfloor=120.

(

\lfloor v\rfloor

means the greatest integer less than or equal to

v.

)

\frac2{25}

\frac{241}{2500}

\frac1{10}

\frac{96}{625}

1

把每个取整条件转化为

x

所在的区间

Convert each floor condition into an interval for

x

条件概率等于两个相关区间长度之比

The conditional probability is the ratio of the two relevant interval lengths

已知条件等价于

144\le x\lt169

，该区间的长度为

25

。所求条件等价于

120^2\le100x\lt121^2\text{，}

即

144\le x\lt146.41

。这个区间的长度为

2.41=\frac{241}{100}

，所以条件概率为

\frac{\frac{241}{100}}{25}=\frac{241}{2500}\text{。}

所以正确答案是 B。

The given condition is

144\le x\lt169,

an interval of length

25.

The desired condition is

120^2\le100x\lt121^2,

or

144\le x\lt146.41.

Its length is

2.41=\frac{241}{100},

so the conditional probability is

\frac{\frac{241}{100}}{25}=\frac{241}{2500}.

Thus the correct answer is B.

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