1985 AMC 12 第 24 题

Question

按如下方式选取一个非零数字：选中数字

d

的概率为

\log_{10}(d+1)-\log_{10}d

。选中数字

2

的概率恰为所选数字属于下列哪个集合的概率的

\frac{1}{2}

？

A non-zero digit is chosen in such a way that the probability of choosing digit

d

is

\log_{10}(d+1)-\log_{10}d.

The probability that the digit

2

is chosen is exactly

\frac{1}{2}

the probability that the digit chosen is in the set

\{2,3\}

\{3,4\}

\{4,5,6,7,8\}

\{5,6,7,8,9\}

\{4,5,6,7,8,9\}

数字

2

的概率的两倍为

2\log_{10}(\frac{3}{2})

Twice the probability of digit

2

is

2\log_{10}(\frac{3}{2})

对连续数字集合的概率求和时，可将对数逐项消去

Sum the probabilities over a consecutive set by telescoping its logarithms

选中

2

的概率为

\log_{10}(\frac{3}{2})

，所以其两倍为

\log_{10}(\frac{9}{4})

。对于从

4

到

8

的数字，概率和逐项消去：

\begin{aligned} &\sum_{d=4}^{8} \bigl(\log_{10}(d+1)\\ &\qquad{}-\log_{10}d\bigr) =\log_{10}\frac94\text{。} \end{aligned}

因此该集合的概率恰为选中数字

2

的概率的两倍。

所以正确答案是 C。

The probability of

2

is

\log_{10}(\frac{3}{2}),

so twice that probability is

\log_{10}(\frac{9}{4}).

For the digits

4

through

8,

the sum telescopes:

\begin{aligned} &\sum_{d=4}^{8} \bigl(\log_{10}(d+1)\\ &\qquad{}-\log_{10}d\bigr) =\log_{10}\frac94. \end{aligned}

Therefore that set has exactly twice the probability of digit

2.

Thus the correct answer is C.

其他年份的第 24 题