1980 AMC 12 第 20 题

Question

一个盒子里有

2

枚一美分硬币、

4

枚五美分硬币和

6

枚十美分硬币。在不放回且每枚硬币被选中的概率相同的情况下抽取六枚硬币。抽到的硬币总值至少为

50

美分的概率是多少？

A box contains

2

pennies,

4

nickels and

6

dimes. Six coins are drawn without replacement, with each coin having an equal probability of being chosen. What is the probability that the value of the coins drawn is at least

50

cents?

\frac{37}{924}

\frac{91}{924}

\frac{127}{924}

\frac{132}{924}

以上都不是

none of these

答案：C

知识点：组合无放回抽样分类讨论

难度评级：2100

小提示：

\binom{12}{6}

种等可能的六枚硬币组合

There are

\binom{12}{6}

equally likely sets of six coins

大提示：

按照抽到六枚、五枚或四枚十美分硬币分类计数有利情形

Classify favorable selections by whether they contain six, five, or four dimes

解答：

\binom{12}{6}=924

种选法。总值至少为

50

美分的情形包括：六枚都是十美分硬币；五枚十美分硬币加任意一枚其他硬币；或四枚十美分硬币加两枚五美分硬币。有利选法数为

\begin{aligned} &\binom66\\ &\quad+\binom65\binom61\\ &\quad+\binom64\binom42\\ &=1+36+90\\ &=127 \end{aligned}\text{。}

因此概率为

\frac{127}{924}

。

因此，正确答案是 C。

There are

\binom{12}{6}=924

selections. A value of at least

50

cents occurs with six dimes; with five dimes and any one other coin; or with four dimes and two nickels. The favorable count is

\begin{aligned} &\binom66\\ &\quad+\binom65\binom61\\ &\quad+\binom64\binom42\\ &=1+36+90\\ &=127. \end{aligned}

Thus the probability is

\frac{127}{924}.

Therefore, the correct answer is C.

1980 AMC 12 第 20 题

答案：C

难度评级：2100

小提示：

大提示：

解答：

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