2011 AMC 10A 第 21 题

Question

两枚重量相同的假币与

8

枚相同的真币混在一起。每枚假币的重量不同于每枚真币。先从

10

枚硬币中随机不放回选出一对硬币，再从剩下

8

枚中随机不放回选出第二对硬币。已知第一对硬币的总重量等于第二对硬币的总重量。所选

4

枚硬币全是真币的概率是多少？

Two counterfeit coins of equal weight are mixed with

8

identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the

10

coins. A second pair is selected at random without replacement from the remaining

8

coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all

4

selected coins are genuine?

\dfrac{7}{11}

\dfrac{9}{13}

\dfrac{11}{15}

\dfrac{15}{19}

\dfrac{15}{16}

答案：D

知识点：条件概率组合分类讨论

难度评级：1990

小提示：

两对重量相等要求每对中假币数量相同。

Equal pair weights require the same number of counterfeit coins in each pair

大提示：

统计全是真币的情况和每对各有一枚假币的情况。

Count the all-genuine case and the one-counterfeit-in-each-pair case

解答：

可能情况只有两类：两对都全是真币，或每一对都含一枚假币。

全是真币时，选第一对有

\binom{8}{2} = 28

种，第二对有

\binom{6}{2} = 15

种。

两对可交换，所以还要除以

2

，得到

28 \cdot 15 \div 2 = 210

种配置。

每对各有一枚假币时，选择两枚真币有

\binom{8}{2} = 28

种，而假币的选择只有一种。

把两枚假币分别与两枚真币配对有两种方法。

因此这一类共有

28 \cdot 2 = 56

种。

因此所求概率为

\dfrac{210}{210 + 56} = \dfrac{210}{266} = \dfrac{15}{19}\text{。}

所以正确答案是 D。

There are two cases: either both selected pairs contain only genuine coins or each selected pair has one counterfeit coin.

For the first case, there are

\binom{8}{2} = 28

ways to choose the coins for the first pair and

\binom{6}{2} = 15

choices for the second pair.

We also have to divide by

2

since we can swap the pairs. This gives us

28 \cdot 15 \div 2 = 210

configurations for this case.

For the second case, there are

\binom{8}{2} = 28

ways to choose the non-counterfeit coins. There is only one choice for the counterfeit coins.

There are two ways to create the two pairs, two choices for which counterfeit coin goes with a genuine coin.

This means that there are

28 \cdot 2 = 56

configurations for this case.

The desired probability is then

\dfrac{210}{210 + 56} = \dfrac{210}{266} = \dfrac{15}{19}.

Thus, D is the correct answer.

2011 AMC 10A 第 21 题

答案：D

难度评级：1990

小提示：

大提示：

解答：

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