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

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 $2$ ，得到种配置。 $28 \cdot 15 \div 2 = 210$

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

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

解答：

其他年份的第 21 题