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2010 AMC 12A 第 16 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

16.

Bernardo 随机选出 33 个不同的数，这些数来自集合 {1,2,3,4,5,6,7,8,9}\{1,2,3,4,5,6,7,8,9\} 并按降序排列形成一个 33 位数。Silvia 也随机选出 33 个不同的数，这些数来自集合 {1,2,3,4,5,6,7,8}\{1,2,3,4,5,6,7,8\} 并按降序排列形成一个 33 位数。Bernardo 的数大于 Silvia 的数的概率是多少？

Bernardo randomly picks 33 distinct numbers from the set {1,2,3,4,5,6,7,8,9}\{1,2,3,4,5,6,7,8,9\} and arranges them in descending order to form a 33-digit number. Silvia randomly picks 33 distinct numbers from the set {1,2,3,4,5,6,7,8}\{1,2,3,4,5,6,7,8\} and also arranges them in descending order to form a 33-digit number. What is the probability that Bernardo’s number is larger than Silvia’s number?

4772\dfrac{47}{72}

3756\dfrac{37}{56}

23\dfrac{2}{3}

4972\dfrac{49}{72}

3956\dfrac{39}{56}

答案：B
知识点：基本概率分类讨论对称性
难度评级：1900
小提示：

按 Bernardo 是否选到 99 分情况。

Separate cases according to whether Bernardo picks 99

大提示：

若没有 99，两人的数对称，除了选到相同集合的情况。

Without a 9,9, the two numbers are symmetric except when the chosen sets match

解答：

分两种情况：Bernardo 选到 99，或者没有选到。

情况 11： Bernardo 选到 99。

固定一个数后，另外两个数有 (82)=28\binom{8}{2} = 28 种选法。

总选法为 (93)=84\binom{9}{3} = 84，所以此情况的概率是 2884=13。 \dfrac{28}{84} = \dfrac{1}{3}\text{。}

注意，如果 Bernardo 选到 99，他一定比 Silvia 的数大。

因此本情况中 Bernardo 一定获胜。

情况 22： Bernardo 没有选到 99。

这种情况发生的概率为 1−13=231 - \frac{1}{3} = \frac{2}{3}。此时两人从同一组数字中选择，获胜机会对称。

我们还需要求出两人选到相同数字的概率。Silvia 与 Bernardo 选到同一组数字的概率为 1(83)=156 \dfrac{1}{\binom{8}{3}} = \dfrac{1}{56} 于是此时 Bernardo 的数更大的概率为 1−1562=55112。 \dfrac{1 - \frac{1}{56}}{2} = \dfrac{55}{112}\text{。}

因此 Bernardo 得到较大数的总概率为 13+23⋅55112=3756。 \dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{55}{112} = \dfrac{37}{56}\text{。}

所以正确答案是 B。

There are two cases: Bernardo picks a 99 or he doesn’t.

Case 1:1: Bernardo picks a 99

Since a number is fixed, there are (82)=28\binom{8}{2} = 28 ways to choose the other two numbers.

There are a total of (93)=84\binom{9}{3} = 84 ways to pick all three numbers. The probability is then 2884=13. \dfrac{28}{84} = \dfrac{1}{3}.

Note that if Bernardo picks a 9,9, he automatically has a greater number than Silvia.

This means that Bernardo always wins in this case.

Case 2:2: Bernardo doesn’t pick a 99

There is a 1−13=231 - \frac{1}{3} = \frac{2}{3} chance of this happening. Since both people are choosing from the same numbers, they have an equal chance of winning.

We still need to find the probability that the numbers are the same. There is a 1(83)=156 \dfrac{1}{\binom{8}{3}} = \dfrac{1}{56} chance that Silvia chooses the same numbers as Bernardo. The probability that Bernardo gets a higher number is then 1−1562=55112. \dfrac{1 - \frac{1}{56}}{2} = \dfrac{55}{112}.

The total probability of Bernardo getting a higher number is then 13+23⋅55112=3756. \dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{55}{112} = \dfrac{37}{56}.

Thus, B is the correct answer.

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