2005 AMC 12B 第 11 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。

11.

一个信封中有八张纸币:22 张一元、22 张五元、22 张十元、22 张二十元。不放回地随机抽出两张。它们面值和至少为 $20\$20 的概率是多少?

An envelope contains eight bills: 22 ones, 22 fives, 22 tens, and 22 twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $20\$20 or more?

14\dfrac{1}{4}

27\dfrac{2}{7}

37\dfrac{3}{7}

12\dfrac{1}{2}

23\dfrac{2}{3}

答案:D
知识点:基本概率组合分类讨论
难度评级:1500
解答:

共有 (82)=28\binom{8}{2} = 28 对等可能的纸币。

总额达到 $20\$20 或更多的情况有:两张二十(11 种),一张二十配六张较小纸币之一(26=122 \cdot 6 = 12 种),以及两张十(11 种)。

有利情况共 1+12+1=141 + 12 + 1 = 14 种,所以概率为 1428=12\dfrac{14}{28} = \dfrac12

所以正确答案是 D

There are (82)=28\binom{8}{2} = 28 equally likely pairs of bills.

The sum is $20\$20 or more in these cases: both twenties (11 way), one twenty with one of the six smaller bills (26=122 \cdot 6 = 12 ways), or both tens (11 way).

That is 1+12+1=141 + 12 + 1 = 14 favorable pairs, so the probability is 1428=12.\dfrac{14}{28} = \dfrac12.

Thus, the correct answer is D.

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