AMC10 2019 A

AMC10 2019 A · Q4

AMC10 2019 A · Q4. It mainly tests Pigeonhole principle.

A box contains 28 red balls, 20 green balls, 19 yellow balls, 13 blue balls, 11 white balls, and 9 black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 15 balls of a single color will be drawn?

一个盒子中有 28 个红球，20 个绿球，19 个黄球，13 个蓝球，11 个白球，9 个黑球。从盒中不放回地抽取最少多少个球，才能保证至少有一种颜色的球抽到 15 个？

(A) 75 75

(B) 76 76

(D) 84 84

(E) 91 91

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The greatest number of balls that can be drawn without getting 15 of one color is obtained if and only if 14 red, 14 green, 14 yellow, 13 blue, 11 white, and 9 black balls are drawn, a total of 75 balls. When another ball is drawn, it will be the 15th ball of one of the colors red, green, or yellow. Thus the requested minimum is $75+1=76$.

答案（B）：在不抽到某一种颜色达到 15 个的情况下，最多能抽到的球数当且仅当抽到 14 个红球、14 个绿球、14 个黄球、13 个蓝球、11 个白球和 9 个黑球时取得，总共 75 个球。再抽取一个球时，它将成为红、绿或黄三种颜色之一的第 15 个球。因此所求的最小值为 $75+1=76$。

Topics

CP.005 Pigeonhole principle