Question

一个盒子里有 $28$ 个红球、 $20$ 个绿球、 $19$ 个黄球、 $13$ 个蓝球、 $11$ 个白球和 $9$ 个黑球。从盒中不放回地取球，最少要取多少个球，才能保证至少取到 $15$ 个同色球？

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?

$75$

$76$

$79$

$84$

$91$

答案：B

知识点：抽屉原理极端原理

难度评级：1070

解答：

在还不能保证有 $15$ 个同色球的最坏情况下，可以取出所有黑球、白球、蓝球，以及红、绿、黄各 $14$ 个。

此时红、绿、黄三种颜色都只有 $14$ 个，而其余三种颜色本来就少于十五个，所以仍可能没有任何颜色达到十五个。

总数为再多取一个球，就必定使某种颜色达到 $15$ 个，因此答案为 $75 + 1 = 76$ 。 $9 + 11 + 13 + 3 \cdot 14$ $= 33 + 42$ $= 75.$

所以正确答案是 B。

Note that we can pull as many as $14$ balls of each color without ensuring that $15$ balls of one color are drawn.

This means that we can draw all of the black, white and blue balls, along with $14$ red, green, and yellow balls.

This gives us a total of $9 + 11 + 13 + 3 \cdot 14$ $= 33 + 42$ $= 75.$ We need to add one at the end, however, to ensure that we get that $15$ th ball of some color, $75 + 1 = 76.$

Thus, B is the correct answer.

2019 AMC 10A 第 4 题

答案：B

难度评级：1070

解答：

其他年份的第 4 题