There are 30 students each of which has one pen, one pencil, and one eraser. Then 26 students lost their pens, 23 students lost their erasers, and 21 students lost their pencils. What is the minimal possible number of students who lost all three things?
题目要求的是minimal possible number of students who lost all three things,遗失三种东西的学生数的最小值。
这不是概率问题,是容斥极值问题,即求三个集合公共部分的数量最小值。这种问题有容斥极值公式直接求解:
min(A∩B∩C) = A+B+C-2I
其中A,B,C分别代表三个集合元素数量,I代表全集的数量
即丢失全部三种东西的学生数量最少为26+23+21-30×2=10个