Deriving the number of students reading three newspapers based on information regarding readership of each

Question

There are $3$ newspapers $A , B , C$ and $400$ students. $180$ students read $A$; $270$ read $B$; and $371$ read $C$. I want to figure out at least how many are there who read all three.

I tried with Venn diagrams but ended up having two many variables and too less information. May be I am not doing it right. How to do this?

Answer 1

Asssume that we have a Venn diagram like below:

We will have

${\begin{cases}a_1+a_4+a_6+a_7=180\\a_2+a_4+a_5+a_7=270\\a_3+a_5+a_6+a_7=371\\a_1+a_2+a_3+a_4+a_5+a_6+a_7=400\end{cases}}$ and we need to find the minimum value of $a_7$.

We know that:

• At least zero students only read one newspaper $A$.

• At least zero students only read one newspaper $B$.

• At least zero students only read one newspaper $C$.

$\Rightarrow a_1+a_2+a_3\ge 0$.

We also have

$a_1+a_2+a_3\ge 0$

$\Rightarrow a_7-a_7+a_1+a_2+a_3\ge 0$

$\Rightarrow a_7\ge -a_1-a_2-a_3+a_7$

$\Rightarrow a_7\ge (a_1-2a_1)+(a_2-2a_2)+(a_3-2a_3)+(a_4+a_4-2a_4)+(a_5+a_5-2a_5)+(a_6+a_6-2a_6)+(a_7+a_7+a_7-2a_7)$

$\Rightarrow a_7\ge a_1+a_4+a_6+a_7+a_2+a_4+a_5+a_7+a_3+a_5+a_6+a_7-2a_1-2a_2-2a_3-2a_4-2a_5-2a_6-2a_7$

$\Rightarrow a_7\ge A+B+C-2(a_1+a_2+a_3+a_4+a_5+a_6+a_7)$

$\Rightarrow a_7\ge 180+270+371-2\times 400$

$\Rightarrow a_7\ge 21$

The equality holds if $a_1=a_2=a_3=0;a_4=29;a_5=220;a_6=130;a_7=21$.

So at least $21$ students read all three newspapers. This method can be generalized for most cases.

Answer 2

Maximize the number of people who read $2$ newspapers in order to find the least amount of people who read all three newspapers.

Answer 3

Assuming everyone reads at least one newspaper, the addition formula: $$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C) \Rightarrow\\ 400=180+270+371-(P(A\cap B)+P(A\cap C)+P(B\cap C))+P(A\cap B\cap C).$$ To minimize $P(A\cap B\cap C)$, we must maximize the number inside brackets, for which we must minimize the number of people reading only $A$, $B$, $C$ (i.e. set them equal to $0$): $$P(A\cap B\cap C)=P(A\cap B)+P(A\cap C)+P(B\cap C)-421=\\ 400+2P(A\cap B\cap C)-421 \Rightarrow \\ P(A\cap B\cap C)=21.$$