Question

A survey of $500$ television watches produced the following information : $285$ watch football, $195$ watch hockey, $115$ watch basketball, $45$ watch football and basketball, $70$ watch football and hockey, $50$ watch hockey and basketball, $50$ do not watch any of the three games. How many watch all the three games?

Answer 1

Given , $N=500,n\left(F\right)=285,n\left(H\right)=195,n\left(B\right)=115,n(F\cap H)=70,n(H\cap B)=50,n(F\cap B)=45,n\left(\right(F\cup B\cup H{)}^{\prime })=50$
Now, $n(F\cup B\cup H)=N-n\left(\right(F\cup H\cup B{)}^{\prime })$
$\Rightarrow n(F\cup B\cup H)=450$
$\Rightarrow n\left(F\right)+n\left(B\right)+n\left(H\right)-n(F\cap B)-n(B\cap H)-n(H\cap F)+n(F\cap B\cap H)=450$
$\Rightarrow n(F\cap B\cap H)=20$