Question

A survey of $500$ television watchers produced the following information; $285$ watch foot-ball, $195$ 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 ? How many watch exactly one of the three games ?

Answer 1

20, Only football 190 etc.
Given $N=500,n\left(F\right)=285,$
$n\left(H\right)=195,n\left(B\right)=115,n(F\cap B)=45,$
$n(F\cap H)=70,n(H\cap B)=50,$
$n(F^{\prime }\cap H^{\prime }\cap B^{\prime })=50.$
To find $n(F\cap H\cap B),n(F\cap H^{\prime }\cap B^{\prime }),n(F^{\prime }\cap H\cap B^{\prime })andn(F^{\prime }\cap H^{\prime }\cap B)$
we have
$50=n(F^{\prime }\cap H^{\prime }\cap B^{\prime })=n{(F\cup H\cup B)}^{{}^{\prime }},$
By De-Morgan law
$=N-n(F\cup H\cup B)=N-\{S_1-S_2+S_3\}$
$=N-\{n\left(F\right)+n\left(H\right)+n\left(B\right)-n(F\cap H)-n(H\cap B)-n(B\cap F)+n(F\cap H\cap B)\}$
$=500-285-195-115+70+50+45-n(F\cap H\cap B)$
$=665-595--n(F\cap H\cap B)$
$=70-n(F\cap H\cap B)$
$\therefore n(F\cap H\cap B)=20$
Again, $n(F\cap H^{\prime }\cap B^{\prime })=\{F\cap {(H\cup B)}^{{}^{\prime }}\}$
By De-Morgan law
$=n\left(F\right)-\{F\cap (H\cup B)\}$
$=n\left(F\right)-n\{(F\cap H)\cup (F\cap B)\}$
$=N\left(F\right)-n\{n(F\cap H)+n(F\cap B)-n(F\cap H\cap B)\}$
$=285-70-45+20=190$