Question

A survey of 500 television viewers 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 ? How many watch exactly one of the three games ?

Answer 1

Let,

n(P) , denote total number of television viewers,

n(F) be the number of people who watch football,

n(H) be the number of people who watch hockey and

n(B) be the number of people who watch basket ball.

Then, n (P) = 500, n (F) = 285, n(H) = 195, n (B) = 115, $n(F\cap B)=45,n(F\cap H)=70.$

$n(H\cap B)=50andn(F\cup H\cup B)=50$

Now,

n{$(F\cup H\cup B{)}^{\prime }$} = $n\left(P\right)-n(F\cup H\cup B)$

$\Rightarrow 50=500-$ {n (F)+n(H)+n(B) - $n(F\cap H)-n(H\cap B)-n(F\cap H\cap B)$}

$\Rightarrow$ 50= 500- {285+195+115-70-50-45+$n(F\cap H\cap B)$ }

$\Rightarrow$ 50 = 500-430 - $n(F\cap H\cap B)$

$\Rightarrow$ 50 = 70 - $n(F\cap H\cap B)$

$\Rightarrow$ $n(F\cap H\cap B)$ = 70-50 = 20

Hence, 20 people watch all the 3 games

Number of people who watch only football

= 285- (50+20+25)

= 285 - 95

= 190

Number of people who watch only hockey

= 195 - (50+20+30)

= 195-100

= 95

And, Number of people who watch only basket ball

= 115- (25+20+30)

= 115-75

= 40

Number of people who watch exactly one of the three games

= Number of people who watch either football only or hockey only or basket only

= 190+95+40 [$\because$ they are pairwise disjoint]

= 325

Hence, 325 people watch exactly one of the three games.

Number of people who watch exactly one of the three games.

= number of people who watch either football only or hockey only or basketball only.

= 190+95+40 [$\because$ they are pairwise disjoint]

= 325

Hence, 325 people who watch exactly one of the three games.