A survey of 500 television viewers produced the following information;

Solution:

Let F, H B denote the sets of students who watch football, hockey and basketball, respectively.

Also, let U be the universal set.

We have:

n(F) = 285, n(H) = 195, n(B) = 115, n(F∩">∩∩B) = 45, n(F∩">∩∩H) = 70 and n(H∩">∩∩B) = 50

Also, we know:

$n\left(\mathrm{~F}^{\prime} \cap \mathrm{H}^{\prime} \cap \mathrm{B}^{\prime}\right)=50$

 

$\Rightarrow n(\mathrm{~F} \cup \mathrm{H} \cup \mathrm{B})^{\prime}=50$

$\Rightarrow n(\mathrm{U})-n(\mathrm{~F} \cup \mathrm{H} \cup \mathrm{B})=50$

 

$\Rightarrow 500-n(\mathrm{~F} \cup \mathrm{H} \cup \mathrm{B})=50$

$\Rightarrow n(\mathrm{~F} \cup \mathrm{H} \cup \mathrm{B})=450$

Number of students who watch all three games $=n(\mathrm{~F} \cap \mathrm{H} \cap \mathrm{B})$

$\Rightarrow n(\mathrm{~F} \cup \mathrm{H} \cup \mathrm{B})-n(\mathrm{~F})-n(\mathrm{H})-n(\mathrm{~B})+n(\mathrm{~F} \cap \mathrm{B})+n(\mathrm{~F} \cap \mathrm{H})+n(\mathrm{H} \cap \mathrm{B})$

$\Rightarrow 450-285-195-115+45+70+50$

 

$\Rightarrow 20$

Number of students who watch exactly one of the three games

$=n(\mathrm{~F})+n(\mathrm{H})+n(\mathrm{~B})-2\{n(\mathrm{~F} \cap \mathrm{B})+n(\mathrm{~F} \cap \mathrm{H})+n(\mathrm{H} \cap \mathrm{B})\}+3\{n(\mathrm{~F} \cap \mathrm{H} \cap \mathrm{B})\}$

$=285+195+115-2(45+70+50)+3(20)$

 

$=325$