# A survey of 500 television viewers produced the following information;

Question:

$\cap$ 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?

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

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

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

$=325$