A survey of 500 television viewers produced the following information, 285 watch foot ball, 195 watch hockey, 115 watch basket ball, 45 watch foot ball and basket ball, 70 watch foot ball and hockey, 50 watch hockey and basket ball, 50 do not watch any of the three games. The number of viewers, who watch exactly one of the three games is

$\begin{array}{l}\mathrm{n}(\cup )=500,\mathrm{n}\left(\mathrm{F}\right)=285;\mathrm{n}\left(\mathrm{H}\right)=195,\mathrm{n}\left(\mathrm{B}\right)=115\\ \mathrm{n}(\mathrm{F}\cap \mathrm{B})=45,\mathrm{n}(\mathrm{F}\cap \mathrm{H})=70\\ \mathrm{n}(\mathrm{H}\cap \mathrm{B})=50,\mathrm{n}(\overline{\mathrm{F}}\cap \overline{\mathrm{B}}\cap \overline{\mathrm{C}})=50\end{array}$

$$\begin{gathered} n\left\{(F\cup H\cup B{)}^{\mathrm{\prime }}\right\}=n\left(P\right)-n(F\cup H\cup B) \\ \Rightarrow 50=500-\left\{n\right(F)+n(H)+n(B)-n(F\cap H)-n(H\cap B)-n(F\cap H\cap B\left)\right\} \\ \Rightarrow 50=500-\{285+195+115-70-50-45+n(F\cap H\cap B\left)\right\} \\ \Rightarrow 50=500-430-n(F\cap H\cap B) \\ \Rightarrow 50=70-\mathrm{n}(\mathrm{F}\cap \mathrm{H}\cap \mathrm{B}) \\ \Rightarrow \mathrm{n}(\mathrm{F}\cap \mathrm{H}\cap \mathrm{B})=70-50=20 \\ \text{ Hence, }20\text{ people watch all the }3\text{ games } \\ \text{ Number of people who watch only football }=285-(50+20+25) \\ =285-95 \\ =190 \\ \text{ Number of people who watch only hockey }=195-(50+20+30) \\ =195-100 \\ =95 \\ \text{ And, Number of people who watch only basket ball }=115-(25+20+30) \\ =115-75 \\ =40 \\ \text{ Number of people who watch exactly one of the three games }=\text{ Number of people who watch either football only or hockey only or basket only} \\ =190+95+40[\because \text{ they are pairwise disjoint }] \\ =325 \\ \text{ Hence, }325\text{ people watch exactly one of the three games. } \end{gathered}$$