Let ,

Total number of People n(P) = 500.

People who watch Basketball n(B) =115.

People who watch Football n(F) = 285.

People who watch Hockey n(H) = 195.

People who watch Basketball and Hockey n(B ∩ H) = 50

People who watch Football and Hockey n(H ∩ F) = 70

People who watch Basketball and Football n(B ∩ F) = 45

People who do not watch any games. n(H∪B∪F)= 50

Now,

n(H∪B∪F)’ = n(P) – n(H∪B∪F)

50 = 500–( n(H)+n(B)+n(F) – n (H ∩ B)– n (H ∩ F)– n (B ∩ F)+ n (H ∩B ∩ F))

50 = 500–(285+195+115–70–50–45 +n (H ∩B ∩ F))

50 = 500–430 + n (H ∩B ∩ F))

n (H ∩B ∩ F) = 70–50

n (H ∩B ∩ F)) = 20

∴ 20 People watch all three games.

Number of people who only watch football

= 285–(50+20+25)

= 285–95

= 190.

Number of people who only watch Hockey

= 195–(50+20+30)

= 195–100

= 95.

Number of people who only watch Basketball

= 115–(25+20+30)

= 115–75

= 40.

Number of people who watch exactly one of the three games

As the sets are pairwise disjoint we can write

= number of people who watch either football only or hockey only or Basketball only

=190+95+40

=325

∴ 325 people watch exactly one of the three games.