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.
