Q. 12

# A survey of 500 t

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(HBF)= 50

Now,

n(HBF)’ = n(P) – n(HBF)

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.

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