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∩ 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 - Mathematics

\[\cap\]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?

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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\[\cap\]B) = 45, n(F\[\cap\]H) = 70 and n(H\[\cap\]) = 50
Also, we know:
n(F\['\]\[\cap\]H\['\]\[\cap\]B\['\]= 50

\[\Rightarrow\]n(F\[\cup\]H\[\cup\]B)'= 50

\[\Rightarrow\]n(U) \[-\]n(F\[\cup\]H\[\cup\]B) = 50

\[\Rightarrow\]500 \[-\]n(F\[\cup\]H\[\cup\]B) = 50 

\[\Rightarrow\]n(F\[\cup\]H\[\cup\] = 450

Number of students who watch all three games = n(F\[\cap\]H\[\cap\]B)

\[\Rightarrow\]n(F\[\cup\]H\[\cup\]B) \[-\]n(F) \[-\]n(H)\[-\]n(B) + n(F\[\cap\]B) + n(F\[\cap\]H) + n(H\[\cap\]B)

\[\Rightarrow\]450 \[-\] 285 \[-\]195 \[-\]115 + 45 + 70 + 50

\[\Rightarrow\]20 

Number of students who watch exactly one of the three games
n(F) + n(H) + n(B) 

\[-\] 2{n(F\[\cap\]B) + n(F\[\cap\]H) + n(H\[\cap\]B)} + 3{n(F\[\cap\]H\[\cap\]B)}

= 285 + 195 + 115 

\[-\]2(45 + 70 + 50) + 3(20)
= 325

 

Concept: The Empty Set
  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 1 Sets
Exercise 1.8 | Q 12 | Page 47
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