Question

A survey of 500 television watchers produces 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 doesn’t watch any of the game? How many watch all the three games? How many watch exactly one of the games?

Answer 1

Hint: In this question, we are given the total number of television watchers i.e. 500. In which 285 watch football, 195 watch hockey, and 115 watch basketball. Also 45 watch football and basketball, 70 watch football and hockey, 50 watch hockey and basketball. 50 doesn’t watch any of the games. We have to find watchers who watch all the three games and also who watch exactly one of the games i.e. either watches only basketball or only football or only hockey. You can solve this by the Venn diagram.

Complete step by step answer:

The Venn diagram for this question is given below:

Let us assume n(F) be the number of football watchers, n(H) be the number of hockey watchers and n(B) be the number of basketball watchers.
$\begin{gathered}
   \Rightarrow n(F) = 285 \\
   \Rightarrow n(H) = 195 \\
   \Rightarrow n(B) = 115 \\
\end{gathered} $
The total number of television watchers i.e. n(T) is 500. Out of which 50 watchers doesn’t watch any game. So, number of watchers who watch games is given by $n(F \cup H \cup B)$ so we have,
$ \Rightarrow n(F \cup H \cup B) = 500 - 50 = 450$
The number of watchers who watch two games are given by:
$\begin{gathered}
   \Rightarrow n(F \cap H) = 70 \\
   \Rightarrow n(F \cap B) = 45 \\
   \Rightarrow n(H \cap B) = 50 \\
\end{gathered} $
The numbers of watchers who watches all the three games is given by $n(F \cap H \cap B)$ and formula is as follows:
$ \Rightarrow n(F \cup H \cup B) = n(F) + n(H) + n(B) - n(F \cap H) - n(H \cap B) - n(B \cap F) + n(F \cap H \cap B)$
$\begin{gathered}
   \Rightarrow 450 = 285 + 195 + 115 - 70 - 50 - 45 + n(F \cap H \cap B) \\
   \Rightarrow 450 = 595 - 165 + n(F \cap H \cap B) = 430 + n(F \cap H \cap B) \\
   \Rightarrow n(F \cap H \cap B) = 450 - 430 = 20 \\
\end{gathered} $
The number of watchers who watch all the three games is 20.
Now we have to find who watches only one game i.e. only football or only basket ball or only hockey we can do it with the help of Venn diagram
$\begin{gathered}
   \Rightarrow n(onlyF) = n(F) - n(F \cap H) - n(F \cap B) + n(F \cap H \cap B) \\
   \Rightarrow n(onlyF) = 285 - 70 - 45 + 20 = 305 - 115 = 190 \\
\end{gathered} $
$\begin{gathered}
   \Rightarrow n(onlyB) = n(B) - n(F \cap B) - n(H \cap B) + n(F \cap H \cap B) \\
   \Rightarrow n(onlyB) = 115 - 45 - 50 + 20 = 135 - 95 = 40 \\
\end{gathered} $
$\begin{gathered}
   \Rightarrow n(onlyH) = n(H) - n(F \cap H) - n(H \cap B) + n(F \cap H \cap B) \\
   \Rightarrow n(onlyH) = 195 - 70 - 50 + 20 = 215 - 120 = 195 \\
\end{gathered} $
So, the watchers who watches only one game $ = 190 + 40 + 195 = 325$.

Note: There is a difference between only one and one of these game viewers. Here students make the mistake of taking all viewers who watch one game including watches both also or watches all the three games. Students may get wrong while taking total game watcher people they can take 500 but in 500 there are 50 people who didn’t watch any game. So, 50 will be subtracted from 500 to calculate the total game watcher people. Take care of these types of mistakes.