Question

In a survey of $500$ TV viewers, it was found that $285$ watch cricket, $195$ watch football and $115$ watch tennis. Also, $45$ watch both cricket and football, $70$ watch both cricket and tennis and $50$ watch both football and tennis. If $50$ do not watch any game on TV, then the number of viewers who watch all the three games are:
A)$10$
B)$20$
C)$30$
D)$50$

Answer 1

Hint: This question can be easily solved by applying
$n\left( {A \cup B \cup C} \right) = n\left( A \right) + n\left( B \right) + n\left( C \right) - n\left( {A \cap B} \right) - n\left( {B \cap C} \right) - n\left( {C \cap A} \right) + n\left( {A \cap B \cap C} \right)$
Before using this formula, make sure to calculate the number of TV viewers who watch at least one game.

Complete step-by-step answer:
Total number of viewers on television $ = 500$
People who watch cricket denoted by
$n\left( C \right) = 285$ (1)
People watching football $n\left( F \right) = 195$ (2)
People watching tennis $n\left( T \right) = 115$ (3)
People watching both cricket and football would be denoted by intersection of C and F
$n\left( {C \cap F} \right) = 45$ (4)
Similarly, people watching both cricket and tennis
$n\left( {C \cap T} \right) = 70$ (5)
People watching football and tennis
$n\left( {F \cap T} \right) = 50$ (6)
Let the number of viewers watching all the three games,
$n\left( {C \cap F \cap T} \right) = x$ (7)
We know that
$n\left( {A \cup B \cup C} \right) = n\left( A \right) + n\left( B \right) + n\left( C \right) - n\left( {A \cap B} \right) - n\left( {B \cap C} \right) - n\left( {C \cap A} \right) + n\left( {A \cap B \cap C} \right)$
But here $n\left( {A \cup B \cup C} \right)$ means $n\left( {C \cup F \cup T} \right)$,i.e., viewers watching at least one game which is equal to:
Total viewers $ - $viewers who do not watch any game
$
   = 500 - 50 \\
    \\
 $
$n\left( {C \cup F \cup T} \right) = 450$ (8)
According to our notation, the formula becomes
$n\left( {C \cup F \cup T} \right) = n\left( C \right) + n\left( F \right) + n\left( T \right) - n\left( {C \cap F} \right) - n\left( {F \cap T} \right) - n\left( {T \cap C} \right) + n\left( {C \cap F \cap T} \right)$ (9)
Substituting values from equation (1) to (8) in (9)
We get
$
  450 = 285 + 195 + 115 - 45 - 70 - 50 + x \\
  450 = 595 - 165 + x \\
  x = 450 - 430 \\
  x = 20 \\
 $
Viewers watching all games $ = 20$
Option B is correct.

Note: This problem could also be solved by using a venn diagram. The diagram should be made carefully by overlapping circles accurately otherwise the answer would be wrong.