Question

In a survey of $450$ people, it was found that $110$ play cricket, $160$ play tennis and $70$ play both cricket as well as tennis. How many plays neither cricket nor tennis?

Answer 1

Hint: In the given question, we have to work out the no. of people neither playing cricket nor tennis. There are two ways to solve this question –
By using own diagram
By using set formula approach
For Venn diagrams, simply draw two circles, one for cricket and other for tennis, intersect because there are some people who play both cricket and tennis. After doing so, just work out the answer required. For the set theorem, use the formula $n(A \cup B) = n(A) + n\left( B \right) - n\left( {A \cap B} \right)$. Put values given in it and solve.

Complete step-by-step answer:
Let’s consider cricket with set $A$, So, individuals play cricket,
$n(A) = 110$
Consider, tennis with set $B$, So, individuals playing tennis,
$n\left( B \right) = 160$
No. of individuals playing both cricket and tennis are,
$n\left( {A \cap B} \right) = 70$
So, individuals playing either of the sport can be calculated using formula,
$
  P(A) + P(B) - P\left( {A \cap B} \right) = P(A \cup B) \\
   \Rightarrow P(A \cup B) = 110 + 160 - 70 = 200 \\
$
$\therefore 200$ Individuals are playing sports.
Therefore, individuals do not play any sports are,
$
   = 450 - 200 \\
   = 250 \\
$
Using Venn Diagram:
     

Set $A$ indicates, people playing cricket i.e. $ = 100$
Set $B$ indicates, people playing Tennis i.e. $ = 160$
$\left( {A \cap B} \right)$ indicates, people playing both i.e. $ = 70$
$\therefore $ Peoples playing only cricket $ = (110 - 70) = 40$
Peoples playing only Tennis $ = \left( {160 - 70} \right) = 90$
$\therefore $ People playing any game $ = 40 + 90 + 70$$ = 200$
$\therefore $ Peoples do not play any game are $ = 450 - 200$ $ = 250$

Note: This question is asked from the topic-set theory. It can be solved by logical ideas. One can do this question in any way according to their wish. Set theory is a branch of mathematical logic, which actually is simple and interesting, if they are solved by using Venn diagrams.