Question

A and B are two subsets of universal set U such that $n\left( U \right) = 700,n\left( A \right) = 200,n\left( B \right) = 300$ and $n\left( {A \cap B} \right) = 100$. Find $n\left( {A' \cap B'} \right)$.

Answer 1

Hint: Use formulae $n\left( {A' \cap B'} \right) = n{\left( {A \cup B} \right)^\prime }$ and $n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$ and substitute values to find the required value.

Complete step by step answer:
According to question, $n\left( U \right) = 700,n\left( A \right) = 200,n\left( B \right) = 300$ and $n\left( {A \cap B} \right) = 100$.
We know from set theory that:
$ \Rightarrow n\left( {A' \cap B'} \right) = n{\left( {A \cup B} \right)^\prime } .....(i)$
Further, we also know that:
$ \Rightarrow n{\left( {A \cup B} \right)^\prime } = n\left( U \right) - n\left( {A \cup B} \right)$ and $n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$.
So putting the value of $n\left( {A \cup B} \right)$ in the formula of $n{\left( {A \cup B} \right)^\prime }$. We will get:
$ \Rightarrow n{\left( {A \cup B} \right)^\prime } = n\left( U \right) - \left[ {n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)} \right]$
Putting the values, $n\left( U \right) = 700,n\left( A \right) = 200,n\left( B \right) = 300$ and $n\left( {A \cap B} \right) = 100$, we’ll get:
$
   \Rightarrow n{\left( {A \cup B} \right)^\prime } = 700 - \left( {200 + 300 - 100} \right), \\
   \Rightarrow n{\left( {A \cup B} \right)^\prime } = 700 - 400, \\
   \Rightarrow n{\left( {A \cup B} \right)^\prime } = 300 \\
$
Putting the value of $n{\left( {A \cup B} \right)^\prime }$ in equation $(i)$, we’ll get
$ \Rightarrow n\left( {A' \cap B'} \right) = 300$

Thus, the value of $n\left( {A' \cap B'} \right)$ is 300.

Note: In case for set A and B, if $n\left( {A \cap B} \right) = 0$ then set A and B doesn’t have any common element.
$ \Rightarrow \left( {A \cap B} \right) = \phi $
So we have
$ \Rightarrow n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right)$