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Question: A and B are two subsets of universal set U such that \(n\left( U \right) = 700,n\left( A \right) = 2...

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

Explanation

Solution

Hint: Use formulae n(A′∩B′)=n(A∪B)′n\left( {A' \cap B'} \right) = n{\left( {A \cup B} \right)^\prime } and n(A∪B)=n(A)+n(B)−n(A∩B)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(U)=700,n(A)=200,n(B)=300n\left( U \right) = 700,n\left( A \right) = 200,n\left( B \right) = 300 and n(A∩B)=100n\left( {A \cap B} \right) = 100.
We know from set theory that:
⇒n(A′∩B′)=n(A∪B)′.....(i)\Rightarrow n\left( {A' \cap B'} \right) = n{\left( {A \cup B} \right)^\prime } .....(i)
Further, we also know that:
⇒n(A∪B)′=n(U)−n(A∪B)\Rightarrow n{\left( {A \cup B} \right)^\prime } = n\left( U \right) - n\left( {A \cup B} \right) and n(A∪B)=n(A)+n(B)−n(A∩B)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(A∪B)n\left( {A \cup B} \right) in the formula of n(A∪B)′n{\left( {A \cup B} \right)^\prime }. We will get:
⇒n(A∪B)′=n(U)−[n(A)+n(B)−n(A∩B)]\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(U)=700,n(A)=200,n(B)=300n\left( U \right) = 700,n\left( A \right) = 200,n\left( B \right) = 300 and n(A∩B)=100n\left( {A \cap B} \right) = 100, we’ll get:
⇒n(A∪B)′=700−(200+300−100), ⇒n(A∪B)′=700−400, ⇒n(A∪B)′=300  \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(A∪B)′n{\left( {A \cup B} \right)^\prime } in equation (i)(i), we’ll get
⇒n(A′∩B′)=300\Rightarrow n\left( {A' \cap B'} \right) = 300

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

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