Question

Let $A$ and $B$ be the two sets such that $n\left(A\right)=0.16$, $n\left(B\right)=0.14$, $n\left(A\cup B\right)=0.25$.Then $n\left(A\cap B\right)$ is equal to

• A. $0.3$
• B. $0.5$
• C. $0.05$
• D. None of these

Answer 1

The correct option is C

$0.05$

Determine $n\left(A\cap B\right)$.

The union of two sets $A$and $B$ is the set of all elements which are either in $A$ or in $B$.It is denoted by $A\cup B$.

$A\cup B=\left\{x:x\in A\text{or}x\in B\right\}$.

The intersection of two sets $A$and $B$ is the set of all elements common to both $A$ and $B$.It is denoted by $A\cap B$.

$A\cap B=\left\{x:x\in A\text{and}x\in B\right\}$

The cardinality of set operations is given by: $n\left(A\cup B\right)=n\left(A\right)+n\left(B\right)-n(A\cap B)$

Therefore, substituting the given values of $n\left(A\right)=0.16$, $n\left(B\right)=0.14$ and $n\left(A\cup B\right)=0.25$, we have

$$\begin{gathered} \Rightarrow 0.25=0.16+0.14-n(A\cap B) \\ \Rightarrow n(A\cap B)=0.30-0.25 \\ \Rightarrow n(A\cap B)=0.05 \end{gathered}$$

Hence, option (C) is the correct answer.