Question

If $n\left(\mu \right)=48,n\left(A\right)=28$, $n\left(B\right)=33$ and $n(B-A)=12$, then $n(A\cap B{)}^C$ is

• A. $27$
• B. $28$
• C. $29$
• D. $30$

Answer 1

The correct option is A $27$

Given $n\left(\mu \right)=48,n\left(A\right)=28,n\left(B\right)=33,n(B-A)=12$

As we know that

$n(B-A)=n\left(B\right)-n(A\cap B)=33-n(A\cap B)$

$⟹33-n(A\cap B)=12$

$⟹n(A\cap B)=33-12=21$

$n(A\cap B{)}^C=n(\mu )-n(A\cap B)=48-21=27$