Question

Let $n\left(U\right)=700,n\left(A\right)=200,n\left(B\right)=300$ and $n(A\cap B)=100$, then $n(A^c\cap B^c)=$

• A. 400
• B. 600
• C. 300
• D. 200

Answer 1

The correct option is C 300

Given that $n\left(U\right)=700,n\left(A\right)=200,n\left(B\right)=300,n(A\cap B)=100$

we know that $A^c\cap B^c=(A\cup B{)}^c$ by demorgan's law

Therefore, $n(A^c\cap B^c)=n(A\cup B{)}^c$

$=n\left(U\right)-n(A\cup B)$

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

$=700-200-300+100=300$

Therefore, $n(A^c\cap B^c)=300$