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

Answer: (C)

$300$

$n(\cup )=700,n\left(A\right)=200,n\left(B\right)=300$

$n(A\cap B)=100$ then $n(A^c\cup B^c)=?$

$n\left(A^c\right)=700-200=500$

$n\left(B^c\right)=700-300=40$

$\Rightarrow A^c\cap B^c=(A\cup B{)}^c$

$\Rightarrow n(A^c\cap B^c)=n\left(U\right)-n(A\cup B)$

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

$=700-\{200+300-100\}$

$\Rightarrow A^c\cap B^c\Rightarrow 700-400=300$