Question

If $n(A\cap B^C)=5,n(B\cap A^C)=6,n(A\cap B)=4$ then the value of $n(A\cup B)$ is

• A. $18$
• B. $15$
• C. $16$
• D. $17$

Answer 1

The correct option is B $15$

$n(A\cap B^c)=n(A-B)=5=n\left(A\right)-n(A\cap B)$

$n(B\cap A^c)=n(B-A)=6=n\left(B\right)-n(A\cap B)$

$n(A\cap B)=4$

$n(A\cup B)=n(A-B)+n(B-A)+n(A\cap B)$

$=5+6+4$

$=15$