Question

If $n\left(A\right)=43, n\left(B\right)=51\quad and \quad n\left(A\cup B\right)=75, then\quad n\left(A-B\right)\cup\left(B-A\right)$ is equal to

• A. $53$
• B. $45$
• C. $56$
• D. $66$

Answer 1

Answer: (C)

$56$

Answer 2

Given, $n(A)=43, \,n(B)=51$ and $n(A \cup B)=75$
Now, by addition theorem of probability,
$n(A \cap B) =n(A)+n(B)-n(A \cup B)$
$=43+51-75=19$
Now, $n[(A-B) \cup(B-A)]$
$=n(A \cup B)-n(A \cap B)$
$=75-19$
$=56$

So, the correct option is (C): $56$