Question

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

• A. $53$
• B. $45$
• C. $56$
• D. $66$
• E. $46$

Answer 1

Answer: (C)

$56$

Given, $n\left(A\right)=43,n\left(B\right)=51$ and $n(A\cup B)=75$

Now, by addition theorem of probability,

$n(A\cap B)=n\left(A\right)+n\left(B\right)-n(A\cup B)$
$=43+51-75=19$

Now, $n\left[\right(A-B)\cup (B-A\left)\right]$

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

$=75-19=56$