If $n(A)=30, n(B) = 45, n(A \cup B) = 65$ then find $n(A \cap B)$, $n(A-B)$ and $n(B-A)$.

Given :

The given terms are $n(A)=30, n(B) = 45, n(A \cup B) = 65$.

To do :

We have to find $n(A \cap B)$, $n(A-B)$ and $n(B-A)$.

Solution :

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

$65 = 30 + 45- n(A \cap B) $

$65 = 75 -  n(A \cap B) $

$ n(A \cap B) = 75-65$

$ n(A \cap B) =10$

$n(A-B) = n(A)  - n(A \cap B)$

$n(A-B) = 30 -10 = 20$

$n(A-B) =20$

$n(B-A) = n(B)  - n(A \cap B)$

$n(B-A) = 45 - 10=35$

$n(B-A) = 35$

Therefore, $ n(A \cap B) =10$, $n(A-B) =20$ and $n(B-A) = 35$.