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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 $n(B-A) = 35$.
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