Given $n(U) = 20,\ n(A) = 12,\ n(B) = 9,n(A \cap B) = 4$, where $U$ is the universal set, $A$ and $B$ are subsets of $U$, then $n\lbrack(A \cup B)^{c}\rbrack$ equals to

Question

Dr. Abubacker · Accepted Answer

(d)
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 12 + 9 - 4 = 17$
Hence, $n\left\lbrack (AUB \right)^{c}\rbrack = n(U) - n(A \cup B)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 20 - 17 = 3$

Q. Given $n(U) = 20,\ n(A) = 12,\ n(B) = 9,n(A \cap B) = 4$, where $U$ is the universal set, $A$ and $B$ are subsets of $U$, then $n\lbrack(A \cup B)^{c}\rbrack$ equals to

Explanation

Reviewed by Dr. Abubacker

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