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14.Probability
easy
If $P(A) = \frac{1}{2},\,\,P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{7}{{12}},$ then the value of $P\,(A' \cap B')$ is
A
$\frac{7}{{12}}$
B
$\frac{3}{4}$
C
$\frac{1}{4}$
D
$\frac{1}{6}$
Solution
(b) $P(A' \cap B') = 1 – P(A \cup B)$
$ = 1 – \left( {\frac{1}{2} + \frac{1}{3} – \frac{7}{{12}}} \right) = 1 – \frac{1}{4} = \frac{3}{4}$.
Standard 11
Mathematics