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Q. If $A$ and $B$ are two events such that $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$ and $P(\frac{A}{B})=\frac{1}{4}$, then $P\left(A^{\prime} \cap B^{\prime}\right)$ is

KCETKCET 2022Probability - Part 2

Solution:

$P ( A )=\frac{1}{2}, P ( B )=\frac{1}{3}, P ( A \cap B )=\frac{1}{4} \times \frac{1}{3}=\frac{1}{12}$
$P ( A \cap B )= P (\overline{ A \cup B })$
$P ( A \cap B )=1- P ( A \cup B )$
$P ( A \cap B )=1-\left[\frac{1}{2}+\frac{1}{3}-\frac{1}{12}\right]$
$P ( A \cap B )=1-\left[\frac{6+4-1}{12}\right]$
$P ( A \cap B )=1-\frac{9}{12}$
$P ( A \cap B )=\frac{1}{4}$