If A and B are events such that P(A cup B) = 3/4, P(A cap B) = 1/4, P(bar A) = 2/3, then P(bar A cap

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14.Probability

medium

If $A$ and $B$ are events such that $P(A \cup B) = 3/4,$ $P(A \cap B) = 1/4,$ $P(\bar A) = 2/3,$ then $P(\bar A \cap B)$ is

A

$\frac{5}{{12}}$

B

$\frac{3}{8}$

C

$\frac{5}{8}$

D

$\frac{1}{4}$

(AIEEE-2002)

Solution

(a) $P\,(A \cup B)\, = \frac{3}{4},$ $P(A \cap B) = \frac{1}{4}$

$P(\bar A)\, = \frac{2}{3}\,\, \Rightarrow $ $P(A) = \,\frac{1}{3}$

$\therefore $$P(A \cap B)\, = \,P(A)\, + P\,(B)\, – P(A \cup B)$

$\frac{1}{4} = \frac{1}{3} + P(B)\, – \frac{3}{4} ⇒ P(B)\, = \frac{2}{3}$

$P(\bar A \cap \,B)$= $P(B) – \,P\,(A \cap B)$

$ = \frac{2}{3} – \frac{1}{4} = \frac{{8 – 3}}{{12}} = \frac{5}{{12}}.$

Standard 11

Mathematics

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