If P,(A) = 0.4,,,P,(B) = x,,,P,(A cup B) = 0.7 and events A and B are mutually exclusive, then x =

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14.Probability

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If $P\,(A) = 0.4,\,\,P\,(B) = x,\,\,P\,(A \cup B) = 0.7$ and the events $A$ and $B$ are mutually exclusive, then $x = $

A

$\frac{3}{{10}}$

B

$\frac{1}{2}$

C

$\frac{2}{5}$

D

$\frac{1}{5}$

Solution

(a) Since events are mutually exclusive, therefore

$P(AB) = 0$ i.e., $P(A \cup B) = P(A) + P(B)$

$ \Rightarrow 0.7 = 0.4 + x \Rightarrow x = \frac{3}{{10}}.$

Standard 11

Mathematics

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