Let A and B are two events and P(A') = 0.3 , P(B) = 0.4,,P(A cap B') = 0.5 , then P(A cup B&

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14.Probability

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Let $A$ and $B$ are two events and $P(A') = 0.3$, $P(B) = 0.4,\,P(A \cap B') = 0.5$, then $P(A \cup B')$ is

A

$0.5$

B

$0.8$

C

$1$

D

$0.1$

Solution

(b) $P(A') = 0.3$, $\therefore$ , $P(A) = 0.7$

$P(B') = 0.6$, $\therefore $ $P(B) = 0.4$ and $P(A \cap B') = 0.5$

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

$= 0.7 + 0.6 -0.5 = 0.8$.

Standard 11

Mathematics

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