If P(A cup B) = 0.8 and P(A cap B) = 0.3, then P(bar A) + P(bar B) =

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14.Probability

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If $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3,$ then $P(\bar A) + P(\bar B) = $

A

$0.3$

B

$0.5$

C

$0.7$

D

$0.9$

Solution

(d) $P(A'\, \cap B')\, = 1 – P(A \cup B)\, = 1 – 0.8 = 0.2$

$P(A' \cup B') = 1 – P(A \cap B) = 1 – 0.3 = 0.7$

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

$⇒ \,0.7\, = P(A') + P(B') – 0.2$ $⇒$ $ P(A')+ P(B')=09 $.

Standard 11

Mathematics

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