If A and B are arbitrary events, then

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14.Probability

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If $A$ and $B$ are arbitrary events, then

A

$P(A \cap B) \ge P(A) + P(B)$

B

$P(A \cup B) \le P(A) + P(B)$

C

$P(A \cap B) = P(A) + P(B)$

D

None of these

Solution

(b) $P(A \cup B) = P(A) + P(B) – P(A \cap B) \le P(A) + P(B)$,$(\because \,\,P(A \cap \,B) \geqslant \,0)$.

Standard 11

Mathematics

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