If A and B are two mutually exclusive events then $P (A \cup B) =$

P (A) + P (B) - P (A \cap B)

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P (A) - 2 P (A \cap B)

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P (A) + P (B)

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P (A) . P (B)

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Solution

The correct option is B $P (A) + P (B)$
Two exclusive events signify that they have nothing in common between them. so $P (A \cap B) = 0$
$P (A \cup B) = P (A) + P (B) - P (A \cap B) s o P (A \cup B) = P (A) + P (B)$

Suggest Corrections

Similar questions

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P (A / B) = \frac{P (A \cap B)}{P (B)}

P (B / A) = \frac{P (A \cap B)}{P (A)}

If A and B are mutually exclusive events then
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(c) PA<PB
(d) None of these

A

B

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(i) $P (A \cup B)$

(ii) $P (A \cap B)$

(iii) $P (A \cap ¯ ¯¯ ¯ B)$

(iv) $P (¯ ¯¯ ¯ A \cap ¯ ¯¯ ¯ B)$

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