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

0.3

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0.5

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0.7

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0.9

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Solution

The correct option is D $0.9$
$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
$\Rightarrow P (A) + P (B) = 0.8 + 0.3 = 1.1$
And $P (¯ ¯¯ ¯ A) + P (¯ ¯¯ ¯ B) = 1 - P (A) + 1 - P (B) = 2 - 1.1 = 0.9$

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Similar questions

(A)

(B)

P (A) = 0.7, P (B) = 0.55, P (C) = 0.5, P (A \cap B) = x, P (A \cap C) = 0.45, P (B \cap C) = 0.3

P (A \cap B \cap C) = 0.2

P (A \cup B) = 0.8

P (A \cap B) = 0.3

P (A^{'}) + P (B^{'})

P (A) = 0.5, P (B) = 0.4, P (A \cap B) = 0.3,

P (A^{'} / B^{'})

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