If PA - 0-25- PB - 0-50- PA- B - 0-14- then P neither A nor B -

The correct option is A 0.39 P(A∪ B)=P(A)+P(B) P(A∩ B) ⇒ P(A∪ B)=0.25+0.50 0.14=0.61 P(neither A nor B)=P(A̅∩B̅)=1 P(A ∪ B)=1 0.61=0.39 Opt ...

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Standard XII

Mathematics

Conditional Probability

If PA = 0.2...

Question

If $P (A) = 0.25, P (B) = 0.50, P (A \cap B) = 0.14$ , then $P$ (neither $A$ nor $B) =$

0.39

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0.25

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0.11

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0.24

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Solution

The correct option is A $0.39$
$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
$\Rightarrow P (A \cup B) = 0.25 + 0.50 - 0.14 = 0.61$
$P (neither A nor B) = P (¯ A \cap ¯ B) = 1 - P (A \cup B) = 1 - 0.61 = 0.39$
Option A is correct.

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If P(A)=0.25,P(B)=0.50,P(A∩B)=0.14, then P (neither A nor B)=

The correct option is A 0.39P(A∪B)=P(A)+P(B)−P(A∩B)⇒P(A∪B)=0.25+0.50−0.14=0.61P(neither A nor B)=P(¯A∩¯B)=1−P(A∪B)=1−0.61=0.39Option A is correct.

If $P (A) = 0.25, P (B) = 0.50, P (A \cap B) = 0.14$ , then $P$ (neither $A$ nor $B) =$

The correct option is A $0.39$
$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
$\Rightarrow P (A \cup B) = 0.25 + 0.50 - 0.14 = 0.61$
$P (neither A nor B) = P (¯ A \cap ¯ B) = 1 - P (A \cup B) = 1 - 0.61 = 0.39$
Option A is correct.