P(A∩B)+P(A∩B¯)= P(A∩B)+P(A)−P(A∩B) = P(A)

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$P (A \cap B) + P (A \cap \bar{B}) =$

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P(A)

P(A – B)

P(A $\cap$ B)

P(B)

answer is C.

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Detailed Solution

$P (A \cap B) + P (A \cap \bar{B}) =$ $P (A \cap B) + P (A) - P (A \cap B)$

= $P (A)$

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