A and B are two events on a sample space S such that P(A) = 0.8, P(B) = 0.6 and (A $\cup$ B) = 0.9 find P(A $\cap$ B).

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Solution

Given:
P(A) = 0.8, P(B) = 0.6 and P(A $\cap$ B) = 0.9
On applying the addition theorem of probability, we get:
P(A $\cup$ B) = P(A) + P(B) $-$ P(A $\cap$ B)
= 0.8 + 0.6 $-$ 0.9
= 0.5

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P (A) = 0.8, P (B) = 0.6

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P (A \cap B) = \frac{1}{3}, P (B) = \frac{1}{3} . Find P (A) and P (A') .

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