If p(A) = \(\frac{1}{3}\), P(B) = \(\frac{2}{3}\) and P(A ∩ B) =�

1.	If p(A) = \(\frac{1}{3}\), P(B) = \(\frac{2}{3}\) and P(A ∩ B) = \(\frac{1}{6}\), then find P(A’ ∩ B’).
Answer» According to the law of addition of probability, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = \(\frac{1}{3}+\frac{2}{3}-\frac{1}{6}\) = \(\frac{2+4−1}{6}\) = \(\frac{5}{6}\) Now, P(A’ ∩ B’) = P(A ∪ B)’ = 1 – P(A ∪ B) = \(1 – \frac{5}{6}\) = \(\frac{6−5}{6}\) = \(\frac{1}{6}\)

If p(A) = \(\frac{1}{3}\), P(B) = \(\frac{2}{3}\) and P(A ∩ B) = \(\frac{1}{6}\), then find P(A’ ∩ B’).

Answer»

According to the law of addition of probability,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{1}{3}+\frac{2}{3}-\frac{1}{6}\)

= \(\frac{2+4−1}{6}\)

= \(\frac{5}{6}\)

Now, P(A’ ∩ B’) = P(A ∪ B)’

= 1 – P(A ∪ B)

= \(1 – \frac{5}{6}\)

= \(\frac{6−5}{6}\)

= \(\frac{1}{6}\)

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If p(A) = \(\frac{1}{3}\), P(B) = \(\frac{2}{3}\) and P(A ∩ B) = \(\frac{1}{6}\), then find P(A’ ∩ B’).

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