If PA- B-PA- B- then show that relation between PA and PB is PA-PB-

P( A∪ B ) =P( A ) +P( B ) P( A∩ B ) If P( A∪ B ) =P( A∩ B ) Then P( A ) and P( B ) are equalSince P( A∪ B ) =P( A∩ B ) ⇒ A and ...

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Probability

If PA∪ B=PA...

Question

If $P (A \cup B) = P (A \cap B) \Leftrightarrow$ then show that relation between $P (A)$ and $P (B)$ is $P (A) = P (B)$ .

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P (A \cup B) = P (A \cap B)

P (A)

P (B)

P (A \cup B) = P (A \cap B),

P (A \cup B) = P (A \cap B)

P (A)

P (B)

P (A) = P (B)

(i) P (A) = \frac{1}{3}, P (B) = \frac{1}{4} and P (A \cup B) = \frac{5}{12}, then find P (A | B) and P (B | A) . (ii) P (A) = \frac{6}{11}, P (B) = \frac{5}{11} and P (A \cup B) = \frac{7}{11}, then find P (A \cap B), P (A | B) and P (B | A) . (iii) P (A) = \frac{7}{13}, P (B) = \frac{9}{13} and P (A \cap B) = \frac{4}{13}, then find P (A | B) . (iv) P (A) = \frac{1}{2}, P (B) = \frac{1}{3} and P (A \cap B) = \frac{1}{4}, then find P (A | B), P (B | A), P (A | B) and P (A | B) .

A, B, C

P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A) + P (A \cap B \cap C)

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