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let A and B be two events such that P(B|A)=P(B|A^c), A^c is A complement. Are A and B independent? Please Give reasons!

Note by Sourav Agarwal 3 years ago

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For \(P(A) \in (0,1) \), we can write \(\displaystyle P(B|A) = \frac{P(B \cap A)}{P(A)} \) and \( \displaystyle P(B|A^c) = \frac{P(B \cap A^c)}{P(A^c)} \) (Using Bayes' Theorem ).

Since, \(\displaystyle P(B|A) = P(B|A^c) \), therefore, \(\displaystyle \frac{P(B \cap A)}{P(A)} = \frac{P(B \cap A^c)}{P(A^c)} \) \( \displaystyle \Rightarrow P(B \cap A)P(A^c) = P(A)P(B \cap A^c) \Rightarrow P(B \cap A)( 1- P(A)) = P(A) (P(B) - P(B \cap A) ) \) Simplifying, we obtain, \(\displaystyle P(A \cap B) = P(A)P(B) \) which is sufficient to prove that \(A\) and \(B\) are independent. – Sudeep Salgia · 3 years ago

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TopNewestFor \(P(A) \in (0,1) \), we can write \(\displaystyle P(B|A) = \frac{P(B \cap A)}{P(A)} \) and \( \displaystyle P(B|A^c) = \frac{P(B \cap A^c)}{P(A^c)} \) (Using Bayes' Theorem ).

Since, \(\displaystyle P(B|A) = P(B|A^c) \), therefore, \(\displaystyle \frac{P(B \cap A)}{P(A)} = \frac{P(B \cap A^c)}{P(A^c)} \) \( \displaystyle \Rightarrow P(B \cap A)P(A^c) = P(A)P(B \cap A^c) \Rightarrow P(B \cap A)( 1- P(A)) = P(A) (P(B) - P(B \cap A) ) \) Simplifying, we obtain, \(\displaystyle P(A \cap B) = P(A)P(B) \) which is sufficient to prove that \(A\) and \(B\) are independent. – Sudeep Salgia · 3 years ago

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