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WBJEE · Maths · Probability

Let \(A\) and \(B\) be two events with \(P\left(A^{C}\right)=0.3\) \(P(B)=0.4 \quad\) and \(\quad P\left(A \cap B^{C}\right)=0.5 . \quad\) Then
\(P\left(B \mid A \cup B^{c}\right)\) is equal to

  1. A \(\frac{1}{4}\)
  2. B \(\frac{1}{3}\)
  3. C \(\frac{1}{2}\)
  4. D \(\frac{2}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{4}\)

Step-by-step Solution

Detailed explanation

\(P\left(\frac{B}{A \cup B^{c}}\right)=\frac{P\left(B \cap\left(A \cup B^{e}\right)\right)}{P\left(A \cup B^{c}\right)}\) \(=\frac{P(A \cap B)}{P(A)+P\left(B^{c}\right)-P\left(A \cap B^{c}\right)}\)…