KCET · Maths · Probability
If \(A\) and \(B\) are events, such that \(P(A)=\frac{1}{4}\) \(P(A / B)=\frac{1}{2}\) and \(P(B / A)=\frac{2}{3}\), then \(P(B)\) is
- A \(\frac{1}{3}\)
- B \(\frac{2}{3}\)
- C \(\frac{1}{2}\)
- D \(\frac{1}{6}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
Given,
\(P(A)=\frac{1}{4}, P\left(\frac{A}{B}\right)=\frac{1}{2}\) and \(P\left(\frac{B}{A}\right)=\frac{2}{3}\)
We know that, \(P\left(\frac{E_1}{E_2}\right)=\frac{P\left(E_1 \cap E_2\right)}{P\left(E_2\right)}\)
\(\begin{aligned} & \Rightarrow \quad \frac{2}{3}=\frac{P(A \cap B)}{1 / 4} \quad[\because A \cap B=B \cap A] \\ & \Rightarrow \quad P(A \cap B)=\frac{2}{3} \times \frac{1}{4}=\frac{1}{6}\end{aligned}\)
\(\because \quad P\left(\frac{A}{B}\right)=\frac{1}{2}\)
\(\therefore \quad \frac{P(A \cap B)}{P(B)}=\frac{1}{2}\)
\(\Rightarrow \quad P(B)=2 P(A \cap B)=2 \times \frac{1}{6}=\frac{1}{3}\)
\(P(A)=\frac{1}{4}, P\left(\frac{A}{B}\right)=\frac{1}{2}\) and \(P\left(\frac{B}{A}\right)=\frac{2}{3}\)
We know that, \(P\left(\frac{E_1}{E_2}\right)=\frac{P\left(E_1 \cap E_2\right)}{P\left(E_2\right)}\)
\(\begin{aligned} & \Rightarrow \quad \frac{2}{3}=\frac{P(A \cap B)}{1 / 4} \quad[\because A \cap B=B \cap A] \\ & \Rightarrow \quad P(A \cap B)=\frac{2}{3} \times \frac{1}{4}=\frac{1}{6}\end{aligned}\)
\(\because \quad P\left(\frac{A}{B}\right)=\frac{1}{2}\)
\(\therefore \quad \frac{P(A \cap B)}{P(B)}=\frac{1}{2}\)
\(\Rightarrow \quad P(B)=2 P(A \cap B)=2 \times \frac{1}{6}=\frac{1}{3}\)
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