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For two events A and B, if P(A)=P\left(\frac{A}{B}\right)=\frac{1}{4} and P\left(\frac{B}{A}\right)=\frac{1}{2}, then which of the following is not true ?

Option: 1

A and B are independent 


Option: 2

\begin{aligned} & P\left(\frac{\dot{A}^{\prime}}{B}\right)=\frac{3}{4} \\ \end{aligned}


Option: 3

P\left(\frac{B^{\prime}}{A^{\prime}}\right)=\frac{1}{2}


Option: 4

None of these 


Answers (1)

best_answer

\begin{gathered} P\left(\frac{B}{A}\right)=\frac{1}{2} \Rightarrow \frac{P(B \cap A)}{P(A)}=\frac{1}{2} \Rightarrow P(B \cap A)=\frac{1}{8} \\ \\P\left(\frac{A}{B}\right)=\frac{1}{4} \Rightarrow \frac{P(A \cap B)}{P(B)}=\frac{1}{4} \Rightarrow P(B)=\frac{1}{2} \end{gathered}

P(A \cap B)=\frac{1}{8}=P(A) \cdot P(B)

\therefore Events A and B are independent 

\begin{aligned} & \text { Now, } P\left(\frac{A^{\prime}}{B}\right)=\frac{P\left(A^{\prime} \cap B\right)}{P(B)}=\frac{P\left(A^{\prime}\right) P(B)}{P(B)}=\frac{3}{4} \\ \\& \text { and } P\left(\frac{B^{\prime}}{A^{\prime}}\right)=\frac{P\left(B^{\prime} \cap A^{\prime}\right)}{P\left(A^{\prime}\right)}=\frac{P\left(B^{\prime}\right) P\left(A^{\prime}\right)}{P\left(A^{\prime}\right)}=\frac{1}{2} \end{aligned}

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Devendra Khairwa

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