KCET · Maths · Probability
If \(A\) and \(B\) are two non-mutually exclusive events such that \(P(A \mid B)=P(B \mid A)\), then
- A \(A \subset B\) but \(A \neq B\)
- B \(A=B\)
- C \(\mathrm{A} \cap \mathrm{B}=\phi\)
- D \(\mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B})\)
Answer & Solution
Correct Answer
(D) \(\mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B})\)
Step-by-step Solution
Detailed explanation
\(A\) and \(B\) are non mutually exclusive
\(\begin{aligned} & \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{B} \mid \mathrm{A}) \\ & \Rightarrow \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})} \\ & \Rightarrow \mathrm{P}(\mathrm{B})=\mathrm{P}(\mathrm{A})[\because \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \neq 0]\end{aligned}\)
\(\begin{aligned} & \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{B} \mid \mathrm{A}) \\ & \Rightarrow \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})} \\ & \Rightarrow \mathrm{P}(\mathrm{B})=\mathrm{P}(\mathrm{A})[\because \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \neq 0]\end{aligned}\)
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