A) I only
B) I and II
C) I and III
D) II and III
Correct Answer: A
Solution :
\[P\left( \frac{B}{A} \right)=\frac{P(A\cap B)}{P(A)}\,\Rightarrow \,\frac{1}{2}=\frac{P(A\cap B)}{1/4}\] \[\Rightarrow P(A\cap B)=\frac{1}{8}\] Hence events A and B are not mutually exclusive. \Statement II is incorrect. \[P\,\left( \frac{A}{B} \right)=\frac{P(A\cap B)}{P(B)\,}\Rightarrow \,P(B)=\frac{1}{2}\] \[\because \] \[P(A\cap B)=\frac{1}{8}=P(A)\,.\,P(B)\] \ events A and B are independent events. \[P\,\left( \frac{{{A}^{c}}}{{{B}^{c}}} \right)=\frac{P({{A}^{c}}\cap {{B}^{c}})}{P({{B}^{c}})}=\frac{P({{A}^{c}})\,P({{B}^{c}})}{P\,({{B}^{c}})}=\frac{3}{4}.\,\frac{1}{2}.\,\frac{2}{1}=\frac{3}{4}\] Hence statement I is correct. Again \[P\left( \frac{A}{B} \right)+\,P\left( \frac{A}{{{B}^{c}}} \right)\] \[=\frac{1}{4}+\frac{P(A\cap {{B}^{c}})}{P({{B}^{c}})}\] \[=\frac{1}{4}+\frac{P(A)-P(A\cap B)}{P({{B}^{c}})}\]\[=\frac{1}{4}+\frac{\frac{1}{4}-\frac{1}{8}}{\frac{1}{2}}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\] Hence statement III is incorrect.
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