A) \[\frac{3}{8}\]
B) \[\frac{1}{5}\]
C) \[\frac{3}{4}\]
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
Let E be the event that a six occurs and A be the event that man reports that It is a six. \[\therefore \]\[P(E)=\frac{1}{6},P(E)=\frac{5}{6},P\left( \frac{A}{E} \right)=\frac{3}{4}\] and \[P\left( \frac{A}{E} \right)=\frac{1}{4}\] Using Bays theorem \[P\left( \frac{E}{A} \right)=\frac{P(E).P\left( \frac{A}{E} \right)}{P(E).P\left( \frac{A}{E} \right)+P(E)P\left( \frac{A}{E} \right)}\] \[=\frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}}=\frac{3}{8}\]
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