A) \[\frac{24}{29}\]
B) \[\frac{1}{4}\]
C) \[\frac{3}{4}\]
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
| [a] Let \[{{E}_{1}}\] be the event that the answer is guessed, \[{{E}_{2}}\] be the event that the answer is copied, \[{{E}_{2}}\] be the event that the examinee knows the answer end E be the event that the examinee answer correctly. |
| Given \[P({{E}_{1}})=\frac{1}{3},P({{E}_{2}})=\frac{1}{6}\] |
| Assume that events \[{{E}_{1}},{{E}_{2}}\And {{E}_{3}}\] are exhaustive. |
| \[\therefore P({{E}_{1}})+P({{E}_{2}})+P({{E}_{3}})=1\] |
| \[\therefore P({{E}_{3}})=1-P({{E}_{1}})-P({{E}_{2}})=1-\frac{1}{3}-\frac{1}{6}=\frac{1}{2}\] |
| Now, \[P\left( \frac{E}{{{E}_{1}}} \right)=\] Probability of getting correct answer by guessing \[=\frac{1}{4}\] (Since 4 alternatives)\[P\left( \frac{E}{{{E}_{2}}} \right)\]= Probability of answering correctly by copying \[=\frac{1}{8}\] |
| And \[P\left( \frac{E}{{{E}_{3}}} \right)=\]Probability of answering correctly be knowing = 1 |
| Clearly, \[\left( \frac{{{E}_{3}}}{E} \right)\] is the event he knew the answer to the question given that he correctly answered it. Using Baye?s theorem \[P\left( \frac{{{E}_{3}}}{E} \right)\] |
| \[=\frac{P({{E}_{3}}).P\left( \frac{E}{{{E}_{3}}} \right)}{P({{E}_{1}}).P\left( \frac{E}{{{E}_{1}}} \right)+P({{E}_{2}}).P\left( \frac{E}{{{E}_{2}}} \right)+P({{E}_{3}}).P\left( \frac{E}{{{E}_{3}}} \right)}\] |
| \[=\frac{\frac{1}{2}\times 1}{\frac{1}{3}\times \frac{1}{4}+\frac{1}{6}\times \frac{1}{8}+\frac{1}{2}\times }=\frac{24}{29}\] |
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