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# A student answers a multiple choice question with 5 alternatives, of which exactly one is correct. The probability that he knows the correct answer is p, $0< p <1$. If he does not know the correct answer, he randomly ticks one answer. Given that he has answered the question correctly, then what is the probability that he did not tick the answer randomly.${\text{A}}{\text{. }}\dfrac{{3p}}{{4p + 3}} \\ {\text{B}}{\text{. }}\dfrac{{5p}}{{3p + 2}} \\ {\text{C}}{\text{. }}\dfrac{{5p}}{{4p + 1}} \\ {\text{D}}{\text{. }}\dfrac{{4p}}{{3p + 1}} \\$  Hint- Here, we will be assuming event A as the student answered the question correctly, event ${{\text{E}}_1}$ as the student ticked the answer randomly and event ${{\text{E}}_2}$ as the student did not tick the answer randomly (or he knows the answer) and then will be using Bayes Theorem i.e., ${\text{P}}\left( {\dfrac{{{{\text{E}}_2}}}{{\text{A}}}} \right) = \dfrac{{{\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_2}}}} \right){\text{P}}\left( {{{\text{E}}_2}} \right)}}{{{\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_1}}}} \right){\text{P}}\left( {{{\text{E}}_1}} \right) + {\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_2}}}} \right){\text{P}}\left( {{{\text{E}}_2}} \right)}}$.

Given, probability that the student knows the correct answer = p where 0Since, the question is a multiple choice question containing 5 alternatives.
If ${\text{P}}\left( {\dfrac{{{{\text{E}}_2}}}{{\text{A}}}} \right)$ is the probability of occurrence of event ${{\text{E}}_2}$ when event A had already occurred.
If ${\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_{\text{1}}}}}} \right)$ is the probability of occurrence of event A when event ${{\text{E}}_1}$ had already occurred.
If ${\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_{\text{2}}}}}} \right)$ is the probability of occurrence of event A when event ${{\text{E}}_2}$ had already occurred.
If ${\text{P}}\left( {{{\text{E}}_1}} \right)$ is the probability of occurrence of event ${{\text{E}}_1}$ and if ${\text{P}}\left( {{{\text{E}}_2}} \right)$ is the probability of occurrence of event ${{\text{E}}_2}$.
According to Bayes Theorem, we can write
${\text{P}}\left( {\dfrac{{{{\text{E}}_2}}}{{\text{A}}}} \right) = \dfrac{{{\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_2}}}} \right){\text{P}}\left( {{{\text{E}}_2}} \right)}}{{{\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_1}}}} \right){\text{P}}\left( {{{\text{E}}_1}} \right) + {\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_2}}}} \right){\text{P}}\left( {{{\text{E}}_2}} \right)}}{\text{ }} \to {\text{(1)}}$
Also we know that the general formula for the probability of occurrence of an event is given by
Probability of occurrence of an event$= \dfrac{{{\text{Number of favourable cases}}}}{{{\text{Total number of possible cases}}}}$
Let ${{\text{E}}_1}$ be the event that the student ticked the answer randomly and ${{\text{E}}_2}$ be the event that the student did not tick the answer randomly (or he knows the answer) and A be the event that the student answered the question correctly.
Probability that the student did not tick the answer randomly (or probability that he knows the correct answer) is ${\text{P}}\left( {{{\text{E}}_2}} \right) = p$
Probability that the student ticked the answer randomly is ${\text{P}}\left( {{{\text{E}}_1}} \right) = 1 - {\text{P}}\left( {{{\text{E}}_2}} \right) = 1 - p$.
Probability that the student answered the question correctly, given that he did not tick the answer randomly (or he knows the answer) is ${\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_2}}}} \right) = 1$
Probability that the student answered the question correctly, given that he did not tick the answer randomly (or he knows the answer) is ${\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_1}}}} \right) = \dfrac{{{\text{Number of correct answers}}}}{{{\text{Total number of alternatives}}}} = \dfrac{1}{5}$
Substituting all the obtained values in equation (1), we get
Probability that the student did not tick the answer randomly (or he knows the answer), given that he answered the question correctly is
${\text{P}}\left( {\dfrac{{{{\text{E}}_2}}}{{\text{A}}}} \right) = \dfrac{{1 \times p}}{{\left[ {\dfrac{1}{5} \times \left( {1 - p} \right)} \right] + \left[ {1 \times p} \right]}} \\ \Rightarrow {\text{P}}\left( {\dfrac{{{{\text{E}}_2}}}{{\text{A}}}} \right) = \dfrac{p}{{\left[ {\dfrac{{\left( {1 - p} \right)}}{5} + p} \right]}} \\ \Rightarrow {\text{P}}\left( {\dfrac{{{{\text{E}}_2}}}{{\text{A}}}} \right) = \dfrac{p}{{\left[ {\dfrac{{\left( {1 - p} \right) + 5p}}{5}} \right]}} \\ \Rightarrow {\text{P}}\left( {\dfrac{{{{\text{E}}_2}}}{{\text{A}}}} \right) = \dfrac{{5p}}{{4p + 1}} \\$
Therefore, the required value of the probability is $\dfrac{{5p}}{{4p + 1}}$.
Hence, option C is correct.

Note- In this particular problem, ${\text{P}}\left( {{{\text{E}}_1}} \right) + {\text{P}}\left( {{{\text{E}}_2}} \right) = 1$ because there are only two possibilities that either the student knows the correct answer (i.e., event ${{\text{E}}_2}$) or the student answers the question randomly (i.e., event ${{\text{E}}_1}$). Also, ${\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_2}}}} \right) = 1$ because if the student knows the correct answer then obviously he will answer it correctly.
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