Question

In a test, an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is \[\dfrac{1}{3}\] and the probability that he copies the answer is \[\dfrac{1}{6}\]. The probability that his answer is correct given that he copied it is \[\dfrac{1}{8}\]. The probability that he knew the answer to the question given that he correctly answered it is \[\dfrac{4k}{29}\]. Find the value of k.

Answer 1

Hint: In this question, we first need to find the probability for the different possibilities when the answer is correct. Then from the formula of Bayes’ theorem we can get the probability that he knew the answer to the question that he correctly answered.
\[P\left( \dfrac{{{E}_{i}}}{A} \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( \dfrac{A}{{{E}_{i}}} \right)}{\sum\limits_{i=1}^{n}{P\left( {{E}_{i}} \right)P\left( \dfrac{A}{{{E}_{i}}} \right)}},i=1,2,3,..,n\]

Complete step-by-step answer:

Let us first define the events
E1 be the event that he guesses the answer to the question.
E2 is the event that he copies the answer to the question.
E3 is the event that he knows the answer to the question.
E be the event that the answer is correct.
Here,
\[\begin{align}
  & P\left( {{E}_{1}} \right)=\dfrac{1}{3} \\
 & P\left( {{E}_{2}} \right)=\dfrac{1}{6} \\
 & P\left( {{E}_{3}} \right)=1-\left( \dfrac{1}{3}+\dfrac{1}{6} \right)=\dfrac{1}{2} \\
\end{align}\]
As the question that he needs to attempt is a multiple choice he can guess any one of the given 4 options which can be written in terms of probability as
The probability for the answer to be correct when he guessed is
\[P\left( \dfrac{E}{{{E}_{1}}} \right)=\dfrac{1}{4}\]
As already given in the question that the probability for answering the question right provided he copied is given by
\[P\left( \dfrac{E}{{{E}_{2}}} \right)=\dfrac{1}{8}\]
The event of choosing the correct provided he knew the answer is a sure event whose probability is given by
\[P\left( \dfrac{E}{{{E}_{3}}} \right)=1\]
BAYES THEOREM: Let S be the sample space and let E1, E2,....., En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E1 or E2 or .. or En, then the probability of occurrence of \[{{E}_{i}}\], when A occurred, is
 \[P\left( \dfrac{{{E}_{i}}}{A} \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( \dfrac{A}{{{E}_{i}}} \right)}{\sum\limits_{i=1}^{n}{P\left( {{E}_{i}} \right)P\left( \dfrac{A}{{{E}_{i}}} \right)}},i=1,2,3,..,n\]
Now, the probability that he knew the answer to the question given that he correctly answered can be obtained by substituting the respective values in the above formula.
\[\Rightarrow P\left( \dfrac{{{E}_{3}}}{E} \right)=\dfrac{P\left( {{E}_{3}} \right)P\left( \dfrac{E}{{{E}_{3}}} \right)}{\sum\limits_{i=1}^{3}{P\left( {{E}_{i}} \right)P\left( \dfrac{E}{{{E}_{i}}} \right)}}\]
Now, by substituting the respective values we get,
\[\begin{align}
  & \Rightarrow P\left( \dfrac{{{E}_{3}}}{E} \right)=\dfrac{1\times \dfrac{1}{2}}{\dfrac{1}{4}\times \dfrac{1}{3}+\dfrac{1}{8}\times \dfrac{1}{6}+\dfrac{1}{2}\times 1} \\
 & \Rightarrow P\left( \dfrac{{{E}_{3}}}{E} \right)=\dfrac{\dfrac{1}{2}}{\dfrac{1}{12}+\dfrac{1}{48}+\dfrac{1}{2}} \\
\end{align}\]
Now, on further simplification we get,
\[\therefore P\left( \dfrac{{{E}_{3}}}{E} \right)=\dfrac{24}{29}\]
Let us now compare this value with the given value in the question to get the value of k.
\[\begin{align}
  & \Rightarrow \dfrac{4k}{29}=\dfrac{24}{29} \\
 & \Rightarrow 4k=24 \\
 & \therefore k=6 \\
\end{align}\]

Note: It is important to note that the probability that he knew the answer to the question he answered correctly cannot be found directly because the questions that he answered correctly may have different possibilities. So, we need to use the Bayes’ theorem to solve it further as he already answered the question correctly. While calculating the probabilities of different possibilities it is important to note the number of favourable outcomes and the total number of possible outcomes because neglecting any of the possibilities there changes the probability accordingly and then when substituted in the formula gives the wrong result.