Question

In a multiple-choice test, an examinee either knows the correct answer with probability p, or guesses with probability 1−p. The probability of answering a question correctly is $\dfrac{1}{m}$ ​, if he or she merely guesses. If the examinee answers a question correctly, the probability that he or she really knows the answer is

(a)$\dfrac{mp}{1+mp}$
(b)$\dfrac{mp}{1+\left( m-1 \right)p}$
(c)$\dfrac{\left( m-1 \right)p}{1+\left( m-1 \right)p}$
(d)$\dfrac{\left( m-1 \right)p}{1+mp}$

Answer 1

The formula and definition of conditional probability which will be used in this question is given as follows, Conditional probability is the probability of an event (A), given that another (B) has already occurred. If A and B are any events in S then the conditional probability of B relative to A is given as follows$P({}^{B}/{}_{A})=\dfrac{P(B\bigcap{A})}{P(A)}$, where $P(B\bigcap{A})$ is the probability for event A and B happening together and P (A) is the probability of happening of the event A and also it should not be equal to zero.

Complete step-by-step answer:
In the question, we are asked to calculate the probability that the examinee really knows the answer of the question in multiple choice questions.
Let the probability that the examinee knows the answer be P (A) and let the probability that the answer is correct be P (B).
P (B) is calculated by taking the two cases as when the examinee makes a guess as well as when the examinee knows the correct answer.
Hence,
P(B)=P(examinee knows the correct answer)P(getting the answer right)+P(examinee guesses correct answer)P(answering a question correctly if guessed)

& P(B)\ =p\cdot 1+(1-p)\cdot \dfrac{1}{m} \\\ & \ \ \ \ \ \ \ \ \ =p\left( 1-\dfrac{1}{m} \right)+\dfrac{1}{m} \\\ & \ \ \ \ \ \ \ \ \ =p\left( \dfrac{m-1}{m} \right)+\dfrac{1}{m}\ \ \ \ \ ...(i) \\\ \end{aligned}$$ This is because for the first part of the sum, if a person knows an answer, then he will definitely get the right answer and for the second part, if that person guesses then the probability of him getting the right answer is given in the question and in these two terms the probabilities are combined by the fundamental theorem in which two events whose occurrence are to be checked together and hence, their respective probabilities are to be multiplied. Now, using the formula for conditional probability which is as follows to get the answer is $$\begin{aligned} & P({}^{A}/{}_{B})=\dfrac{P(B\bigcap{A})}{P(B)} \\\ & \ \ \ \ \ \ \ \ \ \ =\dfrac{p\cdot 1}{p\left( \dfrac{m-1}{m} \right)+\dfrac{1}{m}} \\\ &\ \ \ \ \ \ \ \ \ \ \ =\dfrac{p}{\dfrac{pm-p+1}{m}} \\\ & \ \ \ \ \ \ \ \ \ \ =\dfrac{pm}{pm-p+1} \\\ & \ \ \ \ \ \ \ \ \ \ =\dfrac{pm}{p\left( m-1 \right)+1} \end{aligned}$$ Using the equation (i) which gives us the value of P(B) and for $$P(A\cap B)$$, we can see that if the examinee knows the answer, then he will definitely get the correct answer. Hence, the probability of the examinee knowing an answer and getting it correct $$\dfrac{pm}{p\left( m-1 \right)+1}$$. Hence, option ( b ) is correct. **So, the correct answer is “Option (b)”.** **Note:** The students can make an error if they don’t know about the formula and the definition of conditional probability which is given in the hint as follows. Always remember that if A and B are any events in S then the conditional probability of B relative to A is given as follow $$P({}^{B}/{}_{A})=\dfrac{P(B\bigcap{A})}{P(A)}$$ , where $$P(B\bigcap{A})$$ is the probability for event A and B happening together and P (A) is the probability of happening of the event A and also it should not be equal to zero.