Q91 / 180 · COMEDK 2021 · Mathematics
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, the probability that he did not tick the answer randomly, is
- A$\dfrac{3 p}{4 p+3}$
- B$\dfrac{5 p}{3 p+2}$
- Correct answerC$\dfrac{5 p}{4 p+1}$
- D$\dfrac{4 p}{3 p+1}$
Solution
Let $K$ be the event that the student knows the correct answer, and $R$ be the event that the student ticks the answer randomly. Let $C$ be the event that the student answers the question correctly.
The probability that the student knows the answer is $P(K) = p$. The probability that the student does not know the answer is $P(K^{c}) = 1 - p$.
If the student knows the answer, the probability of answering correctly is $P(C|K) = 1$. If the student does not know the answer, the student ticks randomly among 5 alternatives, so the probability of answering correctly is $P(C|K^{c}) = \dfrac{1}{5}$.
Using the law of total probability, the probability of answering correctly is $P(C) = P(C|K)P(K) + P(C|K^{c})P(K^{c}) = 1 \times p + \dfrac{1}{5} \times (1 - p) = p + \dfrac{1 - p}{5} = \dfrac{5p + 1 - p}{5} = \dfrac{4p + 1}{5}$.
We want to find the probability that the student knew the answer given that the answer is correct, which is $P(K|C)$. Using Bayes theorem, $P(K|C) = \dfrac{P(C|K)P(K)}{P(C)} = \dfrac{1 \times p}{\dfrac{4p + 1}{5}} = \dfrac{5p}{4p + 1}$.
Answer: $\dfrac{5 p}{4 p+1}$
The probability that the student knows the answer is $P(K) = p$. The probability that the student does not know the answer is $P(K^{c}) = 1 - p$.
If the student knows the answer, the probability of answering correctly is $P(C|K) = 1$. If the student does not know the answer, the student ticks randomly among 5 alternatives, so the probability of answering correctly is $P(C|K^{c}) = \dfrac{1}{5}$.
Using the law of total probability, the probability of answering correctly is $P(C) = P(C|K)P(K) + P(C|K^{c})P(K^{c}) = 1 \times p + \dfrac{1}{5} \times (1 - p) = p + \dfrac{1 - p}{5} = \dfrac{5p + 1 - p}{5} = \dfrac{4p + 1}{5}$.
We want to find the probability that the student knew the answer given that the answer is correct, which is $P(K|C)$. Using Bayes theorem, $P(K|C) = \dfrac{P(C|K)P(K)}{P(C)} = \dfrac{1 \times p}{\dfrac{4p + 1}{5}} = \dfrac{5p}{4p + 1}$.
Answer: $\dfrac{5 p}{4 p+1}$
Hence, the correct answer is option: 'C'.