How do I proceed for this probability problem? The question is — In an entrance examination with multiple choice questions, with each question having four options and a single correct answer, suppose that only 20% candidates think they know the answer to one difficult question and only half of them know it correctly and the other half get it wrong. The remaining candidates pick one option out of the four randomly and tick the same. If a candidate has correctly answered the question, what is the (conditional) probability that she knew the answer?
Edit:
Let $K$ be the event that a candidate knows the answer (correctly or not), and $R$ is the event that a candidate answers the question randomly.
So, $P(K) = 0.2$, $P(R) = 0.8$.
Let $C$ be the event that a candidate correctly answers the question.
We want to find $P(K|C)$.
By Bayes’ Theorem,
$P(K|C) = \frac{P(K).P(C|K)}{P(K).P(C|K) + P(R).P(C|R)}$
We have $P(C|K) = 0.1$ and $P(C|R) = 0.25$.
On substituting, $P(K|C) = \frac{1}{11}$
Is this correct?
Edit #2: This is wrong. My mistake was interpreting the fact that half of the 20% of candidates knowing the answer correctly meant that 10% of all candidates knew it correctly. Instead, it means that the probability that any of these 20% candidates knowing the answer is 0.5.
So $P(C|K) = 0.5$ and not $0.1$
 A: Let $C$ be the event of correctly answering the question, $K$ be the event of knowing the answer, and $NK$ be the event of not knowing the answer.
We need $P(K|C)$, this can happen in two ways, either she knew the answer or she didn't. Now the answer can be known in two ways, either correctly or incorrectly. (Here, she knew the answer means that she thought she knew the answer)
By Bayes' Theorem:
$$P(K|C)=\dfrac{P(K)\cdot P(C|K)}{P(C)}$$
Also $P(C)=P(K).P(C|K)+P(NK).P(C|NK)$
$$P(C)=(0.2)(0.5)+(0.8)(0.25)=0.3$$ because if she knew the answer, then she either knew it correct or incorrect, the probability of both being $0.5$, whereas if she guessed, then she had a $0.25$ probability of correctly answering the question.
$$\therefore P(K|C)=\dfrac{0.1}{0.3}=0.33$$
A: 10% of the candidates give the wrong answer
10% of the candidates give the right answer (they knew the answer)
80% of the candidates give the right answer with probability $0.25$ (they guessed right)
Thus the right answer is given by $0.8\times0.25+0.1$
Applying Bayes' theorem you get
$$\frac{0.1}{0.8\times0.25+0.1}=\frac{1}{3}$$
