A student appears for a quiz consisting of only true-false type questions and answers all the questions. The student knows the answers of some questions and guesses the answers for the remaining questions. Whenever the student knows the answer of a question, he gives the correct answer. Assume that the probability of the student giving the correct answer for a question, given that he has guessed it, is 1/2. Also assume that the probability of the answer for a question being guessed, given that the student’s answer is correct, is 1/6. Then the probability that the student knows the answer of a randomly chosen question is

The correct option is 5/7.

Let us define the events for a randomly chosen question:
Let $K$ be the event that the student knows the answer.
Let $G$ be the event that the student guesses the answer. Since a question is either known or guessed, we have $P\left(G\right)=1-P\left(K\right)$.
Let $C$ be the event that the student's answer is correct.

Based on the problem description, we are given the following:
1. If the student knows the answer, he always answers correctly. Thus:
$P\left(C\right|K)=1$
2. The probability of giving the correct answer given that he guessed it is 1/2:
$P\left(C\right|G)=\frac{1}{2}$
3. The probability that the answer was guessed, given that the student's answer is correct, is 1/6:
$P\left(G\right|C)=\frac{1}{6}$

We want to find $P\left(K\right)$, the probability that the student knows the answer. Let $P\left(K\right)=p$. Then, $P\left(G\right)=1-p$.

Using Bayes' theorem, the conditional probability $P\left(G\right|C)$ is given by:
$P\left(G\right|C)=\frac{P\left(G\right)P\left(C\right|G)}{P\left(K\right)P\left(C\right|K)+P(G\left)P\right(C\left|G\right)}$

Substituting the known values into the equation:
$\frac{1}{6}=\frac{(1-p)\cdot \frac{1}{2}}{p\cdot 1+(1-p)\cdot \frac{1}{2}}$

Simplify the fraction in the denominator:
$\frac{1}{6}=\frac{\frac{1-p}{2}}{p+\frac{1-p}{2}}$
Multiply the numerator and denominator by 2 to clear the fraction:
$\frac{1}{6}=\frac{1-p}{2p+1-p}$
$\frac{1}{6}=\frac{1-p}{p+1}$

Now, cross-multiply to solve for $p$:
$1\cdot (p+1)=6\cdot (1-p)$
$p+1=6-6p$
Combine the terms containing $p$ on one side:
$p+6p=6-1$
$7p=5$
$p=\frac{5}{7}$

Therefore, the probability that the student knows the answer to a randomly chosen question is $\frac{5}{7}$.