Question

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 \(\frac{1}{2}\) . Also assume that the probability of the answer for a question being guessed, given that the student’s answer is correct, is \(\frac{1}{6}\) . Then the probability that the student knows the answer of a randomly chosen question is

• A. \(\frac{1}{12}\)
• B. \(\frac{1}{7}\)
• C. \(\frac{5}{7}\)
• D. \(\frac{5}{12}\)

Answer 1

The Correct Option is C

To solve the problem, we need to find the probability that the student knows the answer to a randomly chosen question given the information about guessing and correctness.

1. Define the events:
Let:
- \( K \) = event that the student knows the answer
- \( G \) = event that the student guesses the answer
- \( C \) = event that the student's answer is correct

2. Given probabilities:
- \( P(C \mid G) = \frac{1}{2} \) (Probability of correct answer given guessed)
- \( P(G \mid C) = \frac{1}{6} \) (Probability of guessing given answer is correct)

3. Using Bayes' Theorem to find \( P(K \mid C) \):
Note that the student either knows or guesses the answer, so:
\[ P(K) + P(G) = 1 \] and
\[ P(C) = P(C \mid K) P(K) + P(C \mid G) P(G) \] Since the student always answers correctly when he knows the answer:
\[ P(C \mid K) = 1 \]

4. Express \( P(G \mid C) \) using Bayes' Theorem:
\[ P(G \mid C) = \frac{P(C \mid G) P(G)}{P(C)} \] Given \( P(G \mid C) = \frac{1}{6} \), so:
\[ \frac{1}{6} = \frac{P(C \mid G) P(G)}{P(C)} = \frac{\frac{1}{2} P(G)}{P(C)} \] which implies:
\[ P(C) = 3 P(G) \]

5. Express \( P(C) \) using total probability:
\[ P(C) = P(C \mid K) P(K) + P(C \mid G) P(G) = 1 \cdot P(K) + \frac{1}{2} P(G) = P(K) + \frac{1}{2} P(G) \]

6. Using \( P(K) = 1 - P(G) \), substitute in \( P(C) = 3 P(G) \):
\[ 3 P(G) = (1 - P(G)) + \frac{1}{2} P(G) = 1 - P(G) + \frac{1}{2} P(G) = 1 - \frac{1}{2} P(G) \] So, \[ 3 P(G) + \frac{1}{2} P(G) = 1 \] \[ \frac{7}{2} P(G) = 1 \implies P(G) = \frac{2}{7} \]

7. Calculate \( P(K) \):
\[ P(K) = 1 - P(G) = 1 - \frac{2}{7} = \frac{5}{7} \]

Final Answer:
The probability that the student knows the answer to a randomly chosen question is \( \boxed{\frac{5}{7}} \).

Answer 2

To determine the probability that the student knows the answer to a randomly chosen question, we can use Bayes' Theorem. Let us define the events first:

• K: The student knows the answer to a question.
• G: The student guesses the answer to a question.
• C: The student's answer is correct.

We are provided with the following probabilities:

• The probability of a correct answer given a guess is \(P(C|G)=\frac{1}{2}\).
• The probability of a guessed answer given a correct answer is \(P(G|C)=\frac{1}{6}\).

We want to find \(P(K|C)\), which is the probability that the student knows the answer given that the student's answer is correct.

Using Bayes' Theorem, we have:

\(P(K|C)=\frac{P(C|K)\cdot P(K)}{P(C)}\)

Since the student always gives the correct answer if they know it, \(P(C|K)=1\).

Considering the complementary event, \(P(K)+P(G)=1\), and we also need to find \(P(C)\), the total probability of a correct answer. This can be expressed as:

\(P(C)=P(C|K)\cdot P(K)+P(C|G)\cdot P(G)\)

\(P(C)=1\cdot P(K)+\frac{1}{2}\cdot P(G)\)

We know \(P(G|C)=\frac{1}{6}\), which can be expanded with Bayes' Theorem:

\(P(G|C)=\frac{P(C|G)\cdot P(G)}{P(C)}\)

\(\frac{1}{6}=\frac{\frac{1}{2}\cdot P(G)}{P(C)}\)

Rearranging this, we get:

\(P(C)=3\cdot P(G)\)

Now substitute back in the equation for \(P(C)\):

\(P(C)=P(K)+\frac{1}{2}\cdot P(G)=3\cdot P(G)\)

\(P(K)+\frac{1}{2}P(G)=3P(G)\)

\(P(K)=3P(G)-\frac{1}{2}P(G)\)

\(P(K)=\frac{5}{2}P(G)\)

Using \(P(K)+P(G)=1\):

\(\frac{5}{2}P(G)+P(G)=1\)

\(\frac{7}{2}P(G)=1\)

\(P(G)=\frac{2}{7}\)

Thus:

\(P(K)=1-P(G)=1-\frac{2}{7}=\frac{5}{7}\)

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