In a test an examinee either guesses or copies or knows the 
answer in a multiple choice question with four choices. The 
probability that he makes a guess is 1
3
and the probability that he 
copies the answer is 1
6
. The probability that his answer is correct 
given that he copied it is 1
8
. Find the probability that he knew the answer to the question given that he correctly answered it.

We are given the problem in which an examinee answers a multiple-choice question with four choices. The probabilities associated with the three possible actions are as follows:

1. Probability of guessing (\( P(G) \)) = \( \frac{1}{3} \)
2. Probability of copying (\( P(C) \)) = \( \frac{1}{6} \)
3. Probability of knowing the answer (\( P(K) \)) = \( 1 - P(G) - P(C) \)
   
    Since the sum of probabilities must be 1, we can calculate:
    \[
    P(K) = 1 - \frac{1}{3} - \frac{1}{6} = \frac{1}{2}
    \]

The problem also gives us the following conditional probabilities:

- Probability that the answer is correct given that the examinee guessed (\( P(Correct|G) \)) = \( \frac{1}{4} \), since in guessing, there is a 1 in 4 chance of choosing the correct answer.
- Probability that the answer is correct given that the examinee copied (\( P(Correct|C) \)) = \( \frac{1}{8} \), as stated in the problem.
- Probability that the answer is correct given that the examinee knew the answer (\( P(Correct|K) \)) = 1, because if the examinee knows the answer, they will answer correctly with certainty.

### Question:
We are tasked with finding the probability that the examinee knew the answer given that they answered correctly. This is expressed as \( P(K|Correct) \).

We can use **Bayes' Theorem** to calculate this probability:
\[
P(K|Correct) = \frac{P(Correct|K) \cdot P(K)}{P(Correct)}
\]
To solve this, we need to first calculate \( P(Correct) \), the total probability of answering the question correctly. This can be done using the law of total probability:

\[
P(Correct) = P(Correct|G) \cdot P(G) + P(Correct|C) \cdot P(C) + P(Correct|K) \cdot P(K)
\]

Substituting the known values:
\[
P(Correct) = \frac{1}{4} \cdot \frac{1}{3} + \frac{1}{8} \cdot \frac{1}{6} + 1 \cdot \frac{1}{2}
\]
\[
P(Correct) = \frac{1}{12} + \frac{1}{48} + \frac{1}{2}
\]
\[
P(Correct) = \frac{4}{48} + \frac{1}{48} + \frac{24}{48} = \frac{29}{48}
\]

Now, we can use Bayes' Theorem:
\[
P(K|Correct) = \frac{1 \cdot \frac{1}{2}}{\frac{29}{48}} = \frac{\frac{1}{2}}{\frac{29}{48}} = \frac{1}{2} \cdot \frac{48}{29} = \frac{24}{29}
\]

### Final Answer:
The probability that the examinee knew the answer given that they answered correctly is \( \frac{24}{29} \).

Would you like further details on any part of the solution, or do you have any questions?

Here are 5 related questions to explore:

1. How does the total probability formula help in such problems?
2. What is the role of Bayes' Theorem in conditional probability?
3. How can we interpret the difference between guessing, copying, and knowing in probabilistic terms?
4. How does changing the probability of copying affect the final answer?
5. What happens if the number of choices in the multiple-choice question changes?

**Tip:** Bayes' Theorem is particularly useful when you need to update the probability of an event based on new information, such as knowing that an answer is correct in this case.

In a test an examinee either guesses or copies or knows the answer in a multiple choice question with four choices. The probability that he makes a guess is 1/3 and the probability that he copies the answer is 1/6. The probability that his answer is correct given that he copied it is 1/8. Find the probability that he knew the answer to the question given that he correctly answered it.

Learn how to calculate the probability that an examinee knew the answer to a multiple choice question given they answered it correctly. This problem uses Bayes' Theorem and conditional probability with a detailed step-by-step solution. Suitable for students in Grades 10-12.

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