In a competitive examination, an examinee either guesses or copies or knows the answer to 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 the answer is correct, given that he copies it, is `1/8`. Find the probability that he knows the answer to the question, given that he correctly answered

Answer

To solve the problem, we will use the concept of conditional probability and the law of total probability. Here is the step-by-step solution:

### Step 1: Define the Events
Let:
- \( G \): the event that the examinee guesses the answer.
- \( C \): the event that the examinee copies the answer.
- \( K \): the event that the examinee knows the answer.
- \( A \): the event that the answer is correct.

### Step 2: Given Probabilities
From the problem, we have:
- \( P(G) = \frac{1}{3} \)
- \( P(C) = \frac{1}{6} \)
- \( P(K) = 1 - P(G) - P(C) = 1 - \frac{1}{3} - \frac{1}{6} = \frac{1}{2} \)

### Step 3: Find the Probability of Correct Answers
Next, we need to find the probability of getting the answer correct given each event:
- If the examinee guesses, the probability of getting the answer correct is \( P(A|G) = \frac{1}{4} \) (since there are 4 choices).
- If the examinee copies, the probability of getting the answer correct is \( P(A|C) = \frac{1}{8} \).
- If the examinee knows the answer, the probability of getting the answer correct is \( P(A|K) = 1 \).

### Step 4: Use the Law of Total Probability
Now, we can find the total probability of answering correctly, \( P(A) \):
\[ P(A) = P(A|G)P(G) + P(A|C)P(C) + P(A|K)P(K) \]
Substituting the values:
\[ P(A) = \left(\frac{1}{4} \cdot \frac{1}{3}\right) + \left(\frac{1}{8} \cdot \frac{1}{6}\right) + \left(1 \cdot \frac{1}{2}\right) \]
Calculating each term:
- \( P(A|G)P(G) = \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12} \)
- \( P(A|C)P(C) = \frac{1}{8} \cdot \frac{1}{6} = \frac{1}{48} \)
- \( P(A|K)P(K) = 1 \cdot \frac{1}{2} = \frac{1}{2} = \frac{24}{48} \)

Now combine these:
\[ P(A) = \frac{1}{12} + \frac{1}{48} + \frac{24}{48} \]
Finding a common denominator (48):
\[ P(A) = \frac{4}{48} + \frac{1}{48} + \frac{24}{48} = \frac{29}{48} \]

### Step 5: Find the Desired Probability
We want to find \( P(K|A) \), the probability that the examinee knows the answer given that they answered correctly. Using Bayes' theorem:
\[ P(K|A) = \frac{P(A|K)P(K)}{P(A)} \]
Substituting the known values:
\[ P(K|A) = \frac{1 \cdot \frac{1}{2}}{\frac{29}{48}} = \frac{\frac{1}{2}}{\frac{29}{48}} = \frac{24}{29} \]

### Final Answer
Thus, the probability that the examinee knows the answer given that they answered correctly is:
\[ \boxed{\frac{24}{29}} \]