A man is known to speak the truth 3 out of 4 times. He throws a dice and reports that it is a six. Find the probability that it is actually a six.

Explanation

To solve the problem, we will use Bayes' theorem. We need to find the probability that the dice actually shows a six given that the man reports it as a six.

### Step 1: Define the events
Let:
- \( A \) be the event that the dice shows a six.
- \( B \) be the event that the man reports a six.

### Step 2: Find the probabilities
1. **Probability of reporting the truth**: The man speaks the truth 3 out of 4 times.
\[ P(\text{Truth}) = P(A | B) = \frac{3}{4} \]

2. **Probability of lying**: The probability that he lies is:
\[ P(\text{Lie}) = 1 - P(\text{Truth}) = 1 - \frac{3}{4} = \frac{1}{4} \]

3. **Probability of rolling a six**: The probability of rolling a six on a fair die is:
\[ P(A) = \frac{1}{6} \]

4. **Probability of rolling a non-six**: The probability of rolling a number that is not six (1, 2, 3, 4, or 5) is:
\[ P(A') = 1 - P(A) = 1 - \frac{1}{6} = \frac{5}{6} \]

### Step 3: Calculate the probabilities of reporting a six
1. **Probability of reporting a six when the dice shows a six**:
\[ P(B | A) = P(\text{Truth}) = \frac{3}{4} \]

2. **Probability of reporting a six when the dice does not show a six**:
- If the dice shows any number from 1 to 5, he will lie and say it is a six. The probability of lying is \( P(\text{Lie}) = \frac{1}{4} \).
\[ P(B | A') = P(\text{Lie}) = \frac{1}{4} \]

### Step 4: Use the law of total probability to find \( P(B) \)
\[ P(B) = P(B | A) \cdot P(A) + P(B | A') \cdot P(A') \]
Substituting the values:
\[ P(B) = \left(\frac{3}{4} \cdot \frac{1}{6}\right) + \left(\frac{1}{4} \cdot \frac{5}{6}\right) \]
\[ = \frac{3}{24} + \frac{5}{24} = \frac{8}{24} = \frac{1}{3} \]

### Step 5: Apply Bayes' theorem to find \( P(A | B) \)
Using Bayes' theorem:
\[ P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} \]
Substituting the known values:
\[ P(A | B) = \frac{\left(\frac{3}{4}\right) \cdot \left(\frac{1}{6}\right)}{\frac{1}{3}} \]
\[ = \frac{\frac{3}{24}}{\frac{1}{3}} = \frac{3}{24} \cdot \frac{3}{1} = \frac{9}{24} = \frac{3}{8} \]

### Final Answer
The probability that the dice actually shows a six given that the man reports it as a six is:
\[ \boxed{\frac{3}{8}} \]