A man is known to speak the truth on an average 3 out of 4 times. He throws a fair die and reports that it is a six. The probability the it is actually a six is

Options

A. `(3)/(5)`
B. `(3)/(8)`
C. `(3)/(4)`
D. `(1)/(5)`

Answer

To solve the problem, we will use Bayes' theorem. Let's break down the steps:

### Step 1: Define Events
Let:
- \( S_1 \): The event that the die shows a six.
- \( S_2 \): The event that the die does not show a six.
- \( E \): The event that the man reports that the die shows a six.

### Step 2: Calculate Probabilities
1. The probability that the die shows a six:
\[ P(S_1) = \frac{1}{6} \]

2. The probability that the die does not show a six:
\[ P(S_2) = 1 - P(S_1) = 1 - \frac{1}{6} = \frac{5}{6} \]

3. The probability that the man reports a six given that it is actually a six:
\[ P(E | S_1) = \frac{3}{4} \]

4. The probability that the man reports a six given that it is not a six:
\[ P(E | S_2) = 1 - P(E | S_1) = 1 - \frac{3}{4} = \frac{1}{4} \]

### Step 3: Use Bayes' Theorem
We want to find \( P(S_1 | E) \), the probability that the die shows a six given that the man reports it as a six. According to Bayes' theorem:
\[ P(S_1 | E) = \frac{P(E | S_1) \cdot P(S_1)}{P(E)} \]

### Step 4: Calculate \( P(E) \)
To find \( P(E) \), we use the law of total probability:
\[ P(E) = P(E | S_1) \cdot P(S_1) + P(E | S_2) \cdot P(S_2) \]
Substituting the known values:
\[ P(E) = \left(\frac{3}{4} \cdot \frac{1}{6}\right) + \left(\frac{1}{4} \cdot \frac{5}{6}\right) \]
Calculating each term:
\[ P(E) = \frac{3}{24} + \frac{5}{24} = \frac{8}{24} = \frac{1}{3} \]

### Step 5: Substitute Back into Bayes' Theorem
Now substituting back into Bayes' theorem:
\[ P(S_1 | E) = \frac{P(E | S_1) \cdot P(S_1)}{P(E)} = \frac{\left(\frac{3}{4}\right) \cdot \left(\frac{1}{6}\right)}{\frac{1}{3}} \]
Calculating the numerator:
\[ = \frac{3}{24} \]
Now dividing by \( \frac{1}{3} \):
\[ P(S_1 | E) = \frac{3}{24} \cdot 3 = \frac{9}{24} = \frac{3}{8} \]

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