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

Answer

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

Let:
- \( S_1 \): The event that the man speaks the truth.
- \( S_2 \): The event that the man lies.
- \( E \): The event that the die shows a six.

We know:
- \( P(S_1) = \frac{3}{4} \) (the probability that he speaks the truth)
- \( P(S_2) = 1 - P(S_1) = \frac{1}{4} \) (the probability that he lies)

Now, we need to find:
- \( P(E | S_1) \): The probability that the die shows a six given that he speaks the truth. Since the die is fair, this is \( \frac{1}{6} \).
- \( P(E | S_2) \): The probability that the die shows a six given that he lies. If he lies, he will say it is a six when it is not. The probability that he lies and says it is a six is \( \frac{5}{6} \) (because there are 5 other outcomes).

Now, we apply Bayes' theorem:
\[ P(S_1 | E) = \frac{P(S_1) \cdot P(E | S_1)}{P(S_1) \cdot P(E | S_1) + P(S_2) \cdot P(E | S_2)} \]

Substituting the values:
\[ P(S_1 | E) = \frac{\left(\frac{3}{4}\right) \cdot \left(\frac{1}{6}\right)}{\left(\frac{3}{4}\right) \cdot \left(\frac{1}{6}\right) + \left(\frac{1}{4}\right) \cdot \left(\frac{5}{6}\right)} \]

Calculating the numerator:
\[ \text{Numerator} = \frac{3}{4} \cdot \frac{1}{6} = \frac{3}{24} \]

Calculating the denominator:
\[ \text{Denominator} = \frac{3}{4} \cdot \frac{1}{6} + \frac{1}{4} \cdot \frac{5}{6} = \frac{3}{24} + \frac{5}{24} = \frac{8}{24} = \frac{1}{3} \]

Now substituting back into the equation:
\[ P(S_1 | E) = \frac{\frac{3}{24}}{\frac{8}{24}} = \frac{3}{8} \]

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