A person is known to speak the truth 4 times out of 5. He throws a die and reports that it is an SIX. The probability that it is actually an six, is

Options

A. `(1)/(3)`
B. `(2)/(9)`
C. `(4)/(9)`
D. `(5)/(9)`

Answer

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

### Step-by-step Solution:

1. **Define Events**:
- Let \( A \) be the event that the die shows a six.
- Let \( E \) be the event that the person reports that the die shows a six.
- Let \( B \) be the event that the die does not show a six.

2. **Find Probabilities**:
- The probability of getting a six when a die is thrown:
\[ P(A) = \frac{1}{6} \]
- The probability of not getting a six:
\[ P(B) = 1 - P(A) = \frac{5}{6} \]

3. **Determine Conditional Probabilities**:
- The probability that the person reports a six given that it is actually a six (truth-telling):
\[ P(E|A) = \frac{4}{5} \]
- The probability that the person reports a six given that it is not a six (lying):
\[ P(E|B) = \frac{1}{5} \]

4. **Apply Bayes' Theorem**:
Bayes' theorem states:
\[ P(A|E) = \frac{P(E|A) \cdot P(A)}{P(E|A) \cdot P(A) + P(E|B) \cdot P(B)} \]

5. **Substitute Values**:
Substitute the values we have found into Bayes' theorem:
\[ P(A|E) = \frac{\left(\frac{4}{5}\right) \cdot \left(\frac{1}{6}\right)}{\left(\frac{4}{5} \cdot \frac{1}{6}\right) + \left(\frac{1}{5} \cdot \frac{5}{6}\right)} \]

6. **Calculate the Numerator**:
\[ \text{Numerator} = \frac{4}{5} \cdot \frac{1}{6} = \frac{4}{30} \]

7. **Calculate the Denominator**:
\[ \text{Denominator} = \frac{4}{30} + \frac{5}{30} = \frac{9}{30} \]

8. **Final Calculation**:
\[ P(A|E) = \frac{\frac{4}{30}}{\frac{9}{30}} = \frac{4}{9} \]

Thus, the probability that the die actually shows a six given that the person reports it as a six is \( \frac{4}{9} \).

### Final Answer:
The required probability is \( \frac{4}{9} \).

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