Question

A person speaks truth 4 out of 5 times. A die is tossed. He reports that there is a six. What is the chance that actually there was six?
1. \(\frac{4}{9}\)
2. \(\frac{5}{9}\)
3. \(\frac{1}{9}\)
4. \(\frac{1}{3}\)

• A. \(\frac{4}{9}\)
• B. \(\frac{5}{9}\)
• C. \(\frac{1}{9}\)
• D. \(\frac{1}{3}\)

Hint:

To solve this quickly, think in terms of frequency:
Out of 30 rolls:
- Six occurs 5 times. The person reports it truthfully 4 times.
- Six does not occur 25 times. The person lies and reports six 5 times.
Total times reported six = \( 4 + 5 = 9 \).
Actually a six = 4.
Probability = \( 4/9 \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a classic conditional probability problem that requires using Bayes' Theorem to find the posterior probability of an event after receiving a report.
Key Formula or Approach:
Bayes' Theorem is given by:
\[ P(A|B) = \frac{P(A) \cdot P(B|A)}{P(A) \cdot P(B|A) + P(A^c) \cdot P(B|A^c)} \]

Step 2: Detailed Explanation:
Let us define the events systematically:
- Let \( S \) be the event that a six actually occurs when the die is rolled.
The probability of rolling a six is:
\[ P(S) = \frac{1}{6} \] - Let \( S^c \) be the event that a six does not occur.
The probability of not rolling a six is:
\[ P(S^c) = \frac{5}{6} \] - Let \( R \) be the event that the person reports rolling a six.
Now, let us determine the conditional probabilities based on the person's truthfulness:
- The person speaks the truth with probability \( 4/5 \).
- If a six actually occurred (\( S \)), the person reports a six (\( R \)) only if they are telling the truth:
\[ P(R|S) = \frac{4}{5} \] - If a six did not occur (\( S^c \)), the person reports a six (\( R \)) only if they are lying:
\[ P(R|S^c) = 1 - P(\text{Truth}) = 1 - \frac{4}{5} = \frac{1}{5} \] We want to find \( P(S|R) \), the probability that a six actually occurred given that the person reported a six.
Apply Bayes' Theorem:
\[ P(S|R) = \frac{P(S) \cdot P(R|S)}{P(S) \cdot P(R|S) + P(S^c) \cdot P(R|S^c)} \] Substitute the probabilities into the formula:
\[ P(S|R) = \frac{\frac{1}{6} \times \frac{4}{5}}{\left(\frac{1}{6} \times \frac{4}{5}\right) + \left(\frac{5}{6} \times \frac{1}{5}\right)} \] \[ P(S|R) = \frac{\frac{4}{30}}{\frac{4}{30} + \frac{5}{30}} \] \[ P(S|R) = \frac{4}{4 + 5} = \frac{4}{9} \]

Step 3: Final Answer:
The probability that there actually was a six is \( 4/9 \).
Therefore, the correct choice is Option (A).