Question

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 six.

• A. \(\frac{1}{8}\)
• B. \(\frac{2}{8}\)
• C. \(\frac{3}{8}\)
• D. \(\frac{1}{2}\)

Hint:

Whenever information is reported by a person whose statements may be true or false, Bayes' Theorem is usually the correct approach.

Answer 1

The Correct Option is C

Step 1: Find the prior probabilities. Since the die is fair, \[ P(E)=\frac16, \qquad P(E')=\frac56. \]

Step 2: Find the truth and lie probabilities. The man speaks truth with probability \[ \frac34. \] Hence, \[ P(A|E)=\frac34. \] Therefore, \[ P(A|E')=\frac14. \]

Step 3: Apply Bayes' Theorem. \[ P(E|A) = \frac{\frac16\cdot\frac34} {\frac16\cdot\frac34+\frac56\cdot\frac14}. \] \[ = \frac{\frac{3}{24}} {\frac{3}{24}+\frac{5}{24}}. \] \[ = \frac{3}{8}. \] Conclusion: \[ {\frac{3}{8}} \]