Question

A person is known to speak the truth in 3 out of 4 occasions. If he throws a die and reports that it is six, then the probability that it is actually six is:

• A. \(\frac{3}{8}\)
• B. \(\frac{2}{7}\)
• C. \(\frac{1}{9}\)
• D. \(\frac{4}{5}\)

Hint:

Apply Bayes theorem to update probabilities based on new information.

Answer 1

The Correct Option is A

Step 1: Define events
Let \(T\) be the event that person tells the truth, \(S\) be the event the die shows six. \[ P(T) = \frac{3}{4}, \quad P(\overline{T}) = \frac{1}{4} \] Step 2: Calculate probability of reporting six
\[ P(\text{reports six}) = P(\text{reports six} | S) P(S) + P(\text{reports six} | \overline{S}) P(\overline{S}) \] \[ = P(T) \times \frac{1}{6} + P(\overline{T}) \times \frac{5}{6} = \frac{3}{4} \times \frac{1}{6} + \frac{1}{4} \times \frac{5}{6} = \frac{3}{24} + \frac{5}{24} = \frac{8}{24} = \frac{1}{3} \] Step 3: Use Bayes theorem to find \(P(S | \text{reports six})\)
\[ P(S | \text{reports six}) = \frac{P(\text{reports six} | S) P(S)}{P(\text{reports six})} = \frac{\frac{3}{4} \times \frac{1}{6}}{\frac{1}{3}} = \frac{\frac{3}{24}}{\frac{1}{3}} = \frac{3}{24} \times 3 = \frac{3}{8} \]