Let's solve the problem step-by-step. --- ## **Step 1: Assign Probabilities** Let: - \( E \): The event actually occurred. - \( A \): A says the event occurred. - \( B \): B says the event occurred. Given: - Probability A speaks truth = \( P(A_T) = .75 \) - Probability B speaks truth = \( P(B_T) = .60 \) Assume both speak independently. --- ## **Step 2: Conditional Probabilities** - If the event **actually occurred** (\( E \)): - Probability both say it occurred: \( .75 \times .60 = .45 \) - If the event **did NOT occur** (\( \bar{E} \)): - Probability both lie (i.e., both say it **did** occur when it did **not**): - Probability A lies: \( 1 - .75 = .25 \) - Probability B lies: \( 1 - .60 = .40 \) - Both lie: \( .25 \times .40 = .10 \) --- ## **Step 3: Required Probability** We want: **Probability that the event actually occurred, given both say it occurred.** This is \( P(E \mid \text{both say event occurred}) \). By Bayes' theorem: \[ P(E \mid \text{both say occurred}) = \frac{P(\text{both say occurred} \mid E) \cdot P(E)}{P(\text{both say occurred})} \] Assume **no bias** about whether the event occurs, so \( P(E) = .5 \) (if not given, it's standard to assume this). --- ### **Calculate Numerator** \[ P(\text{both say occurred} \mid E) \cdot P(E) = .45 \times .5 = .225 \] --- ### **Calculate Denominator** \[ P(\text{both say occurred}) = P(\text{both say occurred} \mid E)P(E) + P(\text{both say occurred} \mid \bar{E})P(\bar{E}) \] \[ = (.45 \times .5) + (.10 \times .5) = .225 + .05 = .275 \] --- ### **Final Calculation** \[ P(E \mid \text{both say occurred}) = \frac{.225}{.275} = \frac{225}{275} = \frac{9}{11} \] --- ## **Final Answer** \[ \boxed{\frac{9}{11}} \] or approximately \[ \boxed{.818} \] --- ### **Explanation** - **Both speaking truth** when the event occurs: .75 × .60 = .45. - **Both lying** when the event does **not** occur: .25 × .40 = .10. - Use Bayes to compute the probability that the event really happened, given both say so. Let me know if you'd like a deeper breakdown on any step!