Answer: What is the probability that this manager is a superstar?, What is the probability that this manager will beat the market next year?

## Explanation This is a **Bayesian probability** problem involving conditional probabilities and Bayes' theorem. ### Given Information: - **P(S) = 0.16** (Probability a manager is a superstar) - **P(O) = 0.84** (Probability a manager is ordinary) - **P(B|S) = 0.70** (Probability of beating market given superstar) - **P(B|O) = 0.50** (Probability of beating market given ordinary) - Manager has beaten market for 3 consecutive years ### Part A: Probability manager is a superstar given 3 years of beating market Using **Bayes' Theorem**: \[ P(S|B^3) = \frac{P(B^3|S) \cdot P(S)}{P(B^3|S) \cdot P(S) + P(B^3|O) \cdot P(O)} \] Where: - \[ P(B^3|S) = (0.70)^3 = 0.343 \] - \[ P(B^3|O) = (0.50)^3 = 0.125 \] \[ P(S|B^3) = \frac{0.343 \times 0.16}{0.343 \times 0.16 + 0.125 \times 0.84} = \frac{0.05488}{0.05488 + 0.105} = \frac{0.05488}{0.15988} \approx 0.343 \] **Answer: Approximately 34.3%** ### Part B: Probability manager beats market next year This is the **weighted average** of the conditional probabilities: \[ P(B_{next}) = P(S|B^3) \cdot P(B|S) + P(O|B^3) \cdot P(B|O) \] Where: - \[ P(O|B^3) = 1 - P(S|B^3) = 1 - 0.343 = 0.657 \] \[ P(B_{next}) = 0.343 \times 0.70 + 0.657 \times 0.50 = 0.2401 + 0.3285 = 0.5686 \] **Answer: Approximately 56.9%** ### Key Insights: 1. **Bayesian updating**: The manager's track record increases our belief they might be a superstar from 16% to 34.3% 2. **Future performance**: The probability of beating the market next year is higher than the unconditional probability but not as high as a confirmed superstar 3. **Independence assumption**: Each year's performance is independent given the manager type

Author: Tanishq Prabhu