Answer: 0.3433

## Explanation Let's define the following events: - **S**: the manager is a superstar - **O**: the manager is ordinary - **B**: the manager beats the market in a given year - **U**: the manager underperforms the market in a given year From the information given: - P(S) = 0.16 (the prior probability of being a superstar) - P(O) = 0.84 (the prior probability of being ordinary) - P(B|S) = 0.7 (probability of beating the market given the manager is a superstar) - P(B|O) = 0.5 (probability of beating the market given the manager is ordinary) We want to find P(S|B³), the probability that the manager is a superstar today, given that he has beaten the market for three consecutive years. Using Bayes' theorem: \[ P(S|B^3) = \frac{P(B^3|S) \cdot P(S)}{P(B^3)} \] Calculate each term: 1. **P(B³|S)** = P(B|S)³ = 0.7³ = 0.343 2. **P(B³|O)** = P(B|O)³ = 0.5³ = 0.125 3. **P(B³)** = P(B³|S) × P(S) + P(B³|O) × P(O) = (0.343 × 0.16) + (0.125 × 0.84) = 0.05488 + 0.105 = 0.15988 4. **P(S|B³)** = \[ \frac{0.343 \times 0.16}{0.15988} = \frac{0.05488}{0.15988} = 0.3433 \] Therefore, the probability that the manager is a superstar given three consecutive years of beating the market is **0.3433**. This demonstrates how Bayesian updating works - even though the prior probability of being a superstar was only 16%, three consecutive years of strong performance significantly increases this probability to about 34.33%.

Author: Tanishq Prabhu