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Answer: 0.66
This is a Bayesian probability problem. We need to find P(O | 3B) = probability that the manager is ordinary given they beat the market for 3 consecutive years. **Given:** - P(S) = 0.16 (probability of being a superstar) - P(O) = 1 - 0.16 = 0.84 (probability of being ordinary) - P(B|S) = 0.7 (probability of beating market given superstar) - P(B|O) = 0.5 (probability of beating market given ordinary) **Step 1: Calculate P(3B|S)** Since performance is independent year to year: P(3B|S) = (0.7)^3 = 0.343 **Step 2: Calculate P(3B|O)** P(3B|O) = (0.5)^3 = 0.125 **Step 3: Calculate P(3B)** (total probability of beating market 3 years) P(3B) = P(3B|S) × P(S) + P(3B|O) × P(O) = 0.343 × 0.16 + 0.125 × 0.84 = 0.05488 + 0.105 = 0.15988 ≈ 0.16 **Step 4: Apply Bayes' Theorem to find P(S|3B)** P(S|3B) = [P(3B|S) × P(S)] / P(3B) = (0.343 × 0.16) / 0.15988 = 0.05488 / 0.15988 ≈ 0.343 **Step 5: Find P(O|3B)** P(O|3B) = 1 - P(S|3B) = 1 - 0.343 = 0.657 ≈ 0.66 Therefore, the probability that the manager is NOT a superstar is approximately 0.66.
Author: Nikitesh Somanthe
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You are an analyst at a large mutual fund. After examining historical data, you establish that all fund managers fall into 2 categories: superstars (S) and ordinaries (O).
Superstars are by far the best managers. The probability that a superstar will beat the market in any given year stands at 70%. Ordinaries, on the other hand, are just as likely to beat the market as they are to underperform it. Regardless of the category in which a manager falls, the probability of beating the market is independent of year to year. Superstars are rare diamonds because only a meager 16% of all recruits turn out to be superstars.
During the analysis, you stumble upon the profile of a manager recruited 3 years ago, who has since gone on to beat the market every year.
What is the probability that the manager is NOT a superstar?
A
0.66
B
0.7
C
0.45
D
0.64