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Answer: 0.34
We need to determine P(S | 3B): The probability that the manager is a superstar given that they have managed to beat the market in three consecutive years. As such, we need to apply Bayes' theorem. P(S | 3B) = P(S) * P(3B | S)/P(3B) Now, we already have P(S) = 16% = 4/25 P(3B | S) = (7/10)^3 since performance is independent from one year to the next = 343/1000 P(3B) = unconditional probability of beating the market in 3 consecutive years = weighted average probability of 3 marketing-beating years over both superstars and ordinaries = P(3B | S) * P(S) + P(3B | O) * P(O) = [(7/10)^3 * 4/25] + [(1/2)^3 * 21/25] = (343/1000 * 4/25) + (1/8 * 21/25) = 1372/25000 + 21/200 = 16% Therefore, 16% * 34.3% / 16% = 34.3% or 0.343
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 a superstar as at present?
A
0.46
B
0.34
C
0.84
D
0.16
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