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Answer: 0.359
This is a conditional probability problem that can be solved using Bayes' theorem. Let: - X, Y, Z be events of choosing routes X, Y, Z respectively - O be the event of arriving on time Given: - P(O|X) = 0.60 - P(O|Y) = 0.65 - P(O|Z) = 0.70 - P(X) = P(Y) = P(Z) = 1/3 (equally likely routes) We want P(Z|O): $$P(Z|O) = \frac{P(Z) \cdot P(O|Z)}{P(Z) \cdot P(O|Z) + P(Y) \cdot P(O|Y) + P(X) \cdot P(O|X)}$$ $$= \frac{(\frac{1}{3} \cdot 0.7)}{(\frac{1}{3} \cdot 0.7) + (\frac{1}{3} \cdot 0.65) + (\frac{1}{3} \cdot 0.6)}$$ $$= \frac{0.2333}{(0.2333 + 0.2167 + 0.2)}$$ $$= \frac{0.2333}{0.65} = 0.3589 \approx 0.359$$ The calculation shows that given she arrived on time, the probability she took route Z is approximately 0.359.
Author: Tanishq Prabhu
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A financial risk manager has three routes to get to the office. The probability that she gets to the office on time using routes X, Y, and Z are 60%, 65%, and 70%. She does not have a preferred route and is therefore equally likely to choose any of the three routes. Calculate the probability that she chose route Z given that she arrives to work on time.
A
0.359
B
0.233
C
0.216
D
0.2