Answer-first summary for fast verification
Answer: 0.359
This is a Bayes' Theorem problem. We need to find P(Z|O) - the probability she chose route Z given that she arrived on time. **Given:** - P(O|X) = 0.60 (probability of arriving on time given route X) - P(O|Y) = 0.65 (probability of arriving on time given route Y) - P(O|Z) = 0.70 (probability of arriving on time given route Z) - P(X) = P(Y) = P(Z) = 1/3 (equally likely to choose any route) **Using Bayes' Theorem:** P(Z|O) = [P(Z) × P(O|Z)] / [P(X) × P(O|X) + P(Y) × P(O|Y) + P(Z) × P(O|Z)] **Calculation:** Numerator: (1/3) × 0.70 = 0.23333 Denominator: (1/3 × 0.60) + (1/3 × 0.65) + (1/3 × 0.70) = 0.20 + 0.21667 + 0.23333 = 0.65 P(Z|O) = 0.23333 / 0.65 = 0.3589 ≈ 0.359 **Verification:** The total probability of arriving on time is 0.65. Route Z contributes 0.23333 to this total probability, so the conditional probability is 0.23333/0.65 = 0.3589.
Author: Nikitesh Somanthe
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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
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