Financial Risk Manager Part 1

Financial Risk Manager Part 1

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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.

TTanishq

Explanation:

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)=P(Z)⋅P(O∣Z)P(Z)⋅P(O∣Z)+P(Y)⋅P(O∣Y)+P(X)⋅P(O∣X)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)}

=(13⋅0.7)(13⋅0.7)+(13⋅0.65)+(13⋅0.6)= \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)}

=0.2333(0.2333+0.2167+0.2)= \frac{0.2333}{(0.2333 + 0.2167 + 0.2)}

=0.23330.65=0.3589≈0.359= \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.

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