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FRM Part I · FRM Exam Part I · Fundamentals of Probability

A credit officer partitions next year's outcomes for a corporate borrower into three mutually exclusive and exhaustive events: E1 = 'no change in rating', E2 = 'upgrade', E3 = 'downgrade or default'. The probabilities satisfy P(E2) = 0.5 × P(E3) and P(E1) = 0.70. What is P(E3)?

P(E3) is 0.20. The three events partition the sample space, so their probabilities sum to 1, leaving 0.30 for E2 and E3. Since E2 is half of E3, E3 takes two thirds of 0.30, which is 0.20.

  1. A0.150
  2. B0.200Correct
  3. C0.100
  4. D0.300

Explanation

Exhaustive and mutually exclusive events sum to 1, so P(E2) + P(E3) = 0.30. With P(E2) = 0.5 P(E3), 1.5 P(E3) = 0.30, so P(E3) = 0.20 and P(E2) = 0.10. The value 0.150 results from splitting 0.30 equally, ignoring the stated ratio.

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