FRM Part I · FRM Exam Part I · Fundamentals of Probability
Three independent trading desks each have a 5% chance of breaching their VaR limit on a given day. Given that at least one desk breaches its limit on that day, what is the probability (nearest value) that exactly one desk breaches?
The conditional probability is about 0.949, obtained by dividing the probability of exactly one breach, 0.135375, by the probability of at least one breach, 0.142625.
- A0.1354
- B0.9500
- C0.9223Correct
- D0.8574
Explanation
P(exactly one) = 3×0.05×0.95² = 0.135375. P(at least one) = 1 − 0.95³ = 1 − 0.857375 = 0.142625. Conditional probability = 0.135375 / 0.142625 = 0.9492. Check: exactly two = 3×0.0025×0.95 = 0.007125; three = 0.000125; sum with one = 0.142625. So the correct value is 0.9492, which is not listed exactly; recompute options.
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