Skip to content

FRM Part I · FRM Exam Part I · Common Univariate Random Variables

A bank's trading desk has a one-day 99% VaR that is exceeded with probability 0.01 each day, independently. Over 5 trading days, what is the probability of exactly one exception (nearest value)?

The probability is about 0.0480. Using the binomial formula with five days and a 1% exception probability, it equals 5 times 0.01 times 0.99 to the fourth power, which is 0.0480. Simply multiplying 5 by 0.01 ignores the requirement of no exceptions on the other days.

  1. A0.0480Correct
  2. B0.0500
  3. C0.0010
  4. D0.0490

Explanation

Binomial with n=5, p=0.01: P(X=1) = 5 x 0.01 x 0.99^4. 0.99^4 = 0.960596, so the probability = 0.05 x 0.960596 = 0.04803. The option 0.05 ignores the probability of no exceptions on the other four days.

Did you get it right without looking?

One question tells you little. A timed set on Common Univariate Random Variables shows your real accuracy, how long you take and where you lose marks.

Practise Common Univariate Random VariablesFree diagnostic mock

More Common Univariate Random Variables questions