Charta is on Google Play: practise on the go. Get the app
Charta

Define a conditional probability, explain it and work it out.

FRM® Part I · Quantitative Analysis · Fundamentals of Probability · FRM1.QA.1.e

The idea

The conditional probability of A given B is the probability that A occurs when we know that B has occurred: P(A | B) = P(A and B) ÷ P(B). It is defined only when P(B) is positive. Conditioning shrinks the sample space to B, so the joint probability is rescaled by dividing by the probability of B. It is not symmetric: P(A | B) divides by P(B), whereas P(B | A) divides by P(A), and the two usually differ. Rearranged, the same idea gives the multiplication rule for the probability that both occur: P(A and B) = P(A | B) × P(B). In sampling without replacement, the second draw is conditional on the first, so the pool changes between draws. For a bond that has survived so far, default in a later period is conditional on that survival, so divide the unconditional probability of default in that period by the survival probability, not by one. The usual error is dividing by the wrong event. On the exam, identify the given event and put its probability in the denominator.

Formulas

Conditional probability

P(A | B) = P(A and B) ÷ P(B)

P(A and B)
the probability that both A and B occur
P(B)
the probability of the given event B, which must be positive

The given event goes in the denominator. Dividing by P(A) instead gives P(B | A), which usually differs.

A practice question

From the question bank, with the reason each wrong option is wrong.

P(A and B) = 0.12, P(A) = 0.20 and P(B) = 0.30. What is P(A | B)?

  1. 0.60Why not: Divides by P(A) and so computes P(B | A).
  2. 0.06Why not: Multiplies the marginal probabilities, which is not a conditional probability.
  3. 0.18Why not: Subtracts the joint probability from P(B) instead of dividing by P(B).
  4. 0.40 - correct

P(A | B) = P(A and B) / P(B) = 0.12 / 0.30 = 0.40. Note that P(B | A) = 0.12/0.20 = 0.60 is different.

A harder one

A bond portfolio holds 1,000 issuers: 600 are investment grade, of which 6 defaulted this year, and 400 are high yield, of which 40 defaulted. An issuer is picked at random from those that defaulted. What is the probability that it is high yield?

  1. 0.100Why not: Computes P(default | high yield) instead of P(high yield | default); the conditioning event is reversed.
  2. 0.040Why not: Reports the joint probability of being high yield and defaulting out of all 1,000 issuers.
  3. 0.870 - correct
  4. 0.400Why not: Reports the unconditional share of high-yield issuers and ignores the information that the issuer defaulted.

Given default, the sample space is the 46 defaulted issuers, of which 40 are high yield: 40/46 = 0.870. Equivalently P(HY and default)/P(default) = 0.040/0.046.

The hardest tier

A bond's cumulative default probability is 2% by the end of year 1 and 5% by the end of year 2. Given that it has not defaulted by the end of year 1, what is the probability that it defaults during year 2?

  1. 3.00%Why not: Uses the unconditional year-2 default probability of 3% and does not divide by the year-1 survival probability.
  2. 3.06% - correct
  3. 5.10%Why not: Divides the cumulative probability by the survival probability instead of the year-2 increment.
  4. 3.16%Why not: Divides by the survival probability to the end of year 2 rather than to the end of year 1.

Default in year 2 has unconditional probability 0.05 - 0.02 = 0.03. Conditioning on survival to year 1 divides by 0.98: 0.03/0.98 = 3.06%.

In the app: 6 practice questions on this outcome, spaced review that brings it back before you forget it.

Start free →

More in Fundamentals of Probability