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Use a discrete probability function to find how likely a given event is.

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

The idea

A discrete probability function, or probability mass function, gives f(x), the probability that a discrete random variable X takes the value x, for each value that X can take. It has two properties: every f(x) lies between 0 and 1, and the probabilities over all possible values sum to 1. The probability of any event is the sum of f(x) over the values that make up the event. So P(X ≥ 2) adds f(2), f(3) and so on, while P(a ≤ X ≤ b) adds every value from a to b inclusive. When the function contains an unknown constant, set the total to 1 and solve for it first. Be careful with strict and non-strict inequalities, which change which end values are included. Never use a function whose probabilities over all possible values do not sum to 1. On the exam, list the values, check the total, then add only the values in the event.

A practice question

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

Which of the following is a valid probability mass function for a random variable X that takes the values 1, 2, 3 and 4?

  1. f(1) = 0.30, f(2) = 0.30, f(3) = 0.30, f(4) = 0.20Why not: The probabilities sum to 1.10, not 1.
  2. f(1) = 0.60, f(2) = 0.50, f(3) = -0.20, f(4) = 0.10Why not: The total is 1 but f(3) is negative; no probability can be below zero.
  3. f(1) = 0.20, f(2) = 0.30, f(3) = 0.40, f(4) = 0.10 - correct
  4. f(1) = 0.25, f(2) = 0.25, f(3) = 0.25, f(4) = 0.15Why not: The probabilities sum to 0.90, not 1.

A valid probability mass function has every f(x) between 0 and 1 and a total of exactly 1. Only the first option satisfies both: 0.20 + 0.30 + 0.40 + 0.10 = 1.

A harder one

A discrete random variable X takes the values 1, 2, 3, 4 and 5 with P(X = x) = k x for some constant k. What is P(X >= 4)?

  1. 0.33Why not: Computes P(X > 4) = 5/15, excluding the value 4.
  2. 0.40Why not: Treats the five values as equally likely and takes 2 out of 5, ignoring that probabilities rise with x.
  3. 0.90Why not: Sets k = 0.1 without normalising; those probabilities would sum to 1.5.
  4. 0.60 - correct

The total must be 1: k x (1 + 2 + 3 + 4 + 5) = 15k = 1, so k = 1/15. Then P(X >= 4) = (4 + 5)/15 = 0.60.

The hardest tier

A random variable X takes the values -2, -1, 0, 1 and 2 with P(X = x) = k (3 - |x|) for a constant k. What is P(-1 <= X <= 1)?

  1. 0.778 - correct
  2. 0.333Why not: Uses strict inequalities and counts only x = 0, dropping both end values -1 and 1.
  3. 0.556Why not: Includes only x = -1 and x = 0 and forgets x = 1.
  4. 0.889Why not: Computes P(X >= -1), dropping the upper limit and including x = 2.

The weights are 1, 2, 3, 2, 1, which total 9, so k = 1/9. P(-1 <= X <= 1) = (2 + 3 + 2)/9 = 7/9 = 0.778.

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