## **Learning Outcomes**

This article explains how core probability rules connect unconditional, conditional, and joint probabilities, and how Bayes’ theorem updates beliefs when new information arrives. After reading, you will be able to:

- Apply probability rules for unions, intersections, complements, and conditional events
- Compute joint, marginal (unconditional), and conditional probabilities using tables or scenario partitions
- Use the total probability rule and Bayes’ theorem to update risk estimates such as default probabilities

## **GARP FRM Part 1 Syllabus**

For the GARP FRM Part 1 exam, you are expected to understand probability rules and Bayesian updating with a focus on the following syllabus points:

- Define event spaces; distinguish independent, mutually exclusive, and conditionally independent events
- Calculate and interpret unconditional, conditional, and joint probabilities; apply the total probability rule
- Explain and apply Bayes’ rule to update probabilities when new information is observed

## **Test Your Knowledge**

_Attempt these questions before reading this article. If you find some difficult or cannot remember the answers, remember to look more closely at that area during your revision._

1. If $P(A)=0.30$, $P(B)=0.50$, and $P(A \cap B)=0.20$, then $P(A \cup B)$ is:
   a) 0.15  
   b) 0.60  
   c) 0.80  
   d) 1.00

2. A portfolio breaches its VaR on 2% of days unconditionally. A stress indicator triggers on 10% of days, and $P(\text{breach} \mid \text{trigger})=8\%$. What is the joint probability $P(\text{breach} \cap \text{trigger})$?

3. An economy outperforms with probability 60%. If it outperforms, a stock rises with probability 70%; if it underperforms, the stock rises with probability 20%. Compute $P(\text{outperform} \mid \text{stock rises})$.

4. True/False (explain briefly): If two events are mutually exclusive, they must be independent.

## **What Do You Know Already?**

- Can you translate word statements (e.g., “given that”, “either/or”, “both”) into $P(A\mid B)$, $P(A\cup B)$, and $P(A\cap B)$ correctly?
- Can you distinguish $P(A\mid B)$ from $P(B\mid A)$ and explain why they are usually different?
- Can you compute $P(B)$ in Bayes’ theorem using the total probability rule when it is not given directly?

Use these prompts to gauge your starting point; then focus your revision where your answers were least confident.

## **Introduction**

Probability is the language of uncertainty. In financial risk management, uncertainty is not an abstract idea: it is the day-to-day reality behind credit defaults, operational losses, market moves, and model outcomes. Whether you are estimating a one-day 99% VaR, forecasting a default rate, or interpreting a risk signal from a monitoring system, you are using probability rules—often implicitly.

For FRM Part 1, the goal is not only to memorize formulas, but to recognize which probability object you are working with: an unconditional (marginal) probability, a conditional probability, or a joint probability. The total probability rule ties together scenario-based thinking (“which regime are we in?”), and Bayes’ theorem formalizes updating (“given what I just observed, what should I believe now?”).

> **Key Term: Event and event space**  
> An **event** is an outcome (or set of outcomes) from a random experiment. The **event space** (often called the sample space) is the set of all possible outcomes, including the empty event.

## **Core Probability Rules for Risk Problems**

Risk questions often involve combining events: “default and downgrade,” “profit or loss,” “breach given trigger.” These map directly to set operations.

### Events as sets

Let $A$ and $B$ be events.

- Intersection: $A \cap B$ means **both** occur.
- Union: $A \cup B$ means **at least one** occurs (A or B or both).
- Complement: $A^c$ means **A does not occur**.

### Probability axioms and useful identities

From basic probability axioms (probabilities are between 0 and 1, and total probability is 1), we obtain working rules used throughout FRM.

1. Complement rule  
   $$P(A^c) = 1 - P(A)$$

2. General addition rule  
   $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

3. Special case: mutually exclusive events  
   If $A$ and $B$ cannot occur together, then $P(A \cap B)=0$ and:
   $$P(A \cup B) = P(A) + P(B)$$

> **Key Term: Mutually exclusive events**  
> Two events are **mutually exclusive** if they cannot occur at the same time, so $P(A \cap B)=0$.

### Multiplication rule and the chain form

The bridge from “and” to “given” is the multiplication rule:
$$P(A \cap B) = P(A\mid B)P(B) = P(B\mid A)P(A)$$

This identity is the starting point for Bayes’ theorem (later), and it is also how you build joint probabilities from conditional and unconditional components.

> **Key Term: Conditional probability**  
> $P(A\mid B)$ is the probability of event $A$ occurring **given** that event $B$ has occurred. It is computed as $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ when $P(B)>0$.

## **Unconditional, Joint, and Conditional Probabilities in Practice**

A common FRM pattern is: you are given (i) a partition of the world into scenarios and (ii) conditional probabilities inside each scenario. From this, you compute joint probabilities and then unconditional probabilities.

### Unconditional (marginal) vs joint vs conditional

- Unconditional (marginal): $P(A)$, probability of $A$ with no “given.”
- Joint: $P(A \cap B)$, probability that both occur.
- Conditional: $P(A\mid B)$, probability of $A$ once you know $B$ occurred.

### Worked Example 1.1

Acme, Inc. employees earn more than \$40,000 depending on education level

Employees fall into exactly one of three education categories: no degree (ND), bachelor’s degree (BD), and higher-than-bachelor’s degree (HBD). The unconditional probabilities are:

- $P(ND)=60\%$
- $P(BD)=30\%$
- $P(HBD)=10\%$

The conditional probabilities of earning more than \$40,000 are:

- $P(40+\mid ND)=10\%$
- $P(40+\mid BD)=70\%$
- $P(40+\mid HBD)=100\%$

Compute (i) the joint probabilities and (ii) the unconditional probability $P(40+)$.

> **Answer:**  
> Step 1: Compute joint probabilities using $P(A \cap B)=P(A\mid B)P(B)$.
>
> - $P(40+ \cap ND)=P(40+\mid ND)P(ND)=0.10\times0.60=0.06=6\%$
> - $P(40+ \cap BD)=P(40+\mid BD)P(BD)=0.70\times0.30=0.21=21\%$
> - $P(40+ \cap HBD)=P(40+\mid HBD)P(HBD)=1.00\times0.10=0.10=10\%$
>
> Step 2: Sum the joint probabilities across the mutually exclusive and exhaustive education categories to get $P(40+)$:
>
> $$
> P(40+)=6\%+21\%+10\%=37\%
> $$
>
> Interpretation: 37% of Acme employees earn more than \$40,000 per year.

This example also illustrates a general theorem used constantly in scenario-based risk work: the total probability rule.

## **Total Probability Rule and Scenario Partitions**

Many risk models start by partitioning the world into regimes: expansion vs recession, normal vs stressed volatility, solvent vs distressed counterparties. If the regimes form a partition, you can compute an unconditional probability by summing across regimes.

### The total probability rule

Let $\{B_1,\dots,B_n\}$ be **mutually exclusive and exhaustive** events (exactly one occurs). Then:

$$P(A) = \sum_{i=1}^{n} P(A\mid B_i)P(B_i)$$

In Acme, the partition is $\{ND, BD, HBD\}$ and the sum of joint probabilities gives $P(40+)$.

> **Key Term: Total probability rule**  
> If $\{B_i\}$ form a mutually exclusive and exhaustive partition, then an unconditional probability can be computed as $P(A)=\sum_i P(A\mid B_i)P(B_i)$.

### Probability tables and “marginals”

FRM questions may present a probability matrix (a table of joint probabilities). From such a table:

- Row sums and column sums are **marginal** (unconditional) probabilities.
- A conditional probability is a joint cell divided by the marginal of the conditioning event.

For example, if a table implies $P(\text{Negative earnings} \cap \text{Return}=-3\%)=25\%$ and $P(\text{Return}=-3\%)=30\%$, then:
$$P(\text{Negative earnings}\mid \text{Return}=-3\%)=\frac{25\%}{30\%}=83.33\%$$

This is the same conditional probability definition, just computed from a table rather than from a tree.

## **Independence, Mutual Exclusivity, and Conditional Independence**

FRM questions frequently test whether you can distinguish these concepts.

### Independence

Events $A$ and $B$ are independent if knowing one does not change the probability of the other:

$$P(A\mid B)=P(A)$$

An equivalent condition is:

$$P(A\cap B)=P(A)P(B)$$

### Mutual exclusivity is different from independence

If $A$ and $B$ are mutually exclusive and both have positive probability, then they cannot be independent. Why? Mutual exclusivity implies $P(A\cap B)=0$, but independence would require $P(A\cap B)=P(A)P(B)>0$.

This is a common exam trap: “cannot happen together” (mutually exclusive) is not the same as “unrelated” (independent).

### Conditional independence

Two events may be independent overall but not independent once you condition on a third event, or the other way around. Formally, $A$ and $B$ are conditionally independent given $C$ if:

$$P(A\cap B\mid C)=P(A\mid C)P(B\mid C)$$

Risk interpretation: two drivers can look unrelated in aggregate, but once you hold the macro regime fixed, they may become related (or unrelated). For example, defaults in two sectors may appear weakly related over a full cycle, but highly related during recession.

## **Bayes’ Theorem for Updating Risk Estimates**

Bayes’ theorem turns probability into an updating rule: start with a base rate (prior), observe a signal, then compute an updated probability (posterior).

### Deriving Bayes’ theorem from the multiplication rule

From the multiplication rule:

$$P(A\cap B)=P(A\mid B)P(B)=P(B\mid A)P(A)$$

Rearrange to solve for $P(A\mid B)$:

$$P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)}$$

If $P(B)$ is not given, it is often computed using the total probability rule across a partition of $A$ and $A^c$:

$$P(B)=P(B\mid A)P(A)+P(B\mid A^c)P(A^c)$$

> **Key Term: Bayes’ theorem**  
> Bayes’ theorem updates the probability of a hypothesis $A$ after observing evidence $B$: $P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)}$.

### Worked Example 2.1

Updating the probability of an outperforming economy given a stock increase

Given:

- $P(O)=60\%$ where $O$ = economy outperforms
- $P(U)=40\%$ where $U$ = economy underperforms
- $P(S\mid O)=70\%$ where $S$ = stock goes up
- $P(S\mid U)=20\%$

Compute $P(O\mid S)$.

> **Answer:**  
> Step 1: Compute $P(S)$ using the total probability rule over $\{O,U\}$:
>
> $$
> P(S)=P(S\mid O)P(O)+P(S\mid U)P(U)
> $$
>
> $$
> P(S)=0.70(0.60)+0.20(0.40)=0.42+0.08=0.50
> $$
>
> Step 2: Apply Bayes’ theorem:
>
> $$
> P(O\mid S)=\frac{P(S\mid O)P(O)}{P(S)}=\frac{0.70\times0.60}{0.50}=\frac{0.42}{0.50}=0.84
> $$
>
> Result: $P(O\mid S)=84\%$.
>
> Interpretation: observing the stock increase raises the probability you assign to an outperforming economy from 60% to 84%, given the stated conditional relationships.

### Worked Example 2.2

Credit screening and the base rate effect

A bank monitors obligors for possible default over the next year.

- Base default probability: $P(D)=2\%$
- The monitoring system triggers an alert $T$:
  - Sensitivity: $P(T\mid D)=90\%$
  - False positive rate: $P(T\mid D^c)=10\%$

Compute $P(D\mid T)$, the probability of default given an alert.

> **Answer:**  
> Step 1: Compute $P(T)$ using the total probability rule over $\{D, D^c\}$:
>
> $$
> P(T)=P(T\mid D)P(D)+P(T\mid D^c)P(D^c)
> $$
>
> $$
> P(T)=0.90(0.02)+0.10(0.98)=0.018+0.098=0.116
> $$
>
> Step 2: Apply Bayes’ theorem:
>
> $$
> P(D\mid T)=\frac{P(T\mid D)P(D)}{P(T)}=\frac{0.90(0.02)}{0.116}=\frac{0.018}{0.116}\approx0.1552
> $$
>
> Result: $P(D\mid T)\approx15.5\%$.
>
> Interpretation: even with a seemingly “good” trigger, the posterior default probability is only about 15.5% because defaults are rare to begin with (low base rate).

This type of calculation is common in model risk management: a signal can look accurate, but the posterior probability depends heavily on how often the event actually occurs.

### **Exam Warning**

> The most common Bayes errors are:
>
> 1. Swapping $P(A\mid B)$ and $P(B\mid A)$ (they are usually very different).
> 2. Forgetting that $P(B)$ must reflect all ways that $B$ can happen (use the total probability rule).
> 3. Ignoring base rates: a high $P(\text{signal}\mid \text{event})$ does not imply a high $P(\text{event}\mid \text{signal})$ when the event is rare.

## **How These Rules Show Up in FRM Risk Context**

Probability rules are not isolated math; they are the structure behind several FRM topics.

### Conditional risk measures and backtesting language

- VaR backtesting uses events such as $B$ = “VaR breach.” A conditional probability $P(B\mid \text{stress})$ formalizes the idea that breaches cluster in stress.
- Operational risk uses event categories (fraud, system failure, legal) that form partitions; the total probability rule converts category-conditional frequencies into an overall frequency.

### Scenario weights and mixture thinking

When you weight conditional distributions by regime probabilities, you are applying the total probability rule at the distribution level. For example, “low volatility 75% of the time, high volatility 25% of the time” corresponds to weighting outcomes by regime probabilities, the same logic as $P(A)=\sum_i P(A\mid B_i)P(B_i)$.

### Model monitoring as Bayesian updating

Many monitoring tasks are naturally Bayesian:

- A downgrade, widening CDS spread, or covenant breach is “evidence” $B$.
- The bank’s current PD is the “prior” $P(D)$.
- Historical behavior provides $P(B\mid D)$ and $P(B\mid D^c)$.
- The updated PD after observing the evidence is $P(D\mid B)$.

Bayes’ theorem provides the quantitative link between “signals” and “updated risk.”

## **Summary**

Probability rules translate finance language (“and”, “or”, “given that”) into mathematical statements that can be computed consistently. The addition, complement, and multiplication rules allow you to move between unions, intersections, and conditionals. In FRM questions, most calculations reduce to identifying the right event relationships and then applying these identities carefully.

The total probability rule formalizes regime-based thinking: if the world is partitioned into mutually exclusive and exhaustive scenarios, unconditional probabilities are weighted sums of conditional probabilities. Bayes’ theorem then builds on the same joint/conditional identity to reverse conditioning and update beliefs after observing new information—an idea used widely in credit monitoring, model validation, and risk signal interpretation.

## **Key Point Checklist**

_This article has covered the following key knowledge points:_

- Events, complements, unions, intersections, and the basic probability identities
- The general addition rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ and its mutually exclusive special case
- The multiplication rule $P(A\cap B)=P(A\mid B)P(B)=P(B\mid A)P(A)$
- How to compute joint probabilities from conditional and unconditional probabilities
- The total probability rule for mutually exclusive and exhaustive partitions
- Independence vs mutual exclusivity, and the meaning of conditional independence
- Bayes’ theorem $P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)}$ and computing $P(B)$ via total probability
- Common exam mistakes: reversing conditionals, missing the denominator, and ignoring base rates

## **Key Terms and Concepts**

- Event
- Event space (sample space)
- Union $A\cup B$
- Intersection $A\cap B$
- Complement $A^c$
- Unconditional (marginal) probability
- Joint probability
- Conditional probability
- Independence
- Mutually exclusive events
- Conditional independence
- Total probability rule
- Bayes’ theorem
- Prior probability
- Posterior probability
- Likelihood


Probability, random variables, and distributions - Probability rules and Bayes' theorem