# Probability, conditional probability and independent events

> IB Mathematics Analysis and Approaches SL · IB AA SL
> Source: https://www.owlsprep.com/study/ib-math-aa-sl-u4-probability-conditional-probability-and-independent/

This module covers core probability rules for IB AA SL, including how to calculate conditional probabilities, identify independent events, and apply the addition and multiplication rules to solve common exam probability problems.

**Prerequisites:** [Basic set theory and Venn diagrams](https://www.owlsprep.com/study/ib-math-aa-sl-u4-sets-and-venn-diagrams/)

## Learning objectives

- Apply basic probability rules to single and multi-event problems
- Calculate conditional probability using the definition formula
- Identify and verify independent events using probability rules
- Distinguish between independent and mutually exclusive events

## Basic Probability Rules

**Probability of an Event** — For a finite sample space with equally likely outcomes, probability of event $A$ is the ratio of favorable outcomes to total outcomes in the sample space, with $0 \leq P(A) \leq 1$.

*Notation:* P(A)

*Example:* Rolling a fair 6-sided die: $P(\text{even number}) = 3/6 = 0.5$

The addition rule describes the probability of either of two events occurring, accounting for overlap where both events happen at the same time:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

**Worked example:** In a class of 30 students, 12 study biology, 15 study chemistry, and 5 study both. What is the probability a randomly selected student studies biology or chemistry?

1. Define events: $A = \text{studies biology}$, $B = \text{studies chemistry}$
2. Calculate base probabilities:
3. $$P(A) = 12/30 = 0.4, \quad P(B) = 15/30 = 0.5, \quad P(A \cap B) = 5/30 \approx 0.167$$
4. Substitute into the addition rule:
5. $$P(A \cup B) = 0.4 + 0.5 - 0.167 = 0.733$$
6. The probability is $11/15 \approx 0.73$.

> **Exam tip:** In IB exams, "or" always means inclusive or (A or B or both), so always subtract the intersection term once.

## Conditional Probability

**Conditional Probability** — The probability of event $A$ occurring given that event $B$ has already occurred. It restricts the sample space to only outcomes in $B$.

*Notation:* P(A|B)

From the definition, we get the core formula for conditional probability, which holds when $P(B) > 0$:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

**Worked example:** A bag has 5 red marbles and 3 blue marbles. Two marbles are drawn without replacement. Find the probability the second marble is red given the first was blue.

1. Define events: $A = \text{second marble red}$, $B = \text{first marble blue}$
2. After drawing one blue marble, we have 5 red and 2 blue marbles left, so 7 total marbles remaining.
3. Since all marbles are equally likely, the probability is the ratio of favorable red marbles to total remaining marbles:
4. $$P(A|B) = \frac{5}{7} \approx 0.714$$
5. Verifying with the formula: $P(A \cap B) = (3/8)(5/7) = 15/56$, $P(B) = 3/8 = 21/56$, so $P(A|B) = (15/56)/(21/56) = 5/7$, which matches.

**Check your understanding**

What is $P(B|A)$ (probability first is blue given second is red) for the example above?

1. What is the value of $P(B|A)$?

   - $3/7$
   - $3/8$
   - $2/7$
   - $5/8$

   *Why:* Correct! After drawing a red marble first, 4 red and 3 blue remain, so 3 blue out of 7 total marbles.

## Independent Events

**Independent Events** — Two events are independent if the occurrence of one does not change the probability of the other. This means $P(A|B) = P(A)$ and $P(B|A) = P(B)$.

> **warning**
>
> Independent events are NOT the same as mutually exclusive events. Mutually exclusive events have $P(A \cap B) = 0$, meaning if one occurs the other cannot, so they are always dependent.

Rearranging the conditional probability formula gives the key test for independence:

$$P(A \cap B) = P(A) \times P(B) \quad \text{for independent events } A, B$$

**Worked example:** A fair coin is flipped twice. Let $A = \text{first flip heads}$, $B = \text{second flip tails}$. Verify that $A$ and $B$ are independent.

1. Calculate base probabilities: $P(A) = 1/2$, $P(B) = 1/2$
2. The probability that the first is heads and the second is tails is 1 out of 4 equally likely outcomes, so $P(A \cap B) = 1/4$
3. Check the independence condition:
4. $$P(A) \times P(B) = (1/2)(1/2) = 1/4 = P(A \cap B)$$
5. The condition holds, so $A$ and $B$ are independent.

## Common pitfalls

- **Wrong:** Treating sampling without replacement as independent events.
  - Why it fails: Drawing without replacement changes the sample size and outcome counts for subsequent draws, so events are dependent.
  - Correct: Use conditional probability for subsequent draws and adjust counts after each draw.
- **Wrong:** Confusing $P(A|B)$ with $P(B|A)$.
  - Why it fails: The order of events in conditional probability matters; swapping the condition gives a different result.
  - Correct: Always clearly label which event is the given condition, and remember the given event is in the denominator of the formula.
- **Wrong:** Assuming events are independent without verification.
  - Why it fails: Most problems do not state dependence, so students often incorrectly assume independence.
  - Correct: Only use $P(A \cap B) = P(A)P(B)$ if the question explicitly says events are independent, or you verify it.
- **Wrong:** Adding probabilities without subtracting the intersection.
  - Why it fails: Overlap where both events occur is counted twice if the intersection is not subtracted.
  - Correct: Always apply the full addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

## Cheatsheet

| Concept | Formula |
| --- | --- |
| Addition Rule | P(A \cup B) = P(A) + P(B) - P(A \cap B) |
| Conditional Probability | P(A\|B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) > 0 |
| Independence Condition | P(A \cap B) = P(A) \cdot P(B) |
| Independence Definition | P(A\|B) = P(A), \quad P(B\|A) = P(B) |
| Multiplication Rule (Dependent) | P(A \cap B) = P(A\|B) \cdot P(B) |

## What's next

This topic is the foundation for all advanced probability work in IB AA SL. Mastery of conditional probability and independence is essential for solving multi-step problems with tree diagrams, and for working with discrete probability distributions that model random outcomes. These concepts are also the core of the binomial distribution, one of the most frequently tested distributions on the IB AA SL exam. Building a solid understanding here will make more complex topics much easier to master.

- [Discrete probability distributions, expectation and variance](https://www.owlsprep.com/study/ib-math-aa-sl-u4-discrete-probability-distributions-expectation-and/)
- [Binomial probability distribution](https://www.owlsprep.com/study/ib-math-aa-sl-u4-binomial-probability-distribution/)
- [Normal probability distribution](https://www.owlsprep.com/study/ib-math-aa-sl-u4-normal-probability-distribution/)

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