Probability, conditional probability and independent events

IB Mathematics Analysis and Approaches SL· Unit 4: Statistics & Probability, Topic 5· 15 min read

1. Basic Probability Rules★★☆☆☆⏱ 5 min

📘 Definition

Probability of an Event

P(A)

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 $0 \leq P (A) \leq 1$ .

Example:

Rolling a fair 6-sided die: $P (even number) = 3/6 = 0.5$

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: $A = studies biology$ , $B = studies chemistry$
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.

2. Conditional Probability★★★☆☆⏱ 6 min

📘 Definition

Conditional Probability

P(A|B)

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

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: $A = second marble red$ , $B = first marble blue$
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.

✓ Quick check

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

What is the value of $P (B ∣ A)$ ?
- $3/7$
- $3/8$
- $2/7$
- $5/8$
Reveal answer
3/7 —
Correct! After drawing a red marble first, 4 red and 3 blue remain, so 3 blue out of 7 total marbles.

3. Independent Events★★★☆☆⏱ 4 min

📘 Definition

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 $P (B ∣ A) = P (B)$ .

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 $A = first flip heads$ , $B = second flip tails$ . 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.

4. Common Pitfalls

Wrong move:

Treating sampling without replacement as independent events.

Why:

Drawing without replacement changes the sample size and outcome counts for subsequent draws, so events are dependent.

Correct move:

Use conditional probability for subsequent draws and adjust counts after each draw.

Wrong move:

Confusing $P (A ∣ B)$ with $P (B ∣ A)$ .

Why:

The order of events in conditional probability matters; swapping the condition gives a different result.

Correct move:

Always clearly label which event is the given condition, and remember the given event is in the denominator of the formula.

Wrong move:

Assuming events are independent without verification.

Why:

Most problems do not state dependence, so students often incorrectly assume independence.

Correct move:

Only use $P (A \cap B) = P (A) P (B)$ if the question explicitly says events are independent, or you verify it.

Wrong move:

Adding probabilities without subtracting the intersection.

Why:

Overlap where both events occur is counted twice if the intersection is not subtracted.

Correct move:

Always apply the full addition rule: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ .

5. Quick Reference 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)

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 1
Conditional probability calculation
2022 · 2
Identify independent events
2021 · 1
Multi-event probability problem

Going deeper

syllabusIB AA SL Official Syllabus

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.