$P(D) = 0.02$, $P(\text{positive} \mid D) = 0.95$, $P(\text{positive} \mid D') = 0.10$. Given a positive test, find $P(D)$ (posterior).

1. Priors : $P(D) = 0.02$, $P(D') = 0.98$. 2. Total probability of positive : $P(P) = (0.02)(0.95) + (0.98)(0.10) = 0.019 + 0.098 = 0.117$. 3. Bayes : $P(D \mid P) = \frac{P(D \cap P)}{P(P)} = \frac{0.019}{0.117} = 0.16239\ldots \implies \mathbf{0.162}$.

Two events $C$ and $D$ are such that $P(C) = 0.55$ and $P(D) = 0.4$ . The probability of neither event occurring is $P(C' \cap D') = 0.27$ . Find $P(C \cup D)$ . Find $P(C \cap D)$ . Find the probability that event $C$ occurs, but event $D$ does not.

$P(C' \cap D') = P((C \cup D)') = 0.27$ , so $P(C \cup D) = 1 - 0.27 = 0.73$ . A1 $P(C \cup D) = P(C) + P(D) - P(C \cap D)$ . (M1) $0.73 = 0.55 + 0.4 - P(C \cap D) \implies P(C \cap D) = 0.95 - 0.73 = 0.22$ . A1 $P(C \cap D') = P(C) - P(C \cap D)$ . (M1) $= 0.55 - 0.22 = 0.33$ . A1

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IB Maths AA SL · Unit 4: Statistics and Probability

Probability (IB Maths AA SL): notes, exam questions and worked examples

Updated October 2026 · By Pete Bromfield

Probability questions in AA SL use a Venn diagram, a tree diagram or a sample space, and four rules from the formula booklet: the complement, the addition rule, independence and conditional probability. The key question is always "out of what?": a conditional probability $P (A ∣ B)$ restricts you to the outcomes in $B$ .

72 questions
Paper 1: 57
Paper 2: 15
3 worked examples

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Notes
Formulas
Worked examples
Questions
FAQ

Probability notes

Outcomes and complements

With equally likely outcomes, $P (A) = \frac{n (A)}{n (U)}$ . The complement is often quicker: $P (at least one) = 1 - P (none)$ . List the sample space in a table when two things happen together, such as two dice.

Venn diagrams and the addition rule

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$ : subtract the overlap so it is not counted twice. Fill a Venn diagram from the inside out, starting with the intersection. Events are mutually exclusive when they cannot both happen, so $P (A \cap B) = 0$ .

Independent events

$A$ and $B$ are independent when $P (A \cap B) = P (A) P (B)$ : one happening does not change the chance of the other. If $P (A) = 0.4$ , $P (B) = 0.5$ and they are independent, $P (A \cup B) = 0.4 + 0.5 - 0.2 = 0.7$ . Do not assume independence unless the question says so or the numbers show it.

Conditional probability and tree diagrams

$P (A ∣ B) = \frac{P (A \cap B)}{P (B)}$ . On a tree diagram, multiply along the branches and add the paths you need. Without replacement, the second set of branches changes: from 5 red and 3 blue counters, $P (two red) = \frac{5}{8} \times \frac{4}{7} = \frac{5}{14}$ .

Reversing a tree (Bayes)

To find the probability of a first-stage event given a second-stage result, divide the path you want by the total of all paths that give that result. It is the conditional probability formula read from the tree.

Key formulas

Probability In the booklet: $P (A) = \frac{n (A)}{n (U)}, P (A) + P (A^{'}) = 1$
Combined events In the booklet: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$
Mutually exclusive In the booklet: $P (A \cup B) = P (A) + P (B)$
Independent In the booklet: $P (A \cap B) = P (A) P (B)$
Conditional In the booklet: $P (A ∣ B) = \frac{P (A \cap B)}{P (B)}$

The IB formula booklet is provided in every exam paper. In the booklet = printed there, so know where to find it; Learn it = not in the booklet, so learn it by heart. All AA SL formulas →

Explained, with a worked example: Conditional probability · Probability formulas

Probability worked examples

Worked example 1: Independent and combined events · easy

Independent events with $P (A) = 0.3$ , $P (B) = 0.8$ . Find $P (A \cap B)$ and $P (A \cup B)$ .

Solution

2 steps

Examiner tip: Independent (probabilities multiply) is NOT the same as mutually exclusive (intersection is zero). Read the wording carefully.

Worked example 2: Drawing without replacement · medium

Bag has 4 red and 5 blue marbles. Two marbles drawn one after the other without replacement. Find the probability that they are different colours.

Solution

4 steps · 1 to fill in

Examiner tip: Without replacement — both the numerator AND the denominator must decrease for the second draw.

Worked example 3: Applying Bayes' Theorem · hard

$P (D) = 0.02$ , $P (positive ∣ D) = 0.95$ , $P (positive ∣ D^{'}) = 0.10$ . Given a positive test, find $P (D)$ (posterior).

Solution

3 steps · 1 to fill in

Examiner tip: A fully labelled tree diagram is the safest way to tackle Bayes' Theorem — the denominator is always the SUM of all end-branches meeting the given condition.

Probability exam-style questions

3 questions from our AA SL bank, written in the style of IB papers, easiest first. One has its full worked solution open here. Open the worked answer, or check your answer in the practice area.

Question 1 · easy · 4 marks · Paper 2

A biased coin has a probability of landing on tails of $0.38$ . The coin is tossed exactly $250$ times.

Find the probability that the coin lands on heads on any single toss.
Calculate the expected number of times the coin will land on tails.
Calculate the expected number of times the coin will land on heads.

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Question 2 · medium · 5 marks · Paper 2

Two events $C$ and $D$ are such that $P (C) = 0.55$ and $P (D) = 0.4$ . The probability of neither event occurring is $P (C^{'} \cap D^{'}) = 0.27$ .

Find $P (C \cup D)$ .
Find $P (C \cap D)$ .
Find the probability that event $C$ occurs, but event $D$ does not.

Full worked solution Free, no sign-in

$P (C^{'} \cap D^{'}) = P ((C \cup D)^{'}) = 0.27$ , so $P (C \cup D) = 1 - 0.27 = 0.73$ . A1
$P (C \cup D) = P (C) + P (D) - P (C \cap D)$ . (M1)
$0.73 = 0.55 + 0.4 - P (C \cap D) ⟹ P (C \cap D) = 0.95 - 0.73 = 0.22$ . A1
$P (C \cap D^{'}) = P (C) - P (C \cap D)$ . (M1)
$= 0.55 - 0.22 = 0.33$ . A1

Every question here has a full worked solution. See them all →

Question 3 · hard · 6 marks · Paper 2

A box contains $n$ green apples and $4$ red apples. Two apples are selected at random without replacement. The probability that both apples are green is exactly $\frac{1}{3}$ .

Formulate an algebraic equation in terms of $n$ for the probability of selecting two green apples.
Use your graphic display calculator to solve the equation and find the exact number of green apples initially in the box.

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All 72 probability questions →

Where students lose marks on probability

Examiner's noteProbability: assuming independence, and conditional probability with complements

Where marks are lost: Candidates multiplied probabilities for events that were not independent, found a union when an intersection was asked, used the independence formulas to work backwards, mishandled P(A | B′), misread "more than" in a discrete setting, and left answers as unsimplified fractions of decimals. When writing the conditional formula, generic P(A | B) with no reference to the context was not credited.

Examiners flagged this in 3 exam series (November 2024, May 2025, November 2025). Summarised in our own words from IB subject reports, IB Programme Resource Centre. Read the full note →

Probability FAQ

Are the probability formulas in the IB formula booklet?

Yes. The AA SL booklet gives the complement, the addition rule, mutually exclusive and independent events, and conditional probability. Knowing which one fits is the skill, so draw a Venn or tree diagram first.

What is the difference between mutually exclusive and independent?

Mutually exclusive events cannot happen together, so P(A ∩ B) = 0. Independent events can happen together, but one does not affect the other, so P(A ∩ B) = P(A) × P(B).

How many IB Maths AA SL probability questions are there?

There are 72 exam-style probability questions in the AA SL question bank (Paper 1: 57 · Paper 2: 15), graded 16 easy, 36 medium, 16 hard, 4 very hard. Every question has a full IB-style mark scheme (M, A and R marks).

Is probability on Paper 1 or Paper 2?

Both. In the bank, Paper 1: 57 · Paper 2: 15. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.

Next steps

More AA SL Unit 4: Bivariate Data and Correlation · Discrete Random Variables · Measures of Central Tendency · Normal Distribution · Presentation of Data

Probability in other courses: AI SL

Teaching this topic? Free ready-to-teach lessons (slides and teacher notes): Conditional probability · Independent & mutually exclusive events · Venn diagrams & combined events.

Keep revising AA SL: AA SL revision notes · IB Maths AA SL past papers · AA SL common exam mistakes · AA SL syllabus checklist · AA SL statistics and probability knowledge organiser

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← All IB Maths AA SL topics

Probability (IB Maths AA SL): notes, exam questions and worked examples

Try it first

Probability notes

Outcomes and complements

Venn diagrams and the addition rule

Independent events

Conditional probability and tree diagrams

Reversing a tree (Bayes)

Key formulas

Probability worked examples

Worked example 1: Independent and combined events · easy

Solution

Worked example 2: Drawing without replacement · medium

Solution

Worked example 3: Applying Bayes' Theorem · hard

Solution

Probability exam-style questions

Question 1 · easy · 4 marks · Paper 2

Question 2 · medium · 5 marks · Paper 2

Full worked solution Free, no sign-in

Question 3 · hard · 6 marks · Paper 2

Where students lose marks on probability

Probability FAQ

Next steps

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