Probability (AA SL)
Probability measures how likely an event is, on a scale from 0 (impossible) to 1 (certain). This topic covers building and reading sample spaces, combining events with "or" and "and", using tree and Venn diagrams for multi-stage or overlapping events, and working out conditional probability - the chance of one event given that another has already happened.
What the syllabus says
This topic maps onto three points in the official IB Analysis & Approaches syllabus.
| Code | Syllabus content |
|---|---|
| SL4.5 | Concepts of trial, outcome, equally likely outcomes, relative frequency, sample space \(U\) and event. \(P(A) = \dfrac{n(A)}{n(U)}\). The complementary events \(A\) and \(A'\) (not \(A\)). Expected number of occurrences. |
| SL4.6 | Use of Venn diagrams, tree diagrams, sample space diagrams and tables of outcomes to calculate probabilities. Combined events: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Mutually exclusive events: \(P(A\cap B)=0\). Conditional probability: \(P(A|B)=\dfrac{P(A\cap B)}{P(B)}\). Probabilities with and without replacement. Independent events: \(P(A\cap B)=P(A)P(B)\). |
| SL4.11 | Formal definition and use of \(P(A|B)=\dfrac{P(A\cap B)}{P(B)}\) for conditional probabilities, and \(P(A|B)=P(A)=P(A|B')\) for independent events, including testing for independence. |
These are core AA SL syllabus points that are also examinable at AA HL.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is a sample space?
A sample space \(U\) is the set of every possible outcome of a trial. Once you know the sample space, the probability of an event \(A\) is simply the number of outcomes in \(A\) divided by the total number of outcomes: \(P(A) = \dfrac{n(A)}{n(U)}\).
e.g. Rolling two dice gives a sample space of \(6\times6=36\) equally likely outcomes.
What are mutually exclusive events?
Two events are mutually exclusive if they can never happen at the same time - they share no outcomes, so \(P(A\cap B)=0\). For mutually exclusive events, the addition rule simplifies to \(P(A\cup B)=P(A)+P(B)\).
e.g. Rolling a 2 and rolling a 5 on the same die roll are mutually exclusive, so \(P(2\text{ or }5)=\tfrac16+\tfrac16=\tfrac13\).
What is conditional probability?
Conditional probability \(P(A|B)\) is the probability that \(A\) happens, given that \(B\) is already known to have happened. It's found by restricting your sample space to just the outcomes where \(B\) occurred: \(P(A|B)=\dfrac{P(A\cap B)}{P(B)}\).
e.g. Of 50 women surveyed, 20 own a bicycle, so \(P(\text{owns}\mid\text{woman})=\dfrac{20}{50}=0.4.\)
What does it mean for events to be independent?
Events \(A\) and \(B\) are independent if the occurrence of one doesn't affect the probability of the other. Formally, \(P(A\cap B)=P(A)\times P(B)\), which is equivalent to \(P(A|B)=P(A)\).
e.g. Two separate coin tosses are independent, so \(P(\text{two heads})=0.5\times0.5=0.25.\)
What is a tree diagram used for?
A tree diagram shows the outcomes of two or more stages happening in sequence, with each branch labelled by its probability. To find the probability of a specific path, multiply the probabilities along the branches; to combine several paths, add their probabilities.
e.g. Drawing red then red without replacement from 3 red and 5 blue marbles: \(\dfrac{3}{8}\times\dfrac{2}{7}=\dfrac{6}{56}=\dfrac{3}{28}.\)
Key formulas
Five formulas cover almost every question on this topic. The tables below summarise them at a glance - the explanations underneath go into more depth on each one.
Formula reference
The combined-events and conditional probability formulas are on the official formula booklet; the basic probability and independence definitions are assumed prior knowledge.
| Formula | Used for | Booklet? |
|---|---|---|
| \(P(A) = \dfrac{n(A)}{n(U)}\) | Basic probability from a sample space | Not in booklet - prior knowledge |
| \(P(A') = 1 - P(A)\) | Complementary events | Not in booklet - prior knowledge |
| \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\) | Combined ("or") events | ✓ Yes |
| \(P(A|B)=\dfrac{P(A\cap B)}{P(B)}\) | Conditional probability | ✓ Yes |
| \(P(A\cap B)=P(A)P(B)\) | Independent events | Not in booklet - definition, not a listed formula |
Mutually exclusive vs independent
These two ideas are often confused, but they describe completely different situations - mixing them up is one of the most common errors on this topic.
| Feature | Mutually exclusive | Independent |
|---|---|---|
| Meaning | Events cannot happen together | One event doesn't affect the other |
| Test | \(P(A\cap B)=0\) | \(P(A\cap B)=P(A)P(B)\) |
| Union rule | \(P(A\cup B)=P(A)+P(B)\) | \(P(A\cup B)=P(A)+P(B)-P(A)P(B)\) |
| Key fact | If \(A,B\) mutually exclusive and both have nonzero probability, they cannot be independent | Independent events with nonzero probability always overlap |
Combining and restricting probabilities
Addition rule
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
Subtract the overlap so it isn't double-counted.
✓ In the formula bookletComplement rule
\[P(A')=1-P(A)\]
"At least one" problems are almost always faster using the complement.
Not in the formula booklet - prior knowledgeMultiplication along a tree
\[P(A\cap B)=P(A)\times P(B|A)\]
Multiply probabilities along a single branch of a tree diagram.
Not in the formula booklet - prior knowledgeConditional probability and independence
Conditional probability
\[P(A|B)=\dfrac{P(A\cap B)}{P(B)}\]
Restrict the sample space to just the outcomes where \(B\) occurs.
✓ In the formula bookletTesting independence
\[P(A|B)=P(A)\]
Compare \(P(A)\) with \(P(A|B)\) - if they're equal, \(A\) and \(B\) are independent.
✓ In the formula bookletWith vs without replacement
Without replacement, later branch probabilities change because the total (or the count of a colour) has decreased.
Always check whether an item is replaced before drawing the next branch.
Not in the formula booklet - key exam ideaWorked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
Two fair dice are rolled and the scores added.
(a) State the number of outcomes.
(b) Find the probability the total is 7.
Worked solution
(a) Number of outcomes. Each die has 6 faces and they are independent: \(6\times6=36.\) A1
(b) Total is 7. The favourable pairs are \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\) - 6 ways. M1
So \(P=\dfrac{6}{36}=\dfrac16.\) A1
A box has 4 white and 6 black balls. Two are drawn without replacement.
(a) Find P(both white).
(b) Find P(one of each colour).
(c) Find P(at least one black).
Worked solution
(a) Both white. Without replacement: \(\dfrac{4}{10}\times\dfrac{3}{9}=\dfrac{12}{90}\) M1
\(=\dfrac{2}{15}.\) A1
(b) One of each. Two routes: \(\dfrac{4}{10}\cdot\dfrac{6}{9}+\dfrac{6}{10}\cdot\dfrac{4}{9}=\dfrac{48}{90}\) M1
\(=\dfrac{8}{15}.\) A1
(c) At least one black. Complement of both white: \(1-\dfrac{2}{15}\) M1
\(=\dfrac{13}{15}.\) A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Forgetting to subtract the overlap in the addition rule. \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\) - leaving out the \(-P(A\cap B)\) term double-counts the outcomes in both events.
- Not adjusting probabilities when sampling without replacement. Once an item is removed, both the total and the count of that type decrease for the next draw - the second branch probability is not the same as the first.
- Confusing mutually exclusive with independent. Mutually exclusive events cannot happen together (\(P(A\cap B)=0\)); independent events can happen together, just without affecting each other's probability. Two mutually exclusive events with nonzero probability are never independent.
- Restricting the wrong sample space in conditional probability. \(P(A|B)\) means: given \(B\) has happened, find the chance of \(A\) - the denominator must be \(P(B)\), not \(P(A)\) or the full sample space.
Using your GDC
Every step below is a real button sequence, not a vague "use your calculator" hint - covering the TI-84 Plus, TI-Nspire, and Casio fx-9860/fx-CG50. Pick your model to filter down to just the steps that apply to you.
Several probability questions on this topic feed into normal-model problems later in the unit - this is the reverse of a normal probability: given a percentage, find the cut-off value.
- Work out the area to the LEFT of the value you want.
- 2nd → VARS (DISTR) → invNorm(area, μ, σ). Newer OS lets you pick the tail.TI-84
- menu → Probability → Distributions → Inverse Normal; enter the area, μ and σ.Nspire
- Main menu → Statistics → DIST → NORM → InvN; set the tail and enter area, σ, μ.Casio
Tip: invNorm needs the area to the LEFT. For "top 10%", use area = 0.90; for "bottom 25%", use area = 0.25.
Given a probability and one parameter, work back to the missing mean or standard deviation - a standard twist that builds on basic probability reasoning.
- Turn the probability into a \(z\)-value using the inverse normal with \(\mu = 0\), \(\sigma = 1\).
- invNorm(area-to-left, 0, 1) gives \(z\); then solve \(z = (x - \mu)/\sigma\) for the unknown.TI-84
- Use Inverse Normal with \(\mu = 0\), \(\sigma = 1\) to get \(z\), then solve the standardising equation for \(\mu\) or \(\sigma\).Nspire
- DIST → NORM → InvN with \(\mu = 0\), \(\sigma = 1\) gives \(z\); substitute into \(z = (x - \mu)/\sigma\).Casio
- If two probabilities are given, form two equations and solve them simultaneously for \(\mu\) and \(\sigma\).
Tip: Sketch the curve and shade the area on the correct side before using invNorm.
See the full GDC guide for more calculator models and topics.
Ready to practise properly?
Probability questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
What's the difference between P(A or B) and P(A and B)?
P(A or B), written P(A∪B), is the probability that at least one of A or B happens. P(A and B), written P(A∩B), is the probability that both happen together. The addition rule P(A∪B) = P(A) + P(B) − P(A∩B) connects the two, subtracting the overlap so it isn't counted twice.
When do I use a tree diagram versus a Venn diagram?
Use a tree diagram for events that happen in stages or sequence, like drawing balls one after another - each branch is a stage and you multiply along paths. Use a Venn diagram for events that overlap within a single population, like people who play two different sports.
How do I know if two events are independent?
Two events A and B are independent exactly when P(A∩B) = P(A) × P(B), which is equivalent to P(A|B) = P(A) - knowing B happened doesn't change the probability of A. If these don't hold, the events are dependent.
Do I need a calculator for probability questions?
Many probability questions can be done without a calculator, since they're often fraction arithmetic from a tree, Venn diagram or sample space. But some involve binomial or normal probabilities from later in this unit, where technology is expected. See the GDC guide for model-specific instructions.
Sub-topics
Probability broken down into its individual skills, each with its own focused page.
Related topics
More Statistics & Probability topics from the same AA SL syllabus unit, in case you want to keep going.