Probability (AA HL)
Probability measures how likely an event is, on a scale from 0 to 1, and this topic is about combining and re-interpreting those likelihoods correctly. You'll work with Venn diagrams and tree diagrams, combine events with "or" and "and", condition one event on another happening, test whether two events are independent, and at HL use Bayes' theorem to reverse a conditional probability. Getting the right formula for the situation matters more here than heavy calculation.
What the syllabus says
This topic maps onto three points in the official IB Analysis & Approaches syllabus, including one that's HL-only.
| Code | Syllabus content |
|---|---|
| SL4.6 | Use of Venn diagrams, tree diagrams, sample space diagrams and tables of outcomes. Combined events: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Mutually exclusive events: \(P(A\cap B)=0\). Conditional probability: \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\). Probabilities with and without replacement. Independent events: \(P(A\cap B)=P(A)P(B)\). |
| SL4.11 | Formal definition and use of the conditional probability formula, and \(P(A\mid B)=P(A)=P(A\mid B')\) for independent events - used to test whether two events are independent. |
| AHL4.13 | Use of Bayes' theorem for a maximum of three events, building on independent events from SL4.6. |
SL4.6 and SL4.11 are core AA content also examinable at HL; AHL4.13 is additional HL-only content.
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?
The sample space, written \(U\), is the set of every possible outcome of a trial. An event \(A\) is a subset of the sample space - some particular collection of outcomes you're interested in. The probability of \(A\) is \(P(A) = \dfrac{n(A)}{n(U)}\) when outcomes are equally likely.
e.g. Rolling a die, \(U=\{1,2,3,4,5,6\}\); for "even", \(A=\{2,4,6\}\), so \(P(A)=\dfrac36=\dfrac12\).
What is conditional probability?
Conditional probability \(P(A\mid B)\) is the probability of \(A\) happening, given that you already know \(B\) has happened. Knowing \(B\) has occurred can shrink the sample space you're really working within, changing the probability of \(A\).
e.g. If \(P(A\cap B)=0.2\) and \(P(B)=0.5\), then \(P(A\mid B)=\dfrac{0.2}{0.5}=0.4\).
What are independent events?
Two events are independent if one happening doesn't change the probability of the other - formally, \(P(A\cap B)=P(A)P(B)\), which is equivalent to \(P(A\mid B)=P(A)\). This is different from mutually exclusive, where the events can't both happen.
e.g. \(P(A)=0.5\), \(P(B)=0.4\), independent \(\Rightarrow P(A\cap B)=0.5\times0.4=0.2\).
What are mutually exclusive events?
Mutually exclusive (disjoint) events cannot both happen in the same trial, so \(P(A\cap B)=0\). This makes the addition rule simplify to \(P(A\cup B)=P(A)+P(B)\), with no overlap to subtract.
e.g. \(P(A)=0.3\), \(P(B)=0.25\), mutually exclusive \(\Rightarrow P(A\cup B)=0.3+0.25=0.55\).
What is Bayes' theorem?
Bayes' theorem reverses a conditional probability - it lets you find \(P(B\mid A)\) when you only know \(P(A\mid B)\) and the individual probabilities of \(A\) and \(B\). It's built entirely from the conditional probability formula applied twice.
e.g. If \(P(A\mid B)=0.8\), \(P(B)=0.1\), \(P(A)=0.17\): \(P(B\mid A)=\dfrac{0.8\times0.1}{0.17}\approx0.47\).
Key formulas
Five formulas cover almost every probability question at this level - the tables below summarise them, and the sections underneath explain each in more depth.
Formula reference
The combined-events, conditional probability and independence formulas are all on the official formula booklet. Bayes' theorem is built from them but isn't written out separately.
| Formula | Used for | Booklet? |
|---|---|---|
| \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\) | Combined ("or") events | ✓ Yes |
| \(P(A\cap B)=0\) | Mutually exclusive events | ✓ Yes |
| \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\) | Conditional probability | ✓ Yes |
| \(P(A\cap B)=P(A)P(B)\) | Independent events | ✓ Yes |
| \(P(B\mid A)=\dfrac{P(A\mid B)P(B)}{P(A)}\) | Bayes' theorem | Not in the formula booklet - derived from the conditional probability formula |
Independent vs mutually exclusive
These two terms are the single most confused pair on this topic - they describe opposite kinds of relationship between events.
| Feature | Independent events | Mutually exclusive events |
|---|---|---|
| Definition | \(P(A\cap B)=P(A)P(B)\) | \(P(A\cap B)=0\) |
| Can both happen? | Yes - occurring together is normal | No - they can never occur together |
| Does knowing \(B\) change \(P(A)\)? | No: \(P(A\mid B)=P(A)\) | Yes, completely: \(P(A\mid B)=0\) |
| Can both hold at once (\(P(A),P(B)>0\))? | No - if two events with positive probability are mutually exclusive, they cannot also be independent. | |
Combining events
These rules tell you how to build the probability of "\(A\) or \(B\)" from the probabilities of the individual events.
Addition rule
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
Subtracting the overlap avoids double-counting outcomes that are in both events.
✓ In the formula bookletMutually exclusive events
\[P(A\cup B)=P(A)+P(B)\]
When events can't overlap, there's nothing to subtract - the addition rule simplifies.
✓ In the formula bookletComplementary events
\[P(A')=1-P(A)\]
Useful whenever "at least one" appears - it's often faster to find the probability of "none" and subtract from 1.
Not in the formula booklet - prior knowledgeConditional probability & independence
These describe how one event affects (or doesn't affect) the probability of another.
Conditional probability
\[P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\]
Rearranges to \(P(A\cap B)=P(B)\,P(A\mid B)\) - the rule behind every tree diagram.
✓ In the formula bookletTesting for independence
Check whether \(P(A\mid B)=P(A)\), or equivalently \(P(A\cap B)=P(A)P(B)\). If either holds, the events are independent - if not, they're dependent.
✓ In the formula bookletBayes' theorem (HL)
Bayes' theorem reverses the direction of a known conditional probability, using the law of total probability to rebuild \(P(A)\) from its pieces.
Bayes' theorem
\[P(B\mid A)=\dfrac{P(A\mid B)P(B)}{P(A)}\]
Given a test result, find the probability of the underlying cause - the classic "positive test, actually has the condition" question.
Not in the formula booklet - prior knowledgeFinding \(P(A)\) with a tree
When \(P(A)\) isn't given directly, build it from a tree diagram: sum \(P(B_i)P(A\mid B_i)\) over every branch that leads to \(A\), for up to three events \(B_1,B_2,B_3\).
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
Events \(A\) and \(B\) satisfy \(P(A)=\dfrac{1}{2}\), \(P(B)=\dfrac{1}{3}\), \(P(A\cup B)=\dfrac{2}{3}\).
(a) Find \(P(A\cap B)\).
(b) Find \(P(A\mid B)\).
Worked solution
(a) \(P(A\cap B)=P(A)+P(B)-P(A\cup B)=\dfrac12+\dfrac13-\dfrac23\) M1
\(=\dfrac16.\) A1
(b) \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{1/6}{1/3}\) M1
\(=\dfrac12.\) A1
Events \(A\) and \(B\) satisfy \(P(A\mid B)=\dfrac34\), \(P(B)=\dfrac25\) and \(P(A)=\dfrac12\).
(a) Find \(P(A\cap B)\).
(b) Find \(P(A\cup B)\).
(c) Find \(P(B\mid A)\).
Worked solution
(a) \(P(A\cap B)=P(A\mid B)\,P(B)=\dfrac34\times\dfrac25\) M1
\(=\dfrac{3}{10}.\) A1
(b) \(P(A\cup B)=\dfrac12+\dfrac25-\dfrac{3}{10}=\dfrac{5+4-3}{10}=\dfrac{6}{10}\) M1
\(=\dfrac35.\) A1
(c) \(P(B\mid A)=\dfrac{3/10}{1/2}\) M1
\(=\dfrac35.\) 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.
- Confusing independent with mutually exclusive. These describe opposite relationships - independent events can happen together freely, mutually exclusive events never can. Two events with positive probability can never be both.
- Forgetting "without replacement" changes the second probability. If a card, ball or item isn't put back, the total and the count of favourable outcomes both shrink for the next draw - each branch of the tree needs a fresh denominator.
- Adding \(P(A)+P(B)\) without subtracting the overlap. This only works when the events are mutually exclusive. If they can occur together, \(P(A\cap B)\) must be subtracted or you'll double-count outcomes.
- Mixing up \(P(A\mid B)\) and \(P(B\mid A)\). These are generally different numbers - "given it rained, the probability the match was cancelled" is not the same as "given the match was cancelled, the probability it rained". Bayes' theorem exists specifically to convert between them.
Using your GDC
Most probability questions on this topic are worked through by hand with a tree diagram, Venn diagram or the conditional probability formula rather than a calculator routine. The one place your GDC genuinely speeds things up is counting arrangements and selections.
Count arrangements and selections quickly - useful whenever a question involves choosing a committee, dealing cards, or arranging items without listing every outcome by hand.
- Type n, then MATH → PROB → nCr (or nPr), then r, then ENTER.TI-84
- Use nCr(n, r) or nPr(n, r) from the catalog.Nspire
- Type n, then OPTN → PROB → nCr (or nPr), then r.Casio
Tip: nCr ignores order (choosing a team); nPr counts order (ranking places). nCr(n, r) is also the binomial coefficient.
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 independent and mutually exclusive events?
Independent events don't affect each other's probability - knowing one happened tells you nothing about the other. Mutually exclusive events can never both happen at once. In fact, two events with non-zero probability can never be both independent and mutually exclusive at the same time.
How do I know whether to add or multiply probabilities?
Add probabilities for "or" situations (\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)). Multiply probabilities for "and" situations involving independent events, or use \(P(A\cap B)=P(B)\,P(A\mid B)\) when events depend on each other.
Is Bayes' theorem examined at HL?
Yes. AHL4.13 requires using Bayes' theorem for up to three events, usually to reverse a conditional probability - for example, given a positive test result, finding the probability the person actually has the condition.
Can I use my GDC for probability questions?
Most probability questions are worked through by hand using tree diagrams, Venn diagrams or the conditional probability formula. Your GDC is mainly useful for evaluating nCr and nPr quickly when counting arrangements or selections.
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 HL syllabus unit, in case you want to keep going.