Probability (AI SL)
Probability measures how likely an event is, from 0 (impossible) to 1 (certain). This topic covers reading Venn diagrams and tree diagrams to build up probabilities from raw counts, combining events with "or" and "and", conditioning one event on another, and using expected value to judge whether a game or decision is worthwhile.
What the syllabus says
This topic maps onto two points in the official IB Applications & Interpretation 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: \(P(A\cap B)=P(A)P(B)\). |
Problems can be solved using a Venn diagram, tree diagram, sample space diagram or table of outcomes without needing the formulae explicitly.
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, \(U\), is the set of every possible outcome of an experiment. Once you know the sample space, the probability of any event \(A\) is just the fraction of outcomes belonging to \(A\): \(P(A)=\dfrac{n(A)}{n(U)}\).
e.g. Rolling a die: \(U=\{1,2,3,4,5,6\}\), so \(P(\text{even})=\dfrac{3}{6}=0.5\).
What is conditional probability?
Conditional probability is the chance of one event given that another has already happened. It restricts the sample space to just the outcomes where the condition is true, then asks what fraction of those also satisfy the event you want.
e.g. If \(P(A\cap B)=0.2\) and \(P(B)=0.5\), then \(P(A|B)=\dfrac{0.2}{0.5}=0.4\).
What are independent events?
Two events are independent if one happening doesn't change the probability of the other. Algebraically, \(A\) and \(B\) are independent exactly when \(P(A\cap B)=P(A)\times P(B)\) - this is the test you check, not an assumption you can just declare.
e.g. Two coin flips: \(P(\text{HH})=0.5\times0.5=0.25\).
What is a tree diagram?
A tree diagram shows a sequence of events as branches, with each branch labelled by its probability. The probability of following a particular path is the product of the probabilities along it; the probability of an outcome that can happen several ways is the sum of those paths.
e.g. \(P(\text{white then black})=0.6\times0.4=0.24\) along one branch.
What is expected value?
The expected value \(E(X)\) of a discrete random variable is the long-run average outcome, found by multiplying each possible value by its probability and adding the results. It's not a value \(X\) can actually take - it's a weighted average over many repeats.
e.g. \(E(X)=5(0.2)+1(0.3)+0(0.5)=1.3\).
Key formulas
Four 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, conditional probability, independence and expected-value formulas are all in the official formula booklet.
| Formula | Used for | Booklet? |
|---|---|---|
| \(P(A\cup B) = P(A)+P(B)-P(A\cap B)\) | Combined events ("or") | ✓ Yes |
| \(P(A|B) = \dfrac{P(A\cap B)}{P(B)}\) | Conditional probability | ✓ Yes |
| \(P(A\cap B) = P(A)\,P(B)\) | Independent events | ✓ Yes |
| \(E(X) = \sum x\,P(X=x)\) | Expected value (discrete) | ✓ Yes |
| \(P(A) = \dfrac{n(A)}{n(U)}\) | Basic probability from a sample space | Not in booklet - prior knowledge |
Mutually exclusive vs independent
These two ideas are easy to mix up because both describe a relationship between two events - but they mean opposite things.
| Feature | Mutually exclusive | Independent |
|---|---|---|
| Meaning | The events can never both happen | One event doesn't affect the other |
| Test | \(P(A\cap B)=0\) | \(P(A\cap B)=P(A)P(B)\) |
| Union formula | \(P(A\cup B)=P(A)+P(B)\) | \(P(A\cup B)=P(A)+P(B)-P(A)P(B)\) |
| Example | Rolling a 2 and rolling a 5 on one die | Two separate coin flips |
Combining events
These rules let you build up the probability of a compound event from simpler pieces - use a Venn diagram to see the overlap clearly.
Union ("or")
\[P(A\cup B) = P(A)+P(B)-P(A\cap B)\]
Subtract the overlap once so it isn't double-counted.
✓ In the formula bookletComplement
\[P(A') = 1-P(A)\]
Useful whenever "at least one" is easier to find as 1 minus "none".
Not in the formula booklet - prior knowledgeConditional probability
\[P(A|B) = \dfrac{P(A\cap B)}{P(B)}\]
Restrict the sample space to \(B\) first, then ask what fraction is also \(A\).
✓ In the formula bookletIndependence and expected value
Independence is a property you test for; expected value is a single summary number for an entire probability distribution.
Independent events
\[P(A\cap B) = P(A)\,P(B)\]
Multiply along the branches of a tree diagram when draws are with replacement.
✓ In the formula bookletExpected value
\[E(X) = \sum x\,P(X=x)\]
A fair game has \(E(X)=0\) for the player's net gain.
✓ In the formula bookletExpected number of occurrences
\[n \times P(\text{event})\]
Multiply the probability of one occurrence by the number of trials.
Not in the formula booklet - prior knowledgeWorked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
In a class of 30: 18 study French, 14 study Spanish, 6 study both.
(a) Complete the Venn diagram to show this information.
(b) Find \(P(\text{French or Spanish})\).
Worked solution
(a) French only \(=18-6=12.\) A1 Both \(=6.\) Spanish only \(=14-6=8.\) A1 In at least one subject \(=12+6+8=26,\) so neither \(=30-26=4.\)
(b) \(P(F\cup S).\) \(\dfrac{26}{30}=\dfrac{13}{15}\) M1
\(\approx 0.867.\) A1 (Check via the formula: \(P(F\cup S)=\tfrac{18}{30}+\tfrac{14}{30}-\tfrac{6}{30}=\tfrac{26}{30}.\))
A factory uses two machines. Machine A makes 60% of items, with 3% defective. Machine B makes 40%, with 5% defective.
(a) Find the probability a random item is defective.
(b) Given an item is defective, find the probability it came from Machine A.
Worked solution
(a) \(P(D)\) - law of total probability. Sum over both machines: M1
\(P(D)=0.6(0.03)+0.4(0.05)=0.018+0.020\) A1
\(=0.038.\) A1
(b) \(P(A\mid D)\) - reverse the tree (Bayes). Of the defective probability, the share from A is: R1
\(P(A\mid D)=\dfrac{P(A\cap D)}{P(D)}=\dfrac{0.018}{0.038}\) M1
\(\approx 0.474.\) 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 mutually exclusive with independent. Mutually exclusive events can't both happen (\(P(A\cap B)=0\)); independent events just don't affect each other (\(P(A\cap B)=P(A)P(B)\)). Two events with nonzero probability can't be both at once.
- Forgetting that "without replacement" changes later branches. If an item is removed and not put back, the totals and probabilities on the next branch of a tree diagram must be updated - they are not the same as the first draw.
- Adding along a branch instead of multiplying. Multiply probabilities along a single path through a tree diagram; only add the probabilities of separate paths that lead to the same outcome.
- Mixing up \(P(A|B)\) with \(P(B|A)\). These are generally different numbers - always check which event is the "given" condition, since it fixes which probability goes on the bottom of the fraction.
Using your GDC
Probability questions on this topic are usually solved by setting up the right expression from a Venn diagram, tree diagram or table - the GDC's job is just to evaluate the arithmetic accurately.
Once you've written the correct expression from a Venn diagram, tree diagram or table, your GDC just needs to evaluate it - no special probability mode is needed for this topic.
- Write out the full expression first, e.g. \(0.6\times0.03+0.4\times0.05\), rather than rounding intermediate values.
- Type the expression directly on the home screen and press ENTER.TI-84
- Type the expression in Run-Matrix and press EXE.Casio
- Type the expression on a Calculator page and press ENTER.Nspire
- Keep several decimal places until the final answer, then round only at the end.
Tip: Store an intermediate value (like \(P(D)\) in a Bayes' question) so you can reuse it exactly in the next part, instead of retyping a rounded version.
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 "or" and "and" in probability?
"Or" means the union - either event happening - and uses \(P(A\cup B) = P(A) + P(B) - P(A\cap B)\). "And" means the intersection - both events happening together - and is the \(P(A\cap B)\) term you subtract to avoid double-counting the overlap.
How do I know if two events are independent or mutually exclusive?
Test independence with \(P(A\cap B) = P(A)\times P(B)\) - if that's true, the events don't affect each other. Mutually exclusive events can never happen together, so \(P(A\cap B) = 0\). Two events with a nonzero probability can't be both independent and mutually exclusive at once.
Is probability tested with or without a GDC?
Both. Straightforward Venn diagram and tree diagram questions often appear on the non-calculator Paper 1, but anything with messy decimals or several combined events is just as likely on the calculator papers - your GDC evaluates the arithmetic once you've set up the right expression.
What's the difference between P(A|B) and P(B|A)?
\(P(A|B)\) is the probability of \(A\) given that \(B\) has already happened; \(P(B|A)\) is the reverse. They're generally different numbers unless \(P(A) = P(B)\) - always check which event is the "given" condition before dividing.
Sub-topics
Probability broken down into its individual skills, each with its own focused page.
Related topics
More Statistics & Probability topics from the same AI SL syllabus unit, in case you want to keep going.