Every core rule of chance — the probability scale, complement, addition and conditional rules, independence, two-way tables, Venn and tree diagrams — plus HL Bayes' theorem, for IB Mathematics Applications & Interpretation HL. Built by an IBO-certified Singapore tutor, with a print-ready PDF to download.
Probability is the engine room of the Statistics & Probability strand in IB Mathematics Applications & Interpretation HL, and it is pure AI in spirit: you are asked to model uncertainty — a game, a survey, a medical test, a bag of counters — and then read a sensible number out of it. A calculator is always allowed, so almost no marks are for arithmetic. They live in choosing the right rule (add or multiply? condition on which event?), in picturing the situation with a Venn diagram, a two-way table or a tree, and in interpreting the answer in context.
This cheatsheet condenses SL 4.5–4.6 onto one page — the probability scale and sample space, complementary and combined events, the crucial difference between mutually exclusive and independent, conditional probability, two-way tables, Venn and tree diagrams, and the HL capstone of reversing a conditional with Bayes' theorem — and flags the traps that quietly cost method marks. The random-variable toolkit that builds on these rules (expected value, the binomial distribution and combining variables) lives on the companion Probability Distributions sheet. The print-ready PDF is at the bottom, free to download.
§1 — The probability scale & sample space SL 4.5
Every probability sits on a scale from $0$ (impossible) to $1$ (certain). The sample space $U$ is the list of all possible outcomes; for equally likely outcomes the theoretical probability is a simple count.
Scale:$0\le P(A)\le 1$ — and all outcomes in a sample space sum to $1$.
Theoretical:$P(A)=\dfrac{n(A)}{n(U)}$ — favourable outcomes over total equally likely outcomes.
Experimental:$P(A)\approx\dfrac{\text{number of times }A\text{ occurs}}{\text{number of trials}}$ — the relative frequency.
Expected number:Expected occurrences of $A$ in $n$ trials $=n\,P(A)$.
NoteAs the number of trials grows, the experimental (relative-frequency) probability settles towards the theoretical value — the law of large numbers. A short run of data can look nothing like the true probability.
TrickAdd every category to get the total $n(U)$ — the outcome you want is always included in that total. If probabilities are given as a table, they must sum to $1$, which lets you find any missing value.
§2 — Complementary events SL 4.5
Every event $A$ pairs with its complement $A'$ ("not $A$"); together they fill the whole sample space, so their probabilities add to $1$.
Complement:$P(A')=1-P(A)$
"At least one":$P(\text{at least one})=1-P(\text{none})$ — usually far quicker than adding cases.
TrapSubtract from $1$, not from $100$, when the probabilities are decimals or fractions. A "percentage" answer still comes from a probability between $0$ and $1$.
TrickThe moment a question says at least one, reach for the complement: find $P(\text{none})$ and subtract. This turns a long "or" calculation into a single step.
§3 — Combined events: the addition rule SL 4.6
For the union "$A$ or $B$ (or both)", the addition rule stops the overlap being counted twice.
Addition rule:$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
Mutually exclusive:If $A$ and $B$ cannot both occur, $P(A\cap B)=0$, so $P(A\cup B)=P(A)+P(B)$.
TrapSubtract the overlap $P(A\cap B)$ exactly once — forget it and the shared part is double-counted, giving a probability that can even exceed $1$.
NoteRead the symbols carefully: "$\cup$" (union) means or; "$\cap$" (intersection) means and. In IB, "or" is the inclusive or — it includes the case where both happen.
§4 — Mutually exclusive vs independent SL 4.6
These two ideas are constantly confused, yet they describe completely different situations. Mutually exclusive is about events that cannot coincide; independent is about events that do not influence each other.
TrapIndependent $\Rightarrow$ multiply for "and"; mutually exclusive $\Rightarrow$ add for "or". Do not mix them up — two events with non-zero probability are almost never both at once.
TrickTo test independence, compare $P(A)\,P(B)$ with the given $P(A\cap B)$: if they are equal the events are independent, if not they are dependent.
§5 — Conditional probability SL 4.6
Conditional probability rescales to a smaller world — the event you are told has already happened.
Conditional:$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$ — restrict attention to $B$ only.
Independent case:If $A,B$ are independent, $P(A\mid B)=P(A)$ — the condition tells you nothing new.
TrapThe denominator is the given event. Conditioning on "boy" means dividing by the number of boys, not by everyone in the sample.
NoteThe multiplication rule $P(A\cap B)=P(A)\,P(B\mid A)$ is exactly what you do when you multiply along the branches of a tree diagram (§8) — conditional probability and trees are the same idea.
§6 — Two-way (contingency) tables SL 4.5
A two-way table cross-classifies a group by two features; every probability is just a count read straight from the grid. Fill in the margin totals first.
Travel
Boy
Girl
Total
Bus
12
8
20
Walk
6
4
10
Total
18
12
30
Unconditional:$P(A)=\dfrac{\text{row or column total}}{\text{grand total }N}$ — e.g. $P(\text{Bus})=\tfrac{20}{30}=\tfrac{2}{3}$.
Conditional:$P(A\mid B)=\dfrac{n(A\cap B)}{n(B)}$ — one cell over its row/column total, e.g. $P(\text{Girl}\mid\text{Walk})=\tfrac{4}{10}$.
TrapFor a conditional, the denominator is the total of the given row or column — never the grand total $N$.
NoteIndependence straight from the table: $P(\text{Boy})\,P(\text{Bus})=\tfrac{18}{30}\cdot\tfrac{20}{30}=\tfrac{12}{30}=P(\text{Boy}\cap\text{Bus})$, so here travel mode and gender are independent — the test from §4 in action.
§7 — Venn diagrams SL 4.5
A Venn diagram shows the sample space as regions inside a box. Fill it from the overlap outwards: write the intersection first, then the single regions, then whatever is left over sits in "neither".
A two-set Venn diagram with $n(A)=18$, $n(B)=14$ and overlap $8$: the single regions are $18-8=10$ and $14-8=6$, leaving $30-24=6$ in "neither". The four regions add to $N=30$.
Only A / only B:$n(A)-n(A\cap B)$ and $n(B)-n(A\cap B)$ — strip out the shared middle.
TrapSubtract the "both" count once from each circle before you do anything else, or you will double-count the overlap and the regions will not add up to $N$.
TrickAlways write the intersection number in first, then work outwards. The four regions (only $A$, both, only $B$, neither) must total $N$ — a built-in check.
§8 — Tree diagrams: with & without replacement SL 4.5
A tree diagram lays out a sequence of stages. The rule is simple: multiply along the branches of one path, then add between the paths that give your event.
A two-stage tree for two draws without replacement from 5 red and 3 blue counters. Multiply along each path; the four outcomes $\tfrac{20}{56},\tfrac{15}{56},\tfrac{15}{56},\tfrac{6}{56}$ add to $1$. Because a counter is not replaced, every second-stage denominator drops to $7$.
Along a path:$P(\text{path})=$ product of that path's branch probabilities (an "and").
Between paths:$P(\text{event})=$ sum of every path that produces it (an "or").
With replacement:The counts reset, so every stage uses the same probabilities (independent draws).
Without replacement:Reduce that outcome's count and the total by $1$ for the next stage (dependent draws).
TrapWithout replacement the second-stage denominator is $n-1$. For "both the same colour" from $r$ red and $g$ green, $P=\dfrac{r}{n}\cdot\dfrac{r-1}{n-1}+\dfrac{g}{n}\cdot\dfrac{g-1}{n-1}$ — reduce both the colour count and the total.
NoteHL synthesis. Repeated independent trials that each succeed with probability $p$ (and fail with $q=1-p$) give a geometric pattern: the first success on trial $n$ has probability $q^{\,n-1}p$. In a turn-taking game the player who goes first wins with $\dfrac{p}{1-q^{2}}=\dfrac{1}{2-p}$ — a geometric series to infinity of ratio $q^{2}$, where probability meets Sequences & Series.
§9 — Reversing a conditional: Bayes' theorem AHL 4.13
A tree runs forwards — cause, then evidence. Often you want to run it backwards: given the evidence, how likely is the cause? Bayes' theorem does exactly this.
Bayes:$P(A\mid D)=\dfrac{P(A)\,P(D\mid A)}{P(D)}$
Total probability:$P(D)=P(A)\,P(D\mid A)+P(A')\,P(D\mid A')$ — sum every branch that reaches $D$.
TrapBuild the denominator $P(D)$ from all routes to the evidence first, then divide by the single route you want. Forgetting the second branch is the classic Bayes error.
NoteFor a rare condition, even an accurate test can give a surprisingly low posterior $P(A\mid D)$ — most positive results come from the huge healthy group, not the few true cases. Always compute $P(D)$ before judging the answer.
§10 — Exam attack plan All sections
Question cue
What to do
Watch for
"Find $P(A\cup B)$ / $A$ or $B$"
Addition rule $P(A)+P(B)-P(A\cap B)$
Subtract the overlap once; mutually exclusive $\Rightarrow$ it is $0$
Bayes $\dfrac{P(A)P(D\mid A)}{P(D)}$, $P(D)=$ all routes
Posterior can be low for a rare cause
Worked Example — Two Counters Without Replacement
Question (AI HL Paper 2 style — 7 marks)
A bag contains $5$ red and $3$ blue counters. Maya draws two counters at random, one after the other, without replacement. (a) Find the probability that both counters are red. (b) Find the probability that the two counters are different colours. (c) Given that the second counter drawn is red, find the probability that the first counter was blue.
Solution
(a) There are $8$ counters; after a red is removed, $4$ of the remaining $7$ are red: $P(\text{RR})=\dfrac{5}{8}\times\dfrac{4}{7}=\dfrac{20}{56}=\dfrac{5}{14}\approx0.357$. (M1)(A1)
(b) "Different colours" is red-then-blue or blue-then-red: $P=\dfrac{5}{8}\times\dfrac{3}{7}+\dfrac{3}{8}\times\dfrac{5}{7}=\dfrac{15}{56}+\dfrac{15}{56}=\dfrac{30}{56}=\dfrac{15}{28}\approx0.536$. (M1)(A1)(Check: it also equals $1-P(\text{RR})-P(\text{BB})=1-\tfrac{20}{56}-\tfrac{6}{56}$.)
(c) First find $P(\text{2nd red})$ from every path ending in red: $P(\text{2nd red})=\dfrac{20}{56}+\dfrac{3}{8}\times\dfrac{5}{7}=\dfrac{20}{56}+\dfrac{15}{56}=\dfrac{35}{56}$. (M1) Then $P(\text{1st blue}\mid\text{2nd red})=\dfrac{P(\text{1st blue}\cap\text{2nd red})}{P(\text{2nd red})}=\dfrac{15/56}{35/56}=\dfrac{15}{35}=\dfrac{3}{7}\approx0.429$. (M1)(A1)
Examiner's note: keep exact fractions or round to 3 s.f., and make sure every second-draw denominator drops to $7$ — that is what "without replacement" means. Part (c) is a reverse conditional: the denominator is $P(\text{2nd red})$, built from all the paths reaching a red second draw (total probability), not from a single branch.
Common Student Questions
What is the difference between mutually exclusive and independent events?
They are completely different. Mutually exclusive events cannot both happen, so $P(A\cap B)=0$ and you add their probabilities for "or". Independent events do not affect each other, so $P(A\cap B)=P(A)P(B)$ and you multiply for "and". In fact two events with non-zero probability cannot be both mutually exclusive and independent at the same time — if one rules the other out, they must influence each other.
How do I know whether to add or to multiply?
Look at the joining word. "$A$ or $B$" is a union, so use the addition rule $P(A)+P(B)-P(A\cap B)$ (drop the overlap if they are mutually exclusive). "$A$ and $B$", or a sequence of stages, is an intersection, so you multiply — along the branches of a tree. If the second event depends on the first, multiply by the conditional $P(B\mid A)$ rather than $P(B)$.
In a conditional probability, which number goes on the bottom?
The given event. $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$, so you divide by $P(B)$ — the event you have been told has happened. In a two-way table this means the total of the given row or column, not the grand total. Getting the wrong denominator is the single most common conditional-probability mistake.
With replacement or without replacement — what actually changes?
With replacement, the item goes back, so the counts reset and every draw uses the same probabilities — the draws are independent. Without replacement, you reduce the chosen outcome's count and the total by $1$ before the next draw, so the second-stage denominators shrink (from $n$ to $n-1$) and the draws become dependent. On a tree, only the without-replacement second-stage branches change.
When do I need Bayes' theorem instead of a normal tree?
When the question runs the tree backwards: you are given the evidence and asked for the probability of the cause, such as "given the test is positive, find the probability the person has the condition". Use $P(A\mid D)=\dfrac{P(A)P(D\mid A)}{P(D)}$, building $P(D)$ from every branch that reaches the evidence (the law of total probability), then divide by the branch you want. For a rare condition the answer can be surprisingly small.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
Notes PDF — every rule and definition in full, with worked Venn, two-way-table, tree and Bayes examples.
Tutorial booklet — 25+ AI HL-style questions from probability fluency to full applications (conditional probability, without-replacement trees, Bayes).
Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations for every probability question type.
Predicted-paper questions — the probability question types most likely to appear in the next AI HL session.
