When should I use a Venn diagram versus a tree diagram?

Use a Venn diagram when you are sorting a single population by overlapping labels — students who study Physics, Chemistry or both, for instance. Use a tree diagram when events happen in sequence — drawing two counters one after another, or a two-stage test. A quick tell: if the question describes 'first... then...', reach for a tree; if it describes one group with several attributes, reach for a Venn diagram.

What is the difference between 'mutually exclusive' and 'independent'?

Mutually exclusive events cannot happen at the same time, so $P(A\cap B)=0$; if one occurs the other cannot. Independent events are ones where the outcome of one does not change the probability of the other, tested by $P(A\cap B)=P(A)P(B)$ — and these events CAN happen together. They are opposite in spirit: mutually exclusive events are strongly linked (one rules out the other), while independent events ignore each other. Any two events with non-zero probability cannot be both at once.

How do I handle 'at least one' questions?

Use the complement. 'At least one' is the opposite of 'none', so $P(\text{at least one})=1-P(\text{none})$. On a tree diagram, find the single path where the event never happens, multiply along it, and subtract from 1 — far quicker than adding up every other path.

How do tree diagrams change when there is no replacement?

Without replacement, each draw removes an item, so the total falls by one at every stage and the count of the drawn colour falls too. The second-stage branches then depend on what the first stage produced — that dependence is exactly conditional probability. With replacement the item returns, the totals reset, and the stages are independent, so the second-stage branches repeat the first.

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Probability, Venn diagrams and tree diagrams…

Revise Probability, Venn diagrams and tree diagrams for IB Mathematics: Applications and Interpretation (4.4)

Looking for free IB Mathematics: Applications and Interpretation notes on Probability, Venn diagrams and tree diagrams? This Higher Level lesson covers official syllabus point 4.4 on Maths AI HL. Use the visual steps, key terms, and quick checks below — then practise with IB criterion marking on Probability, Venn diagrams and tree diagrams. Built for students searching "IB Maths AI 4.4", "Probability, Venn diagrams and tree diagrams IB revision", or a free alternative to paid IB note sites.

Probability, Venn diagrams and tree diagrams

Probability is the language we use to measure uncertainty, turning words like 'likely' and 'even chance' into numbers between 0 and 1. This lesson covers the whole of 4.4: the sample space and the probability of a single event, the complement, combined events and the addition rule, mutually exclusive and independent events, conditional probability, and the two diagrams that make all of it manageable — Venn diagrams for classifying a population and tree diagrams for events in sequence. Because 4.4 is examined on Paper 2, the GDC does the arithmetic; the skill being marked is choosing the right diagram, applying the right rule, and writing down the method. As in every calculation topic, the worked examples below are laid out exactly the way our engine marks your own practice.

BY THE END, YOU CAN…

1Describe a sample space, calculate the probability of an event, and use the complement $P(A')=1-P(A)$ .
2Apply the addition rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ and recognise mutually exclusive events where $P(A\cap B)=0$ .
3Represent a population with a Venn diagram and read unions, intersections, complements and conditional probabilities from it.
4Test independence with $P(A\cap B)=P(A)P(B)$ , and find conditional probabilities using $P(A|B)=\frac{P(A\cap B)}{P(B)}$ .
5Construct tree diagrams for successive events, with and without replacement, and present working so that method marks and follow-through protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Mapping Out Chance

Probability turns a vague sense of 'likely' into a number between 0 and 1. Venn diagrams and tree diagrams are the two maps that make that number easy to find: one sorts a group by overlapping labels, the other traces a sequence of choices step by step.

THINK OF IT LIKE…

Think about a class survey. A Venn diagram is like sorting everyone into overlapping hoops on the floor — one hoop for students who play football, one for those who play chess, and the overlap for the few who do both; the probability of a random student playing football is just the fraction of people standing in that hoop. A tree diagram is different: it maps a story that unfolds in stages, like drawing two sweets from a bag one after another, where each branch remembers what already happened and adjusts the odds for what comes next.

1
Name the sample space — the complete set of equally likely outcomes — and count it.
2
Decide the shape of the problem: a Venn diagram for one population with overlapping labels, a tree diagram for events that happen in sequence.
3
Fill the diagram carefully. For a Venn diagram start from the innermost overlap; for a tree diagram check whether the second stage is with or without replacement.
4
Read off the answer with the right rule — add along a union, multiply along a branch, divide for a conditional probability — and show the method that earns the marks.

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Step 1

Name the sample space — the complete set of equally likely outcomes — and count it.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$P(A)=\dfrac{n(A)}{n(U)}$

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Full topic notes

Formal explanation with the rigour you need for the exam.

Sample space, probability and the complement

An experiment's sample space, written $U$ , is the set of all possible outcomes. When those outcomes are equally likely, the probability of an event $A$ is the fraction of the sample space it fills: count the favourable outcomes and divide by the total. Every probability lies between 0 (impossible) and 1 (certain). The complement of $A$ , written $A'$ , is the event that $A$ does not happen — and since $A$ and $A'$ together make up the whole sample space, their probabilities add to 1.

Probability of an event: $P(A)=\dfrac{n(A)}{n(U)}$ , with $0 \le P(A) \le 1$ . \n Complement: $P(A')=1-P(A)$ .

Sample space $U$ : every possible outcome. On a Venn diagram it is the surrounding rectangle.
Probability: for equally likely outcomes, favourable outcomes over total outcomes.
Complement $A'$ : 'not $A$ '. Use $P(A')=1-P(A)$ , especially for 'at least one' questions where the complement is 'none'.

Combined events, the addition rule and mutual exclusivity

Two events can be combined in two ways. Their intersection $A\cap B$ is 'A and B' — both happening. Their union $A\cup B$ is 'A or B or both'. If you simply add $P(A)$ and $P(B)$ you count the overlap twice, so the addition rule subtracts it back off. When two events cannot occur together they are mutually exclusive: their intersection is empty, $P(A\cap B)=0$ , and the addition rule reduces to a plain sum.

Addition rule: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ . \n Mutually exclusive events: $P(A\cap B)=0$ , so $P(A\cup B)=P(A)+P(B)$ .

EXAM TIP

The single most common addition-rule slip is forgetting the $-P(A\cap B)$ term. Only drop it when you have checked the events are mutually exclusive. If the events overlap and you leave it out, your union exceeds its true value — and often exceeds 1, which should ring an alarm.

Venn diagrams: classifying a population

A Venn diagram is the right tool when one population is labelled by several attributes. A rectangle holds the sample space $U$ ; circles inside it hold the events; the overlap is the intersection and the region outside every circle is the complement of the union — 'neither'. The reliable way to fill one is to start at the innermost overlap and work outward, subtracting as you go so nothing is double-counted, then check that every region sums to the population total.

The rectangle is the sample space $U$ ; each circle is an event.
The overlap is the intersection $A\cap B$ ('A and B'); the combined region is the union $A\cup B$ ('A or B').
The region outside all circles is $(A\cup B)'$ ('neither A nor B').
Fill from the innermost overlap outward, subtracting so each region is counted once.

Independent events

Two events are independent when one occurring does not change the probability of the other — successive coin flips, for instance. The defining test is that the probability of both equals the product of the separate probabilities. This is genuinely different from mutual exclusivity: mutually exclusive events are strongly linked because one rules the other out, whereas independent events ignore each other entirely. Two events with non-zero probability can never be both at once.

Test for independence: $A$ and $B$ are independent if and only if $P(A\cap B)=P(A)\,P(B)$ . \n Equivalently, $P(A|B)=P(A)$ .

Tree diagrams for successive events

When an experiment happens in stages — draw one, then another — a tree diagram lays out every path. Each stage sprouts a set of branches whose probabilities sum to 1, and the probability of a full sequence is the product along its path. The crucial detail is replacement. WITH replacement, the item returns and the second stage repeats the first, so the draws are independent. WITHOUT replacement, the total falls by one and so does the count of whatever was drawn, so the second-stage branches are conditional on the first — which is conditional probability in visual form.

Branches leaving a single point always sum to 1.
Multiply along a path to find the probability of that whole sequence.
Add the probabilities of the separate paths that give the required overall result.
Check with vs without replacement before writing the second-stage probabilities.

EXAM TIP

Read the words 'with replacement' or 'without replacement' — or infer it from the context — before you fill the second stage. Without replacement, the second-stage denominator drops by one and the drawn colour's numerator drops too. Getting this wrong is the single biggest source of lost marks in tree-diagram questions.

Common mistakes examiners penalise

Confusing mutually exclusive with independent — mutually exclusive means $P(A\cap B)=0$ (cannot co-occur); independent means $P(A\cap B)=P(A)P(B)$ (no influence). They are different, and non-trivial events cannot be both.
Dropping the $-P(A\cap B)$ term — only omit it after confirming the events are mutually exclusive. Otherwise the union is overstated, sometimes above 1.
Inverting the conditional formula — it is $P(A|B)=\dfrac{P(A\cap B)}{P(B)}$ , dividing by the given event's probability. Writing $\dfrac{P(A\cap B)}{P(A)}$ answers a different question.
Not changing the second-stage probabilities without replacement — reduce the total (and the relevant count) by one; reusing the first-stage fractions treats the draw as with replacement.
Forgetting the second order in 'one of each' — 'one green and one yellow' includes both G–Y and Y–G, so add both paths; using only one halves the answer.
Filling a Venn diagram from the outside in — start at the innermost overlap and subtract outward, or regions get double-counted and the total will not match the population.
Adding probabilities that should be multiplied — multiply ALONG a path (a sequence), add ACROSS separate paths (alternatives); mixing these is a frequent tree-diagram error.

Model answer — marked the way our engine marks it

On Paper 2 the marks are analytic: each is tied to a specific line of working — a method mark (M) or an accuracy mark (A) — and an accuracy mark depends on the method mark it follows. Follow-through (FT) means an earlier slip need not cost the marks that build on it, provided the later step is correct on your own figures. The engine also ignores subsequent working (ISW) once a correct answer appears, and accepts any equivalent form and any correctly-rounded value. That protection only exists if the method is on the page. Study how each mark below is earned by a specific line.

Where this leads

The ideas in 4.4 are the foundation of everything probabilistic that follows. The multiplication and addition you do along tree branches become the mechanics of discrete random variables and expected value; conditional probability and independence underpin the binomial and normal models, where 'trials' are assumed independent; and the habit of naming the sample space, choosing the right diagram and showing each rule is exactly the discipline every Paper 2 probability question rewards. Master the two maps — a Venn diagram for one population with overlapping labels, a tree diagram for events in sequence — and the rest of the probability course becomes variations on a picture you already know how to draw.

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

In a year group of 120 students, 65 study Mathematics (M), 52 study Physics (P), and 20 study both. A student is chosen at random. \n (a) Draw a Venn diagram to represent this information. \n (b) Find the probability that the student studies Mathematics but not Physics. \n (c) Find the probability that the student studies neither subject. \n (d) Given that a student studies Mathematics, find the probability that they also study Physics. [7]

1
(a) Venn diagram. Draw the rectangle $U$ with two overlapping circles, M and P. Start at the overlap: $n(M\cap P)=20$ . [M1]\nMaths only: $65-20=45$ . Physics only: $52-20=32$ . In circles: $45+20+32=97$ , so neither $=120-97=23$ . [A1 for all four regions: 45, 20, 32, and 23 outside]\n\n**(b) Maths but not Physics.** That is the 45 students in the Maths-only region: $P=\dfrac{45}{120}=\dfrac{3}{8}=0.375$ . [A1]\n\n**(c) Neither subject.** The 23 students outside both circles: $P=\dfrac{23}{120}=0.192$ (3 s.f.). [A1]\n\n**(d) Given Maths, also Physics.** This is conditional: the sample space shrinks to the 65 Maths students, of whom 20 also study Physics.\n $P(P|M)=\dfrac{n(M\cap P)}{n(M)}$ [M1]\n $=\dfrac{20}{65}=\dfrac{4}{13}=0.308$ (3 s.f.). [A1]

✓ FULLY WORKED

EXAMPLE 2

For two events, $P(A)=0.6$ , $P(B)=0.5$ and $P(A\cup B)=0.8$ . \n (a) Find $P(A\cap B)$ . \n (b) Determine whether $A$ and $B$ are independent. \n (c) Find $P(A|B)$ . [6]

1
(a) Intersection from the addition rule. Rearrange $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ : [M1]\n $P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.6+0.5-0.8=0.3$ . [A1]\n\n**(b) Test for independence.** Compare $P(A\cap B)$ with $P(A)P(B)$ : [M1]\n $P(A)P(B)=0.6\times0.5=0.3$ , which equals $P(A\cap B)=0.3$ . Since they are equal, $A$ and $B$ ARE independent. [A1]\n\n**(c) Conditional probability.** $P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.3}{0.5}=0.6$ . [M1][A1]\nAs a check, this equals $P(A)=0.6$ , exactly what independence predicts.

✓ FULLY WORKED

EXAMPLE 3

A bag contains 7 green balls and 5 yellow balls. David picks a ball at random and does not replace it, then picks a second ball at random. \n (a) Draw a tree diagram to show the possible outcomes. \n (b) Find the probability that David picks two green balls. \n (c) Find the probability that David picks one ball of each colour. [6]

1
(a) Tree diagram. Total $=12$ . First pick: $P(G_1)=\dfrac{7}{12}$ , $P(Y_1)=\dfrac{5}{12}$ . After one ball is removed, 11 remain (without replacement). [M1: first-stage probabilities]\nSecond pick if first was Green: $P(G_2|G_1)=\dfrac{6}{11}$ , $P(Y_2|G_1)=\dfrac{5}{11}$ .\nSecond pick if first was Yellow: $P(G_2|Y_1)=\dfrac{7}{11}$ , $P(Y_2|Y_1)=\dfrac{4}{11}$ . [A1: correct second-stage conditional probabilities]\n\n**(b) Two green.** Multiply along the G–G path: [M1]\n $P(G_1\cap G_2)=\dfrac{7}{12}\times\dfrac{6}{11}=\dfrac{42}{132}=\dfrac{7}{22}=0.318$ (3 s.f.). [A1]\n\n**(c) One of each colour.** This happens two ways, G–Y or Y–G, so add the two paths: [M1]\n $\dfrac{7}{12}\times\dfrac{5}{11}+\dfrac{5}{12}\times\dfrac{7}{11}=\dfrac{35}{132}+\dfrac{35}{132}=\dfrac{70}{132}=\dfrac{35}{66}=0.530$ (3 s.f.). [A1]

✓ FULLY WORKED

EXAMPLE 4

A bag has 6 red and 4 blue counters. Two are drawn without replacement. Draw a tree diagram and find the probability that the two counters are different colours. [5]

1
Model answer — full working.\n\nThere are $6+4=10$ counters. First draw: $P(R_1)=\dfrac{6}{10}$ , $P(B_1)=\dfrac{4}{10}$ . Since the first counter is NOT replaced, 9 counters remain for the second draw.\n\nSecond-draw (conditional) branches:\nAfter red: $P(R_2|R_1)=\dfrac{5}{9}$ , $P(B_2|R_1)=\dfrac{4}{9}$ .\nAfter blue: $P(R_2|B_1)=\dfrac{6}{9}$ , $P(B_2|B_1)=\dfrac{3}{9}$ .\n\nDifferent colours happens two ways — red then blue (RB), or blue then red (BR):\n $P(RB)=\dfrac{6}{10}\times\dfrac{4}{9}=\dfrac{24}{90}, \qquad P(BR)=\dfrac{4}{10}\times\dfrac{6}{9}=\dfrac{24}{90}.$ \n\nAdd the two paths:\n $P(\text{different})=\dfrac{24}{90}+\dfrac{24}{90}=\dfrac{48}{90}=\dfrac{8}{15}=0.533$ (3 s.f.).\n\n---\nHow our marking engine awards the 5 marks:\n\n- M1 — tree with correct without-replacement probabilities. Awarded for a tree whose second stage uses a denominator of 9 with the reduced counts ( $\tfrac{5}{9}$ , $\tfrac{4}{9}$ after red; $\tfrac{6}{9}$ , $\tfrac{3}{9}$ after blue). Reusing $/10$ denominators (treating it as with replacement) loses this mark.\n- M1 — identify the two paths RB and BR. A method mark for recognising that 'different colours' is achieved by red-then-blue OR blue-then-red, not a single path.\n- M1 — add the two products. Awarded for combining the paths by addition, $\tfrac{6}{10}\cdot\tfrac{4}{9}+\tfrac{4}{10}\cdot\tfrac{6}{9}$ . This is the structural step the engine checks: multiply along, add across.\n- A1 — $\dfrac{48}{90}$ . The first accuracy mark, for the correct unsimplified sum. It depends on the method marks above and follows through: a candidate with slightly different (but correctly conditional) branches who adds their own products correctly still earns it.\n- A1 — $\dfrac{8}{15}$ . The final accuracy mark for the simplified value (equivalently $0.533$ ). FT applies — a correctly simplified version of the candidate's own $\tfrac{48}{90}$ keeps this mark.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts $\dfrac{48}{90}$ , $\dfrac{8}{15}$ , $0.533$ or $53.3\%$ as the final answer, and accepts the two products written in either order. Once a correct final value appears, ISW means later restatements do not lose marks.\n\nBottom line: three of the five marks are method marks that survive an arithmetic slip, and the two accuracy marks are shielded by follow-through. A candidate who writes only ' $\tfrac{8}{15}$ ' with no tree and no paths risks losing up to 3 marks; a candidate who draws the tree, shows RB and BR, and adds the products keeps the method regardless of a slip in the final number.

✓ FULLY WORKED

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEAProbability, Venn diagrams and tree diagrams

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

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QUESTION · 1 / 11

Sample space $U$ and probability of an event

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ANSWER

The sample space is the set of all possible outcomes. For equally likely outcomes, $P(A)=\dfrac{n(A)}{n(U)}$ — the number of outcomes in $A$ divided by the total. Every probability satisfies $0 \le P(A) \le 1$ .

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Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
Sample space $U$ : every possible outcome. On a Venn diagram it is the surrounding rectangle.
✓
Probability: for equally likely outcomes, favourable outcomes over total outcomes.
✓
Complement $A'$ : 'not $A$ '. Use $P(A')=1-P(A)$ , especially for 'at least one' questions where the complement is 'none'.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

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Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Get a Paper 2 probability question marked: draw the tree, show every path, and see the marks awarded on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

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Probability, Venn diagrams and tree diagrams

BY THE END, YOU CAN…

1Describe a sample space, calculate the probability of an event, and use the complement $P(A')=1-P(A)$ .
2Apply the addition rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ and recognise mutually exclusive events where $P(A\cap B)=0$ .
3Represent a population with a Venn diagram and read unions, intersections, complements and conditional probabilities from it.
4Test independence with $P(A\cap B)=P(A)P(B)$ , and find conditional probabilities using $P(A|B)=\frac{P(A\cap B)}{P(B)}$ .
5Construct tree diagrams for successive events, with and without replacement, and present working so that method marks and follow-through protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Mapping Out Chance

THINK OF IT LIKE…

1
Name the sample space — the complete set of equally likely outcomes — and count it.
2
Decide the shape of the problem: a Venn diagram for one population with overlapping labels, a tree diagram for events that happen in sequence.
3
Fill the diagram carefully. For a Venn diagram start from the innermost overlap; for a tree diagram check whether the second stage is with or without replacement.
4
Read off the answer with the right rule — add along a union, multiply along a branch, divide for a conditional probability — and show the method that earns the marks.

Explore the concept

Use the live diagram, PhET or GeoGebra sim, and synced steps — play it, drag controls, or tap a step.

GEOGEBRASTEP 1 / 4 · Step 1

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Use “Interactive Venn Diagram Tool” — shift curves or drag points to see equilibrium and slopes change.

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Step 1

Name the sample space — the complete set of equally likely outcomes — and count it.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$P(A)=\dfrac{n(A)}{n(U)}$

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Full topic notes

Formal explanation with the rigour you need for the exam.

Sample space, probability and the complement

Probability of an event: $P(A)=\dfrac{n(A)}{n(U)}$ , with $0 \le P(A) \le 1$ . \n Complement: $P(A')=1-P(A)$ .

Sample space $U$ : every possible outcome. On a Venn diagram it is the surrounding rectangle.
Probability: for equally likely outcomes, favourable outcomes over total outcomes.
Complement $A'$ : 'not $A$ '. Use $P(A')=1-P(A)$ , especially for 'at least one' questions where the complement is 'none'.

Combined events, the addition rule and mutual exclusivity

Addition rule: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ . \n Mutually exclusive events: $P(A\cap B)=0$ , so $P(A\cup B)=P(A)+P(B)$ .

EXAM TIP

Venn diagrams: classifying a population

The rectangle is the sample space $U$ ; each circle is an event.
The overlap is the intersection $A\cap B$ ('A and B'); the combined region is the union $A\cup B$ ('A or B').
The region outside all circles is $(A\cup B)'$ ('neither A nor B').
Fill from the innermost overlap outward, subtracting so each region is counted once.

Independent events

Test for independence: $A$ and $B$ are independent if and only if $P(A\cap B)=P(A)\,P(B)$ . \n Equivalently, $P(A|B)=P(A)$ .

Tree diagrams for successive events

Branches leaving a single point always sum to 1.
Multiply along a path to find the probability of that whole sequence.
Add the probabilities of the separate paths that give the required overall result.
Check with vs without replacement before writing the second-stage probabilities.

EXAM TIP

Common mistakes examiners penalise

Confusing mutually exclusive with independent — mutually exclusive means $P(A\cap B)=0$ (cannot co-occur); independent means $P(A\cap B)=P(A)P(B)$ (no influence). They are different, and non-trivial events cannot be both.
Dropping the $-P(A\cap B)$ term — only omit it after confirming the events are mutually exclusive. Otherwise the union is overstated, sometimes above 1.
Inverting the conditional formula — it is $P(A|B)=\dfrac{P(A\cap B)}{P(B)}$ , dividing by the given event's probability. Writing $\dfrac{P(A\cap B)}{P(A)}$ answers a different question.
Not changing the second-stage probabilities without replacement — reduce the total (and the relevant count) by one; reusing the first-stage fractions treats the draw as with replacement.
Forgetting the second order in 'one of each' — 'one green and one yellow' includes both G–Y and Y–G, so add both paths; using only one halves the answer.
Filling a Venn diagram from the outside in — start at the innermost overlap and subtract outward, or regions get double-counted and the total will not match the population.
Adding probabilities that should be multiplied — multiply ALONG a path (a sequence), add ACROSS separate paths (alternatives); mixing these is a frequent tree-diagram error.

Model answer — marked the way our engine marks it

Where this leads

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

1
(a) Venn diagram. Draw the rectangle $U$ with two overlapping circles, M and P. Start at the overlap: $n(M\cap P)=20$ . [M1]\nMaths only: $65-20=45$ . Physics only: $52-20=32$ . In circles: $45+20+32=97$ , so neither $=120-97=23$ . [A1 for all four regions: 45, 20, 32, and 23 outside]\n\n**(b) Maths but not Physics.** That is the 45 students in the Maths-only region: $P=\dfrac{45}{120}=\dfrac{3}{8}=0.375$ . [A1]\n\n**(c) Neither subject.** The 23 students outside both circles: $P=\dfrac{23}{120}=0.192$ (3 s.f.). [A1]\n\n**(d) Given Maths, also Physics.** This is conditional: the sample space shrinks to the 65 Maths students, of whom 20 also study Physics.\n $P(P|M)=\dfrac{n(M\cap P)}{n(M)}$ [M1]\n $=\dfrac{20}{65}=\dfrac{4}{13}=0.308$ (3 s.f.). [A1]

✓ FULLY WORKED

EXAMPLE 2

For two events, $P(A)=0.6$ , $P(B)=0.5$ and $P(A\cup B)=0.8$ . \n (a) Find $P(A\cap B)$ . \n (b) Determine whether $A$ and $B$ are independent. \n (c) Find $P(A|B)$ . [6]

1
(a) Intersection from the addition rule. Rearrange $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ : [M1]\n $P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.6+0.5-0.8=0.3$ . [A1]\n\n**(b) Test for independence.** Compare $P(A\cap B)$ with $P(A)P(B)$ : [M1]\n $P(A)P(B)=0.6\times0.5=0.3$ , which equals $P(A\cap B)=0.3$ . Since they are equal, $A$ and $B$ ARE independent. [A1]\n\n**(c) Conditional probability.** $P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.3}{0.5}=0.6$ . [M1][A1]\nAs a check, this equals $P(A)=0.6$ , exactly what independence predicts.

✓ FULLY WORKED

EXAMPLE 3

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(a) Tree diagram. Total $=12$ . First pick: $P(G_1)=\dfrac{7}{12}$ , $P(Y_1)=\dfrac{5}{12}$ . After one ball is removed, 11 remain (without replacement). [M1: first-stage probabilities]\nSecond pick if first was Green: $P(G_2|G_1)=\dfrac{6}{11}$ , $P(Y_2|G_1)=\dfrac{5}{11}$ .\nSecond pick if first was Yellow: $P(G_2|Y_1)=\dfrac{7}{11}$ , $P(Y_2|Y_1)=\dfrac{4}{11}$ . [A1: correct second-stage conditional probabilities]\n\n**(b) Two green.** Multiply along the G–G path: [M1]\n $P(G_1\cap G_2)=\dfrac{7}{12}\times\dfrac{6}{11}=\dfrac{42}{132}=\dfrac{7}{22}=0.318$ (3 s.f.). [A1]\n\n**(c) One of each colour.** This happens two ways, G–Y or Y–G, so add the two paths: [M1]\n $\dfrac{7}{12}\times\dfrac{5}{11}+\dfrac{5}{12}\times\dfrac{7}{11}=\dfrac{35}{132}+\dfrac{35}{132}=\dfrac{70}{132}=\dfrac{35}{66}=0.530$ (3 s.f.). [A1]

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EXAMPLE 4

A bag has 6 red and 4 blue counters. Two are drawn without replacement. Draw a tree diagram and find the probability that the two counters are different colours. [5]

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Model answer — full working.\n\nThere are $6+4=10$ counters. First draw: $P(R_1)=\dfrac{6}{10}$ , $P(B_1)=\dfrac{4}{10}$ . Since the first counter is NOT replaced, 9 counters remain for the second draw.\n\nSecond-draw (conditional) branches:\nAfter red: $P(R_2|R_1)=\dfrac{5}{9}$ , $P(B_2|R_1)=\dfrac{4}{9}$ .\nAfter blue: $P(R_2|B_1)=\dfrac{6}{9}$ , $P(B_2|B_1)=\dfrac{3}{9}$ .\n\nDifferent colours happens two ways — red then blue (RB), or blue then red (BR):\n $P(RB)=\dfrac{6}{10}\times\dfrac{4}{9}=\dfrac{24}{90}, \qquad P(BR)=\dfrac{4}{10}\times\dfrac{6}{9}=\dfrac{24}{90}.$ \n\nAdd the two paths:\n $P(\text{different})=\dfrac{24}{90}+\dfrac{24}{90}=\dfrac{48}{90}=\dfrac{8}{15}=0.533$ (3 s.f.).\n\n---\nHow our marking engine awards the 5 marks:\n\n- M1 — tree with correct without-replacement probabilities. Awarded for a tree whose second stage uses a denominator of 9 with the reduced counts ( $\tfrac{5}{9}$ , $\tfrac{4}{9}$ after red; $\tfrac{6}{9}$ , $\tfrac{3}{9}$ after blue). Reusing $/10$ denominators (treating it as with replacement) loses this mark.\n- M1 — identify the two paths RB and BR. A method mark for recognising that 'different colours' is achieved by red-then-blue OR blue-then-red, not a single path.\n- M1 — add the two products. Awarded for combining the paths by addition, $\tfrac{6}{10}\cdot\tfrac{4}{9}+\tfrac{4}{10}\cdot\tfrac{6}{9}$ . This is the structural step the engine checks: multiply along, add across.\n- A1 — $\dfrac{48}{90}$ . The first accuracy mark, for the correct unsimplified sum. It depends on the method marks above and follows through: a candidate with slightly different (but correctly conditional) branches who adds their own products correctly still earns it.\n- A1 — $\dfrac{8}{15}$ . The final accuracy mark for the simplified value (equivalently $0.533$ ). FT applies — a correctly simplified version of the candidate's own $\tfrac{48}{90}$ keeps this mark.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts $\dfrac{48}{90}$ , $\dfrac{8}{15}$ , $0.533$ or $53.3\%$ as the final answer, and accepts the two products written in either order. Once a correct final value appears, ISW means later restatements do not lose marks.\n\nBottom line: three of the five marks are method marks that survive an arithmetic slip, and the two accuracy marks are shielded by follow-through. A candidate who writes only ' $\tfrac{8}{15}$ ' with no tree and no paths risks losing up to 3 marks; a candidate who draws the tree, shows RB and BR, and adds the products keeps the method regardless of a slip in the final number.

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MAIN IDEAProbability, Venn diagrams and tree diagrams

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QUESTION · 1 / 11

Sample space $U$ and probability of an event

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Sample space $U$ : every possible outcome. On a Venn diagram it is the surrounding rectangle.
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Probability: for equally likely outcomes, favourable outcomes over total outcomes.
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Complement $A'$ : 'not $A$ '. Use $P(A')=1-P(A)$ , especially for 'at least one' questions where the complement is 'none'.

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Khan Academy: Conditional Probability and Independence↗

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Revise Probability, Venn diagrams and tree diagrams for IB Mathematics: Applications and Interpretation (4.4)

In simple terms

Explore the concept

Key formulas

Full topic notes

Sample space, probability and the complement

Combined events, the addition rule and mutual exclusivity

Venn diagrams: classifying a population

Independent events

Tree diagrams for successive events

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Get a Paper 2 probability question marked: draw the tree, show every path, and see the marks awarded

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

In simple terms

Explore the concept

Key formulas

Full topic notes

Sample space, probability and the complement

Combined events, the addition rule and mutual exclusivity

Venn diagrams: classifying a population

Independent events

Tree diagrams for successive events

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Get a Paper 2 probability question marked: draw the tree, show every path, and see the marks awarded

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Probability, Venn diagrams and tree diagrams