Free Year 11 Probability: Tree Diagrams and Conditional Probability
An HSC/VCE-style study guide to multi-stage probability: the addition and multiplication rules, tree diagrams with and without replacement, conditional probability and testing for independence. Worked examples include a two-draw bag problem, a two-way class survey, an independence check and a factory defect problem where the tree runs backwards, with a six-question self-check and answers.
Instant download · No sign-up · Free for personal, classroom and homeschool use
- Year
- Year 11
- Subject
- Mathematics
- Topic
- Probability
- Difficulty
- Level 2 · Standard
- Estimated time
- 25 minutes
- Curriculum
- Australian Curriculum
- Answers
- Not applicable
- Format
- PDF (A4) + print
Students will practise
- applying the addition rule and the complement to find probabilities
- drawing tree diagrams for events with and without replacement and multiplying along branches
- calculating conditional probabilities using P(A | B) = P(A ∩ B) ÷ P(B)
- testing whether two events are independent
Curriculum: Australian Curriculum. We show specific outcome codes only where they have been verified against the official curriculum document.
What's next?
Completed: Probability: Tree Diagrams and Conditional Probability
Ready for more? Move on to Year 12 Mathematics.
How to use this study guide
- Read it together first, pausing at each worked example to try the step before reading the answer.
- Attempt the "Check yourself" questions at the end without looking back.
- Then practise with a worksheet from the pathway above and finish with the topic test.
Common questions
Who is this study guide for?
Year 11 students (typically ages 16–17) working on probability. It is pitched at level 2 · standard.
Are the answers included?
This is a study guide, so there is no separate answer sheet; the 'Check yourself' questions include answers.
How long does it take?
About 25 minutes. Short, regular sessions work best: two or three a week beats one long one.
Do I need to sign up to download?
No. Click Download Free PDF and it opens immediately. It is free for personal, classroom and homeschool use.
What should we do next?
Explore more Year 11 mathematics resources.
Preview
Year 11 · Mathematics · Probability
Probability: Tree Diagrams and Conditional Probability
Student name:
Date:
What you need to know
For equally likely outcomes, P(A) = [frac number of favourable outcomes/total number of outcomes], and the complement rule says P(not A) = P(A′) = 1 − P(A). The addition rule for "A or B" is P(A ∪ B) = P(A) + P(B) − P(A ∩ B); the overlap is subtracted so it is not counted twice. If A and B cannot both happen (mutually exclusive), P(A ∩ B) = 0.
For two stages, draw a tree diagram: multiply along a path to find the probability of that sequence, and add across the paths that match the question. Events are independent when one does not change the probability of the other, and then P(A ∩ B) = P(A) × P(B). When one event does change the other (for example drawing without replacement), use conditional probability: P(B | A) means "the probability of B given that A has happened".
| Rule | Formula | When to use it |
|---|---|---|
| Complement | P(A′) = 1 − P(A) | "at least one" questions: 1 − P(none) |
| Addition | P(A ∪ B) = P(A) + P(B) − P(A ∩ B) | A or B (or both) |
| Multiplication (general) | P(A ∩ B) = P(A) × P(B | A) | A and then B — the second branch of a tree |
| Independence | P(A ∩ B) = P(A) × P(B) | only when A and B are independent (test it!) |
| Conditional | P(A | B) = | "given that", "of those who…", "if it is known that…" |
A bag holds 3 red and 2 blue marbles. Two are drawn without replacement.
- First draw: P(R) = , P(B) = . Second draw has only 4 marbles left, and the colours depend on the first draw.
- P(both red) = × = = [frac 3/10]
- P(one of each) = P(RB) + P(BR) = × + × = + = = [frac 3/5]
- P(at least one blue) = 1 − P(both red) = 1 − = [frac 7/10]
- With replacement instead, every branch stays or , so P(both red) would be × = .
In a class of 30, 18 study Chemistry, 12 study Physics and 8 study both.
- Chemistry only: 18 − 8 = 10. Physics only: 12 − 8 = 4. Neither: 30 − (10 + 8 + 4) = 8.
- P(Chemistry or Physics) = = = [frac 11/15]
- P(neither) = = [frac 4/15]
- P(Physics | Chemistry) = = = [frac 4/9]. (Of the 18 Chemistry students, 8 also do Physics — the condition shrinks the total to 18.)
P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2. Are A and B independent?
- Test: P(A) × P(B) = 0.4 × 0.5 = 0.2
- This equals P(A ∩ B) = 0.2, so A and B are independent.
- Equivalently P(A | B) = = 0.4 = P(A): knowing B happened does not change the chance of A.
Machine A makes 60% of a factory's parts and 2% of them are faulty. Machine B makes the other 40% and 5% are faulty.
- Tree: first branch is the machine (0.6 / 0.4), second branch is faulty or not (0.02 / 0.98 after A, 0.05 / 0.95 after B).
- P(faulty) = 0.6 × 0.02 + 0.4 × 0.05 = 0.012 + 0.020 = 0.032
- A faulty part is found. P(it came from A | faulty) = = = 0.375 (= )
- Even though A makes most parts, less than half of the faulty parts come from A because its fault rate is lower.
Check yourself
- P(A) = 0.3. Find P(A′).
- A fair die is rolled twice. Find the probability of two sixes.
- A bag holds 4 green and 6 yellow counters. Two are drawn without replacement. Find P(both green).
- P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.1. Find P(A ∪ B).
- Using the values in question 4, find P(A | B).
- Using the values in question 4, are A and B independent? Explain.
Answers: 1. 0.7 2. × = 3. × = = 4. 0.5 + 0.4 − 0.1 = 0.8 5. = 0.25 6. No: P(A) × P(B) = 0.2, which is not equal to P(A ∩ B) = 0.1
Preview — the PDF contains the full resource.
About this resource
- Created by
- Success Tutoring
- Last reviewed
- 1 October 2026
- How it was made
- Written by our team
Free to print and share for personal, classroom and homeschool use. Please don't resell. Spotted an error? Let us know.