With replacement: red, red
Bag: 3 red, 2 blue. Take one, put it back, take another.
P(red, red) = 3/5 × 3/5 = 9/25
Tree diagrams When do you multiply probabilities, and when do you add them?
Foundation and Higher · AQA · 45 minutes
Students test the idea that two rolls give twice the chance, then count it on a grid of outcomes. Trees follow: multiply along a path, add the paths you want, change the branches when nothing is put back, and use 1 − P(none) for at least one. Slides, a vote that shows who's stuck, practice and homework.
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Roll a fair dice with k faces x times. The curve is the real chance of at least one 1. The straight line adds 1/k for every roll, and the flat line is 1: certain. Move k. Where does adding go wrong? The numbers below are the curve's chances for 0, 1, 2 … rolls.
At x = 0, 1, 2 …: 0, 0.17, 0.31, 0.42, 0.52, 0.6, 0.67, 0.72
The idea in one minute
A tree diagram lists the outcomes of two or more events in order, with a probability on each branch. Along a path the events all happen (this and then that), so you multiply. When several paths answer the question (this or that), you add their results. The branches from one point always add up to 1.
The lesson starts with a grid of the 36 outcomes of two dice, so students see that adding 1/6 + 1/6 counts one outcome twice. The tree is then the same outcomes, organised. Students read each path aloud before they multiply, change the second branches when an item is not put back, and find at least one as 1 − P(none). Independent events are sorted from those that are not. The stretch, for Higher, is a conditional probability from a tree.
By the end, students can
- list outcomes in a sample space grid and find probabilities from it
- draw and complete a tree diagram for two events
- multiply along a path and add the paths a question needs
- change the branches when items are not replaced
- find the probability of at least one as 1 − P(none)
Key words
- outcome
- sample space
- tree diagram
- independent events
- with and without replacement
- at least one
- conditional probability (Higher)
Before this lesson
Students should multiply and add fractions and decimals, know that probabilities go from 0 to 1, and find a probability for one event.
The lesson, step by step
45 minutes in total. The stretch is optional.
5 min 1 · Hook · students think
Twice the chance?
Is a six on at least one of two dice 1/6 + 1/6? Students count on a grid of outcomes.
Say“Roll two dice. The chance of a six on one dice is 1/6. So the chance of at least one six is 1/6 + 1/6. True or false? Vote, then count.”
Ask- With six dice, adding gives 6/6. Is a six then certain?
- Which dot on the grid did adding count twice?
Watch forCounting the double six twice: count dots, not rows.
For the teacher
The grid has a dot for each of the 36 outcomes. Eleven have a six, so the chance is 11/36, a little less than 12/36.
Adding counted the double six twice, once for each dice. With more rolls, adding goes past 1, which no probability can. We need a better way: the tree.
Vote first (Ask the class): true or false. Then count the dots with a six together.
Shorter: Show the grid with its 11 dots marked; skip the vote (−2 min).
12 min 2 · Explain · you explain
Branches, paths and putting back
A tree for a bag with replacement, then without; multiply along, add across, and 1 − P(none).
Say“Along a path, this AND that: multiply. Several paths, this OR that: add.”
Ask- Read the top path aloud. What happened first, and then?
- After a red is taken out and kept, how many reds are left? How many counters?
- Which path is 'no red at all'?
Watch forWithout replacement: the top of the fraction drops on the wrong branch, or the bottom doesn't drop.
Add the paths: exactly one red
Red then blue, or blue then red
3/5 × 2/5 + 2/5 × 3/5 = 12/25
Without replacement: red, red
After a red: 2 red and 2 blue left
P(red, red) = 3/5 × 2/4 = 6/20 = 3/10
At least one red, with replacement
Every path but blue, blue
1 − 2/5 × 2/5 = 1 − 4/25 = 21/25
For the teacher
First slide: the tree for a bag of 3 red and 2 blue, the counter put back. Up is red, down is blue, the same at both stages because nothing changed. Each pair of branches adds to 1. The top path is red and then red: 3/5 × 3/5 = 9/25.
Second slide: exactly one red is two paths, red then blue or blue then red. Work each path out, then add: 6/25 + 6/25 = 12/25. All four paths add up to 25/25 = 1.
Third slide: the same bag, the counter kept. After a red there are 2 red and 2 blue, so 2/4 and 2/4; after a blue, 3 red and 1 blue, so 3/4 and 1/4. The second draw depends on the first: the events are not independent.
Last slide: at least one red is every path except blue then blue. One minus that path is quicker than adding three paths, and it can never pass 1.
Have students read the highlighted path aloud on each slide before anyone multiplies.
Shorter: Skip exactly one red; do at least one on the without-replacement tree only (−4 min).
8 min 3 · Your turn · students write
Worked, then yours
Each worked example beside a twin that students solve themselves.
Say“Watch me do one. Then you do its twin, which is almost the same.”
Ask- Is anything put back? Does the first result change the second?
Watch forAt least one: adding the two chances, 0.2 + 0.2.
Independent events
A spinner lands on green with probability 0.3. Spin twice.
P(green, green) = 0.3 × 0.3 = 0.09
Then theirs
A team wins each game with probability 0.4, independently. P(it wins both of two games)?
Without replacement
A bag has 4 red and 3 blue counters. Two are taken, not put back.
P(both red) = 4/7 × 3/6 = 12/42 = 2/7
Then theirs
A bag has 5 green and 3 yellow sweets. Two are eaten. P(both yellow)?
At least one
It rains on a day with probability 0.2, independently. Two days.
P(rain on at least one) = 1 − 0.8 × 0.8 = 0.36
Then theirs
A bus is late with probability 0.1 each day, independently. P(late at least once in two days)?
For the teacher
Work the first example aloud, writing each line. Then students solve its twin while you watch who sketches a tree first.
Do the same for each pair. The sweets twin catches anyone who keeps the 8 on the second branch; the bus twin catches anyone who adds 0.1 + 0.1.
Open the answers slide once most have finished their twins.
Shorter: Do the first two pairs only (−3 min).
15 min 4 · Who's stuck? · students vote
Vote, then practise
One question shows who's stuck; then practice in levels, and a stretch.
Say“Choose an answer on your own. Nobody talks until everyone has chosen.”
Ask- After the first red is taken, how many counters are left in the bag, and how many are red?
Watch forKeeping 10 as the second denominator, or adding along the path.
Sketch a tree or a grid for each question. Give answers as fractions unless the question says otherwise. The stretch is optional.
The question that shows who's stuck
- A bag has 4 red and 6 blue counters. Two are taken out, one after the other, without replacement. What is the probability that both are red?
In your Classroom, as they answer (an example) - A5
- B8
- C4
- D9
Each wrong answer is a different mistake. While they answer, your Classroom shows who chose which, so you know in seconds whom to help with what.
- A fair spinner has 8 equal parts, and 3 of them are red. Drag the marker to P(red) on the probability scale.They drag a marker on a number line from 0 to 1.
- Two fair coins are flipped. What is the probability of two heads? Give it as a fraction.
- Two fair dice are rolled and their scores added. What is the probability of a total of 7? Give it as a fraction.
- A bus is late with probability 0.3 each day, independently. What is the probability that it is late on Monday and not late on Tuesday? Give it as a decimal.
- Are the two events independent (the first doesn't change the chances of the second)? Sort each pair into its group.They sort 6 items into Independent and Not independent.
- Two independent events: P(A) = 0.6 and P(B) = 0.7. Sam says the probability that at least one of them happens is 0.6 + 0.7 = 1.3. What is it?
- A spinner lands on red with probability x, blue with probability 2x and green with probability 0.4. Solve the equation for x, one line at a time.They solve x + 2x + 0.4 = 1 line by line, each line checked as they write it.
- A bag has 5 red and 3 green sweets. Two are eaten, one after the other. What is the probability that they are different colours? Give it as a fraction.
- Stretch (Higher): a bag has 4 red and 6 blue counters. Two are taken without replacement. The second one is red. What is the probability that the first was red too? Give it as a fraction.
Support · Sketch the tree. On the first branch, write red over all counters. Take one red out: on the second branch, one fewer red and one fewer counter. Then multiply along the path.
For the teacher
Everyone chooses before anyone explains. Ask a student for each answer to say how they got it, then ask the class what is in the bag before the second draw.
Questions 1–3 build fluency (one event, two coins, a grid of two dice); 4–5 change the look; 6–7 ask for reasons; 8 is exam-style. The stretch is a conditional probability (Higher).
Students show their working on paper or the board; the quiz keeps their final answers.
If many chose a wrong answer, take two counters out of a real bag, one at a time, and count what is left after each.
Shorter: After the vote, questions 1–4 only (−6 min).
Longer: The stretch, then: why does knowing the second counter change the chances for the first? (+10 min).
5 min 5 · Exit · students show
Place it, then explain
Place a probability on the scale, then say why adding goes wrong.
Say“Flip a fair coin twice. Put the chance of at least one head on the scale. Then answer the second question.”
Watch forPlacing it at 1, from 1/2 + 1/2: that would make a head certain.
Place the probability, then one or two sentences.
- A fair coin is flipped twice. Drag the marker to the probability of at least one head on the probability scale.They drag a marker on a number line from 0 to 1.
- Explain why 1/6 + 1/6 is not the probability of at least one six when two fair dice are rolled.
For the teacher
Place the probability of at least one head in two flips of a fair coin on the probability scale.
Then explain in a sentence or two why 1/6 + 1/6 is not the chance of at least one six in two rolls.
The explanations show who sees why adding fails: start the next lesson with two of them. Set the homework.
Longer: How many rolls until at least one six is more likely than not? (+5 min).
One to one: ask the student to read every path aloud as a sentence (red, and then blue) before writing any number, and to check that each pair of branches and all the end results add up to 1.
Homework: multiply along, add across
Draw a tree or a grid for each question, and write the probability on every branch. Give answers as fractions or decimals, as the question has them.
- A fair coin is flipped and a fair dice is rolled. List all the outcomes. What is the probability of a head and a 5?
- Each of two tests is passed with probability 0.8, independently. Draw the tree. Find P(passes both) and P(passes exactly one).
- A box has 6 red and 4 white beads. Two are taken without replacement. Find P(both white).
- For the same box, find P(at least one red).
- A fair dice is rolled three times. Find the probability of at least one six.
- Jo says rolling a 6 and then a 6 is less likely than rolling a 1 and then a 6. Is she right? Explain.
Print the worksheet with working space and homework. Teacher answers come with your Library copy.
Teaching tree diagrams: common questions
When do you multiply and when do you add on a tree diagram?
Multiply along a path: the path is one event and then another, and both must happen. Add the results of different paths when any of them answers the question (red then blue, or blue then red). Reading each path aloud as a sentence before writing numbers helps students choose.
How do you teach without replacement?
Take counters out of a real bag, one at a time, and count what is left before the second draw: one fewer item in all, and one fewer of the colour taken. Students then write the second branches from the bag as it is, and check that each pair of branches adds to 1.
Why is 'at least one' 1 minus 'none'?
At least one covers every outcome except the one where it never happens. On a tree that is every path but one, so one minus that path is quicker than adding the rest. Adding the single chances instead counts the overlap twice and can even pass 1.
Is this tree diagrams lesson suitable for Foundation GCSE?
Yes. Sample spaces, tree diagrams for independent events and at least one are on both tiers, and simple without-replacement trees appear on Foundation papers too. The stretch, a conditional probability from a tree, is Higher.
Can I use the lesson without creating an account?
Yes. Read the lesson, print the worksheet and open its slides without signing in; nothing you draw there is kept. A free teacher account keeps the whole lesson in your Library to change and teach, with the quizzes, the answers and the answer slides.
Curriculum reference: AQA GCSE Mathematics 8300: P6 and P7 (listing outcomes and sample spaces), P8 (independent and dependent combined events, tree diagrams, when to add and when to multiply) and P4 (probabilities that add up to 1). The conditional stretch is P9, Higher.
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