Maths · Probability
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Teacher view: every answer and mark scheme set out in full.
Tree Diagrams
Combining independent and dependent events with tree diagrams, with and without replacement.
Learning Objectives
- 1Draw and complete a tree diagram for two or more events.
- 2Multiply along the branches for "and", and add the outcomes for "or".
- 3Solve problems with independent events, and with dependent events such as picking without replacement.
- 4Use "at least one" as 1 minus the probability of none.
Tracking two events
When two or more things happen one after the other, a tree diagram shows every possible path, with the probability written on each branch. It stops you missing outcomes, and it gives you a method that always works: multiply the probabilities along a path to get the chance of that path, and add the paths that match what you want. Tree diagram questions are among the most common on a non-calculator paper, usually with fractions that cancel nicely.
Independent events
Two events are independent if the first does not change the probability of the second, such as two throws of a coin, or picking a counter and then putting it back.
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Same probabilities on each branch
The second set of branches has the same probabilities as the first.
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The branches from one point add up to 1
\(0.3 + 0.7 = 1\).
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And means multiply
The probability of rain on both days is \(0.3 \times 0.3 = 0.09\).
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Or means add
Add the end results that match what the question wants.
A tree diagram for two independent events
The four outcomes at the end cover everything that can happen, so their probabilities add up to 1: \(0.09 + 0.21 + 0.21 + 0.49 = 1\). That is a good check.
Using the tree
- Rain on both days \(0.09\).
- Rain on exactly one day Add the two paths: \(0.21 + 0.21 = 0.42\).
- No rain on either day \(0.49\).
- Rain on at least one day \(1 - 0.49 = 0.51\), which is quicker than adding three paths.
At least one
The probability that it rains on any day is 0.3. Assuming the days are independent, work out the probability that it rains on at least one of two days.
Show the solutionHide the solution
- 1 Find the probability of the opposite The opposite of "at least one" is "none": no rain on both days.
- 2 Multiply \(0.7 \times 0.7 = 0.49\).
- 3 Subtract from 1 \(1 - 0.49 = 0.51\).
- 4 Check with the paths \(0.09 + 0.21 + 0.21 = 0.51\).
Answer0.51
Dependent events
When an item is taken and not replaced, the numbers change for the second pick, so the second probabilities depend on the first.
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Reduce the total
After one counter is removed, there is one fewer counter in the bag.
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Reduce the right colour
If the first counter was red, there is one fewer red counter too.
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Branches still add to 1
\(\dfrac{2}{4} + \dfrac{2}{4} = 1\).
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Take care
The second probabilities on each branch are different, not the same as the first.
A tree diagram without replacement
A bag has 3 red and 2 blue counters, and two are taken without replacement. After a red is taken, 4 counters are left, and 2 of them are red. After a blue is taken, 4 counters are left, and 3 of them are red.
Reading the tree
- Two reds \(\dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} = \dfrac{3}{10}\).
- Same colour \(\dfrac{6}{20} + \dfrac{2}{20} = \dfrac{8}{20} = \dfrac{2}{5}\).
- Different colours \(\dfrac{6}{20} + \dfrac{6}{20} = \dfrac{12}{20} = \dfrac{3}{5}\).
- With replacement Two reds would be \(\dfrac{3}{5} \times \dfrac{3}{5} = \dfrac{9}{25}\), a different answer.
Without replacement
A bag contains 3 red counters and 2 blue counters. Two counters are taken at random without replacement. Work out the probability that they are different colours.
Show the solutionHide the solution
- 1 Red then blue \(\dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20}\).
- 2 Blue then red \(\dfrac{2}{5} \times \dfrac{3}{4} = \dfrac{6}{20}\).
- 3 Add the two paths \(\dfrac{6}{20} + \dfrac{6}{20} = \dfrac{12}{20}\).
- 4 Simplify \(\dfrac{3}{5}\).
Answer\(\dfrac{3}{5}\)
Longer trees and finding a missing probability
The same rules work for three events, and for problems where you must find a branch.
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Three events
Multiply three branches. For three coin tosses, each path has probability \(\dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}\).
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Missing branch
The probabilities leaving a point add up to 1. If one branch is 0.6, the other is 0.4.
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Missing number of counters
If \(P(\text{two reds}) = \dfrac{2}{15}\) with 4 red counters in a bag of \(n\), the equation \(\dfrac{4}{n} \times \dfrac{3}{n - 1} = \dfrac{2}{15}\) can be solved (Higher tier).
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Check the end
The outcome probabilities should add up to 1.
Test yourself
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1
What do you do along the branches of a tree diagram?
Show answerHide answer
Multiply the probabilities.
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2
What do you do to combine different paths that give the outcome you want?
Show answerHide answer
Add their probabilities.
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3
What is \(P(\text{at least one})\)?
Show answerHide answer
\(1 - P(\text{none})\).
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4
What changes for the second pick when there is no replacement?
Show answerHide answer
The total number, and the number of the colour picked first.
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5
Two coins are tossed. What is the probability of two heads?
Show answerHide answer
\(\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\).
Exam technique: tree diagrams
Write on every branch, because the working is where the marks are.
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Label every branch
Write the outcome and its probability on each.
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Check the branches
Each pair leaving a point should add to 1.
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Show the multiplication
Write \(\dfrac{3}{5} \times \dfrac{2}{4}\) before simplifying.
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Use the opposite
For "at least one", subtracting from 1 is nearly always quicker.
Summary and exam focus
- A tree diagram shows all the outcomes, with probabilities written on the branches.
- Multiply along a path for "and", and add paths for "or".
- Without replacement, the numbers change on the second set of branches.
- \(P(\text{at least one}) = 1 - P(\text{none})\).
Exam focus
A bag contains 4 red counters and 6 blue counters. Two counters are taken at random without replacement. Work out the probability that both counters are blue. (2 marks) (2 marks)
The first counter is blue with probability \(\dfrac{6}{10}\), and then there are 5 blue counters among 9 left, so \(\dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}\). Show the two fractions before multiplying.
Key terms
The words this lesson expects you to use. Each one is linked from the first place it appears above.
- Tree diagram
- A diagram with branches that shows the outcomes of events in order and their probabilities.
- Branch
- A line on a tree diagram showing one possible outcome and its probability.
- Independent events
- Events where one does not affect the probability of the other.
- Dependent events
- Events where the first changes the probability of the second.
- Replacement
- Putting an item back before the next pick.
- Without replacement
- Not putting an item back, so the next pick has fewer items.
- At least one
- One or more, found as 1 minus the probability of none.
- Outcome probability
- The probability of one complete path through the tree.
- Product
- The result of multiplying.
Questions and answers
15 questions set on this lesson, with the mark schemes and model answers open.
A spinner lands on red with probability 0.4. It is spun twice. Find the probability that it lands on red both times. [2 marks]
Mark scheme — 2 marks available
- \(0.4 \times 0.4\) — M1
- \(0.16\) — A1
Model answer
\(0.4 \times 0.4 = 0.16\).
Zara plays two games of chess against Ahmed. The probability that Zara wins a game is \(\dfrac{2}{3}\). The games are independent. The incomplete tree diagram shows the first game and the second game. (a) Complete the tree diagram. [2 marks] (b) Calculate the probability that Zara wins exactly one of the two games. [2 marks]
Mark scheme — 4 marks available
- (a) \(\dfrac{1}{3}\) on the first lose branch — B1
- (a) \(\dfrac{2}{3}\) and \(\dfrac{1}{3}\) on all second-game branches — B1
- (b) \(\dfrac{2}{3} \times \dfrac{1}{3}\) and \(\dfrac{1}{3} \times \dfrac{2}{3}\) added — M1
- (b) \(\dfrac{4}{9}\) — A1
Model answer
(a) The probability of losing is \(1 - \dfrac{2}{3} = \dfrac{1}{3}\). Every second-game branch is \(\dfrac{2}{3}\) for win and \(\dfrac{1}{3}\) for lose. (b) \(\dfrac{2}{3} \times \dfrac{1}{3} + \dfrac{1}{3} \times \dfrac{2}{3} = \dfrac{4}{9}\).
A bag contains 4 lemon sweets and 3 mint sweets. Two sweets are taken at random without replacement. Calculate the probability that the two sweets have different flavours. [3 marks]
Mark scheme — 3 marks available
- \(\dfrac{4}{7} \times \dfrac{3}{6}\) or \(\dfrac{3}{7} \times \dfrac{4}{6}\) — M1
- Both products added — M1
- \(\dfrac{4}{7}\) or \(\dfrac{24}{42}\) — A1
Model answer
Lemon then mint: \(\dfrac{4}{7} \times \dfrac{3}{6} = \dfrac{12}{42}\). Mint then lemon: \(\dfrac{3}{7} \times \dfrac{4}{6} = \dfrac{12}{42}\). The total is \(\dfrac{24}{42} = \dfrac{4}{7}\).
Two machines work independently. The probability that machine \(A\) works is 0.9 and the probability that machine \(B\) works is 0.8. Calculate the probability that at least one of the machines works. [3 marks]
Mark scheme — 3 marks available
- \(0.1\) and \(0.2\) seen — M1
- \(0.1 \times 0.2\) — M1
- \(0.98\) — A1
Model answer
The probability that neither works is \(0.1 \times 0.2 = 0.02\), so the probability that at least one works is \(1 - 0.02 = 0.98\).
A bag contains 5 blue counters and 3 green counters. Two counters are taken at random without replacement. (a) Show that the probability that both counters are blue is \(\dfrac{5}{14}\). [2 marks] (b) Calculate the probability that at least one counter is green. [2 marks]
Mark scheme — 4 marks available
- (a) \(\dfrac{5}{8} \times \dfrac{4}{7}\) — M1
- (a) \(\dfrac{20}{56} = \dfrac{5}{14}\) shown — A1
- (b) \(1 - \dfrac{5}{14}\) — M1
- (b) \(\dfrac{9}{14}\) — A1
Model answer
(a) \(\dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}\). (b) \(1 - \dfrac{5}{14} = \dfrac{9}{14}\).
Three cards are numbered 1, 2 and 3. Two cards are taken at random without replacement, and the two numbers are added. Calculate the probability that the total is even. [3 marks]
Mark scheme — 3 marks available
- Lists the three possible pairs or totals — M1
- Totals 3, 4 and 5 — A1
- \(\dfrac{1}{3}\) — A1
Model answer
The possible pairs are 1 and 2, 1 and 3, and 2 and 3, which are equally likely. The totals are 3, 4 and 5, so only one total is even. The probability is \(\dfrac{1}{3}\).
The probability of rain on any day is 0.3, and days are independent. What is the probability of rain on both of two days?
Why: \(0.3 \times 0.3 = 0.09\).
The probability of rain on any day is 0.3, and days are independent. What is the probability of rain on at least one of two days?
Why: \(1 - 0.7 \times 0.7 = 1 - 0.49 = 0.51\).
Two fair coins are tossed. What is the probability of two heads?
Why: \(\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\).
A bag has 3 red and 2 blue counters. Two are taken without replacement. What is the probability that both are red?
Why: \(\dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} = \dfrac{3}{10}\).
A bag has 3 red and 2 blue counters. Two are taken without replacement. What is the probability that they are different colours?
Why: \(\dfrac{3}{5} \times \dfrac{2}{4} + \dfrac{2}{5} \times \dfrac{3}{4} = \dfrac{12}{20} = \dfrac{3}{5}\).
A bag has 3 red and 2 blue counters. Two are taken without replacement. What is the probability that they are the same colour?
Why: \(\dfrac{6}{20} + \dfrac{2}{20} = \dfrac{8}{20} = \dfrac{2}{5}\).
On a tree diagram, one branch has probability 0.6. What is the other branch from the same point?
Why: The branches from one point add up to 1.
A bag has 4 red and 6 blue counters. Two are taken without replacement. What is the probability that both are blue?
Why: \(\dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}\).
A bag has 4 red counters and some blue counters, \(n\) in all. Two are taken without replacement and \(P(\text{two reds}) = \dfrac{2}{15}\). How many counters are in the bag?
Why: \(\dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}\), so \(n = 10\).