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Maths · Probability

Conditional probability

Work out probabilities when one event changes the next, such as taking items without replacement, and use two-way tables to find conditional probabilities.

  • Higher
  • 6 key terms
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Teacher resources

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Student handouts

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Warm-up

Answer each one, then check.

  1. 1

    Work out \(\dfrac{5}{8} \times \dfrac{4}{7}\).

    Show answerHide answer

    \(\dfrac{5}{14}\)

  2. 2

    Simplify \(\dfrac{30}{56}\).

    Show answerHide answer

    \(\dfrac{15}{28}\)

  3. 3

    Work out \(1 - \dfrac{5}{14}\).

    Show answerHide answer

    \(\dfrac{9}{14}\)

  4. 4

    What does "independent" mean?

    Show answerHide answer

    One event does not affect the other

  5. 5

    What does "with replacement" mean?

    Show answerHide answer

    The item is put back before the next pick

Learning Objectives

  1. 1Explain what conditional probability means.
  2. 2Draw a tree diagram for events without replacement.
  3. 3Work out probabilities of combined events without replacement.
  4. 4Find a conditional probability from a two-way table.

CONDITIONAL PROBABILITY

The probability of an event happening, given that another event has already happened, is a conditional probability.

Without replacement, the numbers change after each pick.

Without Replacement

A bag has 5 red and 3 blue counters. Two counters are taken at random without replacement. Find the probability that both are red.

Show the solutionHide the solution
  1. 1 First counter red \(\dfrac{5}{8}\)
  2. 2 Now 4 red among 7 remaining \(\dfrac{4}{7}\)
  3. 3 Multiply \(\dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56}\)
  4. 4 Simplify \(\dfrac{5}{14}\)

Answer\(\dfrac{5}{14}\)

One of Each Colour

For the same bag, find the probability that the two counters are different colours.

Show the solutionHide the solution
  1. 1 Red then blue \(\dfrac{5}{8} \times \dfrac{3}{7} = \dfrac{15}{56}\)
  2. 2 Blue then red \(\dfrac{3}{8} \times \dfrac{5}{7} = \dfrac{15}{56}\)
  3. 3 Add \(\dfrac{30}{56}\)
  4. 4 Simplify \(\dfrac{15}{28}\)

Answer\(\dfrac{15}{28}\)

At Least One Red

A bag has 4 red and 6 blue counters. Two are taken without replacement. Find the probability that at least one is red.

Show the solutionHide the solution
  1. 1 At least one red is the opposite of both blue Use \(1 - P(\text{both blue})\)
  2. 2 Both blue \(\dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}\)
  3. 3 Subtract from 1 \(1 - \dfrac{1}{3} = \dfrac{2}{3}\)

Answer\(\dfrac{2}{3}\)

Two-Way Tables

Conditional probabilities can be read from a table by restricting to a group.

  • The idea

    For "given that", use only the people in that group as the total.

  • Example

    60 students: 35 like football, and 20 of those are boys. Given that a student likes football, the probability that they are a boy is \(\dfrac{20}{35}\).

  • Simplify

    \(\dfrac{20}{35} = \dfrac{4}{7}\).

A Two-Way Table

60 students were asked whether they like football.

  • Boys

    Does not like football: 20. Total: 10. 30

  • Girls

    Does not like football: 15. Total: 15. 30

  • Total

    Does not like football: 35. Total: 25. 60

Given That

Use the table. A student is chosen at random. Find the probability that they are a boy given that they like football.

Show the solutionHide the solution
  1. 1 Restrict to students who like football 35 students
  2. 2 Boys among them 20
  3. 3 Probability \(\dfrac{20}{35} = \dfrac{4}{7}\)

Answer\(\dfrac{4}{7}\)

Cards Without Replacement

Three cards are taken from a pack of ten cards numbered 1 to 10, one after another without replacement. Find the probability that the first is even and the second is odd, and that the first two are both greater than 7.

1. Reduce the numbers after each pick.

2. Multiply along the path.

A good answer shows: First even and second odd: \(\dfrac{5}{10} \times \dfrac{5}{9} = \dfrac{25}{90} = \dfrac{5}{18}\). Both greater than 7: \(\dfrac{3}{10} \times \dfrac{2}{9} = \dfrac{6}{90} = \dfrac{1}{15}\).

Can I...?

  1. 1Explain what conditional means.
  2. 2Adjust the numbers after a pick.
  3. 3Draw a tree diagram without replacement.
  4. 4Multiply along the branches.
  5. 5Add paths for combined outcomes.
  6. 6Use "1 minus" for at least one.
  7. 7Find a conditional probability from a table.
  8. 8Simplify my fractions.

Summary & Exam Focus

  • Without replacement, probabilities on the second branches change.
  • Multiply along the branches; add different paths.
  • "Given that" means restrict to that group.
  • At least one \(= 1 -\) none.

Exam focus

A bag has 5 red and 3 blue counters. Two counters are taken at random without replacement. Work out the probability that the counters are different colours. (3 marks) (3 marks)

Both orders count: red then blue AND blue then red. Reduce the total by 1 for the second pick.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Conditional probability
The probability of an event given that another has happened.
Without replacement
The item is not put back, so later probabilities change.
Dependent events
Events where one affects the probability of the other.
Tree diagram
A diagram showing outcomes and probabilities on branches.
Two-way table
A table showing two categories at once.
Complement
The event "not A".

Questions and answers

12 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Non-calculator 3 marks Easier

A bag contains 5 red counters and 3 blue counters. Two counters are taken at random without replacement. Work out the probability that both counters are red.

Mark scheme — 3 marks available

  • \(\dfrac{5}{8}\) — M1
  • \(\dfrac{5}{8} \times \dfrac{4}{7}\) — M1
  • \(\dfrac{5}{14}\) — A1

Model answer

\(\dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}\).

2. Exam question Non-calculator 3 marks Easier

For the same bag, work out the probability that the two counters are different colours.

Mark scheme — 3 marks available

  • One correct product — M1
  • Adding both orders — M1
  • \(\dfrac{15}{28}\) — A1

Model answer

\(\dfrac{5}{8} \times \dfrac{3}{7} + \dfrac{3}{8} \times \dfrac{5}{7} = \dfrac{15}{56} + \dfrac{15}{56} = \dfrac{15}{28}\).

3. Exam question Non-calculator 3 marks Easier

A bag contains 4 red counters and 6 blue counters. Two counters are taken at random without replacement. Complete the tree diagram.

A tree diagram for two counters taken without replacement from 4 red and 6 blue, with the first-stage probabilities given and the second stage blank.

Mark scheme — 3 marks available

  • Denominator 9 on all four branches — B1
  • After red: 3/9, 6/9 — B1
  • After blue: 4/9, 5/9 — B1

Model answer

After red: red 3/9 and blue 6/9. After blue: red 4/9 and blue 5/9.

4. Exam question Non-calculator 3 marks Easier

Using the tree diagram, work out the probability that both counters are blue.

Mark scheme — 3 marks available

  • \(\dfrac{6}{10} \times \dfrac{5}{9}\) — M1
  • \(\dfrac{30}{90}\) — A1
  • \(\dfrac{1}{3}\) — A1

Model answer

\(\dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}\).

5. Exam question Non-calculator 3 marks Easier

Using the tree diagram, work out the probability that at least one of the counters is red.

Mark scheme — 3 marks available

  • \(1 - P(\text{both blue})\) — M1
  • \(1 - \dfrac{1}{3}\) — M1
  • \(\dfrac{2}{3}\) — A1

Model answer

\(1 - P(\text{both blue}) = 1 - \dfrac{1}{3} = \dfrac{2}{3}\).

6. Exam question Non-calculator 3 marks Easier

60 students were asked whether they like football. 35 like football, and 20 of these are boys. 30 of the 60 students are boys. A student is chosen at random. Given that the student likes football, work out the probability that they are a boy.

Mark scheme — 3 marks available

  • 35 as the total — M1
  • \(\dfrac{20}{35}\) — A1
  • \(\dfrac{4}{7}\) — A1

Model answer

Of the 35 who like football, 20 are boys. The probability is \(\dfrac{20}{35} = \dfrac{4}{7}\).

7. Multiple choice 1 mark Easier

A bag has 4 red and 6 blue counters. One is taken and not replaced. How many counters are left?

  1. A 10
  2. B 9 Correct
  3. C 8
  4. D 4

Why: \(10 - 1 = 9\).

8. Multiple choice 1 mark Core

A bag has 5 red and 3 blue counters. A red is taken and not replaced. What is P(second is red)?

  1. A \(\dfrac{5}{8}\)
  2. B \(\dfrac{5}{7}\)
  3. C \(\dfrac{4}{7}\) Correct
  4. D \(\dfrac{4}{8}\)

Why: 4 red remain out of 7: \(\dfrac{4}{7}\).

9. Multiple choice 1 mark Core

Without replacement, the events are...

  1. A Independent
  2. B Impossible
  3. C Certain
  4. D Dependent Correct

Why: The first pick changes what is left, so they are dependent.

10. Multiple choice 1 mark Core

"The probability that a student is a girl, given that they play tennis" means you should...

  1. A Look only at the tennis players Correct
  2. B Look at all students
  3. C Look only at the girls
  4. D Add the tennis and girls totals

Why: Use only the tennis players as the total.

11. Multiple choice 1 mark Core

A bag has 3 red and 2 blue counters. Two are taken without replacement. What is P(both red)?

  1. A \(\dfrac{9}{25}\)
  2. B \(\dfrac{3}{10}\) Correct
  3. C \(\dfrac{6}{25}\)
  4. D \(\dfrac{1}{2}\)

Why: \(\dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} = \dfrac{3}{10}\).

12. Multiple choice 1 mark Stretch

A bag has 2 red and 2 blue counters. Two are taken without replacement. What is P(different colours)?

  1. A \(\dfrac{1}{2}\)
  2. B \(\dfrac{1}{3}\)
  3. C \(\dfrac{2}{3}\) Correct
  4. D \(\dfrac{3}{4}\)

Why: \(\dfrac{2}{4} \times \dfrac{2}{3} + \dfrac{2}{4} \times \dfrac{2}{3} = \dfrac{1}{3} + \dfrac{1}{3} = \dfrac{2}{3}\).