OpenRevise GCSE

Viewing as

Student Teacher

Teaching this? The teacher view adds 2 files, the mark schemes and the model answers.

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
  • All boards
Download the full pack · 3 files

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".

Practice questions

Have a go at each one before you open its answer.

  1. Question 1 Non-calculator 3 marks

    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.

    Show answerHide answer

    Model answer

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

    Mark scheme

    • \(\dfrac{5}{8}\) — M1
    • \(\dfrac{5}{8} \times \dfrac{4}{7}\) — M1
    • \(\dfrac{5}{14}\) — A1
  2. Question 2 Non-calculator 3 marks

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

    Show answerHide answer

    Model answer

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

    Mark scheme

    • One correct product — M1
    • Adding both orders — M1
    • \(\dfrac{15}{28}\) — A1
  3. Question 3 Non-calculator 3 marks

    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.
    Show answerHide answer

    Model answer

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

    Mark scheme

    • Denominator 9 on all four branches — B1
    • After red: 3/9, 6/9 — B1
    • After blue: 4/9, 5/9 — B1
  4. Question 4 Non-calculator 3 marks

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

    Show answerHide answer

    Model answer

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

    Mark scheme

    • \(\dfrac{6}{10} \times \dfrac{5}{9}\) — M1
    • \(\dfrac{30}{90}\) — A1
    • \(\dfrac{1}{3}\) — A1
  5. Question 5 Non-calculator 3 marks

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

    Show answerHide answer

    Model answer

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

    Mark scheme

    • \(1 - P(\text{both blue})\) — M1
    • \(1 - \dfrac{1}{3}\) — M1
    • \(\dfrac{2}{3}\) — A1
  6. Question 6 Non-calculator 3 marks

    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.

    Show answerHide answer

    Model answer

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

    Mark scheme

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

Quick check

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

    1. A10
    2. B9
    3. C8
    4. D4
    Show answerHide answer

    B: 9

    \(10 - 1 = 9\).

  2. 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\(\dfrac{4}{7}\)
    4. D\(\dfrac{4}{8}\)
    Show answerHide answer

    C: \(\dfrac{4}{7}\)

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

  3. Without replacement, the events are...

    1. AIndependent
    2. BImpossible
    3. CCertain
    4. DDependent
    Show answerHide answer

    D: Dependent

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

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

    1. ALook only at the tennis players
    2. BLook at all students
    3. CLook only at the girls
    4. DAdd the tennis and girls totals
    Show answerHide answer

    A: Look only at the tennis players

    Use only the tennis players as the total.

  5. 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\(\dfrac{3}{10}\)
    3. C\(\dfrac{6}{25}\)
    4. D\(\dfrac{1}{2}\)
    Show answerHide answer

    B: \(\dfrac{3}{10}\)

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

  6. 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\(\dfrac{2}{3}\)
    4. D\(\dfrac{3}{4}\)
    Show answerHide answer

    C: \(\dfrac{2}{3}\)

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

Downloads

Free to keep, print and annotate.

Download all 3 as a zip
Something here looks wrong?

Tell us what and we will go and look. It goes to whoever writes these pages, nobody else, and we do not ask who you are — so there is nothing to sign and nothing comes back to you.