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Maths · Probability

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Conditional Probability

Finding probabilities given that something has already happened, from tables, Venn diagrams and tree diagrams.

  • Higher
  • 9 key terms
  • All boards

Learning Objectives

  1. 1Work out conditional probabilities from Venn diagrams, two-way tables and tree diagrams.
  2. 2Use \(P(A \text{ and } B) = P(A \mid B) \times P(B)\) (Higher tier).
  3. 3Decide whether two events are independent.
  4. 4Solve combined-event problems that need more than one method.

Probability when you already know something

A conditional probability is the probability of an event given that something else has already happened, or is known. "Given that" shrinks the group you are looking at, so the bottom of the fraction becomes the size of that smaller group, not the whole total. It is the hardest idea in this chapter, but each of the three tools you know, Venn diagrams, tables and trees, makes it a counting job once you have found the right group. These questions are Higher tier content, so the lesson uses the Higher tier formula as well.

The idea of "given that"

The words "given that" or "knowing that" tell you which group to look at.

  • Shrink the group

    If you know a student plays tennis, you only look at the tennis players.

  • Notation

    \(P(A \mid B)\) means the probability of \(A\) given that \(B\) has happened.

  • Counting

    \(P(A \mid B) = \dfrac{\text{number in both } A \text{ and } B}{\text{number in } B}\).

  • Not symmetric

    \(P(A \mid B)\) is usually different from \(P(B \mid A)\), because the groups are different.

Conditional probability from a table

100 people were asked how they travel to work. Of the 60 males, 36 drive, 14 cycle and 10 take the bus. Of the 40 females, 24 drive, 10 cycle and 6 take the bus. A person who drives is chosen at random. Work out the probability that the person is female.

Show the solutionHide the solution
  1. 1 Find the group "Chosen from those who drive" means \(36 + 24 = 60\) people.
  2. 2 Count the females in the group 24 of those who drive are female.
  3. 3 Write the probability \(\dfrac{24}{60}\).
  4. 4 Simplify \(\dfrac{2}{5}\).

Answer\(\dfrac{2}{5}\)

Conditional probability on trees

On a tree for events without replacement, the second branches are conditional probabilities.

  • Second-stage branches

    After a red counter is picked from 3 red and 2 blue, the probability of another red is \(\dfrac{2}{4}\), which is \(P(\text{red second} \mid \text{red first})\).

  • The multiplication rule

    \(P(A \text{ and } B) = P(A) \times P(B \mid A)\), which is what you do along the branches.

  • Given the second result

    If you are told the second counter was red, find all the paths that end in red and use their total as the bottom of the fraction.

  • Example

    Paths ending in red: RR \(= \dfrac{6}{20}\) and BR \(= \dfrac{6}{20}\), total \(\dfrac{12}{20}\). Given the second counter is red, the probability that the first was red is \(\dfrac{6/20}{12/20} = \dfrac{1}{2}\).

The formula and independence (Higher tier)

These links let you move between the probabilities when you have only some of them.

  • Formula

    \(P(A \text{ and } B) = P(A \mid B) \times P(B)\). It is given on the Higher tier formulae page for some boards and must be learned for others.

  • Rearranged

    \(P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}\).

  • Independent events

    \(A\) and \(B\) are independent if \(P(A \mid B) = P(A)\), which means \(P(A \text{ and } B) = P(A) \times P(B)\).

  • Test

    If \(P(A) = 0.5\), \(P(B) = 0.4\) and \(P(A \text{ and } B) = 0.2\), then \(0.5 \times 0.4 = 0.2\), so they are independent.

Using the formula

(Higher tier) \(P(A) = 0.5\), \(P(B) = 0.4\) and \(P(A \text{ and } B) = 0.1\). Work out \(P(A \mid B)\), and say whether \(A\) and \(B\) are independent.

Show the solutionHide the solution
  1. 1 Rearrange the formula \(P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}\).
  2. 2 Substitute \(\dfrac{0.1}{0.4}\).
  3. 3 Calculate \(\dfrac{1}{4} = 0.25\).
  4. 4 Compare with \(P(A)\) \(0.25 \neq 0.5\), so the events are not independent.

Answer\(P(A \mid B) = 0.25\), and they are not independent

What the AQA exam gives you

AQA gives very few formulae in the exam, so most of the formulae in this lesson have to be remembered.

  • Not given

    AQA does not print \(P(A \text{ and } B) = P(A \mid B) \times P(B)\), so learn it and how to rearrange it to \(P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}\).

  • Also not given

    The test for independence, \(P(A \mid B) = P(A)\), and the method of counting the smaller group from a table, Venn diagram or tree.

  • Practise the counting method

    Most AQA questions can be answered by counting the right group from the diagram, which does not need the formula at all.

Test yourself

  1. 1

    What does \(P(A \mid B)\) mean?

    Show answerHide answer

    The probability of \(A\) given that \(B\) has happened.

  2. 2

    What is the bottom of the fraction in a conditional probability?

    Show answerHide answer

    The number in the group that is given.

  3. 3

    What is the formula for \(P(A \text{ and } B)\) in terms of \(P(A \mid B)\) (Higher tier)?

    Show answerHide answer

    \(P(A \mid B) \times P(B)\).

  4. 4

    What is true of \(P(A \text{ and } B)\) when \(A\) and \(B\) are independent?

    Show answerHide answer

    It equals \(P(A) \times P(B)\).

  5. 5

    Out of 14 tennis players, 6 also play football. What is \(P(F \mid T)\)?

    Show answerHide answer

    \(\dfrac{6}{14} = \dfrac{3}{7}\).

Exam technique: conditional probability

Find the group first, then count inside it.

  • Underline the "given that" part

    It tells you the group for the bottom of the fraction.

  • Count the group carefully

    Add every cell, region or path that belongs to it.

  • Check the denominator is smaller than the total

    A conditional probability usually has fewer than all the items on the bottom.

  • Show a sentence

    "The group is the 14 tennis players" earns a mark for the method.

Summary and exam focus

  • A conditional probability is a probability within a smaller group, so the bottom of the fraction is the size of that group.
  • It can be found from a Venn diagram, two-way table or tree diagram.
  • \(P(A \text{ and } B) = P(A \mid B) \times P(B)\), and independent events have \(P(A \text{ and } B) = P(A) \times P(B)\) (Higher tier).

Exam focus

100 people were asked how they travel to work. Of the 50 who drive, 20 are female. A person who drives is chosen at random. Write down the probability that the person is male. (1 mark) (1 marks)

The group is those who drive, 50 people, and 20 of them are female, so \(50 - 20 = 30\) are male. The answer is \(\dfrac{30}{50} = \dfrac{3}{5}\). The mistake to avoid is dividing by the total of 100.

Key terms

The words 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 event has happened.
Given that
The words that mean you only look at part of the group.
Independent
Events where knowing one has no effect on the probability of the other.
Group
The set of items you are choosing from, after any given information.
Denominator
The number on the bottom of a fraction.
Tree diagram
A diagram showing outcomes in stages with probabilities on branches.
Two-way table
A table of counts sorted by two features.
Venn diagram
A diagram of overlapping sets with the number in each region.
Notation
The symbols used in mathematics, such as \(P(A \mid B)\).

Questions and answers

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

1. Exam question Work out 2 marks Core

A bag contains 4 yellow counters and 6 green counters. One counter is taken at random and not replaced. Another counter is then taken. Given that the first counter is yellow, work out the probability that the second counter is yellow. [2 marks]

Mark scheme — 2 marks available

  • 3 yellow out of 9 remaining — M1
  • \(\dfrac{1}{3}\) — A1

Model answer

After the first counter is taken there are 9 left, 3 of them yellow, so \(\dfrac{3}{9} = \dfrac{1}{3}\).

2. Exam question Work out 5 marks Core

The table shows the drinks chosen by 100 people. (a) A person is chosen at random. Work out the probability that the person chose lemonade. [1 mark] (b) A person under 18 is chosen at random. Work out the probability that the person chose cola. [2 marks] (c) A person who chose water is chosen at random. Work out the probability that the person is aged 18 or over. [2 marks]

A two-way table showing the drinks chosen by 50 people under 18 and 50 people aged 18 and over.

Mark scheme — 5 marks available

  • (a) \(\dfrac{1}{5}\) or \(\dfrac{20}{100}\) — B1
  • (b) \(\dfrac{25}{50}\) — M1
  • (b) \(\dfrac{1}{2}\) — A1
  • (c) \(\dfrac{30}{40}\) — M1
  • (c) \(\dfrac{3}{4}\) — A1

Model answer

(a) \(\dfrac{20}{100} = \dfrac{1}{5}\). (b) \(\dfrac{25}{50} = \dfrac{1}{2}\). (c) 40 people chose water and 30 of them are 18 or over, so \(\dfrac{30}{40} = \dfrac{3}{4}\).

3. Exam question Work out 3 marks Stretch

\(P(A) = 0.6\), \(P(B) = 0.5\) and \(P(A \text{ and } B) = 0.3\). (a) Work out \(P(B \text{ given } A)\). [2 marks] (b) Are \(A\) and \(B\) independent? Give a reason. [1 mark]

Mark scheme — 3 marks available

  • (a) \(\dfrac{0.3}{0.6}\) — M1
  • (a) \(0.5\) — A1
  • (b) Yes, because \(P(B \text{ given } A) = P(B)\) — Q1

Model answer

(a) \(P(B \text{ given } A) = \dfrac{0.3}{0.6} = 0.5\). (b) Yes, because \(P(B \text{ given } A) = 0.5 = P(B)\).

4. Exam question Work out 4 marks Stretch

In a group of 50 students, 30 study Spanish, 25 study French and 10 study both. (a) A student who studies French is chosen at random. Work out the probability that the student also studies Spanish. [2 marks] (b) A student who does not study Spanish is chosen at random. Work out the probability that the student studies French. [2 marks]

Mark scheme — 4 marks available

  • (a) \(\dfrac{10}{25}\) — M1
  • (a) \(\dfrac{2}{5}\) — A1
  • (b) \(\dfrac{15}{20}\) — M1
  • (b) \(\dfrac{3}{4}\) — A1

Model answer

(a) \(\dfrac{10}{25} = \dfrac{2}{5}\). (b) \(50 - 30 = 20\) do not study Spanish. French only is \(25 - 10 = 15\), so the probability is \(\dfrac{15}{20} = \dfrac{3}{4}\).

5. Exam question Work out 4 marks Stretch

A bag contains 6 red counters and 4 white counters. Two counters are taken at random without replacement. Given that the second counter is white, work out the probability that the first counter is also white. [4 marks]

Mark scheme — 4 marks available

  • \(\dfrac{6}{10} \times \dfrac{4}{9} = \dfrac{24}{90}\) — M1
  • \(\dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90}\) — M1
  • \(\dfrac{12}{36}\) or \(\dfrac{12}{90} \div \dfrac{36}{90}\) — M1
  • \(\dfrac{1}{3}\) — A1

Model answer

\(P(\text{red then white}) = \dfrac{6}{10} \times \dfrac{4}{9} = \dfrac{24}{90}\) and \(P(\text{white then white}) = \dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90}\). The probability that the second is white is \(\dfrac{36}{90}\). So the answer is \(\dfrac{12}{36} = \dfrac{1}{3}\).

6. Exam question Work out 4 marks Stretch

1 in 10 people have a medical condition. A test gives a positive result for 90% of the people who have the condition. The test also gives a positive result for 20% of the people who do not have it. A person is chosen at random and has a positive result. Work out the probability that the person has the condition. [4 marks]

Mark scheme — 4 marks available

  • \(0.1 \times 0.9 = 0.09\) — M1
  • \(0.9 \times 0.2 = 0.18\) — M1
  • \(\dfrac{0.09}{0.27}\) — M1
  • \(\dfrac{1}{3}\) — A1

Model answer

Has the condition and positive: \(0.1 \times 0.9 = 0.09\). Does not have it and positive: \(0.9 \times 0.2 = 0.18\). All positives: \(0.27\). The probability is \(\dfrac{0.09}{0.27} = \dfrac{1}{3}\).

7. Multiple choice 1 mark Easier

What does \(P(A \mid B)\) mean?

  1. A The probability of \(A\) and \(B\)
  2. B The probability of \(A\) given that \(B\) has happened Correct
  3. C The probability of \(A\) or \(B\)
  4. D The probability of \(B\) given \(A\)

Why: The vertical line means “given that”.

8. Multiple choice 1 mark Core

14 students play tennis, and 6 of them also play football. What is the probability that a tennis player also plays football?

  1. A \(\dfrac{3}{7}\) Correct
  2. B \(\dfrac{6}{30}\)
  3. C \(\dfrac{1}{3}\)
  4. D \(\dfrac{14}{30}\)

Why: The group is the 14 tennis players: \(\dfrac{6}{14} = \dfrac{3}{7}\).

9. Multiple choice 1 mark Core

Of 50 people who drive to work, 20 are female. A driver is chosen at random. What is the probability that the person is male?

  1. A \(\dfrac{2}{5}\)
  2. B \(\dfrac{30}{100}\)
  3. C \(\dfrac{1}{2}\)
  4. D \(\dfrac{3}{5}\) Correct

Why: \(\dfrac{30}{50} = \dfrac{3}{5}\). The group is the 50 drivers.

10. Multiple choice 1 mark Core

Of 100 people, 36 male drivers and 24 female drivers drive to work. A driver is chosen at random. What is the probability that the driver is female?

  1. A \(\dfrac{24}{100}\)
  2. B \(\dfrac{3}{5}\)
  3. C \(\dfrac{2}{5}\) Correct
  4. D \(\dfrac{24}{40}\)

Why: The group is \(36 + 24 = 60\) drivers, and \(\dfrac{24}{60} = \dfrac{2}{5}\).

11. Multiple choice 1 mark Core

A bag has 3 red and 2 blue counters. One red counter is taken and not replaced. What is the probability that the next counter is red?

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

Why: 2 red counters are left among 4, so \(\dfrac{2}{4} = \dfrac{1}{2}\).

12. Multiple choice 1 mark Core

Is \(P(A \mid B)\) always equal to \(P(B \mid A)\)?

  1. A No, because they use different groups Correct
  2. B Yes, always
  3. C Only for independent events with different probabilities
  4. D Only when \(A\) and \(B\) cannot both happen

Why: The group on the bottom is different in each.

13. Multiple choice 1 mark Stretch

\(P(A) = 0.5\), \(P(B) = 0.4\) and \(P(A \text{ and } B) = 0.2\). Are \(A\) and \(B\) independent?

  1. A No, because \(0.5 + 0.4 \neq 0.2\)
  2. B Yes, because \(0.5 > 0.4\)
  3. C No, because \(0.2 < 0.5\)
  4. D Yes, because \(0.5 \times 0.4 = 0.2\) Correct

Why: Independent events have \(P(A \text{ and } B) = P(A) \times P(B)\).

14. Multiple choice 1 mark Stretch

\(P(A \text{ and } B) = 0.1\) and \(P(B) = 0.4\). What is \(P(A \mid B)\)?

  1. A \(0.04\)
  2. B \(4\)
  3. C \(0.25\) Correct
  4. D \(0.5\)

Why: \(\dfrac{0.1}{0.4} = 0.25\).

15. Multiple choice 1 mark Stretch

\(P(A \mid B) = 0.5\) and \(P(B) = 0.6\). What is \(P(A \text{ and } B)\)?

  1. A \(1.1\)
  2. B \(0.3\) Correct
  3. C \(0.1\)
  4. D \(0.83\)

Why: \(P(A \text{ and } B) = P(A \mid B) \times P(B) = 0.5 \times 0.6 = 0.3\).