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Year 11 · Mathematics · Probability

Conditional probability

Calculate probabilities when one event affects another, using two-way tables, Venn diagrams and tree diagrams, and distinguish independent from dependent events.

Awaiting educator review. This material was drafted by our content team with AI assistance and is waiting for review by a qualified educator. Please check it suits your child before using it. Created 10 October 2026. How we review

These lessons were written around Australian Curriculum strands and matched to UK year groups by age (Year 1 is the year a child turns 6 in England). They are not mapped to the National Curriculum in England or the curricula of Scotland, Wales or Northern Ireland, so check them against your own plans.

Learning goals

Learners will

  • Calculate conditional probabilities from two-way tables and Venn diagrams.
  • Use tree diagrams to find probabilities for two-step experiments without replacement.
  • Distinguish between independent and dependent events and test for independence.

Success looks like

  • I can find P(A | B) by restricting my attention to the cases where B has happened.
  • I can draw a tree diagram for an experiment without replacement and multiply along the branches.
  • I can decide whether two events are independent by checking whether P(A and B) = P(A) × P(B).

The big idea

Conditional probability asks: once we know something has happened, how likely is something else? We write P(A | B), read 'the probability of A given B'. The key idea is that knowing B has happened shrinks the sample space to only the outcomes where B is true.

Two-way tables make this easy. Suppose 50 students are surveyed: 30 play sport, and 18 of those also have a pet; of the 20 who do not play sport, 8 have a pet. So 26 students have a pet. If a student who plays sport is chosen, P(pet | sport) = 1830 = 35. The denominator is 30 (sport players only), not 50.

Order matters: P(sport | pet) = 1826 = 913, which is different from P(pet | sport).

The general formula is P(A | B) = P(A and B) ÷ P(B). With the table above, P(pet and sport) = 1850 and P(sport) = 3050, so P(pet | sport) = 18 ÷ 30 = 35 again.

Venn diagrams work the same way. In a group of 40 people, 22 like tea, 18 like coffee and 7 like both. Given that a person likes tea, P(likes coffee | likes tea) = 722, because we only look inside the tea circle.

Tree diagrams are ideal for two-step experiments. When items are drawn without replacement, the second set of branches changes. With 5 red and 3 blue counters, P(red first) = 58; if a red was taken, only 4 red remain out of 7, so P(red second | red first) = 47. Multiply along the branches: P(red, red) = 58 × 47 = 2056 = 514.

Events are independent if one does not affect the other – such as tossing a coin and rolling a die. Then P(A | B) = P(A), and P(A and B) = P(A) × P(B). If these are not equal, the events are dependent. In the survey, P(pet) = 2650 = 0.52 but P(pet | sport) = 0.6, so having a pet and playing sport are dependent in this group.

Worked examples

From a two-way table

In a survey of 50 students, 30 play sport (18 of these have a pet) and 20 do not play sport (8 of these have a pet). Find P(plays sport | has a pet).

  1. Restrict to students who have a pet: 18 + 8 = 26 students.
  2. Of these 26, the number who play sport is 18.
  3. P(plays sport | has a pet) = 1826 = 913.

Answer: 913 (about 0.69)

A tree diagram without replacement

A bag has 5 red and 3 blue counters. Two counters are taken out one after the other without replacement. Find the probability of getting one of each colour.

  1. First draw: P(R) = 58, P(B) = 38.
  2. Second draw: after a red, P(B) = 37; after a blue, P(R) = 57.
  3. P(R then B) = 58 × 37 = 1556. P(B then R) = 38 × 57 = 1556.
  4. Add the two paths: 1556 + 1556 = 3056 = 1528.

Answer: 1528

Testing for independence

For two events, P(A) = 0.4, P(B) = 0.5 and P(A and B) = 0.2. Are A and B independent?

  1. Calculate P(A) × P(B) = 0.4 × 0.5 = 0.2.
  2. Compare with P(A and B) = 0.2. They are equal.
  3. Alternatively, P(A | B) = 0.2 ÷ 0.5 = 0.4 = P(A), so knowing B happened does not change the probability of A.

Answer: Yes, A and B are independent.

Practice check

Have a go, then check your answers. Each answer comes with an explanation.

  1. 1.What does P(A | B) mean?
  2. 2.In a group of 50 students, 30 play sport, and 18 of those also have a pet. A student who plays sport is chosen at random. What is the probability they have a pet? Give a fraction in simplest form.
  3. 3.A bag holds 4 red and 6 green marbles. Two are taken without replacement. What is the probability that both are green? Give a fraction in simplest form.
  4. 4.Which pair of events is independent?
  5. 5.In a group of 30 people, 12 like cats, 15 like dogs and 5 like both. A person who likes dogs is chosen at random. What is the probability they also like cats? Give a fraction in simplest form.
  6. 6.For two events, P(A) = 0.3, P(B) = 0.6 and P(A and B) = 0.25. Are A and B independent?
  7. 7.If P(A and B) = 0.12 and P(B) = 0.3, find P(A | B).
  8. 8.A bag has 5 red and 3 blue counters. One is taken out and not replaced. Given the first counter was red, what is the probability that the second is blue?
Answer guide for parents

  1. What does P(A | B) mean?

    The probability of A given that B has happened — The vertical line is read 'given'.

  2. In a group of 50 students, 30 play sport, and 18 of those also have a pet. A student who plays sport is chosen at random. What is the probability they have a pet? Give a fraction in simplest form.

    3/5 — Restrict to the 30 sport players: P(pet | sport) = 18/30 = 3/5.

  3. A bag holds 4 red and 6 green marbles. Two are taken without replacement. What is the probability that both are green? Give a fraction in simplest form.

    1/3 — P(green, green) = 6/10 × 5/9 = 30/90 = 1/3.

  4. Which pair of events is independent?

    Tossing a head on a coin and rolling a 6 on a die — The coin has no effect on the die, so the probability of rolling a 6 stays 16 whatever the coin shows.

  5. In a group of 30 people, 12 like cats, 15 like dogs and 5 like both. A person who likes dogs is chosen at random. What is the probability they also like cats? Give a fraction in simplest form.

    1/3 — Restrict to the 15 dog lovers, of whom 5 also like cats: 5/15 = 1/3.

  6. For two events, P(A) = 0.3, P(B) = 0.6 and P(A and B) = 0.25. Are A and B independent?

    No, because P(A) × P(B) = 0.18, which is not equal to 0.25 — Independent events satisfy P(A and B) = P(A) × P(B). Here 0.3 × 0.6 = 0.18 ≠ 0.25, so the events are dependent.

  7. If P(A and B) = 0.12 and P(B) = 0.3, find P(A | B).

    0.4 — P(A | B) = P(A and B) ÷ P(B) = 0.12 ÷ 0.3 = 0.4.

  8. A bag has 5 red and 3 blue counters. One is taken out and not replaced. Given the first counter was red, what is the probability that the second is blue?

    37 — After a red is removed, 7 counters remain and 3 of them are blue, so the probability is 37.

Watch out for

  • Using the whole group as the denominator for a conditional probability. For P(A | B), the denominator is only the cases where B happened.
  • Assuming P(A | B) and P(B | A) are the same. In the survey, P(pet | sport) = 35 but P(sport | pet) = 913.
  • Forgetting to change the second-stage probabilities in a tree diagram when items are not replaced.

Tips for parents

  • Do the counter or sock experiment for real: draw two without replacement many times, tally the results and compare them with the tree diagram.
  • Make a two-way table about your own family or friends (for example, wears glasses / plays an instrument) and ask conditional questions about it.
  • Ask your learner to say each probability in words ('the chance of… given that…') before calculating – it clarifies which group to divide by.

Go further

  • A medical test is 95% accurate and the condition affects 1 in 100 people. Use a two-way table for 10 000 people to find the probability that someone who tests positive actually has the condition. Why is the answer surprisingly low?
  • Draw a full tree diagram for drawing two cards from a standard pack without replacement and find P(two hearts) = 1352 × 1251 = 117.
  • Collect class or family data on two characteristics and test whether they appear to be independent.

About 60 minutes

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You'll need

  • Paper and pencil
  • A calculator
  • A bag with coloured counters or socks (for example 5 red and 3 blue)
  • A pack of playing cards (optional)

Key words

Conditional probability
The probability of an event given that another event has already happened, written P(A | B) and read 'the probability of A given B'.
Two-way table
A table that shows the frequencies of two categorical variables together, with totals for each row and column.
Venn diagram
A diagram of overlapping circles that shows how events or sets overlap.
Tree diagram
A branching diagram showing all outcomes of a multi-step experiment, with probabilities on each branch.
Independent events
Events where the outcome of one does not change the probability of the other.
Dependent events
Events where the outcome of one does change the probability of the other.
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