Short on time? This page is the recap. For explanations, a tree diagram and worked examples, use the full study guide.
Recap
- A conditional probability P(A | B) is the probability that A happens given that B has happened. Read the bar as “given”.
- The condition B shrinks the set of possible outcomes to those in B. You then ask what fraction of B is also in A.
- The general multiplication rule finds a joint probability in stages: first A, then B given A.
- A tree diagram shows the stages. Later branches are conditional on the earlier path.
Key relationships
| Idea | Formula or method |
|---|---|
| Conditional probability | P(A | B) = P(A ∩ B) / P(B), with P(B) > 0 |
| From a two-way table | P(A | B) = (cell for A and B) ÷ (total for B) |
| General multiplication rule | P(A ∩ B) = P(A) · P(B | A) |
| Drawing without replacement | second-draw probability is conditional: reduce the totals after the first draw |
| Tree diagram | branches from one point add to 1; multiply along a path; add paths that make up an event |
| Reversing a condition | P(B | A) = P(A ∩ B) / P(A): find the joint probability, then divide by the new condition |
Assumptions and conventions
- The condition must have a probability above 0. You cannot condition on an impossible event.
- In a two-way table, one individual is chosen at random from the whole table.
- Words like “of the …”, “among the …”, “given that …” and “if … then …” usually name the condition.
- Keep exact values until the final step.
Mistakes to avoid
- Swapping the condition. P(late | van) is not P(van | late).
- Dividing by the grand total when the question gives a condition. That gives the joint probability.
- Ignoring “without replacement”. The second draw has one fewer item.
- Multiplying unconditional probabilities. P(A) · P(B) is only for independent events (Topic 2.7). Use P(A) · P(B | A).
- Rounding early, especially before dividing to reverse a condition.
- Reading cause into a conditional probability from observational data.
Quick self-check
- P(A ∩ B) = 0.12 and P(B) = 0.4. Find P(A | B). (0.12 ÷ 0.4 = 0.3)
- A box has 6 working bulbs and 2 broken bulbs. Two bulbs are taken at random without replacement. Find P(both broken). (2/8 × 1/7 = 1/28 ≈ 0.0357)
- P(A) = 0.5 and P(B | A) = 0.3. Find P(A ∩ B). (0.5 × 0.3 = 0.15)
Next: practice questions.