Maths › Statistics › Conditional probability
Conditional probability
Every probability lives in a sample space, and new information shrinks it. Conditional probability is the arithmetic of that shrinkage. Given that B happened, the world is B now, and everything gets remeasured against it.
Builds on Probability and Venn diagrams.
IN THIS TOPIC
- Calculate conditional probabilities from the formula, from Venn diagrams and from two-way tables.
- Build tree diagrams for successive events, including sampling without replacement.
- Use P(A|B) = P(A) as a test of independence.
- Criticise the assumptions behind a probability model and say which way the answer would move.
COMMON MISCONCEPTION
P(A|B) and P(B|A) are the same thing written two ways.
The shrunken world
P(A|B) reads “the probability of A given B”. Conditioning throws away every outcome outside B and rescales what survives, which is what the formula does. The part of A inside B, divided by B instead of by everything. Take the Venn numbers from last lesson and P(A|B) = 0.2/0.4 = 0.5, while P(A) on its own was also 0.5. That match is no accident, and the last section turns it into a test.
Order matters. P(A|B) and P(B|A) divide the same overlap by different worlds, and they are rarely equal. The probability that a chess player is a grandmaster is tiny. The probability that a grandmaster plays chess is 1.
Multiplying up gives the other useful form, P(A ∩ B) = P(A|B) × P(B), which is what a tree diagram is doing when you multiply along a path.
Do not go looking for the quotient form in the booklet. What the booklet prints, under Probability, is the product rule P(A ∩ B) = P(A)P(B|A) and the full Bayes' theorem. The quotient above is the one-line rearrangement of that product rule, and you have to be able to write it down yourself.
Trees, and drawing without replacement
A tree diagram is conditional probability drawn as branches. Multiply along a path, add between paths. Sampling without replacement is where the conditioning bites, because the second draw's probabilities depend on what the first draw took away.
WORKED EXAMPLE
Two counters, no replacement
A bag holds 3 red and 2 blue counters. Two are drawn without replacement. Find the probability of exactly one red.
Two routes. Red then blue is 3/5 × 2/4 = 6/20, and blue then red is 2/5 × 3/4 = 6/20.
Add the routes: 6/20 + 6/20 = 12/20 = 3/5.
Check against the tree's other ends. Both red is 3/10 and both blue is 1/10, and 3/10 + 1/10 + 3/5 = 1, as it must.
Tables, and the independence test
A two-way table makes conditioning nearly mechanical. Given “the student is in Year 12”, stay inside that row and divide the cell by the row total. The formula and the table are doing one thing. Numerator from the overlap, denominator from the condition.
Independence gets its sharpest statement here. A and B are independent exactly when P(A|B) = P(A), the news of B changing nothing about A. That is last lesson's multiplication test restated, and either version earns full marks provided the comparison is written out.
The model behind the numbers
Every probability question has a model underneath it, and Paper 3 sometimes asks you to poke at it. Treating successive days' weather as independent, or assuming a spinner is fair, or ignoring that people who answer surveys differ from people who do not, are all assumptions doing real work in the arithmetic.
A good criticism names the assumption and says which way the answer moves without it. Rainy days cluster, so assuming independence understates the chance of a wet week. That second half is what separates a scoring answer from a shrug.
ASSESSMENT FOCUS
- Write the conditional formula before you substitute. The method mark is attached to the formula, not to the answer.
- Without replacement means every second-stage denominator drops by one. Forgetting is the single most common tree error in the paper.
- From a table, condition by staying inside one row or column and dividing by its total.
- For “show that A and B are not independent”, compare P(A|B) with P(A), or the overlap with the product, and finish with a sentence.
CHECK YOURSELF
P(A ∩ B) = 0.12 and P(B) = 0.4. Find P(A|B), and state what P(A) would have to equal for A and B to be independent.
Show a hint
Overlap over condition.
Show the answer
P(A|B) = 0.12/0.4 = 0.3.
Independence needs P(A|B) = P(A), so P(A) would have to be 0.3.
Condition means divide: the overlap, measured against the new, smaller world.
On a tree, multiply along and add across, and without replacement the second denominators shrink.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the conditional probability questions page.
CHECK YOUR PROGRESS
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- Calculate conditional probabilities from the formula, from Venn diagrams and from two-way tables.
- Build tree diagrams for successive events, including sampling without replacement.
- Use P(A|B) = P(A) as a test of independence.
- Criticise the assumptions behind a probability model and say which way the answer would move.
Open the full revision checklist to see every objective in the course in one place.