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 is 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(B|A) = P(B) as a test of independence.
WHAT YOU PROBABLY THINK
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”. The condition throws away every outcome outside B and rescales what is left, which is exactly what the formula says: the part of A inside B, measured against B rather than against everything. With the Venn numbers from last lesson, P(A|B) = 0.2/0.4 = 0.5, while P(A) alone was 0.5 too: that match is no accident, and the next section makes it 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 a chess player is a grandmaster is tiny, the probability a grandmaster plays chess is 1.
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 the first draw's result.
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.
The 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 almost mechanical: given “the student is in Year 12”, stay inside that row, and divide the cell by the row total. The formula is doing the same thing the picture does: 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. It is the multiplication test from last lesson in disguise, and either version earns the marks if the comparison is shown.
THE EXAM BIT
- Write the conditional formula before substituting; the method mark attaches to it.
- Without replacement means the second-stage denominators drop by one; forgetting is the single most common tree error.
- 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.
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(B|A) = P(B) as a test of independence.
No animated video for this topic yet; these notes stand alone.