Probability of A, given B has happened: P(A | B) is the probability of A given that B has occurred.
The condition shrinks the sample space to B, so P(A | B) = P(A ∩ B) ÷ P(B).
Conditional probability P(B|A): restrict to the A branch, then read the proportion that is also B.
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IB-style question — straight from the formula
P(A ∩ B) = 0.24 and P(B) = 0.4.
Find P(A | B).
Step by step
- Apply the formula.
- Evaluate.
Final answer
P(A | B) = 0.6.
Divide by the GIVEN event: The event after the '|' is the one you divide by — it's the new, reduced sample space.
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Find the intersection, then condition: If you're given P(A), P(B) and P(A ∪ B), first find P(A ∩ B) from the addition rule, then apply the conditional formula.
IB-style question — two steps
P(A) = 0.6, P(B) = 0.5 and P(A ∪ B) = 0.8.
Find P(A | B).
Step by step
- Addition rule for the intersection.
- Conditional formula.
Final answer
P(A | B) = 0.6.
You need the intersection: The conditional formula needs P(A ∩ B) — recover it from the addition rule before dividing.
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Restrict to the 'given' group: With a table or Venn, the condition restricts you to the given group: count (both) ÷ count (given group).
You only look inside the row/region named after the '|'.
IB-style question — from a table
Of 70 men surveyed, 50 own a car.
(The condition is 'given male'.) Find the probability a randomly chosen man owns a car.
Step by step
- Restrict to the given group (the 70 men).
- Divide.
Final answer
P(car | male) = 5/7 ≈ 0.714.
The given group is your denominator: 'Given male' → divide by the total males, not the whole survey.
Second-stage branches ARE conditional: On a tree, a second-stage branch is already a conditional probability.
To reverse the condition (e.g. P(first stage | second outcome)), use P(A ∩ B) ÷ P(B) with the paths.
IB-style question — reverse the condition
On a rainy day (P = 0.3) the chance of a traffic jam is 0.6; on a dry day it is 0.2.
Given there was a traffic jam, find the probability it was raining.
Step by step
- P(jam) from the two paths.
- Conditional: rainy-and-jam over jam.
Final answer
P(rain | jam) ≈ 0.563.
Numerator = the matching path: The numerator is the single path 'rainy and jam' (0.18); the denominator is the total P(jam) from all paths.