Conditional Probability (AA SL)
Conditional probability asks a narrower question than a plain probability does: given that something has already happened, what's the chance of something else? It's also the tool for testing whether two events are truly independent. This page covers the formula, the independence test, and the mistakes examiners see most, with worked examples. It's part of the broader Probability topic.
21 questions on this sub-topic.
Conditional probability and independence
Covered under IB syllabus reference SL4.6, alongside the combined-events and independence rules it depends on. The conditional-probability formula is in the formula booklet, so the skill being tested is recognising when to use it and substituting the right values.
Conditional probability
\[P(A|B)=\dfrac{P(A\cap B)}{P(B)}\]
Restrict the sample space to just the outcomes where \(B\) occurs, then ask what fraction of those also satisfy \(A\). In the formula booklet.
Independence test
\(P(A\cap B)=P(A)\,P(B)\)
Compare \(P(A\mid B)\) with \(P(A)\) - or equivalently test \(P(A\cap B)\) against \(P(A)P(B)\). If they match, knowing \(B\) hasn't changed the odds of \(A\).
Need basic sample-space probability first, or the full syllabus wording? See Basic Probability or the full Probability topic page.
Worked examples
\(P(A)=0.6\), \(P(B)=0.5\), \(P(A\cap B)=0.3\).
Find \(P(A\mid B)\).
Worked solution
Apply the conditional-probability formula: \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}.\) M1
Substitute: \(P(A\mid B)=\dfrac{0.3}{0.5}=0.6.\) A1
Since \(P(A\mid B)=0.6=P(A)\), knowing \(B\) hasn't changed the chance of \(A\) - the events are in fact independent. R1
Events \(A\) and \(B\) satisfy \(P(A) = 0.4\), \(P(B) = 0.5\). Explain why they cannot be both mutually exclusive and independent (with \(P(A),P(B) > 0\)).
Worked solution
If \(A\) and \(B\) were independent, \(P(A\cap B)=0.4\times0.5=0.2\ne 0.\) M1 A1
If \(A\) and \(B\) were mutually exclusive, \(P(A\cap B)=0.\) A1
Both cannot hold at once, since \(0.2\ne 0\) (and \(P(A),P(B)>0\)) - so the two events cannot be both mutually exclusive and independent. R1
Events \(A\) and \(B\) are mutually exclusive with \(P(A) = 0.35\) and \(P(B) = 0.2\).
Find \(P(A\cup B)\).
Worked solution
They cannot both occur, so \(P(A\cap B)=0.\) R1
\(P(A\cup B)=P(A)+P(B)=0.35+0.2=0.55.\) M1A1
Common mistakes
- Restricting the wrong sample space. \(P(A|B)\) means: given \(B\) has happened, find the chance of \(A\) - the denominator must be \(P(B)\), not \(P(A)\) or the full sample space.
- Not adjusting probabilities when sampling without replacement. Once an item is removed, both the total and the count of that type decrease for the next draw, so a conditional probability calculated from a second draw is not the same fraction as the first.
- Confusing mutually exclusive with independent. Mutually exclusive events cannot happen together (\(P(A\cap B)=0\)); independent events can happen together, just without affecting each other's probability. Two mutually exclusive events with nonzero probability are never independent.
Ready to practise properly?
21 conditional-probability questions, marked instantly like the real exam.
Quick answers
What is the formula for conditional probability?
\(P(A|B) = \dfrac{P(A\cap B)}{P(B)}\). Given that \(B\) has already happened, this restricts the sample space to just the outcomes where \(B\) occurs, then asks what fraction of those also satisfy \(A\).
How do you test whether two events are independent?
Check whether \(P(A\cap B) = P(A)\,P(B)\). If this holds, \(A\) and \(B\) are independent. See the GDC guidance on the Probability topic page for evaluating these expressions on your calculator.