Conditional Probability (AI SL)
Conditional probability asks a narrower question than plain probability: given that something has already happened, how likely is something else? Restricting the sample space this way is also how you test whether two events are independent in the first place. This page covers the formula, the independence test that comes from it, and the places students most often mix up which event is the "given" one. It's part of the broader Probability topic.
22 questions on this sub-topic.
The key formula
Covered under IB syllabus reference SL4.6: conditional probability \(P(A|B)=\dfrac{P(A\cap B)}{P(B)}\) and independent events \(P(A\cap B)=P(A)P(B)\).
Conditional probability
\(P(A|B) = \dfrac{P(A\cap B)}{P(B)}\)
Restrict the sample space to \(B\) first, then ask what fraction is also \(A\). In the formula booklet.
Basic probability
\(P(A) = \dfrac{n(A)}{n(U)}\)
The count you're restricting when you go from plain probability to conditional probability. Not in the formula booklet - prior knowledge.
Need the mutually-exclusive-vs-independent comparison table? See Probability.
Worked examples
Using the class data (30 students: 18 French, 14 Spanish, 6 both), find the probability that a student studies Spanish given that they study French.
Worked solution
\(P(S\mid F)=\dfrac{P(S\cap F)}{P(F)}=\dfrac{6/30}{18/30}=\dfrac{6}{18}\) M1 - restrict attention to the French students only.
\(=\dfrac13\approx 0.333.\) A1 Intuitively: of the 18 French students, 6 also take Spanish.
\(P(A)=0.4,\ P(B)=0.5,\ P(A\cap B)=0.2.\) Determine whether \(A\) and \(B\) are independent.
Worked solution
\(A\) and \(B\) are independent exactly when \(P(A\cap B)=P(A)\times P(B).\) M1
\(P(A)P(B)=0.4\times0.5=0.2.\) A1
This equals the given \(P(A\cap B)=0.2,\) R1
so the events are independent. A1
A coin is tossed twice.
Find the probability of getting two heads.
Worked solution
The two tosses are independent, each with \(P(\text{head})=\dfrac12.\) A1
For independent events multiply: M1
\(P(HH)=\dfrac12\times\dfrac12=\dfrac14.\) A1
For events with \(P(A) = 0.3\) and \(P(B) = 0.5\):
(a) Find \(P(A\cup B)\) if \(A\) and \(B\) are mutually exclusive.
(b) Find \(P(A\cup B)\) if \(A\) and \(B\) are independent.
Worked solution
(a) Mutually exclusive: \(P(A\cap B) = 0\), so \(P(A\cup B) = 0.3 + 0.5\) M1
\(= 0.8.\) A1
(b) Independent: \(P(A\cap B) = 0.15\), so \(P(A\cup B) = 0.3 + 0.5 - 0.15\) M1
\(= 0.65.\) A1
Common mistakes
- Confusing mutually exclusive with independent. Mutually exclusive events can't both happen (\(P(A\cap B)=0\)); independent events just don't affect each other (\(P(A\cap B)=P(A)P(B)\)). Two events with nonzero probability can't be both at once.
- Mixing up \(P(A|B)\) with \(P(B|A)\). These are generally different numbers - always check which event is the "given" condition, since it fixes which probability goes on the bottom of the fraction.
- Dividing by the wrong total. The denominator of \(P(A|B)\) is \(P(B)\), not \(P(A)\) and not \(P(U)\) - it's easy to default back to the full sample space out of habit instead of the restricted one.
Ready to practise properly?
22 conditional-probability questions, marked instantly like the real exam.
Quick answers
What does P(A|B) actually mean?
\(P(A|B)\) is the probability that \(A\) happens, given that you already know \(B\) has happened. It shrinks your sample space down to just the outcomes where \(B\) is true, then asks what fraction of those also satisfy \(A\).
How do I test whether two events are independent?
Compare \(P(A\cap B)\) with \(P(A)\times P(B)\). If they're equal, \(A\) and \(B\) are independent. You can also check whether \(P(A|B)\) equals \(P(A)\) - if conditioning on \(B\) doesn't change the probability of \(A\), they're independent.