Conditional Probability (AA HL)
Conditional probability asks how the chance of one event changes once you know another has already happened. It's also the tool for testing independence - deciding whether knowing about \(B\) tells you anything about \(A\) at all. This page covers the formula, the independence test, and how they interact with combined events. It's part of the broader Probability topic.
13 questions on this sub-topic.
Conditional probability and independence
Covered under IB syllabus reference SL4.6. The conditional probability formula is in the formula booklet, but recognising when a question is testing independence rather than asking for a conditional value directly is the real skill.
Conditional probability
\[P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\]
Rearranges to \(P(A\cap B)=P(B)\,P(A\mid B)\) - the rule behind every tree diagram.
✓ In the formula bookletTesting independence
\[P(A\cap B)=P(A)P(B)\]
Compare the given \(P(A\cap B)\) with the product \(P(A)P(B)\). Equal means independent; unequal means dependent.
Need the full syllabus wording and formula-booklet reference table? See Probability.
Worked examples
Independent events \(A\) and \(B\) have \(P(A) = 0.5\) and \(P(B) = 0.2.\) Find \(P(A\cap B).\)
Worked solution
For independent events, \(P(A\cap B) = P(A)P(B) = 0.5\times0.2\) M1
\(= 0.1.\) A1
Events \(A\) and \(B\) satisfy \(P(A) = 0.4\), \(P(B) = 0.5\), \(P(A\cap B) = 0.2\).
(a) Determine whether \(A\) and \(B\) are independent.
(b) Find \(P(A\cup B)\).
Worked solution
(a) \(P(A)P(B) = 0.4\times0.5 = 0.2\) M1
\(= P(A\cap B)\), so \(A\) and \(B\) are independent. A1
(b) \(P(A\cup B) = 0.4 + 0.5 - 0.2\) M1
\(= 0.7.\) A1
A bag has 5 red and 3 blue counters. Two are drawn without replacement.
(a) Find the probability both are red.
(b) Find the probability they are different colours.
Worked solution
(a) Both red. \(P(RR)=\dfrac{5}{8}\times\dfrac{4}{7}=\dfrac{20}{56}\) M1 \(=\dfrac{5}{14}.\) A1
(b) Different colours. \(P(RB)+P(BR)=\dfrac{5}{8}\cdot\dfrac{3}{7}+\dfrac{3}{8}\cdot\dfrac{5}{7}=\dfrac{30}{56}=\dfrac{15}{28}.\) M1A1
Common mistakes
- Confusing independent with mutually exclusive. These describe opposite relationships - independent events can happen together freely, mutually exclusive events never can. Two events with positive probability can never be both.
- Forgetting "without replacement" changes the second probability. If a card, ball or item isn't put back, the total and the count of favourable outcomes both shrink for the next draw - each branch of the tree needs a fresh denominator.
- Mixing up \(P(A\mid B)\) and \(P(B\mid A)\). These are generally different numbers - "given it rained, the probability the match was cancelled" is not the same as "given the match was cancelled, the probability it rained". Bayes' theorem exists specifically to convert between them.
Ready to practise properly?
13 conditional-probability questions, marked instantly like the real exam.
Quick answers
What is the formula for conditional probability?
\(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\). It rearranges to \(P(A\cap B)=P(B)\,P(A\mid B)\), which is the rule behind every tree diagram.
How do you test whether two events are independent?
Check whether \(P(A)P(B)\) equals \(P(A\cap B)\). If they match, \(A\) and \(B\) are independent; if not, they are dependent. For calculator tips, see the Probability GDC section.