4.6 Combined events and conditional probabilityIB Maths: Applications and Interpretation HL: Revision notes

Probabilities of combined events can be found from a Venn diagram, a tree diagram, a sample space diagram or a table of outcomes. Choose the one that fits the data: Venn diagrams for overlapping groups, tables for two categories, trees for events in sequence. Often a problem can be solved straight from the diagram without any formula. For a class of 30 with 18 studying Spanish, 12 French and 5 both, the Venn diagram has 13 only Spanish, 5 both, 7 only French and 5 neither.

$A\cup B$ means 'A or B or both' and $A\cap B$ means 'A and B'. The word 'or' in mathematics is not exclusive: it includes both. The addition rule is $P(A\cup B)=P(A)+P(B)-P(A\cap B).$ Events are mutually exclusive when they cannot happen together, so $P(A\cap B)=0$ and the rule becomes $P(A\cup B)=P(A)+P(B)$. Example: $\frac{18}{30}+\frac{12}{30}-\frac{5}{30}=\frac{25}{30}=\frac56$.

The probability of $A$ given that $B$ has occurred is $P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad\text{or}\qquad P(A\cap B)=P(B)\,P(A\mid B).$ Knowing $B$ occurred reduces the sample space to $B$. In the class example, $P(F\mid S)=\frac{5}{18}$: of the 18 Spanish students, 5 also study French. With a tree diagram, the second set of branches carries conditional probabilities and you multiply along a branch to get an intersection.

On a tree diagram, multiply along branches for 'and' and add the products of different branches for 'or'. Branches from one point sum to 1. With replacement, the second-stage probabilities are the same as the first. Without replacement, the numbers change. Example: 4 red and 6 yellow pencils, two drawn without replacement: $P(\text{both red})=\frac{4}{10}\times\frac39=\frac{2}{15}$, and $P(\text{different colours})=\frac{4}{10}\times\frac69+\frac{6}{10}\times\frac49=\frac{8}{15}$.

Events $A$ and $B$ are independent if one occurring does not affect the probability of the other: $P(A\cap B)=P(A)\,P(B),\qquad\text{equivalently } P(A\mid B)=P(A).$ To test independence, calculate both sides and compare. For 70 tennis players and 50 badminton players among 120 members with 20 in both, $P(T)P(B)=\frac{35}{144}\approx0.243$ but $P(T\cap B)=\frac16\approx0.167$, so the events are not independent. Do not confuse independent with mutually exclusive: events with non-zero probabilities that are mutually exclusive are never independent.