Elementary probability and sample spacesEdexcel International A Level Maths: Revision notes

An event is a set of outcomes. The sample space is the set of all possible outcomes. When the outcomes are equally likely, $P(A)=\frac{\text{number of outcomes in }A}{\text{total number of outcomes}}$, and $0\leq P(A)\leq1$. Always list or count the sample space carefully, treating the dice or coins as distinguishable. Example: two fair four-sided dice give 16 equally likely ordered pairs. The total 5 comes from $(1,4),(2,3),(3,2),(4,1)$, so $P(\text{total}=5)=\frac{4}{16}=\frac14$. The totals 2 to 8 are not equally likely.

The complement of an event $A$, written $A'$, is the event that $A$ does not happen. Since exactly one of $A$ and $A'$ must occur, $P(A')=1-P(A).$ This is a quick way to find 'at least one' or 'not' probabilities. Example: the probability of at least one head in two coin tosses is $1-P(\text{no heads})=1-\frac14=\frac34$.

Events are mutually exclusive (exclusive) if they cannot happen at the same time, so $P(A\cap B)=0$. For exclusive events the addition rule is $P(A\cup B)=P(A)+P(B).$ Example: scoring 1 or 2 on a die, and scoring 5 or 6, are exclusive. $P(\text{1 or 2 or 5 or 6})=\frac26+\frac26=\frac23$. To show that two events are not exclusive, find one outcome in both, or show that $P(A\cap B)\neq0$.

For any two events, $P(A\cup B)=P(A)+P(B)-P(A\cap B).$ Here $A\cup B$ means '$A$ or $B$ or both' and $A\cap B$ means 'both $A$ and $B$'. The overlap is subtracted because it is counted twice when $P(A)$ and $P(B)$ are added. The rule can be rearranged to find a missing probability, for example $P(A\cap B)=P(A)+P(B)-P(A\cup B)$. Example: for a die with $A$ even and $B$ greater than 3, $P(A\cup B)=\frac12+\frac12-\frac13=\frac23$. The outcomes $\{2,4,5,6\}$ confirm this.

A Venn diagram or a two-way table separates the regions of the sample space. Start with the overlap, then fill in 'only $A$', 'only $B$' and 'neither'. Example: 40 students, 22 study French, 18 study Spanish, 9 study both. Only French is $22-9=13$, only Spanish is $18-9=9$ and both is 9, so 31 study at least one and 9 study neither. $P(\text{neither})=\frac{9}{40}$ and $P(\text{exactly one})=\frac{22}{40}$. Useful results: $P(A\cap B')=P(A)-P(A\cap B)$ and $P(A'\cap B')=1-P(A\cup B)$.