Describe a random experiment by its sample space, write events as subsets, combine them with union, intersection and complement, and recognise mutually exclusive, exhaustive and equally likely events.
Tossing a coin, throwing a die, drawing a card, picking a committee: in each case we know what could happen, but not what will happen on one trial. Such an activity is a random experiment (a trial is one performance of it, the result is an outcome).
The set of all possible outcomes is the sample space $S$, and each outcome is a sample point.
Always say what one outcome is before listing: for two tosses, "the result of toss 1 followed by the result of toss 2" gives ${HH, HT, TH, TT}$ with $n(S) = 4$.
An event is a subset of the sample space. An event occurs when the outcome of the trial is one of its sample points.
Events are sets, so the set operations apply:
| Notation | Meaning |
|---|---|
| $A \cup B$ | $A$ or $B$ or both occur |
| $A \cap B$ | both $A$ and $B$ occur |
| $A'$ | $A$ does not occur ($A'$ is the complement of $A$) |
| $A \cap B'$ | $A$ occurs but $B$ does not |
| $A \subseteq B$ | whenever $A$ occurs, $B$ occurs |
Example: A fair coin is tossed three times. Let $A$ = "at least two heads" and $B$ = "the first toss is a head". List $S$, $A$, $B$, $A \cap B$, $A \cup B$ and $A'$. Writing the three results in order, $S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}$ with $n(S) = 8$. $A = {HHH, HHT, HTH, THH}$ and $B = {HHH, HHT, HTH, HTT}$. Then $A \cap B = {HHH, HHT, HTH}$, $A \cup B = {HHH, HHT, HTH, HTT, THH}$ and $A' = {HTT, THT, TTH, TTT}$ (fewer than two heads).
Example: A fair die is thrown. Let $A = {2, 4, 6}$, $B = {1, 3, 5}$ and $C = {1, 2, 3}$. $A \cap B = \varnothing$ and $A \cup B = S$, so $A$ and $B$ are mutually exclusive and exhaustive ($B = A'$). $A \cap C = {2} \ne \varnothing$, so $A$ and $C$ are not mutually exclusive. $A \cup C = {1, 2, 3, 4, 6} \ne S$, so $A$ and $C$ are not exhaustive either.
For a large sample space we count rather than list, using the multiplication principle, permutations $P(n,r)$ and combinations $\binom{n}{r}$.
Example: A committee of 3 is chosen from 5 men and 4 women. How many outcomes are there, and how many belong to the event $E$ = "exactly two women"? The outcomes are the selections of 3 people from 9 (order does not matter): $n(S) = \binom{9}{3} = 84$. For $E$ choose 2 of the 4 women and 1 of the 5 men: $n(E) = \binom{4}{2}\binom{5}{1} = 6 \times 5 = 30$.
Exam tip: Before any probability is calculated, Paper II questions expect a clear sample space or a counting statement. For two dice, draw the $6 \times 6$ table of $36$ outcomes and shade the event; for selections, say whether order matters and use $\binom{n}{r}$ for "chosen" and $P(n,r)$ or $n!$ for "arranged".