Basic Terminologies

A random experiment is a trial that has more than one possible result: we know in advance every result that could occur, but we cannot say which one will actually happen. Rolling a die, tossing a coin, and drawing a card are all random experiments. Each individual result of such an experiment is called an outcome, and the collection of every possible outcome is the sample space, written $S$ or $\Omega$. The number of outcomes in $S$ is written $n(S)$, and each outcome is also called a sample point.

An event is simply a subset of the sample space - any collection of outcomes we choose to group together, such as 'the die shows an even number'. An outcome that belongs to the event is called a favourable outcome for that event.

Events come in a few standard flavours. An elementary event contains exactly one outcome. The certain event is the whole sample space $S$ itself (every outcome is favourable, so it always occurs). The impossible event is the empty set $\phi$ (no outcome is favourable, so it can never occur).

Because events are sets, the usual set operations let us build new events from old ones. The union $A\cup B$ is the event that at least one of $A$ or $B$ occurs - it contains every outcome belonging to $A$, to $B$, or to both. For example, if $S$ is the positive integers up to 50, $A$ is the multiples of 6 in $S$, and $B$ is the multiples of 9 in $S$, then $A=\{6,12,18,24,30,36,42,48\}$, $B=\{9,18,27,36,45\}$, and $A\cup B=\{6,9,12,18,24,27,30,36,42,45,48\}$ - every element divisible by 6 or by 9. Two events are called exhaustive if their union is the whole sample space, $A\cup B=S$: together they cover every possibility. Throwing a die and letting $A=\{1,2,3,4\}$ (number does not exceed 4) and $B=\{3,4,5,6\}$ (number is not smaller than 3) gives $A\cup B=\{1,2,3,4,5,6\}=S$, so $A$ and $B$ are exhaustive.

The intersection $A\cap B$ is the event that both $A$ and $B$ occur at once - the outcomes common to both sets. With $S$ again the positive integers up to 50, $A$ the multiples of 3 and $B$ the multiples of 5, $A\cap B=\{15,30,45\}$, the numbers divisible by both.

Two events are mutually exclusive (or disjoint) if they share no outcomes at all, i.e. $A\cap B=\phi$: they cannot both happen in the same trial. With $A$ the multiples of 8 and $B$ the multiples of 13 up to 50, $A=\{8,16,24,32,40,48\}$ and $B=\{13,26,39\}$ share no element, so they are mutually exclusive.

When two events are simultaneously mutually exclusive and exhaustive - $A\cap B=\phi$ and $A\cup B=S$ - they are called complementary events: between them they cover the whole sample space with no overlap, so exactly one of the pair always happens. The complement of $A$ is written $A'$ (also $A$-bar or $A^{c}$) and means 'not $A$'.

Combining complement, union and intersection describes richer events: $A'$ means not $A$; $A\cup B$ means at least one of $A$, $B$; $A\cap B$ means both $A$ and $B$; $(A'\cap B)\cup(A\cap B')$ means exactly one of $A$, $B$; and $(A'\cap B')=(A\cup B)'$ means neither $A$ nor $B$.

Worked examples. (1) Tossing a coin and a die together, every outcome pairs a coin face with a die face, so $S=\{(H,1),(H,2),\ldots,(H,6),(T,1),\ldots,(T,6)\}$, 12 outcomes in all. (2) For a two-stage trip - travel to Delhi by car, train or plane, then tour the city by bus or taxi - the sample space lists every (travel mode, tour mode) pair: $S=\{(car,bus),(car,taxi),(train,bus),(train,taxi),(plane,bus),(plane,taxi)\}$. (3) Tossing three coins gives the 8-outcome sample space $S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$. Defining $E_1$ = at least two heads, $E_2$ = at least two tails, $E_3$ = at most one head, and $E_4$ = exactly two heads, we get $E_1=\{HHH,HHT,HTH,THH\}$, $E_2=\{HTT,THT,TTH,TTT\}$, $E_3=\{HTT,THT,TTH,TTT\}$, and $E_4=\{HHT,HTH,THH\}$. Then $E_1\cup E_4=\{HHH,HHT,HTH,THH\}$ (since $E_4\subset E_1$ here) and $E_3'=\{HHH,HHT,HTH,THH\}$. Checking (i) $E_1\cap E_2=\phi$, so $E_1$ and $E_2$ are mutually exclusive; (ii) comparing the two lists shows $E_2=E_3$ exactly, so they are equal events.