How to List Every Possible Outcome of a Random Experiment

Before asking how likely something is, you need every way an experiment can turn out. Tossing coins, rolling dice, drawing balls from a bag and testing products all start with a complete list of outcomes and a precise language for events.

This part covers random experiments and sample spaces, types of events, and sample spaces with tree diagrams when outcomes are not equally likely.

A random experiment has more than one possible outcome, all known in advance, but its actual outcome cannot be predicted, and its sample space S is the set of all possible outcomes, each of which is called a sample point.

Features of a random experiment:

- It can be repeated under identical conditions
- All possible outcomes are known beforehand
- The outcome of any single trial cannot be predicted

Worked examples:

- One coin: $S=\{H,T\}$, with $n(S)=2$
- Two coins: $S=\{HH,HT,TH,TT\}$, with $n(S)=4$
- Two dice: $S=\{(a,b):a,b\in \{1,2,3,4,5,6\}\}$, with $n(S)=36$
- A coin tossed until a head appears: $S=\{H,TH,TTH,\dots \}$, an infinite sample space

Counting with the multiplication principle. Three coins give $2^3=8$ outcomes, and a coin with a die gives $2\times 6=12$.

An everyday example. Noting whether each of two children in a family is a boy or a girl, in order of birth, gives the sample space $\{BB,BG,GB,GG\}$.

The substance. HT and TH are different sample points — writing the two-coin sample space with only three outcomes ruins every probability calculated from it.

An event is a subset of the sample space; the sure event is S itself, the impossible event is the empty set, events are mutually exclusive when they cannot happen together, and events are exhaustive when together they cover the whole sample space.

Types of events:

- Simple event — a single sample point
- Sure event — $E=S$
- Impossible event — $E=\varphi$
- Mutually exclusive — $A\cap B=\varphi$
- Exhaustive — the union of the events is S

Worked example. A die is rolled, so $S=\{1,2,3,4,5,6\}$. Let $A=\{2,4,6\}$ (even), $B=\{1,3,5\}$ (odd) and $C=\{1,2,3\}$ (at most 3).

- A and B are mutually exclusive, since $A\cap B=\varphi$
- A and B are exhaustive, since $A\cup B=S$
- A and C are not mutually exclusive, since $A\cap C=\{2\}$
- 'A number greater than 6' is impossible; 'a number less than 7' is sure

Event language. 'A or B' is $A\cup B$, 'A and B' is $A\cap B$, and 'not A' is $A^{\prime }$; an event and its complement are always mutually exclusive and exhaustive.

An everyday example. The result of a one-day cricket match — win, loss, tie or no result — forms four mutually exclusive and exhaustive events, since exactly one must happen.

The substance. Mutually exclusive and exhaustive are separate ideas — 'even' and 'at most 3' are neither, while 'even' and 'odd' are both.

When outcomes are not equally likely, the sample space still lists every outcome, but each branch of a tree diagram carries its own probability, and the probability of a complete path is the product of the probabilities along it.

Building a tree diagram:

- Draw one set of branches for each stage of the experiment
- Label every branch with its outcome and probability
- Branches leaving one point must have probabilities adding to 1
- Multiply along a path, and add across the paths that make up an event

Worked example. A bag holds 3 red and 2 blue balls. One ball is drawn and not replaced, then a second is drawn.

- First draw: R with probability $\frac{3}{5}$ and B with $\frac{2}{5}$
- After R: R with $\frac{2}{4}$ and B with $\frac{2}{4}$; after B: R with $\frac{3}{4}$ and B with $\frac{1}{4}$
- $S=\{RR,RB,BR,BB\}$, with probabilities $\frac{6}{20}$, $\frac{6}{20}$, $\frac{6}{20}$ and $\frac{2}{20}$

The four path probabilities add to 1, yet the outcomes are not equally likely: BB has probability 0.1, while each of the others has 0.3.

Worked example 2. A biased coin shows heads with probability 0.7. Tossed twice, $P(HH)=0.49$, $P(HT)=P(TH)=0.21$ and $P(TT)=0.09$.

An everyday example. A school bus that is on time with probability 0.8 each day gives a two-day tree in which the path 'late on both days' has probability $0.2\times 0.2=0.04$.

The substance. Listing four outcomes does not make each worth one-quarter — equally likely outcomes must be justified, never assumed.