An experiment (or trial) is an action that produces a result. For example, throwing a six sided die is an experiment (or trial).
A result is an observation from an experiment. For example, the value of the number which is face up after throwing a die.
An event (or outcome) is a set of results. For example, in the die experiment, the event 'throwing an even number' is the set {2, 4, 6}.
The term 'outcome' tends to be used when referring to events in general. The term 'event' tends to be used when referring to a particular event.
The sample space of an experiment is the set of possible results that can occur.
The event space of an experiment is the set of results that satisfy an event.
The probability of an event A occurring is \[ \begin{align} P(A) &= \frac{\text{number of ways event A can occur}} {\text{total number of results}} \\ &= \frac{|\text{Event space}|}{|\text{Sample space}|} \end{align} \ \] where the symbol | | means the 'number of results in'.
Example When throwing a die, to calculate the probability of obtaining an even number:
\[ \text{# of even numbers} = 3 \ \]
\[ \text{# of all possibilities} = 6 \ \]
So
\[ \begin{align}
\text{P(Even number)} &= \frac{3}{6} = \frac{1}{2} \\
\end{align} \ \]
The sets involved in probability can be described in terms of Venn diagrams. The universal set U corresponds to the sample space S and the set of events relating to a particular outcome is the set E.
If there are n possible outcomes \[ (E_1, E_2, \cdots, E_n) \ \] then \[ P(E_1) + P(E_2) + \cdots + P(E_n) = 1 \ \]
Example The probability of drawing a heart or a club or a spade or a diamond from a deck of cards is 1 because these are the only possible outcomes. You can use a special type of Venn diagram to represent this:
If an outcome A is certain it has probability 1
If an outcome A is impossible it has probability 0
From the preceding, probability has a range \[ 0 \le P(A) \le 1 \ \]
Example A card is drawn at random from a standard deck. Then
\[ \begin{align}
\text{P(Seven of Spades)} &= \dfrac{1}{52} \\
\text{P(Heart)} = \dfrac{13}{52} &= \dfrac{1}{4} \\
\text{P(Heart or Diamond)} = \dfrac{13}{52} + \dfrac{13}{52} &= \dfrac{1}{2} \\
\text{P(Heart or Diamond or Club or Spade)} \\
= \dfrac{13}{52} + \dfrac{13}{52} + \dfrac{13}{52} + \dfrac{13}{52} &= 1 \qquad \qquad
\end{align} \ \]
Outcomes that have the same probability of occurring are equally likely outcomes.
For example, when throwing a die, the probability of throwing any of the numbers 1,2,3,4,5,6 is the same, so they are equally likely.
Guided Examples
If A occurs then the notation for A not occurring is \( \bar A \). The associated probabilities add to 1: \[ P(A) + P(\bar A) = 1 \ \] or \[ P(\bar A) = 1 - P(A) \ \] in terms of Venn diagrams this is:
Example If two dice are thrown, what is the probability that the sum of the two numbers is more then three?
The possible outcomes range from two to twelve. In this case it would be easier to calculate the probability of three or less. The total possible outcomes are (1, 1, 1, 2) ... (6, 6). So there are 36 outcomes in the sample space. The possible outcomes for a sum of three or less are (1,1) or (1,2) or (2,1) so \[ \begin{align} \text{P}(\text{Sum}\le 3) &= \frac{3}{36} = \frac{1}{12} \quad \\ \end{align} \ \] and the probability of more than three is \[ \begin{align} \text{P}(\text{Sum}\gt 3) &= 1 - \text{P}(\text{Sum}\le 3) \\ \text{P}(\text{Sum}\gt 3) &= 1 - \frac{1}{12} \\ &= \frac{11}{12} \end{align} \ \]
Outcomes that cannot both occur at the same time are mutually exclusive. This means the event spaces of the two outcomes are disjoint sets.
If A and B are not mutually exclusive, the outcome of both events occurring is denoted by \( A\ and\ B \) and their probability is \( P(AB) \). The event space of both events occurring is the intersection of the individual event spaces.
If A and B are two events, the outcome of A or B occurring is denoted by \( A\ \cup B \).
If the two events are mutually exclusive, then \[ P(A \cup B) = P(A) + P(B) \quad \]
If the two events are not mutually eclusive, then \[ P(A \cup B) = P(A) + P(B) - P(AB) \ \] the last term on the right hand side is to adjust for counting the intersection twice.
Example In a class of high school students, 75% are on Facebook and 45% are on Twitter, and every student is on at least one of these. What is the probability that a student selected at random is on both?
The two possibilities aren't mutually exclusive so \[ \begin{align} P(\text{Facebook} \cup \text{Twitter}) &= P(\text{Facebook}) + P(\text{Twitter}) \\ &\quad- P(\text{Facebook and Twitter}) \\ 1 &= 0.75 + 0.45 \\ &\quad- P(\text{Facebook and Twitter}) \\ P(\text{Facebook and Twitter}) &= 1.2 - 1 \\ &= 0.20 \end{align} \ \] So the probability is 0.20