Textbook
In a finite probability model, $\Omega$ is the set of possible experimental outcomes. Each outcome $\omega$ receives a probability $p_\omega\ge0$, and $\sum_{\omega\in\Omega}p_\omega=1$. An event $A\subseteq\Omega$ is a set of outcomes of interest.
$$P(A):=\sum_{\omega\in A}p_\omega.$$
Consequently, $0\le P(A)\le1$, $P(\varnothing)=0$, and $P(\Omega)=1$. For disjoint events, probabilities add under union. Overlapping outcomes must not be counted twice.
The shortcut $P(A)=|A|/|\Omega|$ applies when the finite outcomes are equally likely. With unequal weights, counting favorable outcomes cannot replace adding their probabilities. A fair coin or die is an assumption of the model, not a reward for convenient arithmetic.
An event’s probability depends on the outcome space and its weights. The same phrase, “greater than 3,” can specify another set in a different experiment. If $A\subseteq B$, summing nonnegative weights over $A$ cannot exceed the sum over $B$, so $P(A)\le P(B)$. This supplies a consistency check: enlarging an event cannot reduce its probability while the underlying model stays unchanged.
Definitions and notation: Mathematics for Computer Science.
In plain language
The whole probability pie has size 1, with a share assigned to each outcome. An event collects the weights of all shares satisfying its condition. Counting alone works only for equal shares; two outcome names do not guarantee equal chances. The complete outcome space and its weights come first: an outcome of two tosses must include both results, and a missing slice cannot be blamed on an unpopular crust.
Example
The constructed model uses a fair six-sided die. The example determines the probability of rolling more than 3.
Event probability in a finite model
All outcomes
Each of the six outcomes has probability 1/6.
Step 1 of 3: All outcomes
Read all steps
All outcomes
Each of the six outcomes has probability 1/6.
- Die
- 123456
Event outcomes
Only 4, 5, and 6 satisfy the condition.
- Greater than 3
- 1excluded2excluded3excluded4included5included6included
Combined weight
Three weights of 1/6 sum to 1/2.
- Selected weights
- 1/6included1/6included1/6included
- Probability
- 1/2current
The scene’s event contains three distinct outcomes: 4, 5, and 6. Their weights sum to half the total. The remaining outcomes 1, 2, and 3 form the other half, so both parts add to one. Counting three out of six works because the die is fair. Keeping the same face names but changing their weights would remove the argument for probability $1/2$.
Computing an event probability requires a defined outcome space, its weights, and the event condition.