The Sample Space is the set of all possible outcomes of a random experiment. In rigorous probability theory, the nature of this set dictates the mathematical tools required for analysis.
| Term | Definition |
|---|---|
| Outcome | A single specific result of an experiment (\(\omega \in \Omega\)). |
| Event | A subset of the sample space (\(A \subseteq \Omega\)). |
| Power Set | The collection of all possible events. For a finite \(\Omega\) of size \(n\), there are \(2^n\) events. |
Events are manipulated using set-theoretic operations. Let \(A\) and \(B\) be events in \(\Omega\):
Modern probability theory is built upon three fundamental axioms proposed by Andrey Kolmogorov in 1933. Any valid probability measure \(P\) must satisfy these conditions.
For any event \(A\), the probability is a non-negative real number:
The probability of the entire sample space is exactly 1, representing certainty:
For any sequence of mutually exclusive events \(A_1, A_2, A_3, \dots\), the probability of their union is the sum of their individual probabilities:
From the fundamental axioms, several critical theorems are derived for calculation:
In advanced measure-theoretic probability, we define a probability space as a triplet \((\Omega, \mathcal{F}, P)\). The \(\sigma\)-algebra \(\mathcal{F}\) is a collection of subsets of \(\Omega\) that is closed under complements and countable unions. This structure ensures that probabilities can be consistently assigned to complex sets without logical contradiction.
Suppose you have a sample space \(\Omega = \{1, 2, 3, \dots\}\) (the set of all positive integers). Is it possible to define a probability measure \(P\) such that every individual outcome has the same probability \(p > 0\)? Justify your answer using the Axioms.
Three points \(X, Y,\) and \(Z\) are chosen independently and uniformly at random on the circumference of a circle. What is the probability that the center of the circle lies inside the triangle formed by these three points?
Using only the three Kolmogorov Axioms, prove the property of monotonicity: if \(A \subseteq B\), then \(P(A) \leq P(B)\). Hint: Express \(B\) as a union of two disjoint sets involving \(A\).