Understanding Probability Laws

Let a random experiment have sample space S. Any assignment of probabilities to events must satisfy three basic laws of probability, called Kolmogorov’s Axioms: 1) For any event A, P(A) ≥ 0. 2) P(S) = 1. 3) If A and B are two mutually exclusive events (i.e., they cannot happen simultaneously), then P(A ( B) = P(A) + P(B).

There are other laws in addition to these three, but Kolmogorov’s Axioms are the foundation for probability theory.

To achieve an understanding of the laws of probability, it helps to have a concrete example in mind.

Consider a single roll of two dice, a red one and a green one. The table below shows the set of outcomes in the sample space, S. Each outcome is a pair of numbers--the number appearing on the red die and the number appearing on the green die. The event that consists of the whole sample space is the event that some one of the outcomes occurs. This event is certain to happen; if we roll the dice, the outcome cannot be something other than one of the 36 outcomes listed in the table. Therefore, the probability associated with the event S is P(S) = 1.

| |1 |(1, 1) |(1,2) |(1, 3) |(1, 4) |(1, 5) |(1, 6) |
| | | | | | | | |
|Number on Green Die| | | | | | | |
| |2 |(2, 1) |(2, 2) |(2, 3) |(2, 4) |(2, 5) |(2, 6) |
| |3 |(3, 1) |(3, 2) |(3, 3) |(3, 4) |(3, 5) |(3, 6) |
| |4