Probability in a finite equally likely model

Content updated: October 22, 2026

The ratio of favorable outcomes to all outcomes is a probability formula only when the outcomes in the chosen model are equally likely. Specify the drawing or tossing procedure before counting.

Learning goals: check the equally likely model; count total and favorable outcomes; distinguish ordered outcomes from unordered groups.

Prerequisites: sample spaces, events, and combinations.

On this page: Definition · Method · Examples · Practice · Further learning

1. Assumptions of the classical formula

Suppose $\Omega$ is finite and nonempty and all outcomes have equal probability. Then for an event $A\subseteq\Omega$,

A favorable outcome is an element of $A$. Each outcome has probability $1/|\Omega|$, and adding these probabilities establishes the formula. If outcomes are not equally likely, sum their probability weights rather than automatically taking a ratio of cardinalities.

From $0\le|A|\le|\Omega|$ it follows that $0\le P(A)\le1$. The impossible event has $P(\varnothing)=0$ and the sure event has $P(\Omega)=1$. These three basic properties follow directly from counting in the finite uniform model.

2. Use a consistent counting model

A fair six-sided die has six equally likely outcomes. For a box of balls, regard each ball as distinct even if some have the same color. A uniformly chosen group of 3 balls from 10, without replacement and without regard to order, gives a sample space with $\binom{10}{3}$ elements.

For sequential draws, use ordered tuples in both numerator and denominator. Each group of 3 distinct balls has exactly $3!$ orders, so the two methods give the same probability for uniform sequential draws without replacement. Do not treat color patterns as equally likely outcomes.

To obtain exactly 2 red balls and 1 white ball, choose 2 from the red balls and then 1 from the white balls. These choices uniquely determine a favorable group, so multiply the two combination counts. Before reducing the fraction, check that the numerator does not exceed the denominator and the result belongs to $[0;1]$.

3. Worked examples

Example 1. A fair die

Roll a fair six-sided die. Here $\Omega=\{1,2,3,4,5,6\}$. The event $A$ of an even result has 3 outcomes; $B$ of at most 4 dots has 4 outcomes. The event $C$ of at least 1 dot is $\Omega$, whereas $D$ of at least 7 dots is empty. Thus

Counting six faces gives the correct denominator because of fairness, not merely because the die has six faces. A biased die has the same possible outcomes but may have different probabilities.

Example 2. Exactly two red balls and one white ball

A box contains 10 balls: 6 red and 4 white. Choose 3 uniformly at random without replacement. For $A$, the event of exactly 2 red balls and 1 white ball, there are $\binom{10}{3}=120$ outcomes and $\binom62\binom41=15\cdot4=60$ favorable outcomes. Hence

Do not use $2/3$ as the probability merely because 2 of the 3 required balls are red. The numerator must count specific groups; the red proportion in a required group is not the probability of selecting such a group.

Example 3. Checking by sequential draws

Use the same box and consider the order red–red–white. Its probability is $(6/10)(5/9)(4/8)=1/6$. There are three positions for the white ball; these cases are mutually exclusive and each has probability $1/6$. Their sum is $1/2$, agreeing with the combination count. Denominators 9 and 8 reflect sampling without replacement, rather than three independent draws.

4. Practice

1. For a fair die, find the probability of a result greater than 4.
2. From the box with 6 red and 4 white balls, choose 3 uniformly without replacement. Find the probability that all 3 are white.
3. Can the four patterns “0,1,2,3 red balls” be treated as four equally likely outcomes of the box experiment?

5. Common mistakes

Retain the equally likely assumption. Do not mix unordered groups in the numerator with ordered tuples in the denominator. Do not assume color patterns are uniform. State whether balls are replaced, since replacement gives a different model.

6. Further learning

Random experiments and events (English lesson) · Permutations and combinations (English lesson) · Conditional probability (English lesson) · The addition rule of probability (English lesson).

Original exercise resource on Scribd.