Equally Likely Outcomes, introduction to probability lesson

A device shows one of the numbers 1 through 12. What is the probability that it shows a multiple of 3? Four of the twelve numbers qualify, but that count alone does not settle the question, because it says nothing about how the device chooses its number. This lesson shows when “favorable outcomes over total outcomes” is a valid rule, why it needs equally likely outcomes at the level you count, and how grouping outcomes or ignoring order can quietly break that assumption.

Key terms

Uniform finite model

Definition Let $\Omega$ be a finite, nonempty sample space with $n$ outcomes. The outcomes are equally likely if every one-outcome event $\{\omega\}$ has the same probability $p$ . A probability model with this property is called the uniform finite model on $\Omega$ . The one-outcome events are pairwise disjoint and their union is $\Omega$ , so additivity and normalization give $np = 1$ . Every outcome $\omega$ therefore has

P(\{\omega\}) = \frac{1}{n}.

Equal likelihood is an assumption about how the outcome is produced, justified for example by a symmetry, a stated selection mechanism or a deliberate design. It does not follow from the fact that the outcomes can be listed.

In plain words Every possible outcome receives the same share of the total probability 1, so with $n$ outcomes each share is $1/n$ . Saying “chosen at random” does not by itself make the shares equal, because a random procedure can favor some outcomes over others.

Example Suppose the device is a generator designed to pick each number from 1 to 12 with the same probability. Then each number has probability $1/12$ . The event $\{3, 6, 9, 12\}$ is the union of four disjoint one-outcome events, so its probability is

4 \cdot \frac{1}{12} = \frac{4}{12} = \frac{1}{3}.

Cardinality

Definition The cardinality of a finite set $A$ , written $\lvert A \rvert$ , is the number of distinct elements of $A$ . The empty event has $\lvert \varnothing \rvert = 0$ . For finite sets $A$ and $B$ ,

\begin{aligned} \lvert A \cup B \rvert &= \lvert A \rvert + \lvert B \rvert \\[0.4em] &\quad - \lvert A \cap B \rvert, \end{aligned}

and the last term is $0$ when $A$ and $B$ are disjoint.

In plain words Cardinality is a count of members, with each distinct element counted once. An element that lies in both $A$ and $B$ is counted in both $\lvert A \rvert$ and $\lvert B \rvert$ but appears only once in $A \cup B$ , which is why the overlap is subtracted once.

Example In $\Omega = \{1, 2, \dots, 12\}$ , let $A = \{3, 6, 9, 12\}$ be the multiples of 3 and let $B$ be the set of the six even numbers. Then $\lvert \Omega \rvert = 12$ , $\lvert A \rvert = 4$ , $\lvert B \rvert = 6$ and $A \cap B = \{6, 12\}$ , so $\lvert A \cup B \rvert = 4 + 6 - 2 = 8$ . Listing the union as $\{2, 3, 4, 6, 8, 9, 10, 12\}$ confirms the count.

Favorable outcomes

Definition For an event $A$ , the favorable outcomes are the outcomes in $A$ . The word simply means that an outcome satisfies the condition being asked about, so when the event of interest is a failure, its favorable outcomes are the failures. The counting rule states that if $\Omega$ is finite and nonempty and its outcomes are equally likely, then every event $A \subseteq \Omega$ has

P(A) = \frac{\lvert A \rvert}{\lvert \Omega \rvert}.

The proof is that $A$ is a disjoint union of $\lvert A \rvert$ one-outcome events, each with probability $1/\lvert \Omega \rvert$ . Without equal likelihood the rule can fail. In any model on a finite sample space, $P(A)$ is the sum of the probabilities of the outcomes in $A$ , and that sum need not equal the ratio of counts.

In plain words Count the outcomes in the event and divide by the number of all outcomes, but first check that all the outcomes of the sample space are equally likely. Both counts must refer to the same experiment and to outcomes described at the same level of detail.

Example Under the uniform model for the 12-number device, the event “at most 6” has six favorable outcomes, so its probability is $6/12 = 1/2$ . Now change the model. A weighted device gives each number from 1 to 6 probability $1/24$ and each number from 7 to 12 probability $1/8$ . These are valid probabilities, since $6 \cdot \tfrac{1}{24} + 6 \cdot \tfrac{1}{8} = \tfrac{1}{4} + \tfrac{3}{4} = 1$ . The event “at most 6” still has six favorable outcomes out of twelve, but its probability is now $6 \cdot \tfrac{1}{24} = \tfrac{1}{4}$ , not $\tfrac{1}{2}$ . The multiples of 3 happen to keep their old probability,

\begin{aligned} P(\{3, 6, 9, 12\}) &= \frac{2}{24} + \frac{2}{8} \\[0.4em] &= \frac{1}{12} + \frac{3}{12} \\[0.4em] &= \frac{1}{3}, \end{aligned}

yet one matching answer does not make the outcomes equally likely, and dividing counts is not a valid method in this model.

Level of detail

Definition The level of detail of a sample space is how much information each outcome records. A coarser description groups the outcomes of a finer one into categories. Each category is an event, and by additivity its probability is the sum of the probabilities of the outcomes it contains. If the finer outcomes are equally likely, a category containing $k$ of the $n$ finer outcomes has probability $k/n$ , so the categories are equally likely exactly when they contain equally many finer outcomes. The counting rule applies at a level of detail where the outcomes are equally likely, with the numerator and the denominator counted at that same level.

In plain words Giving a group of outcomes a single name does not give it an equal share of the probability. The probabilities of the outcomes in a group travel with them.

Example Record only whether the uniform 12-number device shows a small number (1 or 2), a middle number (3 to 6) or a large number (7 to 12). Three category names remain, but they contain 2, 4 and 6 equally likely numbers, so

\begin{gathered} P(\text{small}) = \frac{2}{12} = \frac{1}{6}, \\[0.4em] P(\text{middle}) = \frac{4}{12} = \frac{1}{3}, \\[0.4em] P(\text{large}) = \frac{6}{12} = \frac{1}{2}. \end{gathered}

Giving each category probability $1/3$ would describe a different device, not a different way of computing for this one. Repeated labels raise the same issue. If a bag holds three tokens marked red and one marked green, and a token is drawn so that every physical token is equally likely, then red has probability $3/4$ . There are two color names, but they are not two equally likely outcomes.

Ordered sequences

Definition When an experiment has several stages, an ordered sequence records the result of each stage in the order the stages occur. Two sequences with the same results in different positions, such as HHT and THH, are different outcomes. Three tosses of a coin have $2^3 = 8$ ordered sequences. In the standard model of three fair tosses made independently, these 8 sequences are equally likely, so each has probability $1/8$ . Independence itself is defined precisely later in the course.

In plain words Keeping the order keeps information. A summary such as the number of heads lumps several sequences together, and the lumps can have different sizes.

Example With H for heads and T for tails, the eight sequences are HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. Exactly two heads occurs in HHT, HTH and THH, so the probability of exactly two heads is $3/8$ . Grouping the sequences by their number of heads gives the table below.

Heads	Sequences	Probability
0	TTT	$1/8$
1	HTT, THT, TTH	$3/8$
2	HHT, HTH, THH	$3/8$
3	HHH	$1/8$

There are four possible head counts, but they are not equally likely. Each count has probability $1/8$ or $3/8$ , and none has the probability $1/4$ that dividing by four would suggest.

Check yourself

A bag holds ten tokens numbered 1 through 10, and one token is drawn so that every token is equally likely. Let $E$ be the event that the number is even and $G$ the event that it is greater than 7. Find $P(E)$ , $P(G)$ and $P(E \cup G)$ .

Show the answer

The ten tokens are the equally likely outcomes, so each probability is a number of tokens divided by 10. The even tokens are 2, 4, 6, 8 and 10, which gives $P(E) = 5/10 = 1/2$ . The tokens greater than 7 are 8, 9 and 10, which gives $P(G) = 3/10$ . The union contains the six distinct tokens 2, 4, 6, 8, 9 and 10. Adding $5 + 3$ would count 8 and 10 twice, and the addition rule agrees with the list,

P(E \cup G) = \frac{5 + 3 - 2}{10} = \frac{3}{5}.

Here even and odd numbers each have probability $1/2$ because each category contains five of the ten equally likely tokens, not because there are two category names.

Two fair coins are tossed, modeled as independent fair tosses. A classmate argues that the number of heads is 0, 1 or 2, so the probability of exactly one head is $1/3$ . Find the correct probability and explain the mistake.

Show the answer

The equally likely outcomes are the four ordered sequences HH, HT, TH and TT, each with probability $1/4$ . Exactly one head occurs in HT and TH, so the probability is $2/4 = 1/2$ . The three head counts are categories of different sizes, with probabilities $1/4$ , $1/2$ and $1/4$ for 0, 1 and 2 heads. Dividing by 3 counts at a level of detail where the outcomes are not equally likely.

A die is loaded so that the face 6 has probability $1/4$ and each of the faces 1 to 5 has probability $3/20$ . Find the probability of an even face, and explain why the answer $3/6$ is wrong for this die.

Show the answer

First check that the model is valid. The faces 1 to 5 carry $5 \cdot \tfrac{3}{20} = \tfrac{15}{20}$ , and adding $\tfrac{5}{20}$ for the face 6 gives 1. The even faces are 2, 4 and 6, so

\begin{aligned} P(\text{even}) &= \frac{3}{20} + \frac{3}{20} + \frac{5}{20} \\[0.4em] &= \frac{11}{20}. \end{aligned}

The ratio $3/6$ counts three even faces out of six, which assumes that all six faces are equally likely. That assumption is false here, since the face 6 is more likely than the others, so the counting rule does not apply. The correct value $11/20 = 0.55$ comes from adding the probabilities of the even faces.

Equally Likely Outcomes

Key terms

Uniform finite model

Cardinality

Favorable outcomes

Level of detail

Ordered sequences

Check yourself

How this lesson connects

Builds on

Used later in