﻿ Basic Probability Concepts
BASIC PROBABILITY CONCEPTS
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!

Probability is the area of mathematics devoted to predicting the likelihood of uncertain occurrences.

For example, when you roll a die, it is uncertain whether you'll see a $\,1\,$, $\,2\,$, $\,3\,$, $\,4\,$, $\,5\,$, or $\,6\,$.
However, it is possible to talk about how likely it is for each number to occur.

DEFINITIONS probability, experiment, outcome, sample space, event
A probability is a number between $\,0\,$ and $\,1\,$
that represents the likelihood of occurrence of something.

An experiment is the act of making an observation or taking a measurement.

An outcome is one of the possible things that can occur as a result of an experiment.

The set of all possible outcomes for an experiment is called its sample space,
and is frequently denoted by $\,S\,$.

Any subset of the sample space is called an event, and is frequently denoted by $\,E\,$.

The probability of an event $\,E\,$ is denoted by $\,P(E)\,$.

Given any set $\,A\,$, the notation $\,n(A)\,$ denotes the number of elements in $\,A\,$.

 A die is a cube with the numbers $\,1\,$ through $\,6\,$ represented on its six faces. When it is thrown (‘rolled’), one of these six faces appears on top. For a fair die, each of the numbers $\,1\,$ through $\,6\,$ is equally likely to occur.

There are lots of experiments involving a single fair die.
Here are some of them, with their corresponding outcomes and sample spaces:
 EXPERIMENT OUTCOMES SAMPLE SPACE (1)   Roll once; record the number that appears on the top face. six possible outcomes: $\,1\,$, $\,2\,$, $\,3\,$, $\,4\,$, $\,5\,$, $\,6\,$ $S = \{1,2,3,4,5,6\}$ (2)   Roll once; record the number that appears on the bottom (hidden) face. six possible outcomes: $\,1\,$, $\,2\,$, $\,3\,$, $\,4\,$, $\,5\,$, $\,6\,$ $S = \{1,2,3,4,5,6\}$ (3)   Roll twice; record the numbers on each of these rolls, in order. In this experiment and those below, use the notation $\,1\star5\,$ to denote a roll of $\,1\,$, followed by a roll of $\,5\,$. $36\,$ possible outcomes: $\,1\star1\,$, $\,1\star2\,$, $\,1\star3\,$, $\,1\star4\,$, $\,1\star5\,$, $\,1\star6\,$ $\,2\star1\,$, $\,2\star2\,$, $\,2\star3\,$, $\,2\star4\,$, $\,2\star5\,$, $\,2\star6\,$ $\,3\star1\,$, $\,3\star2\,$, $\,3\star3\,$, $\,3\star4\,$, $\,3\star5\,$, $\,3\star6\,$ $\,4\star1\,$, $\,4\star2\,$, $\,4\star3\,$, $\,4\star4\,$, $\,4\star5\,$, $\,4\star6\,$ $\,5\star1\,$, $\,5\star2\,$, $\,5\star3\,$, $\,5\star4\,$, $\,5\star5\,$, $\,5\star6\,$ $\,6\star1\,$, $\,6\star2\,$, $\,6\star3\,$, $\,6\star4\,$, $\,6\star5\,$, $\,6\star6\,$ $S = \{$ $\,1\star1\,$, $\,1\star2\,$, $\,1\star3\,$, $\,1\star4\,$, $\,1\star5\,$, $\,1\star6\,$, $\,2\star1\,$, $\,2\star2\,$, $\,2\star3\,$, $\,2\star4\,$, $\,2\star5\,$, $\,2\star6\,$, $\,3\star1\,$, $\,3\star2\,$, $\,3\star3\,$, $\,3\star4\,$, $\,3\star5\,$, $\,3\star6\,$, $\,4\star1\,$, $\,4\star2\,$, $\,4\star3\,$, $\,4\star4\,$, $\,4\star5\,$, $\,4\star6\,$, $\,5\star1\,$, $\,5\star2\,$, $\,5\star3\,$, $\,5\star4\,$, $\,5\star5\,$, $\,5\star6\,$, $\,6\star1\,$, $\,6\star2\,$, $\,6\star3\,$, $\,6\star4\,$, $\,6\star5\,$, $\,6\star6\,$ $\}$ (4)   Roll twice; record the sum of the numbers that appear. $\,11\,$ possible outcomes: The smallest sum you can get is $\,2\,$; this results only from rolling a $\,1\,$ followed by a $\,1\,$. The largest sum you can get is $\,12\,$; this results only from rolling a $\,6\,$ followed by a $\,6\,$. Convince yourself that every whole number between $\,2\,$ and $\,12\,$ is also a possible outcome; for example, you could get $\,5\,$ in all these ways: $\,1\star4\,$,   $\,2\star3\,$,   $\,3\star2\,$,   $\,4\star1\,$ $S = \{2,3,4,5,6,7,8,9,10,11,12\}$ (5)   Roll twice; record the greatest number that appears on the two rolls. six possible outcomes: $\,1\,$, $\,2\,$, $\,3\,$, $\,4\,$, $\,5\,$, $\,6\,$ For $\,1\star1\,$, the greatest number is $\,1\,$. This is the only way to get an outcome of $\,1\,$. For $\,1\star2\,$, the greatest number is $\,2\,$. There are three ways to get the outcome $\,2\,$: $\,1\star 2\,$,   $\,2\star 1\,$,   $\,2\star2\,$ For $\,1\star3\,$, the greatest number is $\,3\,$. There are five ways to get the outcome $\,3\,$: $\,1\star 3\,$,   $\,2\star 3\,$,   $\,3\star1\,$,   $\,3\star 2\,$,   $\,3\star 3\,$ $S = \{1,2,3,4,5,6\}$

NOT ALL SAMPLE SPACES ARE CREATED EQUAL

Some sample spaces are much easier to work with than others.

In experiments (1), (2) and (3) above, each member of the sample space is equally likely:

• In experiments (1) and (2), you're equally likely to roll a $\,1\,$, $\,2\,$, $\,3\,$, $\,4\,$, $\,5\,$, or $\,6\,$.
• In experiment (3), each of the $\,36\,$ possible outcomes are equally likely:
for example, the chance of rolling a $\,1\,$ followed by a $\,3\,$ is (say) the same as rolling a $\,5\,$ followed by a $\,4\,$.
In experiments (4) and (5), however, the outcomes are not all equally likely:
• In experiment (4), there's only one way to get a $\,2\,$.
However, there are four ways to get a $\,5\,$.
So, the outcome $\,2\,$ is less likely than the outcome $\,5\,$.
• In experiment (5), there's only one way to get the the outcome $\,1\,$.
However, there are five ways to get the outcome $\,3\,$.
So, the outcome $\,1\,$ is less likely than the outcome $\,3\,$.

When a sample space has equally likely outcomes, then computing probabilities is as easy as counting:

EQUALLY LIKELY OUTCOMES
Let $\,S\,$ be a sample space with a finite number of outcomes.
Suppose each outcome in $\,S\,$ is equally likely to occur.
Then, the probability of an event $\,E\,$ is found by taking the number of outcomes in $\,E\,$
and dividing by the number of outcomes in $\,S\,$.
That is: $$\cssId{s96}{P(E)} \cssId{s97}{= \frac{n(E)}{n(S)}}$$

This is best illustrated by an example.
Let's consider the first (or second) experiment above—a single roll of a fair die.
The example below also clarifies the idea of an ‘event’,
and illustrates notation that is frequently used in connection with probability problems.

EXAMPLE
Consider a single roll of a fair die.
The sample space is $\,S = \{1,2,3,4,5,6\}\,$.
Note that $\,n(S) = 6\,$, since there are $\,6\,$ members in $\,S\,$.

Since this is a fair die, each member of $\,S\,$ is equally likely.
Let $\,x\,$ denote the number that is rolled.

An ‘event’ is a subset of the sample space, so there are many possible events.
Any subset $\,E\,$ of $\,S\,$ will have six or fewer members, so $\,0 \le n(E) \le 6\,$.
The table below illustrates several events, their interpretations, and some conventional language and notation:

 event interpretation of event probability of event some conventional language usedto report the probability $E = \{3\}$ getting a $\,3\,$ on a single roll of a fair die $\displaystyle \cssId{s118}{\frac{n(E)}{n(S)}} \cssId{s119}{= \frac16}$ $\cssId{s120}{P(x = 3)} \cssId{s121}{= \frac16}$ read as: “The probability that $\,x\,$ is $\,3\,$ is $\,\frac16\,$.” $E = \{4\}$ getting a $\,4\,$ on a single roll of a fair die $\displaystyle\frac{n(E)}{n(S)} = \frac16$ $P(x = 4) = \frac16$ $E = \{3,4\}$ getting a $\,3\,$ or a $\,4\,$ on a single roll of a fair die $\displaystyle\frac{n(E)}{n(S)} = \frac26 = \frac13$ $P(x = 3 \text{ or } x=4) = \frac26 = \frac 13$ $E = \{2,4,6\}$ getting an even number on a single roll of a fair die $\displaystyle\frac{n(E)}{n(S)} = \frac36 = \frac12$ $P(x \text{ is even}) = \frac36 = \frac 12$ $E = \{2,3,4,5,6\}$ getting a number greater than $\,1\,$on a single roll of a fair die $\displaystyle\frac{n(E)}{n(S)} = \frac56$ $P(x \gt 1) = \frac56$
Master the ideas from this section