4.2: Terminology and Notation for Probability

Introduction

Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, then the experiment is said to be a chance experiment. Flipping one fair coin twice is an example of an experiment.

A result of an experiment is called an outcome. The sample space of an experiment is the set of all possible outcomes. Three ways to represent a sample space are: to list the possible outcomes either using roster notation or a two-way table, to create a tree diagram, or to create a Venn Diagram. 

The uppercase letter S is often used to denote the sample space. For example, if you flip one fair coin, \(S = \{\text{H, T}\}\) where \(\text{H} =\) heads and \(\text{T} =\) tails are the outcomes.

An event is any combination of outcomes. Upper case letters like \(\text{A}\) and \(\text{B}\) represent events. For example, if the experiment is to flip one fair coin, event \(\text{A}\) might be getting at most one head. The probability of an event \(\text{A}\) is written \(P(\text{A})\).

Definition: probability

The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive (that is, zero and one and all numbers between these values). Also, if we add all the probabilities for each event in the sample space, the sum equals 1.

• \(P(\text{A}) = 0 = 0 \%\) means the event \(\text{A}\) can never happen.
• \(P(\text{A}) = 1 = 100 \%\) means the event \(\text{A}\) always happens.
• \(\sum P(\text{A}) = 1\), where the Greek letter \(\Sigma\) represents "sum".
• For any event E, the probability of E is between 0 and 1 inclusive: \(0 \le P(E) \le 1\)

Probabilities can be expressed as fractions, decimals rounded to at least three places, or percentages. 

• Final answers as fractions are typically reduced but can be more useful in calculations if left unreduced.
• If probability of an event is extremely small, then it is alright to round it to the first nonzero digit after the decimal point. You only want to write 0 as your answer if it is impossible for the event to occur.

Types of Probability

Classical Probability

• Also called Theoretical Probability
• Uses sample spaces to determine the numerical probability that an event will happen.
• An experiment does not actually have to be performed. 
• Assumes all outcomes in the sample space are equally likely to occur.
• The probability of any event \(E\) is \(\dfrac{\text{Number of outcomes in E}}{\text{Total number of outcomes in the sample space}}\)
  • Denoted as \(P(E) = \dfrac{n(E)}{n(S)}\)

Equally likely means that each outcome of an experiment occurs with equal probability. For example, if you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head (\(\text{H}\)) and a Tail (\(\text{T}\)) are equally likely to occur. If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer.

\(P(\text{A}) = 0.5\) means the event \(\text{A}\) is equally likely to occur or not to occur. For example, if you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches 0.5 (the probability of heads)

Example \(\PageIndex{1}\)

Toss a fair dime and a fair nickel. Find the probability of getting one head landing face up. 

Solution

To calculate the probability of an event A when all outcomes in the sample space are equally likely, count the number of outcomes for event \(\text{A}\) and divide by the total number of outcomes in the sample space.

If you toss a fair dime and a fair nickel, the sample space is \(\{\text{HH, TH, HT,TT}\}\) where \(\text{T} =\) tails and \(\text{H} =\) heads. The sample space has four outcomes. Define the event we are interested in as \(\text{A} =\) getting one head. There are two outcomes that meet this condition \(\text{\{HT, TH\}}\), so

\(P(\text{A}) = \frac{2}{4} = 0.5\).

Example \(\PageIndex{2}\)

Suppose you roll one fair six-sided die, with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event \(\text{E} =\) rolling a number that is at least five. Find the probability of that event.

Solution

There are two outcomes {5, 6}, so  \(P(\text{E}) = \frac{2}{6}\).

Empirical Probability

• Relative frequency approximation of probability
• Conduct (or observe) a procedure, and count the number of times that an event actually occurs out of the total number of observations. 
• The probability of any event \(E\) is \(\dfrac{\text{Frequency of the class or classes}}{\text{Total of the frequencies}}\)
  • Denoted as \(P(E) = \dfrac{f}{n}\)

Example \(\PageIndex{3}\)

A store manager gathers some demographic information from the store's customers. The following chart summarizes the age-related information they collected from 152 customers:

Age of Customers

Age	Number of Customers
Younger than 18	37
18 - 21	55
22 - 25	25
26 - 29	20
30 or older	15

 One customer is chosen at random to receive a gift card. What is the probability that the customer is at least 18 but younger than 26?

Solution

There are two classes that fit the event: 18-21 and 22-25, which means we have \(55+25=80\) outcomes for this event, which we will call \(E\). 

So, \(P(E) = \dfrac{80}{152}=0.52631\)...which we will round to 0.526. There is about a 52.6% chance that a customer who is at least 18 but younger than 26 will receive the gift card. 

Consider the earlier example of finding the probability of rolling at least 5 on a fair six sided die. If you were to roll the die only a few times, you should not be surprised if your observed results do not match the probability of \(\frac{2}{6}\). In actual experimentation, many factors affect outcomes. However, if you were to roll the die a very large number of times, the distribution of the data would appraoch the expected probability of \(\frac{2}{6}\). The long-term relative frequency of obtaining this result would approach the theoretical probability of \(\frac{2}{6}\) as the number of repetitions grows larger and larger.

Definition: law of large numbers

This important characteristic of probability experiments is known as the law of large numbers which states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term observed relative frequency will approach the theoretical probability. (The word empirical is often used instead of the word observed.)

It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may be unfair, or biased. Two math professors in Europe had their statistics students test the Belgian one Euro coin and discovered that in 250 trials, a head was obtained 56% of the time and a tail was obtained 44% of the time. The data seem to show that the coin is not a fair coin; more repetitions would be helpful to draw a more accurate conclusion about such bias. Some dice may be biased. Look at the dice in a game you have at home; the spots on each face are usually small holes carved out and then painted to make the spots visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight weight differences due to the different numbers of holes in the faces. Gambling casinos make a lot of money depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice have flat faces; the holes are completely filled with paint having the same density as the material that the dice are made out of so that each face is equally likely to occur. Later we will learn techniques to use to work with probabilities for events that are not equally likely.

Subjective Probability

• Uses a probability value based on an educated guess or estimate using knowledge of the relevant circumstances.
• Based on opinion, inexact data, past experience.

This type of probability calculation is beyond the scope of this course, but some examples include:

• The probability that it will rain during the last week of September.
• The probability that particular horse will win the Kentucky Derby. 
• The probability that an earthquake of magnitude 5.0 or higher will occur in the next 50 years on the San Andreas fault.

All of these events have data associated with them, but are not a simple matter of dividing the number times an event happens by the total number of observations. 

Types of Events 

The "OR" Event

An outcome is in the event \(\text{A OR B}\) if the outcome is in \(\text{A}\) or is in \(\text{B}\) or is in both \(\text{A}\) and \(\text{B}\). 

This type of event is a union of sets. The notation for this is \(\text{A} \cup \text{B}\)

Example \(\PageIndex{4}\)

Let \(\text{A} = \{1, 2, 3, 4, 5\}\) and \(\text{B} = \{4, 5, 6, 7, 8\}\).

Find \(\text{A OR B}\).

Solution

\(\text{A OR B} = \{1, 2, 3, 4, 5, 6, 7, 8\}\). Notice that 4 and 5 are NOT listed twice.

 

The "AND" Event

An outcome is in the event \(\text{A AND B}\) if the outcome is in both \(\text{A}\) and \(\text{B}\) at the same time.

This type of event is an intersection of sets. The notation for this is \(\text{A} \cap \text{B}\)

Example \(\PageIndex{5}\)

Let \(\text{A}\) = {1, 2, 3, 4, 5} and  \(\text{B}\) = {4, 5, 6, 7, 8}.

Find \(\text{A AND B}\).

Solution

 \(\text{A AND B} = {4, 5}\). Notice that only the elements that are listed in both sets appear. 

 

Definition: Complement

The complement of event \(\text{A}\) is denoted \(\text{A'}\) (read "A prime"). \(\text{A'}\) consists of all outcomes that are NOT in \(\text{A}\).

Notice that \(P(\text{A}) + P(\text{A'}) = 1\).

Example \(\PageIndex{6}\)

Let \(\text{S} = \{1, 2, 3, 4, 5, 6\}\) and let \(\text{A} = {1, 2, 3, 4}\).

1. Find \(\text{A'}\).
2. Find \(P(A)\) and \(P(A')\)

Solution

1. \(\text{A′} = {5, 6}\). These are the only outcomes in S that are not already in A. 
2. \(P(A) = \frac{4}{6}\), \(P(\text{A'}) = \frac{2}{6}\)  

Notice: \(P(\text{A}) + P(\text{A'}) = \frac{4}{6} + \frac{2}{6} = 1\). These sets, A and A' contain all possible outcomes of S, so their probabilities when added together are 100% of the possibilities. 

 

Definition: Conditional Events

The conditional probability of \(\text{A}\) given \(\text{B}\) is written \(P(\text{A|B})\). \(P(\text{A|B})\) is the probability that event \(\text{A}\) will occur given that the event \(\text{B}\) has already occurred.

A condition reduces the sample space. We calculate the probability of \(\text{A}\) from the reduced sample space \(\text{B}\).

The formula for \(P(\text{A|B})\) is \(P(\text{A|B} = \frac{n(A \text{and} B)}{n(B)}\)

Alternately, we can use the probabilties calculate \(P(\text{A|B})\) is \(P(\text{A|B}) = \frac{\text{P(A AND B)}}{\text{P(B)}}\) where \(P(\text{B})\) is greater than zero.

Example \(\PageIndex{7}\)

Suppose we toss one fair, six-sided die. Find the probability that a 2 or a 3 is rolled given that we rolled an even number.

Solution

The sample space \(\text{S} = \{1, 2, 3, 4, 5, 6\}\).

Let \(\text{A} =\) face is 2 or 3 and \(\text{B} =\) face is even (2, 4, 6).

To calculate \(P(\text{A|B})\), we count the number of outcomes 2 or 3 in the sample space \(\text{B} = \{2, 4, 6\}\).

Then we divide that by the number of outcomes in \(\text{B}\) rather than \(\text{S}\), so 

\[P(\text{A|B}) =\dfrac{1}{3}\]

We get the same result by using the formula. Remember that \(\text{S}\) has six outcomes.

\[P(\text{A|B}) = \dfrac{ \text{ P(A AND B) } } {P(\text{B})} = \dfrac{\dfrac{\text{(the number of outcomes that are 2 or 3 and even in S)}}{6}}{\dfrac{\text{(the number of outcomes that are even in S)}}{6}} = \dfrac{\dfrac{1}{6}}{\dfrac{3}{6}} = \dfrac{1}{3}\]

 

Practice with Terminology and Symbols

It is important to read each problem carefully to think about and understand what the events are. Understanding the wording is the first very important step in solving probability problems. Reread the problem several times if necessary. Clearly identify the event of interest. Determine whether there is a condition stated in the wording that would indicate that the probability is conditional; carefully identify the condition, if any.

Example \(\PageIndex{8}\)

The sample space \(S\) is the whole numbers starting at one and less than 20.

1. \(S =\) _____________________________

   Let event \(A =\) the even numbers and event \(B =\) numbers greater than 13.

2. \(A =\) _____________________, \(B =\) _____________________
3. \(P(\text{A}) =\) _____________, \(P(\text{B}) =\) ________________
4. \(\text{A AND B} =\) ____________________, \(\text{A OR B} =\) ________________
5. \(P(\text{A AND B}) =\) _________, \(P(\text{A OR B}) =\) _____________
6. \(\text{A′} =\) _____________, \(P(\text{A′}) =\) _____________
7. \(P(\text{A}) + P(\text{A′}) =\) ____________
8. \(P(\text{A|B}) =\) ___________, \(P(\text{B|A}) =\) _____________; are the probabilities equal?

Answer

1. \(\text{S} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19\}\)
2. \(\text{A} = \{2, 4, 6, 8, 10, 12, 14, 16, 18\}, \text{B} = \{14, 15, 16, 17, 18, 19\}\)
3. \(P(\text{A}) = \frac{9}{19}\), \(P(\text{B}) = \frac{6}{19}\)
4. \(\text{A AND B} = \{14,16,18\}\), \(\text{A OR B} = \{2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19\}\)
5. \(P(\text{A AND B}) = \frac{3}{19}\), \(P(\text{A OR B}) = \frac{12}{19}\)
6. \(\text{A′} = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19\); \(P(\text{A′}) = \frac{10}{19}\)
7. \(P(\text{A}) + P(\text{A′}) = 1\left((\frac{9}{19} + \frac{10}{19} = 1\right)\)
8. \(P(\text{A|B}) = \frac{\text{P(A AND B)}}{\text{P(B)}} = \frac{3}{6}, P(\text{B|A}) = \frac{\text{P(A AND B)}}{\text{P(A)}} = \frac{3}{9}\), No

Exercise \(\PageIndex{1}\)

The sample space S is the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)).

1. \(S =\) _____________________________
   Let event \(A =\) the sum is even and event \(B =\) the first number is prime.
2. \(A =\) _____________________, \(B =\) _____________________
3. \(P(\text{A}) =\) _____________, \(P(\text{B}) =\) ________________
4. \(\text{A AND B} =\) ____________________, \(\text{A OR B} =\) ________________
5. \(P(\text{A AND B}) =\) _________, \(P(\text{A OR B}) =\) _____________
6. \(\text{B′} =\) _____________, \(P(\text{B′)} =\) _____________
7. \(P(\text{A}) + P(\text{A′}) =\) ____________
8. \(P(\text{A|B}) =\) ___________, \(P(\text{B|A}) =\) _____________; are the probabilities equal?

Answer

1. \(\text{S} = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)\}\)
2. \(\text{A} = \{(1,1), (1,3), (2,2), (2,4), (3,1), (3,3)\}\)
   \(\text{B} = \{(2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)\}\)
3. \(P(\text{A}) = \frac{1}{2}\), \(P(\text{B}) = \frac{2}{3}\)
4. \(\text{A AND B} = \{(2,2), (2,4), (3,1), (3,3)\}\)
   \(\text{A OR B} = \{(1,1), (1,3), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)\}\)
5. \(P(\text{A AND B}) = \frac{1}{3}, P(\text{A OR B}) = \frac{5}{6}\)
6. \(\text{B′} = \{(1,1), (1,2), (1,3), (1,4)\}, P(\text{B′}) = \frac{1}{3}\)
7. \(P(\text{B}) + P(\text{B′}) = 1\)
8. \(P(\text{A|B}) = \frac{P(\text{A AND B})}{P(\text{B})} = \frac{1}{2}, P(\text{B|A}) = \frac{P(\text{A AND B})}{P(\text{B})} = \frac{2}{3}\), No.

Example \(\PageIndex{9}\)

A fair, six-sided die is rolled. Describe the sample space S, identify each of the following events with a subset of S and compute its probability (an outcome is the number of dots that show up).

1. Event \(\text{T} =\) the outcome is two.
2. Event \(\text{A} =\) the outcome is an even number.
3. Event \(\text{B} =\) the outcome is less than four.
4. The complement of \(\text{A}\).
5. \(\text{A GIVEN B}\)
6. \(\text{B GIVEN A}\)
7. \(\text{A AND B}\)
8. \(\text{A OR B}\)
9. \(\text{A OR B′}\)
10. Event \(\text{N} =\) the outcome is a prime number.
11. Event \(\text{I} =\) the outcome is seven.

Solution

1. \(\text{T} = \{2\}\), \(P(\text{T}) = \frac{1}{6}\)
2. \(A = \{2, 4, 6\}\), \(P(\text{A}) = \frac{1}{2}\)
3. \(\text{B} = \{1, 2, 3\}\), \(P(\text{B}) = \frac{1}{2}\)
4. \(\text{A′} = \{1, 3, 5\}, P(\text{A′}) = \frac{1}{2}\)
5. \(\text{A|B} = \{2\}\), \(P(\text{A|B}) = \frac{1}{3}\)
6. \(\text{B|A} = \{2\}\), \(P(\text{B|A}) = \frac{1}{3}\)
7. \(\text{A AND B} = {2}, P(\text{A AND B}) = \frac{1}{6}\)
8. \(\text{A OR B} = \{1, 2, 3, 4, 6\}\), \(P(\text{A OR B}) = \frac{5}{6}\)
9. \(\text{A OR B′} = \{2, 4, 5, 6\}\), \(P(\text{A OR B′}) = \frac{2}{3}\)
10. \(\text{N} = \{2, 3, 5\}\), \(P(\text{N}) = \frac{1}{2}\)
11. A six-sided die does not have seven dots. \(P(7) = 0\).

Example \(\PageIndex{10}\)

Table describes the distribution of a random sample \(S\) of 100 individuals, organized by gender and whether they are right- or left-handed.

	Right-handed	Left-handed
Males	43	9
Females	44	4

Let’s denote the events \(M =\) the subject is male, \(F =\) the subject is female, \(R =\) the subject is right-handed, \(L =\) the subject is left-handed. Compute the following probabilities:

1. \(P(\text{M})\)
2. \(P(\text{F})\)
3. \(P(\text{R})\)
4. \(P(\text{L})\)
5. \(P(\text{M AND R})\)
6. \(P(\text{F AND L})\)
7. \(P(\text{M OR F})\)
8. \(P(\text{M OR R})\)
9. \(P(\text{F OR L})\)
10. \(P(\text{M'})\)
11. \(P(\text{R|M})\)
12. \(P(\text{F|L})\)
13. \(P(\text{L|F})\)

Answer

1. \(P(\text{M}) = 0.52\)
2. \(P(\text{F}) = 0.48\)
3. \(P(\text{R}) = 0.87\)
4. \(P(\text{L}) = 0.13\)
5. \(P(\text{M AND R}) = 0.43\)
6. \(P(\text{F AND L}) = 0.04\)
7. \(P(\text{M OR F}) = 1\)
8. \(P(\text{M OR R}) = 0.96\)
9. \(P(\text{F OR L}) = 0.57\)
10. \(P(\text{M'}) = 0.48\)
11. \(P(\text{R|M}) = 0.8269\) (rounded to four decimal places)
12. \(P(\text{F|L}) = 0.3077\) (rounded to four decimal places)
13. \(P(\text{L|F}) = 0.0833\)

Independent and Mutually Exclusive Events

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References

1. “Countries List by Continent.” Worldatlas, 2013. Available online at http://www.worldatlas.com/cntycont.htm (accessed May 2, 2013).

Review

In this module we learned the basic terminology of probability. The set of all possible outcomes of an experiment is called the sample space. Events are subsets of the sample space, and they are assigned a probability that is a number between zero and one, inclusive.

Formula Review

\(\text{A}\) and \(\text{B}\) are events

\(P(\text{S}) = 1\) where \(\text{S}\) is the sample space

\(0 \leq P(\text{A}) \leq 1\)

\(P(\text{A|B}) = \frac{\text{P(A AND B)}}{\text{P(B)}}\)

Glossary

Conditional Probability

the likelihood that an event will occur given that another event has already occurred

Equally Likely

Each outcome of an experiment has the same probability.

Event

a subset of the set of all outcomes of an experiment; the set of all outcomes of an experiment is called a sample space and is usually denoted by \(S\). An event is an arbitrary subset in \(S\). It can contain one outcome, two outcomes, no outcomes (empty subset), the entire sample space, and the like. Standard notations for events are capital letters such as \(A, B, C\), and so on.

Experiment

a planned activity carried out under controlled conditions

Outcome

a particular result of an experiment

Probability

a number between zero and one, inclusive, that gives the likelihood that a specific event will occur; the foundation of statistics is given by the following 3 axioms (by A.N. Kolmogorov, 1930’s): Let \(S\) denote the sample space and \(A\) and \(B\) are two events in S. Then:

• \(0 \leq P(\text{A}) \leq 1\)
• If \(\text{A}\) and \(\text{B}\) are any two mutually exclusive events, then \(\text{P}(\text{A OR B}) = P(\text{A}) + P(\text{B})\).
• \(P(\text{S}) = 1\)

Sample Space

the set of all possible outcomes of an experiment

The AND Event

An outcome is in the event \(\text{A AND B}\) if the outcome is in both \(\text{A AND B}\) at the same time.

The Complement Event

The complement of event \(\text{A}\) consists of all outcomes that are NOT in \(\text{A}\).

The Conditional Probability of A GIVEN B

\(P(\text{A|B})\) is the probability that event \(\text{A}\) will occur given that the event \(\text{B}\) has already occurred.

The Or Event

An outcome is in the event \(\text{A OR B}\) if the outcome is in \(\text{A}\) or is in \(\text{B}\) or is in both \(\text{A}\) and \(\text{B}\).

Contributors and Attributions

• Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/30189442-699...b91b9de@18.114.