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

The sample space for a fair six-sided die is \(S = {1,2,3,4,5,6}\).

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. \(P(\text{L|F})\)

Solution

Add columns and rows for their totals for easy calculations:

	Right-handed	Left-handed	Row Totals
Males	43	9	52
Females	44	4	48
Column Totals	87	13	100

1. \(P(\text{M}) = \frac{52}{100}= 0.52\) since there are a total of 52 males when we count both Right and Left-handed people.


2. \(P(\text{F}) = \frac{48}{100} = 0.48\) since there are a total of 48 females when we count both Right and Left-handed people.


3. \(P(\text{R}) = \frac{87}{100} = 0.87\)


4. \(P(\text{L}) = \frac{13}{100} = 0.13\)


5. \(P(\text{M AND R}) = \frac{43}{100} = 0.43\) since we need both qualities to exist in the same selected person, we look at the intersection of those two outcomes.


6. \(P(\text{F AND L}) = \frac{4}{100} = 0.04\)


7. \(P(\text{M OR F}) = \frac{100}{100} = 1\) since those two outcomes account for all of the people in the sample. 


8. \(P(\text{M OR R}) = \frac{96}{100} = 0.96\) since we have 43 right-handed males + 9 left-handed males + 44 right handed females = 96 Male or Right-handed people.


9. \(P(\text{F OR L}) = \frac{57}{100} = 0.57\)


10. \(P(\text{M'}) = \frac{48}{100} = 0.48\) since we want the complement of M we look at the samples that are not M so that is the 48 F samples. We can also use the complement rule that states \(P(\text{M}) + P(\text{M'}) = 1\). So \(P(\text{M'}) = 1 - P(\text{M}) = 1 - 0.52 = 0.48\)


11. For \(P(\text{R|M})\), we first have to identify the sample space based on what was given, which is that we know a male was selected. So, our sample space is just going to be the row with male samples in it. The number of samples that are Right-handed and Male is 43. So, rounded to four decimal places, \[P(\text{R|M})= \frac{n(\text{R AND M})}{n(\text{M})}= \frac{43}{52} =  0.8269 \nonumber\]


12. For \(P(\text{F|L})\), we first has to identify the sample space based on what was given, which is that we know a Left-handed person was selected. So, our sample space is just the column of Left-handed samples. The number of samples that are Female and Left-handed is 4. So, rounded to four decimal places, \[P(\text{F|L}) = \frac{n(\text{F AND L})}{n(\text{L})}=\frac{4}{13} = 0.3077 \nonumber\]


13. For \(P(\text{L|F})\), we first have to identify the sample space based on what was given, which is that we know a female was selected. So, our sample space is just the row of female samples. The number of samples that are Left-handed and female is 4. So, rounded to four decimal places, \[P(\text{L|F}) = \frac{n(\text{L AND F})}{n(\text{F})} = \frac{4}{48} = 0.0833 \nonumber\]

Note: It is tempting to thing that \(P(\text{F|L})=P(\text{L|F})\), but because what is given controls the sample space, or the denominator of the fraction, we have to take that information into account. What is true, however, is that \(P(\text{F AND L}) = P(\text{L AND F})\) since the intersection of L and F doesn't change regardless of which outcome you consider first. 

Independent and Mutually Exclusive Events

Independent and mutually exclusive do not mean the same thing.

Independent Events

Two events are independent if the following are true:

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

Two events \(\text{A}\) and \(\text{B}\) are independent if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are NOT independent, then we say that they are dependent.

Sampling a population

Sampling may be done with replacement or without replacement (Figure \(\PageIndex{1}\)):

• With replacement: If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.
• Without replacement: When sampling is done without replacement, each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.
  • Note: The sample size is 5% or less of the total population, then selections can be treated as independent events since the removal of an outcome doesn't overly affect the final calculation. We will see more of this condition for independence in later chapters. It's known as the "5% rule" or "5% Guideline."

Figure \(\PageIndex{1}\): A visual representation of the sampling process. If the sample items are replaced after each sampling event, then this is "sampling with replacement" if not, then it is "sampling without replacement". (CC BY-SA 4.0; Dan Kernler).

If it is not known whether \(\text{A}\) and \(\text{B}\) are independent or dependent, assume they are dependent until you can show otherwise.

Example \(\PageIndex{11}\): Sampling with and without replacement

You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \(\text{J}\) (jack), \(\text{Q}\) (queen), \(\text{K}\) (king) of that suit.

a. Sampling with replacement:

Suppose you pick three cards with replacement. The first card you pick out of the 52 cards is the \(\text{Q}\) of spades. You put this card back, reshuffle the cards and pick a second card from the 52-card deck. It is the ten of clubs. You put this card back, reshuffle the cards and pick a third card from the 52-card deck. This time, the card is the \(\text{Q}\) of spades again. Your picks are {\(\text{Q}\) of spades, ten of clubs, \(\text{Q}\) of spades}. You have picked the \(\text{Q}\) of spades twice. You pick each card from the 52-card deck.

b. Sampling without replacement:

Suppose you pick three cards without replacement. The first card you pick out of the 52 cards is the \(\text{K}\) of hearts. You put this card aside and pick the second card from the 51 cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining 50 cards in the deck. The third card is the \(\text{J}\) of spades. Your picks are {\(\text{K}\) of hearts, three of diamonds, \(\text{J}\) of spades}. Because you have picked the cards without replacement, you cannot pick the same card twice.

Exercise \(\PageIndex{2}\)

You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \(\text{J}\) (jack), \(\text{Q}\) (queen), \(\text{K}\) (king) of that suit. Three cards are picked at random.

1. Suppose you know that the picked cards are \(\text{Q}\) of spades, \(\text{K}\) of hearts and \(\text{Q}\) of spades. Can you decide if the sampling was with or without replacement?
2. Suppose you know that the picked cards are \(\text{Q}\) of spades, \(\text{K}\) of hearts, and \(\text{J}\) of spades. Can you decide if the sampling was with or without replacement?

Answer a

With replacement

Answer b

No

Example \(\PageIndex{12}\)

You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \(\text{J}\) (jack), \(\text{Q}\) (queen), and \(\text{K}\) (king) of that suit. \(\text{S} =\) spades, \(\text{H} =\) Hearts, \(\text{D} =\) Diamonds, \(\text{C} =\) Clubs.

1. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are \(\text{QS}, 1\text{D}, 1\text{C}, \text{QD}\).
2. Suppose you pick four cards and put each card back before you pick the next card. Your cards are \(\text{KH}, 7\text{D}, 6\text{D}, \text{KH}\).

Which of a. or b. did you sample with replacement and which did you sample without replacement?

Solution a

Without replacement

Solution b

With replacement

Exercise \(\PageIndex{3}\)

You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \(\text{J}\) (jack), \(\text{Q}\) (queen), and \(\text{K}\) (king) of that suit. \(\text{S} =\) spades, \(\text{H} =\) Hearts, \(\text{D} =\) Diamonds, \(\text{C} =\) Clubs. Suppose that you sample four cards without replacement. Which of the following outcomes are possible? Answer the same question for sampling with replacement.

1. \(\text{QS}, 1\text{D}, 1\text{C}, \text{QD}\)
2. \(\text{KH}, 7\text{D}, 6\text{D}, \text{KH}\)
3. \(\text{QS}, 7\text{D}, 6\text{D}, \text{KS}\)

Answer - without replacement

a. Possible; b. Impossible, c. Possible

Answer - with replacement

a. Possible; c. Possible, c. Possible​​​​​

Mutually Exclusive Events

\(\text{A}\) and \(\text{B}\) are mutually exclusive events if they cannot occur at the same time. This means that \(\text{A}\) and \(\text{B}\) do not share any outcomes and \(P(\text{A AND B}) = 0\).

For example, suppose the sample space

\[S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}. \nonumber\]

Let \(\text{A} = \{1, 2, 3, 4, 5\}, \text{B} = \{4, 5, 6, 7, 8\}\), and \(\text{C} = \{7, 9\}\). \(\text{A AND B} = \{4, 5\}\).

\[P(\text{A AND B}) = \dfrac{2}{10} \nonumber\]

and is not equal to zero. Therefore, \(\text{A}\) and \(\text{B}\) are not mutually exclusive. \(\text{A}\) and \(\text{C}\) do not have any numbers in common so \(P(\text{A AND C}) = 0\). Therefore, \(\text{A}\) and \(\text{C}\) are mutually exclusive.

If it is not known whether \(\text{A}\) and \(\text{B}\) are mutually exclusive, assume they are not until you can show otherwise. The following examples illustrate these definitions and terms.

Example \(\PageIndex{13}\)

Flip two fair coins.

The sample space is \(\{HH, HT, TH, TT\}\) where \(T =\) tails and \(H =\) heads. The outcomes are \(HH,HT, TH\), and \(TT\). The outcomes \(HT\) and \(TH\) are different. The \(HT\) means that the first coin showed heads and the second coin showed tails. The \(TH\) means that the first coin showed tails and the second coin showed heads.

• Let \(\text{A} =\) the event of getting at most one tail. (At most one tail means zero or one tail.) Then \(\text{A}\) can be written as \(\{HH, HT, TH\}\). The outcome \(HH\) shows zero tails. \(HT\) and \(TH\) each show one tail.
• Let \(\text{B} =\) the event of getting all tails. \(\text{B}\) can be written as \(\{TT\}\). \(\text{B}\) is the complement of \(\text{A}\), so \(\text{B} = \text{A′}\). Also, \(P(\text{A}) + P(\text{B}) = P(\text{A}) + P(\text{A′}) = 1\).
• The probabilities for \(\text{A}\) and for \(\text{B}\) are \(P(\text{A}) = \dfrac{3}{4}\) and \(P(\text{B}) = \dfrac{1}{4}\).
• Let \(\text{C} =\) the event of getting all heads. \(\text{C} = \{HH\}\). Since \(\text{B} = \{TT\}\), \(P(\text{B AND C}) = 0\). \(\text{B}\) and Care mutually exclusive. \(\text{B}\) and \(\text{C}\) have no members in common because you cannot have all tails and all heads at the same time.)
• Let \(\text{D} =\) event of getting more than one tail. \(\text{D} = \{TT\}\). \(P(\text{D}) = \dfrac{1}{4}\)
• Let \(\text{E} =\) event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) \(\text{E} = \{HT, HH\}\). \(P(\text{E}) = \dfrac{2}{4}\)
• Find the probability of getting at least one (one or two) tail in two flips. Let \(\text{F} =\) event of getting at least one tail in two flips. \(\text{F} = \{HT, TH, TT\}\). \(P(\text{F}) = \dfrac{3}{4}\)

Exercise \(\PageIndex{4}\)

Draw two cards from a standard 52-card deck with replacement. Find the probability of getting at least one black card.

Answer

The sample space of drawing two cards with replacement from a standard 52-card deck with respect to color is \(\{BB, BR, RB, RR\}\).

Event \(A =\) Getting at least one black card \(= \{BB, BR, RB\}\)

\(P(\text{A}) = \dfrac{3}{4} = 0.75\)

Example \(\PageIndex{14}\)

Flip two fair coins. Find the probabilities of the events.

1. Let \(\text{F} =\) the event of getting at most one tail (zero or one tail).
2. Let \(\text{G} =\) the event of getting two faces that are the same.
3. Let \(\text{H} =\) the event of getting a head on the first flip followed by a head or tail on the second flip.
4. Are \(\text{F}\) and \(\text{G}\) mutually exclusive?
5. Let \(\text{J} =\) the event of getting all tails. Are \(\text{J}\) and \(\text{H}\) mutually exclusive?

Solution

Look at the sample space in Example \(\PageIndex{3}\).

1. Zero (0) or one (1) tails occur when the outcomes \(HH, TH, HT\) show up. \(P(\text{F}) = \dfrac{3}{4}\)
2. Two faces are the same if \(HH\) or \(TT\) show up. \(P(\text{G}) = \dfrac{2}{4}\)
3. A head on the first flip followed by a head or tail on the second flip occurs when \(HH\) or \(HT\) show up. \(P(\text{H}) = \dfrac{2}{4}\)
4. \(\text{F}\) and \(\text{G}\) share \(HH\) so \(P(\text{F AND G})\) is not equal to zero (0). \(\text{F}\) and \(\text{G}\) are not mutually exclusive.
5. Getting all tails occurs when tails shows up on both coins (\(TT\)). \(\text{H}\)’s outcomes are \(HH\) and \(HT\).

\(\text{J}\) and \(\text{H}\) have nothing in common so \(P(\text{J AND H}) = 0\). \(\text{J}\) and \(\text{H}\) are mutually exclusive.

Exercise \(\PageIndex{5}\)

A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events:

1. Let \(\text{F} =\) the event of getting the white ball twice.
2. Let \(\text{G} =\) the event of getting two balls of different colors.
3. Let \(\text{H} =\) the event of getting white on the first pick.
4. Are \(\text{F}\) and \(\text{G}\) mutually exclusive?
5. Are \(\text{G}\) and \(\text{H}\) mutually exclusive?

Answer

1. \(P(\text{F}) = \dfrac{1}{4}\)
2. \(P(\text{G}) = \dfrac{1}{2}\)
3. \(P(\text{H}) = \dfrac{1}{2}\)
4. Yes
5. No

Example \(\PageIndex{15}\)

Roll one fair, six-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event \(\text{A} =\) a face is odd. Then \(\text{A} = \{1, 3, 5\}\). Let event \(\text{B} =\) a face is even. Then \(\text{B} = \{2, 4, 6\}\).

• Find the complement of \(\text{A}\), \(\text{A′}\). The complement of \(\text{A}\), \(\text{A′}\), is \(\text{B}\) because \(\text{A}\) and \(\text{B}\) together make up the sample space. \(P(\text{A}) + P(\text{B}) = P(\text{A}) + P(\text{A′}) = 1\). Also, \(P(\text{A}) = \dfrac{3}{6}\) and \(P(\text{B}) = \dfrac{3}{6}\).
• Let event \(\text{C} =\) odd faces larger than two. Then \(\text{C} = \{3, 5\}\). Let event \(\text{D} =\) all even faces smaller than five. Then \(\text{D} = \{2, 4\}\). \(P(\text{C AND D}) = 0\) because you cannot have an odd and even face at the same time. Therefore, \(\text{C}\) and \(\text{D}\) are mutually exclusive events.
• Let event \(\text{E} =\) all faces less than five. \(\text{E} = \{1, 2, 3, 4\}\).

Are \(\text{C}\) and \(\text{E}\) mutually exclusive events? (Answer yes or no.) Why or why not?

Solution

No. \(\text{C} = \{3, 5\}\) and \(\text{E} = \{1, 2, 3, 4\}\). \(P(\text{C AND E}) = \dfrac{1}{6}\). To be mutually exclusive, \(P(\text{C AND E})\) must be zero.

• Find \(P(\text{C|A})\). This is a conditional probability. Recall that the event \(\text{C}\) is {3, 5} and event \(\text{A}\) is {1, 3, 5}. To find \(P(\text{C|A})\), find the probability of \(\text{C}\) using the sample space \(\text{A}\). You have reduced the sample space from the original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5}. So, \(P(\text{C|A}) = \dfrac{2}{3}\).

Exercise \(\PageIndex{6}\)

Let event \(\text{A} =\) learning Spanish. Let event \(\text{B}\) = learning German. Then \(\text{A AND B}\) = learning Spanish and German. Suppose \(P(\text{A}) = 0.4\) and \(P(\text{B}) = 0.2\). \(P(\text{A AND B}) = 0.08\). Are events \(\text{A}\) and \(\text{B}\) independent? Hint: You must show ONE of the following:

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

Answer

\[P(\text{A|B}) = \dfrac{\text{P(A AND B)}}{P(\text{B})} = \dfrac{0.08}{0.2} = 0.4 = P(\text{A})\]

The events are independent because \(P(\text{A|B}) = P(\text{A})\).

Example \(\PageIndex{16}\)

Let event \(\text{G} =\) taking a math class. Let event \(\text{H} =\) taking a science class. Then, \(\text{G AND H} =\) taking a math class and a science class. Suppose \(P(\text{G}) = 0.6\), \(P(\text{H}) = 0.5\), and \(P(\text{G AND H}) = 0.3\). Are \(\text{G}\) and \(\text{H}\) independent?

If \(\text{G}\) and \(\text{H}\) are independent, then you must show ONE of the following:

• \(P(\text{G|H}) = P(\text{G})\)
• \(P(\text{H|G}) = P(\text{H})\)
• \(P(\text{G AND H}) = P(\text{G})P(\text{H})\)

The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information.

1. a. Show that \(P(\text{G|H}) = P(\text{G})\).
2. b. Show \(P(\text{G AND H}) = P(\text{G})P(\text{H})\).

Solution

1. \(P(\text{G|H}) = \dfrac{P(\text{G AND H})}{P(\text{H})} = \dfrac{0.3}{0.5} = 0.6 = P(\text{G})\)
2. \(P(\text{G})P(\text{H}) = (0.6)(0.5) = 0.3 = P(\text{G AND H})\)

Since \(\text{G}\) and \(\text{H}\) are independent, knowing that a person is taking a science class does not change the chance that he or she is taking a math class. If the two events had not been independent (that is, they are dependent) then knowing that a person is taking a science class would change the chance he or she is taking math. For practice, show that \(P(\text{H|G}) = P(\text{H})\) to show that \(\text{G}\) and \(\text{H}\) are independent events.

Exercise \(\PageIndex{7}\)

In a bag, there are six red marbles and four green marbles. The red marbles are marked with the numbers 1, 2, 3, 4, 5, and 6. The green marbles are marked with the numbers 1, 2, 3, and 4.

• \(\text{R} =\) a red marble
• \(\text{G} =\) a green marble
• \(\text{O} =\) an odd-numbered marble
• The sample space is \(\text{S} = \{R1, R2, R3, R4, R5, R6, G1, G2, G3, G4\}\).

\(\text{S}\) has ten outcomes. What is \(P(\text{G AND O})\)?

Answer

Event \(\text{G}\) and \(\text{O} = \{G1, G3\}\)

\(P(\text{G and O}) = \dfrac{2}{10} = 0.2\)

Example \(\PageIndex{17}\)

Let event \(\text{C} =\) taking an English class. Let event \(\text{D} =\) taking a speech class.

Suppose \(P(\text{C}) = 0.75\), \(P(\text{D}) = 0.3\), \(P(\text{C|D}) = 0.75\) and \(P(\text{C AND D}) = 0.225\).

Justify your answers to the following questions numerically.

1. Are \(\text{C}\) and \(\text{D}\) independent?
2. Are \(\text{C}\) and \(\text{D}\) mutually exclusive?
3. What is \(P(\text{D|C})\)?

Solution

1. Yes, because \(P(\text{C|D}) = P(\text{C})\).
2. No, because \(P(\text{C AND D})\) is not equal to zero.
3. \(P(\text{D|C}) = \dfrac{P(\text{C AND D})}{P(\text{C})} = \dfrac{0.225}{0.75} = 0.3\)

Exercise \(\PageIndex{8}\)

A student goes to the library. Let events \(\text{B} =\) the student checks out a book and \(\text{D} =\) the student checks out a DVD. Suppose that \(P(\text{B}) = 0.40\), \(P(\text{D}) = 0.30\) and \(P(\text{B AND D}) = 0.20\).

1. Find \(P(\text{B|D})\).
2. Find \(P(\text{D|B})\).
3. Are \(\text{B}\) and \(\text{D}\) independent?
4. Are \(\text{B}\) and \(\text{D}\) mutually exclusive?

Answer

1. \(P(\text{B|D}) = 0.6667\)
2. \(P(\text{D|B}) = 0.5\)
3. No
4. No

Example \(\PageIndex{18}\)

In a box there are three red cards and five blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card.

Let

• \(\text{R =}\) red card is drawn,
• \(\text{B} =\) blue card is drawn,
• \(\text{E} =\) even-numbered card is drawn.

The sample space \(S = R1, R2, R3, B1, B2, B3, B4, B5\).

\(S\) has eight outcomes.

• \(P(\text{R}) = \dfrac{3}{8}\). \(P(\text{B}) = \dfrac{5}{8}\). \(P(\text{R AND B}) = 0\). (You cannot draw one card that is both red and blue.)
• \(P(\text{E}) = \dfrac{3}{8}\). (There are three even-numbered cards, \(R2, B2\), and \(B4\).)
• \(P(\text{E|B}) = \dfrac{2}{5}\). (There are five blue cards: \(B1, B2, B3, B4\), and \(B5\). Out of the blue cards, there are two even cards; \(B2\) and \(B4\).)
• \(P(\text{B|E}) = \dfrac{2}{3}\). (There are three even-numbered cards: \(R2, B2\), and \(B4\). Out of the even-numbered cards, to are blue; \(B2\) and \(B4\).)
• The events \(\text{R}\) and \(\text{B}\) are mutually exclusive because \(P(\text{R AND B}) = 0\).
• Let \(\text{G} =\) card with a number greater than 3. \(\text{G} = \{B4, B5\}\). \(P(\text{G}) = \dfrac{2}{8}\). Let \(\text{H} =\) blue card numbered between one and four, inclusive. \(\text{H} = \{B1, B2, B3, B4\}\). \(P(\text{G|H}) = \frac{1}{4}\). (The only card in \(\text{H}\) that has a number greater than three is B4.) Since \(\dfrac{2}{8} = \dfrac{1}{4}\), \(P(\text{G}) = P(\text{G|H})\), which means that \(\text{G}\) and \(\text{H}\) are independent.

Exercise \(\PageIndex{9}\)

In a basketball arena,

• 70% of the fans are rooting for the home team.
• 25% of the fans are wearing blue.
• 20% of the fans are wearing blue and are rooting for the away team.
• Of the fans rooting for the away team, 67% are wearing blue.

Let \(\text{A}\) be the event that a fan is rooting for the away team.

Let \(\text{B}\) be the event that a fan is wearing blue.

Are the events of rooting for the away team and wearing blue independent? Are they mutually exclusive?

Answer

• \(P(\text{B|A}) = 0.67\)
• \(P(\text{B}) = 0.25\)

So \(P(\text{B})\) does not equal \(P(\text{B|A})\) which means that \(\text{B} and \text{A}\) are not independent (wearing blue and rooting for the away team are not independent). They are also not mutually exclusive, because \(P(\text{B AND A}) = 0.20\), not \(0\).

Example \(\PageIndex{19}\)

In a particular college class, 60% of the students are female. Fifty percent of all students in the class have long hair. Forty-five percent of the students are female and have long hair. Of the female students, 75% have long hair. Let \(\text{F}\) be the event that a student is female. Let \(\text{L}\) be the event that a student has long hair. One student is picked randomly. Are the events of being female and having long hair independent?

• The following probabilities are given in this example:
• \(P(\text{F}) = 0.60\); \(P(\text{L}) = 0.50\)
• \(P(\text{F AND L}) = 0.45\)
• \(P(\text{L|F}) = 0.75\)

The choice you make depends on the information you have. You could use the first or last condition on the list for this example. You do not know \(P(\text{F|L})\) yet, so you cannot use the second condition.

Solution 1

Check whether \(P(\text{F AND L}) = P(\text{F})P(\text{L})\). We are given that \(P(\text{F AND L}) = 0.45\), but \(P(\text{F})P(\text{L}) = (0.60)(0.50) = 0.30\). The events of being female and having long hair are not independent because \(P(\text{F AND L})\) does not equal \(P(\text{F})P(\text{L})\).

Solution 2

Check whether \(P(\text{L|F})\) equals \(P(\text{L})\). We are given that \(P(\text{L|F}) = 0.75\), but \(P(\text{L}) = 0.50\); they are not equal. The events of being female and having long hair are not independent.

Interpretation of Results

The events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair.

Exercise \(\PageIndex{10}\)

Mark is deciding which route to take to work. His choices are \(\text{I} = \text{the Interstate}\) and \(\text{F} = \text{Fifth Street}\)

• \(P(\text{I}) = 0.44\) and \(P(\text{F}) = 0.55\)
• \(P(\text{I AND F}) = 0\) because Mark will take only one route to work.

What is the probability of \(P(\text{I OR F})\)?

Answer

Because \(P(\text{I AND F}) = 0\),

\(P(\text{I OR F}) = P(\text{I}) + P(\text{F}) - P(\text{I AND F}) = 0.44 + 0.56 - 0 = 1\)

Example \(\PageIndex{20}\)

1. Toss one fair coin (the coin has two sides, \(\text{H}\) and \(\text{T}\)). The outcomes are ________. Count the outcomes. There are ____ outcomes.
2. Toss one fair, six-sided die (the die has 1, 2, 3, 4, 5 or 6 dots on a side). The outcomes are ________________. Count the outcomes. There are ___ outcomes.
3. Multiply the two numbers of outcomes. The answer is _______.
4. If you flip one fair coin and follow it with the toss of one fair, six-sided die, the answer in three is the number of outcomes (size of the sample space). What are the outcomes? (Hint: Two of the outcomes are \(H1\) and \(T6\).)
5. Event \(\text{A} =\) heads (\(\text{H}\)) on the coin followed by an even number (2, 4, 6) on the die.
   \(\text{A}\) = {_________________}. Find \(P(\text{A})\).
6. Event \(\text{B} =\) heads on the coin followed by a three on the die. \(\text{B} =\) {________}. Find \(P(\text{B})\).
7. Are \(\text{A}\) and \(\text{B}\) mutually exclusive? (Hint: What is \(P(\text{A AND B})\)? If \(P(\text{A AND B}) = 0\), then \(\text{A}\) and \(\text{B}\) are mutually exclusive.)
8. Are \(\text{A}\) and \(\text{B}\) independent? (Hint: Is \(P(\text{A AND B}) = P(\text{A})P(\text{B})\)? If \(P(\text{A AND B})\ = P(\text{A})P(\text{B})\), then \(\text{A}\) and \(\text{B}\) are independent. If not, then they are dependent).

Solution

1. \(\text{H}\) and \(\text{T}\); 2
2. 1, 2, 3, 4, 5, 6; 6
3. 2(6) = 12
4. \(T1, T2, T3, T4, T5, T6, H1, H2, H3, H4, H5, H6\)
5. \(\text{A} = \{H2, H4, H6\}\); \(P(\text{A}) = \dfrac{3}{12}\)
6. \(\text{B} = \{H3\}\); \(P(\text{B}) = \dfrac{1}{12}\)
7. Yes, because \(P(\text{A AND B}) = 0\)
8. \(P(\text{A AND B}) = 0\). \(P(\text{A})P(\text{B}) = \left(\dfrac{3}{12}\right)\left(\dfrac{1}{12}\right)\). \(P(\text{A AND B})\) does not equal \(P(\text{A})P(\text{B})\), so \(\text{A}\) and \(\text{B}\) are dependent.

Exercise \(\PageIndex{11}\)

A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let \(\text{T}\) be the event of getting the white ball twice, \(\text{F}\) the event of picking the white ball first, \(\text{S}\) the event of picking the white ball in the second drawing.

1. Compute \(P(\text{T})\).
2. Compute \(P(\text{T|F})\).
3. Are \(\text{T}\) and \(\text{F}\) independent?.
4. Are \(\text{F}\) and \(\text{S}\) mutually exclusive?
5. Are \(\text{F}\) and \(\text{S}\) independent?

Answer

1. \(P(\text{T}) = \dfrac{1}{4}\)
2. \(P(\text{T|F}) = \dfrac{1}{2}\)
3. No
4. No
5. Yes

References

1. “Countries List by Continent.” Worldatlas, 2013. Available online at http://www.worldatlas.com/cntycont.htm (accessed May 2, 2013).
2. Lopez, Shane, Preety Sidhu. “U.S. Teachers Love Their Lives, but Struggle in the Workplace.” Gallup Wellbeing, 2013. http://www.gallup.com/poll/161516/te...workplace.aspx (accessed May 2, 2013).
3. Data from Gallup. Available online at www.gallup.com/ (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.

Two events \(\text{A}\) and \(\text{B}\) are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent.

In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other.

Formula Review

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

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

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

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

If \(\text{A}\) and \(\text{B}\) are independent, \(P(\text{A AND B}) = P(\text{A})P(\text{B}), P(\text{A|B}) = P(\text{A})\) and \(P(\text{B|A}) = P(\text{B})\).

If \(\text{A}\) and \(\text{B}\) are mutually exclusive, \(P(\text{A OR B}) = P(\text{A}) + P(\text{B}) and P(\text{A AND B}) = 0\).

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 \(\text{B}\).

Dependent Events

If two events are NOT independent, then we say that they are dependent.

The OR of Two Events

An outcome is in the event A OR B if the outcome is in A, is in B, or is in both A and B.

The Conditional Probability of One Event Given Another Event

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

Sampling without Replacement

When sampling is done without replacement, each member of a population may be chosen only once.

Sampling with Replacement

If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once.

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.