Chapter Objectives

In this chapter, readers will learn to do the following:

• Define the probability

• Classify classical probability, relative frequency probability, and subjective probability

• Define events and a simple event

• Define and construct a sample space

• Calculate probabilities using simple events and counting rules

• Calculate probabilities of relations

• Construct probability distributions

Probability theory is a branch of mathematics concerned with estimating how likely an event is to happen. Evaluating the probability of an event means estimating the chances of this event occurring on a scale of 0 to 1. If we are confident that the event will never happen, its probability equals 0, and if it would certainly happen, its probability equals 1. Probability theory is primarily used in the natural and social sciences. In some applications, we use the percentage value of probabilities on a scale of 0% to 100%.

In our everyday life, we often refer to probabilities of events. For instance, to strengthen our certainty in expectation of an event, we say, “I am 100% sure”; or to describe the equal chances of two events, we say, “fifty-fifty.”

Imagine tossing a fair coin in the air. We would expect the coin to be either a head or a tail 50% of the time. Hence, the probability of the coin being a head on any toss is 0.5. The notation for declaring the probability of an event is a capital P (denoting probability) followed by an event in parentheses. Therefore, the sentence “the probability of the coin being a head on any toss is 0.5” can be written as P(head) = 0.5.

3.1. Probability with a View Toward Statistics

Recall that in chapter 1, solving Example 1.14, we evaluated the chances of some events happening using the relative frequency histogram (fig. 3.1).

Figure 3.1. Relative frequency histogram constructed from data provided in example 1.14

For instance, we estimated the chances that a randomly selected Oscar winner is 42 years old or older and younger than 46 using the relative frequency histogram 20/96 or 0.21.

This chapter will show that the relative frequency and probability are calculated based on the same concept. Below, we will analyze the relationship between the probability and relative frequency in more detail.

In general, there are three methods that we can use to assign probabilities to events: classical probability, relative frequency probability, and subjective probability.

Classical Probability

One refers to classical probability when the possible outcomes for an event can be mathematically derived. In most cases, but not always, the classical probability is applied if the events are equally likely.

Example 3.1

Indigenous Peoples in Canada and worldwide preserved and still play games based on equal chances to win. Let’s analyze one of those games with different numbers of players.

John, Ann, and George hide a stone in their hands, and Loraine attempts to find out who is holding the stone. It is equally likely that either John, Ann, or George is holding the stone. Therefore, each of them might hold the hidden stone with a chance of 1/3. Using the probability notation, we can write, P(John) = P(Ann) = P(George) = 1/3.

• John, Ann, George, and Meghan hide a stone in their hands, and Loraine attempts to find out who is holding the stone. It is equally likely that either John, Ann, George, or Meghan is holding the stone. Therefore, each of them might hold the hidden stone with a chance of 1/4. Using the probability notation, we can write, P(John) = P(Ann) = P(George) = P(Meghan) = 1/4.

Note: As you know from mathematics courses, it is more convenient to use short notations when writing expressions. So hereafter, we will use P(J) instead of P(John), P(A) instead of P(Ann), and so on.

Example 3.2

Now, let’s analyze an example in which events are not equally likely to happen and evaluate the probability of spinning the sector 2 for the three following spinners (fig. 3.2):

                                                                  (a)                                                 (b)                                              (c)

Figure 3.2. Spinners with various sectors

By inspection, we can determine the portion of sector 2 for cases (a), (b), and (c) as 1/2, 1/4, and 3/4, respectively. Then, using common sense, we can state that these are the values of the probability P(2) for each case.

(a) P(2) = 1/2 = 0.5

(b) P(2) = 1/4 = 0.25

(c) P(2) = 3/4 = 0.75

Relative Frequency Probability

We have already mentioned that probability and relative frequency have a common nature. Usually, this concept is used when the probability is based on observations from many trials or historical records. Analyzing example 1.13, when we evaluated the chances that a randomly selected Oscar winner is 42 years old or older and younger than 46 using the relative frequency histogram 20/96 or 0.21. Using the probability notation, one can write, P(42 ≤ A < 46) = 0.21.

Let us consider another example.

Example 3.3

A landlord asked 140 tenants to complete a demographic questionnaire containing questions about their age and marital status and created the following table:

Table 3.1. Marital status table

Age	Single (S)	Married (M)	Total
30 years old or younger (Y)	77	14	91
Older than 30 years old (O)	28	21	49
Total	105	35	140

What is the probability that a randomly selected tenant is a single individual aged 30 or younger?

What is the probability that a randomly selected tenant is older than 30 years old?

Solution:

From table 3.1, one can see that 77 tenants are single and younger than 30 years old. Therefore, P(Y and S) = 77/140 = 0.55

Similarly, P(O) = 49/140 = 0.35

Subjective Probability

The Great Belt Bridge, which spans the highly trafficked waters separating the Danish islands of Zealand and Funen, is one of the longest bridges in the world. The construction of this unique bridge took place between 1991 and 1998. Before approving the construction, the Danish Parliament asked a group of researchers to evaluate the probability that the bridge would be destroyed by a colliding ship within the next 100 years (Hoffmann, 1994). Since there was no other identical construction in the world, neither of the above-mentioned methods (classical probability or relative frequency probability) could be used to evaluate this probability. We will not provide a detailed explanation of how to evaluate this probability as it requires some theoretical background beyond introductory statistics. However, we can note that evaluating this type of probability includes some subjective information. Hence, it is called subjective probability. The subjective probability is the degree of belief in the person stating the probability. Many everyday probabilities are subjective; for instance, bookmakers’ odds are of this type, as are, often, the exchange rates on the currency market. In this book, we will not provide a detailed analysis of subjective probability; rather, we will complete this topic by offering readers an opportunity to practise the following problem on subjective probability.

A Funny Problem

Evaluate the probability of a tiger winning in the fight against a lion.

Hint: Answers may vary.

Events and Sample Space

At the beginning of this chapter, we defined probability as the chance of an event happening without giving a formal definition of events. Below, we will try to define some basic concepts used in probability theory.

Experiment: In statistics and probability, we consider as an experiment the process of making a measurement or an observation. One must consider that in statistics, the meaning of the measurement is not restricted just by evaluating some physical parameters. For instance, if we record someone’s opinion regarding a political party, this process is also considered a measurement.

Event: In probability theory, an event is a set of outcomes of an experiment to which a probability is assigned. Let’s analyze two popular examples of probability learning.

Tossing a Fair Coin

Assume that we want to estimate our chances of getting a head in the first toss. In other words, we need to evaluate the probability of observation of the head, P(Head). In this case, observation of the head is considered an event.

Tossing a Die

 

Assume that we are tossing a die and would like to determine our chances of getting an even number. That means we need to evaluate the probability of observing a face with 2, 4, or 6, P(2, 4, 6). In this case, our event (observing the face with even numbers) is a set of three single outcomes: a face with 2, a face with 4, and a face with 6. Each of these single outcomes is called a simple event. The events consisting of more than one simple event are called compound events.

The set of all single outcomes is called a sample space. In the example of tossing the coin, the sample space consists of two simple events: head and tail. For example, the sample space of the single die consists of six simple events: faces with 1, 2, 3, 4, 5, and 6. Using the set notation, we can state the following:

Sample space of a coin: S = {head, tail}

Sample space of a die: S = {1, 2, 3, 4, 5, 6}

In this textbook, we will denote a simple event by E_i, where the subscript i specifies the simple event.

3.2. Calculating Probabilities

In mathematics, we can perform calculations on numbers without thinking about what they represent. For instance, by adding 5 and 4, we do not have to specify that 5 or 4 might be an amount of something particular. Calculating probabilities, we assign numerals to chances of actual events, which is why, along with the expressive mathematical formalism, we need to apply a broad range of knowledge and practice to determine probabilities. Below, we will provide two approaches to evaluating the probabilities of events: simple events and counting rules.

Calculating Probabilities Using Simple Events

To evaluate probabilities using simple events, we have to consider the following:

Each probability must lie between 0 and 1.

The sum of all simple events probabilities equals 1.

The probability of a compound event equals the sum of the probabilities of its simple events

Note that similar rules apply to the notion of relative frequencies.

As mentioned above, many Indigenous games focus on finding a hidden object. Below, we will analyze two possible game scenarios.

Example 3.4

Assume that a boy is hiding a stone in his right or left hand, and his opponent has to find the stone. They are given two rounds of attempts, and the boy has no preference for hiding the stone either in his right or left hand. Determine the following:

What is the probability that the boy will hide the stone in his right hand in both rounds?

What is the probability that the boy will hide the stone once in the right and once in the left hand?

Solution:

After two rounds, we can observe one of the following four outcomes (fig. 3.3):

Figure 3.3. Sample space constructed for example 3.4

First round—right; second round—right: RR

First round—right; second round—left: RL

First round—left; second round—right: LR

First round—left; second round—left: LL

The sample space can be determined as S = {RR, RL, LR, LL}. The sample space consists of four equally likely simple events. The total number of outcomes is four.

(a) The simple event we seek (hiding the stone in the right hand in both rounds) is one out of four possible outcomes. Therefore, the probability of this simple event happening is P(RR) = 1/4 or 0.25.

(b) Now let A be an event of hiding the stone once in the right and once in the left hand: Determine the probability of this event. A is a compound event consisting of two simple events, RL and LR. The probabilities of each simple event are P(RL) = P(LR) = 1/4 = 0.25. Therefore, P(A) = P(RL) + P(LR) = 1/4 + 1/4 = 1/2 = 0.5.

Example 3.5

Kihew Iskwew (Cree women’s name that translates as “eagle woman”) painted three red, blue, and green stones. Her daughter randomly selected two stones without returning the previously chosen stone.

Find the probability that first the girl chose red and then blue.

Find the probability that one of the girl's choices was green.

Solution:

This example can be solved in a way similar to the previous one.

Figure 3.4. Sample space constructed for example 3.5

As one can see (fig. 3.4), the sample space contains six equally likely simple events: S = {RB, RG, BR, BG, GR, GB}. Therefore, the probability of each simple event equals 1/6. Now, we can determine the required probabilities.

P(RB) = 1/6 = 0.167

The girl can choose the green stone in four possible outcomes: RG, BG, GR, and GB. In other words, this is a compound event consisting of four simple events. Therefore, P(either RG, BG, GR or GB) = 1/6 + 1/6 + 1/6 + 1/6 = 4/6 = 2/3 or 0.667.

Calculating Probabilities Using Counting Rules

In examples 3.4 and 3.5, we could observe that the total number of outcomes after two rounds of experiments was equal to the product of possible outcomes in the first and second rounds. In example 3.4, the boy had two choices in each round: right or left. Therefore, the total number of outcomes after two rounds was 2×2=4. In example 3.5, the girl had three colours to choose from in the first round. She was left with two colours in the second round, and the total number of outcomes in the two rounds was 2×3=6.

In general, if the experiment consists of m rounds and the numbers of possible outcomes in the 1st, 2nd, . . . , and m^th rounds are n₁, n₂, . . . , and n_m, respectively, the total number of outcomes can be determined as N=n₁×n₂×...×n_m. This way of evaluating the total number of outcomes is called the multiplication counting rule.

Independent Events and Probability Trees

First, we would like to mark one difference between examples 3.4 and 3.5. In example 3.4, the outcomes of the second round did not depend on the outcomes of the first round. In each round, the boy could hide the stone either in his right or left hand. In example 3.5, by contrast, the possible outcomes of the second round depend on the result of the first round. For instance, if the girl chose red in the first round, in the second, she could choose either blue or green, but not red anymore. If her choice in the first round was blue, then in the second round, she could choose either red or green, but not blue, and so on.

The events observed in the first and second rounds of example 3.4 are independent. In general, the events are specified as independent if the occurrence of one does not affect the occurrence of the others. The concept of the independence of events comes from physics. The detailed analysis of this concept is beyond the scope of this textbook. However, we will try to provide a brief explanation.

In 1827, the English botanist Robert Brown noticed that pollen seeds suspended in water moved in an irregular “swarming” motion. Einstein then reasoned that if tiny but visible particles were suspended in a liquid, the invisible atoms in the liquid would bombard the suspended particles and cause them to jiggle. Einstein explained the motion in detail, accurately predicting the irregular, random motions of the particles, which could be directly observed under a microscope. (Einstein and Brownian Motion, APS News, February 2005, 14 [2]). Assume that (1) the motion of each particle is independent of the motions of all the other particles, and (2) the motion of the same particle at a specific time interval does not depend on its motion at different time intervals so long as we think of these time intervals as chosen as a result of their not being too small. In physics, this type of particle motion is known as Brownian motion, and each particle’s motion is considered independent.

The probability tree is one of the methods primarily used to determine the probability of independent events. It is considered a diagram that helps determine the sample space of simple events.

Example 3.6

Hubbub is an Indigenous game played by the Arapaho of Oklahoma (Carlson et al. Moccasins, Chicago Il: Chicago Review Press, 1994, ISBN 1-55652-213-4). To play this game, players need decorated stones. A boy decorated 5 stones: 3 with stars and 2 with circles for playing Hubbub. Then, he randomly selected 2 stones. Determine the probabilities of the following events:

The boy first selected a star and then a circle

The boy first selected a circle and then a star

The boy selected a star and a circle

The boy selected stars both times

The boy selected circles both times

The boy selected a star at least once

The boy selected a star at most once

Solution:

As was done in previous examples, first we need to construct the sample space (fig. 3.5).

 

Figure 3.5. Sample space constructed for example 3.6

Let S1, S2, and S3 denote stones decorated with stars, and C1 and C2 stones decorated with circles. The sample space from this experiment is S = {S1S2, S1S3, S1C1, S1C2, S2S1, S2S3, S2C1, S2C2, S3S1, S3S2, S3C1, S3C2, C1S1, C1S2, C1S3, C1C2, C2S1, C2S2, C2S3, C2C1}. As one can see, there are 20 outcomes.

(a) The event “The boy first selected a star and then a circle” can be observed 6 times out of all 20 outcomes. Therefore, its probability P(either S₁C₁, S₁C₂, S₂C₁, S₂C₂, S₃C₁, or S₃C₂) = 6/20 or 0.3.

(b) The event “The boy first selected a circle and then a star” can be observed 6 times out of all 20 outcomes. Therefore, its probability P(either C₁S₁, C₁S₂, C₁S₃, C₂S₁, C₂S₂, or C₂S₃) = 6/20 or 0.3.

(c) The event “The boy selected a star and a circle” (note that the order does not matter in this event) can be observed 12 times out of all 20 outcomes. Therefore, its probability P(either S₁C₁, S₁C₂, S₂C₁, S₂C₂, S₃C₁, S₃C₂, C₁S₁, C₁S₂, C₁S₃, C₂S₁, C₂S₂, or C₂S₃) = 12/20 or 0.6.

(d) The event “The boy selected stars both times” can be observed 6 times out of all 20 outcomes. Therefore, its probability P(either S₁S₂, S₁S₃, S₂S₁, S₂S₃, S₃S₁, or S₃S₂) = 6/20 or 0.3.

(e) The event “The boy selected circles both times” can be observed 2 times out of all 20 outcomes. Therefore, its probability P(either C₁C₂ or C₂C₁) = 2/20 or 0.1.

(f) The event “The boy selected a star at least once” (either once or twice) can be observed 18 times out of all 20 outcomes. Therefore, its probability P(either S₁S₂, S₁S₃, S₁C₁, S₁C₂, S₂S₁, S₂S₃, S₂C₁, S₂C₂, S₃S₁, S₃S₂, S₃C₁, S₃C₂, C₁S₁, C₁S₂, C₁S₃, C₂S₁, C₂S₂, or C₂S₃) = 18/20 or 0.9.

(g) The event “The boy selected a star at most once” (either once or never) can be observed 14 times out of all 20 outcomes. Therefore, its probability P(either S₁C₁, S₁C₂, S₂C₁, S₂C₂, S₃C₁, S₃C₂, C₁S₁, C₁S₂, C₁S₃, C₁C₂, C₂S₁, C₂S₂, C₂S₃, or C₂C₁) = 14/20 or 0.7.

Although the probability tree diagram is helpful for visualization and counting the numbers of outcomes and simple events, it is not very convenient for large-size samples. Below we will introduce more (mathematical) methods for counting.

Permutation

The permutation is a method of counting that determines the number of possible arrangements in a set of objects when the order of the arrangements does matter.

Example 3.7

Four horses, Alex, Fancy, Sugar, and Tucker, participate in a run. Determine the number of all possible sequences that can arrive at the finish.

Solution:

To make our writing more manageable, we will denote Alex as A, Fancy as F, Sugar as S, and Tucker as T.

Let’s list all possible sequences:

AFST, AFTS, ASFT, ASTF, ATFS, ATSF,

FAST, FATS, FSAT, FSTA, FTAS, FTSA,

SAFT, SATF, SFAT, SFTA, STAF, STFA,

TAFS, TASF, TFAS, TFSA, TSAF, TSFA

Therefore, the horses can arrive at the finish in 24 possible sequences.

Common mathematical problems involve choosing only several items from a set of items in a particular order. Let’s analyze another example.

Example 3.8

Four horses, Alex, Fancy, Sugar, and Tucker, participate in a run. Determine the number of all possible sequences when Alex performs better than Sugar.

Solution:

Now we will list all possible sequences when Alex arrives at the finish before Sugar. To make our task easier, we can refer to the solution of example 3.7, selecting the arrangements where A is listed before S.

AFST, AFTS, ASFT, ASTF, ATFS, ATSF,

FAST, FATS, FTAS, TAFS, TASF, TFAS

As one can see, there are 12 possible ways that Alex performs better than Sugar. Consider that in all these arrangements, the order did matter.

The permutation can be evaluated as a mathematical function using the following formula:

[latex]\displaystyle P(n,k)=\frac{n!}{(n-k)!}\qquad\text{(F3.1)}[/latex]

In this formula, n is the number of given objects and k is the size of ordered subsets. Moreover, n! is a mathematical function defined as n!=n×(n-1)×(n-2)×⋯×3×2×1 and called “n-factorial.”

Note: 0! = 1

Now we can easily solve examples 3.7 and 3.8 using the formula (F3.1).

Example 3. 7 (Revised Solution)

We have to choose four horses out of four, considering that the order does matter: n = 4 and k = 4.                                              Therefore,

[latex]\displaystyle P(4,4)=\frac{4!}{(4-4)!}=\frac{4!}{0!}=\frac{4\times3\times2\times1}{1}=24[/latex]

Example 3. 8 (Revised Solution)

In this example, we have to choose two horses out of four, considering that the order does matter (Alex comes before Sugar): n = 4 and k = 2. Hence,

[latex]\displaystyle P(4,2)=\frac{4!}{(4-2)!}=\frac{4!}{2!}=\frac{4\times3\times2\times1}{2\times1}=4\times3=12[/latex]

 

Combination

Combination is a method of counting that determines the number of possible arrangements in a set of objects when the order of the arrangements does not matter (contrary to the permutation).Let’s use those brave four horses, Alex, Fancy, Sugar, and Tucker, for the following example.

Example 3.9

Rebecca has to choose two out of four horses, Alex, Fancy, Sugar, and Tucker, for watering. How many choices does she have?

Solution:

Although we deal with the same horses, in this case, the order does not matter.

AF, AS, AT, FS, FT, ST

Therefore, there are six ways to choose two horses out of four if the order does not matter.

As in the case of arrangements, there exists a mathematical formula to evaluate the combination:

 

We can use this function to solve the previous example.

Example 3.9 (Revised Solution)

In this example, we must choose two horses out of four, considering that the order does not matter. Here, n = 4, k = 2. Using the formula (F3.2),

[latex]\displaystyle C(n,k)=\frac{n!}{k!(n-k)!}=\frac{4!}{2!(4-2)!}=\frac{4\times3\times2\times1}{2\times1\times2\times1}=6[/latex]

As was marked above, both permutation and combination functions are used mainly to evaluate probabilities for counting simple events. Here it is essential, first, to find out if the order matters in making choices.

Example 3.10

A case lock consists of four gears with ten digits, from 0 to 9 on each. The case owner sets up the key, so there are no repeating digits on the gears, and only the first three digits are odd. What is the probability that a stranger can open the lock after the first attempt?

Solution:

First, we must consider that the order does matter, which is why we will use permutation to count the simple events. The number of total outcomes can be determined as P(10, 4) since we have to arrange a four-digit number without the repetition of digits. There are five odd numbers to use for the first three gears. Therefore, the number of possible choices of first distinct numbers will be P(5, 3). According to the given condition (only the first three digits are odd), we have to choose one out of five even numbers for the fourth gear, P(5, 1). The probability can be determined as follows:

[latex]\displaystyle P(\text{odd, odd, odd, even})=\frac{P(5,3)\cdot P(5,1)}{P(10,4)}=\frac{\frac{5!}{(5-3)!}\cdot\frac{5!}{(5-1)!}}{\frac{10!}{(10-4)!}}=\frac{60\cdot5}{5040}=0.0595[/latex]

Therefore, the stranger has about a 6% chance of opening the key on the first attempt.

Example 3.11

A fishmonger offers 5 guppies and 7 zebrafish for sale. Mathew’s mom allowed him to purchase 5 fish of his choice. What is the probability that Mathew chose 3 guppies and 2 zebrafish?

Solution:

In this example, the order does not matter, so we will use combinations to count the simple events. Matthew has to choose 5 out of 12 fish: C(12, 5). There are C(5, 3) ways to choose 3 guppies out of 5 and C(7, 2) ways to choose 2 zebrafish out of 7. Consequently,

[latex]\displaystyle P(\text{3 guppies and 2 zebrafish})=\frac{C(5,3)\cdot C(2,7)}{C(5,12)}=\frac{\frac{5!}{3!(5-3)!}\cdot\frac{7!}{2!(7-2)!}}{\frac{12!}{5!(12-5)!}}=\frac{10\cdot21}{220}=0.9545[/latex]

Therefore, the probability that Mathew will choose 3 guppies and 2 zebrafish is 0.9545.

3.3. Event Relations and Venn Diagrams

In statistics, we often have to deal with combinations of events. This book will analyze the most useful event relations: union, intersection, and complement. Below, we will provide definitions of each of them. In many cases, diagrams ease understanding of these concepts and solutions to related problems. Generally, an event can be considered a set of simple outcomes. In the 1880s, the English mathematician John Venn (1834–1888), developed a tool to visualize set theory problems. Later this tool became known as a Venn diagram. Venn diagrams are widely used in many fields of mathematics and theoretical physics, including in problems about the probability of event relations.

A Venn diagram of events is an illustration that uses circles to show the relationships among the events. All events are included in a rectangle, representing the sample space. In this textbook, we will denote the sample space by S.

Example 3.12

On March 30, a registrar recorded the names of students registered for algebra and biology classes. Jane, George, Ali, Ann, and Morgan registered for algebra; Christine, Pavlo, and Marian registered for biology. Chris and Matthew are on the waiting list. Construct a Venn diagram for these events and determine the following:

(a) Number of students registered in algebra only

(b) Number of students registered in algebra

(c) Number of students registered in biology only

(d) Number of students registered in biology

(e) Number of students registered in both classes

(f) Number of students on the waiting list

Solution:

Figure 3.6. Venn diagram constructed for example 3.12

 (a) 5

 (b) 5

 (c) 3

 (d) 3

 (e) 8

 (f) 2

Example 3.13

On March 31, a registrar recorded the names of students registered for chemistry and drama classes. Jane, George, Ali, Ann, Morgan, and Pavlo registered for chemistry; Christine, Pavlo, Ali, Ann, and Marian registered for drama; and Chris and Matthew are still on the waiting list. Construct a Venn diagram for these events and determine the following:

(a) Number of students registered in chemistry only

(b) Number of students registered in chemistry

(c) Number of students registered in drama only

(d) Number of students registered in drama

(e) Number of students registered in both classes

(f) Number of students registered in chemistry or drama

(e) Number of students on the waiting list

Solution:

Figure 3.7. Venn diagram constructed for example 3.13

  (a) 3

  (b) 6

  (c) 2

  (d) 5

  (e) 3

  (f) 8

  (g) 2

Later, evaluating the probability of relations, we will see that students’ names are not necessary (with all respect to our students) to solve these questions. To evaluate the probabilities, we need to know the number of students registered in the classes or included on the waiting list. We could use the following Venn diagram instead of the one given in figure 3.7:

Figure 3.8. Venn diagram constructed for counted sets given in example 3.13

Hereafter, we will operate by the numbers of elements without naming them, using the notation |M| to indicate the numbers of elements in a set, M. Evaluating probabilities, we consider each simple event as an element of a set that corresponds to an event. Note that it is important to count the elements in each region, specifying the number of elements in overlapping regions. Using the set notation, the Venn diagram given in figure 3.8 can be described as |C| = 6, |D| = 5, |W| = 2, |C only| = 3, |D only| = 2, |C and D| = 3. Also, note that the number of simple events in the sample space is |S| = 10.

Intersections

In example 3.13, three students registered in the same classes. Using the set theory language, we state that the intersection of events C and D consists of three simple outcomes (registrations of Alli, Ann, and Pavlo). As the reader might notice, the word “and” was used to describe this relation.

The intersection of given events is the collection of all simple outcomes, which are elements of all given events. The intersection is denoted by a symbol, Ç. For example, the expression “Intersection of events A and B” can be written as A Ç B. Using the set theory symbols, we can write the expressions of intersections in examples 3.12 and 3.13 as follows:

A∩B=∅

C∩D={Ali,Ann,Pavlo}

The symbol ∅ denotes the empty set, which does not contain any element. One can express the solutions of parts (e) of examples 3.12 and 3.13 using this symbol as |A Ç B| = 0 and |C Ç D| = 3, respectively. Based on these examples, we can conclude that the number of simple events in an empty set equals zero, |∅|=0.

Unions

The union of given events is the collection of all simple events, which are elements of at least one of the given events. For example, we can state that the union of events C and D in example 3.13 contains eight simple events (registration of all students). The symbol È symbolizes unions. Consequently,

C∪D={Jane,George,Morgan,Ali,Ann,Pavlo,Christine,Marian}

Usually, the words “or” and “either” describes the union of events. Note that the number of students within the union C∪D can be calculated as “number of students in chemistry only” + “number of students in drama only” + “number of students registered in both chemistry and drama.” Using the symbol notations, one can write,

Alternatively, we can count the number of students in this union as

       |C ∪ D| = |C| + |D| − |C ∩ D|                          (F3.4)

Hereafter, we will mainly refer to the last formula. Therefore, for the data provided in example 3.13,

|C ∪ D| = |C| + |D| − |C ∩ D| = 6 + 5 – 3 = 8.

Complements

The complement of a given event is the set of simple events outside of the given event. The complement of event A is denoted by A^C. Let’s consider the data provided in example 3.12 and determine the complement of event A.

A^C = {Chris, Matthew, Christine, Pavlo, Marian}

As mentioned above, we often need to count the number of elements/simple events. Consequently, we might need to determine |A^C|. For the data provided in example 3.12, |A^C| = 5. Note that we could calculate the number of simple events in A^C by subtracting the number of simple events in A from the number of simple events in the sample space, |A^C| = 10 – 5 = 5. In general,

|A^C| = |S| – |A|                                    (F3.5)

3.4. Probabilities of Event Relations.

At the beginning of this chapter, we defined the concept of probability. Then, we analyzed examples within which the probabilities of events were evaluated. However, in real cases, we often need to deal with relations of events. In this section, we will learn how to determine probabilities of various relations: complement, intersection, and union.

Complement Rule

Sometimes it is difficult to determine the probability of an event occurring, but not difficult to determine the probability of an event not occurring. Consider the process of estimating the effectiveness of a vaccine. In reality, this is a very complicated research project. Statistically, this process aims to evaluate the probability that a randomly selected person from the control group is immunized against a virus. In general, to conduct this task, we would first vaccinate all participants and observe their health condition using appropriate medical methods. Then we would evaluate the probability of the effectiveness of the vaccine. Let X and Y denote the effective and ineffective vaccine events, respectively. Consequently, P(X) and P(Y), respectively, denote the probabilities of the effectiveness and ineffectiveness of the vaccine. Assume that the number of immunized participants is x and the number of participants who were not immunized after the vaccination is y. Therefore, the probabilities of the effectiveness and ineffectiveness of the vaccine are P(X) = x/N and P(Y) = y/N, respectively, where N is the number of participants. Note that x + y = N or x = N – y.

Then

[latex]\displaystyle P(X)=\frac{x}{N}=\frac{N-y}{N}=1-\frac{y}{N}[/latex]

Or

Considering that the event X is a complement of the event Y, one can generalize the formula (F3.6) as follows:

[latex]\displaystyle P(A^C)=1-P(A)\qquad\text{(F3.7)}[/latex]

The formula (F3.7) is called the complement rule.

Example 3.14

Rose recorded her and her classmates’ ability to speak Indigenous languages and evaluated the probability of this event, which was equal to 0.7. What is the probability that a randomly selected student in Rose’s class does not speak any Indigenous language?

Solution:

Let I denote the event of the ability to speak at least one Indigenous language. The event of not speaking any Indigenous language is the complement of I. Using the formula (F3.7), we determine [latex]\displaystyle P(I^C)=1-P(I)=1-0.7=0.3[/latex]. Therefore, the probability that a randomly selected student does not speak any Indigenous language equals 0.3 or 30%.

Addition Rule for Probabilities

Another relation that we analyzed above was the union. Let’s return to example 3.12 and evaluate the probability that a randomly selected student is registered in algebra or biology class. In other words, we need to find the probability that a randomly selected student belongs to the union [latex]A\cup B[/latex] . We can still use the Venn diagram to evaluate [latex]\displaystyle P(A\cup B)[/latex] . As discussed earlier, we will operate by the number of simple events in sets. So, the diagram can be modified accordingly (fig. 3.9):

Figure 3.9. Venn diagram constructed for counted sets given in example 3.12

The total number of simple events (students) in the sample space is |S|= 10. The numbers of simple events (students) in A and B are |A|=5 and |B|=3. Note that none of the students are registered in both classes. Consequently, the number of simple events (students) in the union [latex]\displaystyle |A\cup B|\text{ is }|A|+|B|[/latex] . Then [latex]\displaystyle P(A\cup B)=\frac{5}{10}+\frac{3}{10}=\frac{8}{10}=\frac{4}{5}[/latex] or 0.8. Let’s use this solution to generalize the procedure of evaluation of the probability of the union.

[latex]\displaystyle P(A\cup B)=\frac{|A|}{|S|}+\frac{|B|}{|S|}[/latex]

But [latex]\frac{|A|}{|S|}=P(A)[/latex] and [latex]\frac{|B|}{|S|}=P(B)[/latex].

Therefore,

[latex]\displaystyle P(A\cup B)=P(A)+P(B)\qquad\text{(F3.8)}[/latex]

considering that [latex]A\cap B=\emptyset[/latex], i.e., [latex]|A\cap B|=0[/latex]. Events like [latex]A[/latex] and [latex]B[/latex] are mutually exclusive and have no common simple events.

Now let’s evaluate P(A∪B) for the data provided in example 3.13. By definition, the probability that a randomly selected student is registered either in chemistry or drama class can be evaluated as follows:

[latex]\displaystyle P(C\cup D)=\frac{|C\cup D|}{|S|}\qquad\text{(F3.9)}[/latex]

Let’s use the formula (F3.4) in (F3.9):

 

 

One can note that [latex]\frac{|C|}{|S|}=P(C)\text{ and }\frac{|D|}{|S|}=P(D).\ \text{The ratio }\frac{|C\cap D|}{|S|}[/latex] represents the probability that a randomly selected student is registered in both (chemistry and drama) classes. We define this quantity as the probability and denote it as [latex]\displaystyle P(C\cap D)[/latex] . Therefore,

[latex]\displaystyle P(C\cup D)=P(C)+P(D)-P(C\cap D)\qquad\text{(F3.11)}[/latex]

The formula (F3.11) is called the addition rule of the probability. Let’s consider that if the intersection of events C and D are mutually exclusive, then [latex]\displaystyle P(C\cap D)=0[/latex], and the formula (F3.11) turns into the formula (F3.8). In other words, the formula (F3.8) represents the special case of the additional rule of the probability for mutually exclusive events.

Summarising examples 3.12 and 2.13, we can emphasize that events A and B in example 3.12 are mutually exclusive. In contrast, events C and D in example 3.13 overlap and can be classified as not mutually exclusive. In example 3.12, the registered students are in one class only, either algebra or biology; in example 3.13, some students are registered in two classes, chemistry and drama. In general, mutually exclusive events cannot happen at the same time. Therefore, to evaluate the probabilities of unions, first we will have to verify if the events are mutually exclusive or not.

Conditional Probability

Sometimes we need to determine the probability of an event upon a condition. Our knowledge from this chapter allows us to consider this condition as another event. Based on this consideration, we can formalize the definition of conditional probability. The conditional probability is a measure of the chances of an event happening, given that this event takes upon the occurrence of another event. The latter describes the condition.

Using the probability symbols, we denote the probability of event A under the condition B as P(A|B).

Example 3.15

A registrar registered First Nation, Inuit, and Métis students in Cree and Dene language classes and constructed the following table based on her records:

	Cree class	Dene class
First Nation	12	15
Inuit	5	8
Metis	10	6

Determine the probability that a randomly selected Métis student is registered in the Cree class.

Solution:

It is given that we have to look for the Cree class registries among the Métis students. We consider being Métis as an event that defines the condition. There are 10 + 6 = 16 Métis students: |M| = 16. Ten of them are registered in the Cree class. In other words, ten students are Métis and registered in the Cree class. Note that this is the intersection of two events, being Métis and registered in the Cree class. Hence, [latex]|M\cap C|=10[/latex] . Therefore, the probability that a randomly selected Métis student registered in the Cree class can be determined as

 

We can use this solution and derive a formula for evaluating the conditional probability. Assume that S denotes the sample space; hence, |S| is the total number of simple events in the sample space. As we can see from the solution of the example above, the conditional probability of event C, given M, can be evaluated as

[latex]\displaystyle P(C\mid M)=\frac{|M\cap C|}{|M|}\qquad\text{(F3.12)}[/latex]

Let’s divide the numerator and denominator of the fraction in (F3.12) by |S|:

Considering that. [latex]\frac{|M\cap C|}{|S|}=P(M\cap C)\text{ and }\frac{|M|}{|S|}=P(M)[/latex], we can obtain the following general formula for the conditional probability:

[latex]\displaystyle P(C\mid M)=\frac{P(M\cap C)}{P(M)}\qquad\text{(F3.14)}[/latex]

The next example will show that if the conditions defined in (F3.15) are not satisfied, then events A and B are dependent.

Example 3.16

A circus performer has three red and four blue balls in her box (fig. 3.10). For her blind focus, she randomly selects one ball in each of two consecutive attempts, for a total of two balls.

 (a) Determine the probability that in the second attempt, the performer chose a blue ball, given that in the first attempt, her choice             was blue.

(b) Determine the probability that the performer chose the blue ball in the second attempt, given that her choice was red in the first attempt.

Solution:

Figure 3.10. Diagram constructed for example 3.16

(a) If the performer’s first choice is blue, then three red and three blue balls will be left for the second attempt. The probability that            she will choose the blue ball again equals 3/6 = 1/2 or 0.5, or P(B|B) = 0.5.

(b) If the performer’s first choice is red, then two red and four blue balls will be left for the second attempt. Therefore, the probability that she will choose the blue ball on the second attempt equals 4/6 = 2/3 or 0.6̅, or P(B|R) = 0.6̅.

Note that, in this example, the unconditional probability that a randomly selected ball is blue, P(B), is equal to 4/7 = 0.571, while the conditional probabilities P(B|B) = 0.5 and P(B|R) = 0.6̅. In other words, the last two probabilities are determined by given conditions. The next example shows that this is not always the case.

Example 3.17

Consider the spinner shown in figure 3.11.

Figure 3.11. A two-sector spinner from example 3.17

(a) Determine the probability of spinning the blue sector in the second attempt, given that the blue sector was spun in the first attempt.

(b) Determine the probability of spinning the blue sector in the second attempt, given that the red sector was spined in the first attempt.

Solution:

Using common sense, we can conclude that the chance of observing the blue sector in the second attempt does not depend on what colour of the sector was observed in the first attempt. Spinners do not have memory. Let’s confirm our conclusion using a more “statistical” way. The sample space of the two-attempt experiment contains the following:

S = {BB, BR, RB, RR}

(a) The condition for this part was the observation of blue colour in the first attempt. We have two such events, BB and BR. In one of them, the blue colour was observed in the second attempt. Therefore, P(B|B) = 1/2.

(b) The condition for this part was the observation of red colour in the first attempt. We have two such events, RB and RR. In one of them, the blue colour was observed in the second attempt. Therefore, P(B|R) = 1/2.

As one can see, our intuitive conclusion was reasonable. Our statistical analysis shows that the observation of the blue colour in the second attempt does not depend on our observation in the first attempt. At the beginning of this chapter, we defined these types of events as independent. Now we can formalize this definition using conditional probability.

Events A and B are independent if and only if

[latex]\displaystyle P(A\mid B)=P(A)\text{ or }P(B\mid A)=P(A)\qquad\text{(F3.15)}[/latex]

Bayes’ Rule

Example 3.16 (Extended)

Let’s consider an extended version of example 3.16, assuming that the circus performer has three red, four blue, and two green balls in her box. For her blind focus, in her first attempt, she randomly selected a blue ball. What are the chances that the performer would get a green ball in the second one?

Solution:

Using the conditional probability language, we can rephrase the question: Determine the probability that the performer will get the green ball in the second attempt, given that she got the blue ball in her first attempt. After getting the blue ball on the first attempt, she is left with three blue, three red, and two green balls; 3 + 3 + 2 = 8 balls altogether. Therefore, her chances of getting a green ball is 2/8 or 1/4, based on the following:

[latex]\displaystyle P(G\mid B)=\frac{2}{8}=\frac{1}{4}[/latex]

The analysis that we used to solve the example above was generalized by English mathematician and philosopher Thomas Bayes (1701–1761) and is known as Bayes’ Rule.

Assume that S₁,S₂,...S_N represent N mutually exclusive and exhaustive (sample space [latex]S=S_1\cup S_2\cup\ldots\cup S_N[/latex]) partitions with corresponding probabilities [latex]P(S_1),\ P(S_2),\ \ldots,\ P(S_N)[/latex] . The probability of an event Si given that event A occurs can be determined using the following formula:

[latex]\displaystyle P(S_i\mid A)=\frac{P(S_i)\,P(A\mid S_i)}{\sum_{j=1}^{N} P(S_j)\,P(A\mid S_j)}\qquad\text{(F3.14a)}[/latex]

One has to note that the formula F3.14 is a special case of the formula F3.14a.

Now, let’s solve example 3.16 (extended) using the Bayes’ Rule.

[latex]\displaystyle P(G\mid B)=\frac{P(G)P(B\mid G)}{P(B)P(B\mid B)+P(R)P(B\mid R)+P(G)P(B\mid G)}[/latex]

[latex]\displaystyle =\frac{\frac{2}{9}\cdot\frac{4}{8}}{\frac{4}{9}\cdot\frac{3}{8}+\frac{3}{9}\cdot\frac{4}{8}+\frac{2}{9}\cdot\frac{4}{8}}=\frac{8}{32}=\frac{1}{4}[/latex]

As one can see, we got the same answer using Bayes’ Rule, also known as Bayes’ Theorem. Bayes’ Rule is widely used in various areas, including medicine, finance, and machine learning, to make projections and decisions. More detailed explanation of Bayes’ Rule is beyond the scope of this book.

Multiplication Rule for Probabilities

As the reader noticed in the examples analyzed above, we often need to determine the probability of two or more events occurring. For instance, in examples 3.16 and 3.17, we had to find the probability of events occurring in two consecutive attempts. Using the set theory approach, we can consider the occurrence of two or more events as their intersection. To evaluate those probabilities (probabilities of intersection), we used the sample space approach. If the probabilities of certain events are given, the probability of the intersection can be evaluated using the multiplication rule.

For any two events, A and B, the probability of their intersection (the probability that both occur) is equal to

[latex]\displaystyle P(A\cap B)=P(A)\,P(B\mid A)\qquad\text{(F3.16)}[/latex]

According to (F3.15), if A and B are independent events,

[latex]\displaystyle P(A\cap B)=P(A)P(B)\qquad\text{(F3.17)}[/latex]

The formula (F3.17) is known as the special case of the multiplication rule for probabilities.

Example 3.18

According to 2016 census results, 37% of Indigenous language speakers speak Cree (“The Aboriginal languages of First Nations people, Métis and Inuit,” Census of Population, 2016, Statistics Canada, Catalogue no. 98-200-X2016022, ISBN 978-0-660-20371-3). What is the probability that exactly one of the three randomly selected Indigenous speakers speaks Cree?

Solution:

Let C define the event that a randomly selected Indigenous language speaker speaks Cree, hence P(C) = 0.37. Then the event that a randomly selected Indigenous language speaker does not speak Cree is a complement of C and P(C^C) = 1 – 0.37 = 0.63. Either the first, second or third Indigenous language speaker speaks Cree. Consequently,

P(exactly one Indigenous language speaker speaks Cree)

= P(C)P(C^C)P(C^C) + P(C^C)P(C)P(C^C) + P(C^C)P(C^C)P(C) = (0.37)(0.63)(0.63) + (0.63)(0.37)(0.63)

+ (0.63)(0.63)(0.37) = 0.441

Therefore, the probability that exactly one of the three randomly selected Indigenous speakers speaks Cree equals 0.441 or 44.1%.

This type of probability is known as binomial probability, which will be discussed later in this book.

Example 3.19

According to 2016 census results, 37% of Indigenous language speakers speak Cree (“The Aboriginal languages of First Nations people, Métis and Inuit,” Census of Population, 2016, Statistics Canada, Catalogue no. 98-200-X2016022, ISBN 978-0-660-20371-3). In addition to this information, Statistics Canada also reports that 27.8% of Cree speakers live in Saskatchewan. What is the probability that a randomly selected Indigenous language speaker is a Cree-speaking Saskatchewan resident?

Solution:

Let C define the event that a randomly selected Indigenous language speaker speaks Cree and S define the event that a randomly selected person is a Saskatchewan resident. Then P(C) = 0.37 and P(S|C) = 0.278. Consequently,

[latex]\displaystyle P(\text{Cree-speaking Saskatchewan resident})=P(C\cap S)=P(C)P(S\mid C)=0.37\times0.278=0.103[/latex]

Therefore, the probability that a randomly selected Cree speaker is a Saskatchewan resident equals 0.103 or 10.3%.

3.5. Probability Distributions and Random Variables

One of the above-analyzed examples referred to 2016 census results about the Indigenous languages (“The Aboriginal languages of First Nations people, Métis and Inuit,” Census of Population, 2016, Statistics Canada, Catalogue no. 98-200-X2016022, ISBN 978-0-660-20371-3). The concentrations of Indigenous people speaking Algonquian languages in Canadian provinces and territories, as determined within the census, are shown in table 3.2.

Table 3.2. Percentage concentrations of Indigenous People speaking Algonquian languages in Canadian provinces and territories

Provinces and territories	Concentrations of Indigenous People speaking Algonquian languages
Alberta	16.7%
Manitoba	21.7%
Ontario	17.2%
Quebec	21.2%
Saskatchewan	16%
Others	7.2%

Let’s try to interpret the table 3.2 in terms of probabilities. For instance, if the concentration of Indigenous People speaking Algonquian languages in Alberta is 16.7%, we can consider this information as follows: the probability of the event that a randomly selected Algonquian language speaker lives in Alberta is 0.167. So now we can slightly modify table 3.2.

Table 3.3. Probabilities of speaking of Indigenous People Algonquian languages in Canadian provinces

Provinces and territories	Probability
Alberta	0.167
Manitoba	0.217
Ontario	0.172
Quebec	0.212
Saskatchewan	0.16
Others	0.072
Total (all Algonquian language speakers in Canada)	1

Table 3.3 represents the probability distribution over the provinces and territories. The diagram below represents how the probability of speaking Algonquian languages is distributed along the Canadian provinces and territories.

Figure 3.12. Probability distribution diagram constructed for the data provided in table 3.3

Consider that this is a spatial (geographical) probability distribution across Canada. The diagram looks very informative for those who need to compare the knowledge of Algonquian languages in Canadian provinces and territories.

Below we will define the concept of probability distribution in statistics.

Example 3.19

Let’s consider two two-sector spinners with equal areas (fig. 3.12) and determine the probability of spinning the red sector in both of them on the first attempt.

Figure 3.13. Two two-sector spinners with equal areas

Using common sense, we can expect one of the following observations: the red sector can be spun in none of the spinners, in one of them, or in both of them. In other words, the red sector can be observed 0, 1, or 2 times. Let x denote the number of observations of red in the two-trial experiment (table 3.4).

Table 3.4. Probability distribution table for events described in example 3.19

Observation	x	P(x)
BB	0	0.25
BR or RB	1	0.25+0.25=0.5
RR	2	0.25

Mathematically, P(x) can be considered as a function of a variable x. Thereafter, we will call function X, with values x = 0, 1, 2 with probabilities P(x), as a random variable. The function P(x) is called a probability distribution. It is very convenient to represent the probability distribution in a graphical format (fig. 3.14).

Figure 3.14. Probability distribution diagram constructed for events provided in example 3.19

In statistics, we use random variables to quantify outcomes of random occurrences. The concept of random variables was first developed by the Russian mathematician Pafnuty Chebyshev (1821–1894). Chebyshev is known for his fundamental contributions to the fields of probability, statistics, mechanics, and number theory. In previous chapters, we classified variables as discrete and continuous. In general, with discrete random variables we are concerned with counting things; with a continuous random variable we are generally concerned with measuring something.

Once we know all the possible values and the probabilities associated with those values for a discrete random variable, we can construct a discrete probability distribution. The discrete probability distribution describes how the probabilities are spread over the various values that the discrete random variable can take. Three discrete probability distributions—binomial, Poisson, and geometric—have many applications in many areas of science and economy. In the next chapter of this book, we will talk about all three of them, with more details on the binomial distribution.

For discrete random variable distributions, there is always a gap between the distribution points. For instance, the outcome of a roll of a die can only be 1, 2, 3, 4, 5, or 6, not 4.2, 3.5, etc. However, unlike a discrete distribution, there are no gaps between values in a continuous distribution. For example, a height of a randomly selected tree in a forest can be any number within a reasonable interval. Hence the distribution of probabilities of trees’ heights can be considered an example of continuous variable distribution. In this book, we will talk about three types of probability distributions of continuous variables: normal distribution for so-called large samples, Student’s distribution (t-distribution) for small-size samples, and Fisher distribution (F-distribution).

Chapter 3 Summary

• Probability with a view toward statistics
  • o Classical probability
  • o Relative frequency probability
  • o Subjective probability
• Events and sample space
• Calculating probabilities
  • o Subjective probability
  • o Independent events and probability trees
  • o Permutation
  • o Combination
• Event relations and Venn diagrams
  • o Intersections
  • o Unions
  • o Complements
• Probabilities of event relations
  • o Complement rule
  • o Addition rule for probabilities
  • o Conditional probability
  • o Multiplication rule for probabilities
• Probability distributions and random variables

You can also access the presentation file of this chapter. Just click the link to view.

EXERCISES

3.1. Probability with a View Toward Statistics

1. (Introduction to statistics, 2^nd Ed, Test Bank, Anderson, D. R., Sweeney, D.J., Williams, T.A, 1991) The set of all possible sample points (experimental outcomes) is called

a) A sample
b) An event
c) Sample space
d) None of the above answers are correct.

2. (Introduction to statistics, 2^nd Ed, Test Bank, Anderson, D. R., Sweeney, D.J., Williams, T.A, 1991) The probability assigned to each experimental outcome must be

a) Any value larger than zero
b) Smaller than zero
c) At least one
d) Between zero and one

3. (Introduction to statistics, 2^nd Ed, Test Bank, Anderson, D. R., Sweeney, D.J., Williams, T.A, 1991) Probability ranges between

a) 0 to 1.0
b) 1 to infinity
c) Minus infinity to 1
d) Minus infinity to plus infinity
e) None of the above answers are correct

4. (Introduction to statistics, 2^nd Ed, Test Bank, Anderson, D. R., Sweeney, D.J., Williams, T.A, 1991) In statistical experiments, each time the experiment is repeated

a) The same outcome must occur
b) The same outcome cannot occur again
c) A different outcome may occur
d) None of the above answers are correct.

5. (Introduction to statistics, 2^nd Ed, Test Bank, Anderson, D. R., Sweeney, D.J., Williams, T.A, 1991) Since the sun must rise tomorrow, then the probability of the sun rising tomorrow is

a) Much larger than one
b) Zero
c) Less than zero
d) Any value over 100
e) None of the above answers are correct.

3.2. Events and Sample Space

1. The Fast and Furious racers have 3 different sport cars (Mercedes Benz SLR McLaren, Lamborghini Murcielago LP670-4 Superveloce, and Saleen S7 Twin Turbo). Dominic (D), Brian (B) and Mia (M) need to choose a car before they race.  The first person receives the Mercedes, the second one receives the Lamborghini, and the last one receives the Saleen. For example, if Dominic gets the Mercedes, Brian gets the Lamborghini and Mia gets the Saleen, then the outcome is denoted as DBM.

a) List the sample space concerning the outcomes with the notation described above.
b) List the outcomes in each of the following events.
Event A: observe Mia gets the Mercedes
Event B: observe Brian does not get the Lamborghini.
c) Find the probabilities of events A and B. Are A and B mutually exclusive events?
P(A)=
P(B)=
A and B are mutually exclusive events?

2. The Star War light-saber has 3 different colors (Red, Green, and Blue). Anakin (A), Yoda (Y) and Luke (L) need to choose a light-saber before they fight. The first person receives red light-saber, the second one receives green, and the last one receives blue. For example, if Anakin gets red light-saber, Yoda gets green and Luke gets Blue, then the outcome is denoted as AYL.

a) List the sample space concerning the outcomes with the notation described above.
b) List the outcomes in each of the following events.
Event A: observe Yoda gets the red light-saber.
Event B: observe Anakin does not get the blue light-saber.

c) Find the probabilities of events A and B. Are A and B mutually exclusive events?
P(A)=
P(B)=
A and B are mutually exclusive events?  Yes  or   No    ( circle your answer )

 

3. (Introduction to statistics, 2nd Ed, Test Bank, Anderson, D. R., Sweeney, D.J., Williams, T.A, 1991) Which of the following statements is (are) always true?

a)
$$
-1 \leq P(E_i) \leq 1
$$

b)
$$
P(A) = 1 - P(A^c)
$$

c)
$$
P(A) + P(B) = 1
$$

d) Both b and c

e) None of the above answers are correct.

4. (Introduction to statistics, 2nd Ed, Test Bank, Anderson, D. R., Sweeney, D.J., Williams, T.A, 1991) Events that have no sample points in common are

a) Independent events
b) Posterior events
c) Mutually exclusive events
d) Complements
e) None of the above answers are correct

5. (Introduction to Statistics, 2nd Ed, Test Bank, Anderson, D. R., Sweeney, D. J., Williams, T. A., 1991) One of the basic requirements of probability is

a)
$$
P(E_i) \geq 0
$$
for each experimental outcome \(E_i\).

b)
$$
P(A) = P(A^c) - 1
$$

c)
$$
\sum_{i=1}^{k} P(E_i) = 1
$$
if there are \(k\) experimental outcomes.

d) Both b and c

e) None of the above answers are correct

 

3.3. Calculating Probabilities

Using Simple Events
Using Counting Rules

6. In order to survey the quality of tires in Winter time, a researcher decides to select 4 cars from a total of 10 in SIAST parking lot # 15. How many different ways can the selection be made?

7. The Department of Mathematics and Statistics at the University of Regina has 20 faculty members. How many different ways can a Head and an Assistant head be selected?

8. There are 3 Mercedes, 2 Lamborghinis, and 2 Saleens cars on a street near Brian’s garage. Brian and Mia would like to put  2 of them into the garage, one on the left side and another on the right.

a) How many different arrangements of the cars are possible?
b) How many different arrangements of the cars are possible if the left one has to be a Mercedes and the right one must be a Lamborghini or Saleen cars in the garage?

9. There are 2 Stats, 3 Calculus, and 2 Linear Algebra books on the table. Darth Vader would like to arrange (once at a time) 2 of them on the shelf.

a) How many different arrangements of the books are possible?
b) How many different arrangements of the books are possible if the first one has to be a Stats book and the second one should be a Calculus or a Linear Algebra book?

10. There are 3 trails leading to Camp A from your starting position. There are 2 trails from Camp A to Camp B and there are 2 trails from Camp B to Camp C. How many different routes are there from the starting position to Camp C? Draw a tree diagram to illustrate your answer.

3.4. Event Relations and Venn Diagrams

Intersections
Unions
Complements

12.

a)    What is the probability of drawing a club from a deck of cards?

b) What is the probability of drawing either a club or a heart?
c) What is the probability of drawing a 3?
d) What is the probability of drawing either a 7 or a Queen?
e) What is the probability of drawing either a diamond or a 9?

13. In Regina, 48% of all teenagers own a skateboard, 50% own a roller blades, and 39% of all teenagers own a roller blades and skateboards. What is the probability that a teenager owns roller blades or a skateboard?

14. A study was done on individuals who had accidents and came to the emergency ward of a hospital.  Accident severity was ranked as either minor, major or fatal.