Why Is the Statement Incorrect That Probability Can Be Used to Predict the Result of an Activity?

Basic Probability Rules

• Introduction
• Rules of Probability
  • Probability Dominion Ane (For any outcome A, 0 ≤ P(A) ≤ 1)
  • Probability Rule Two (The sum of the probabilities of all possible outcomes is one)
  • Probability Rule Iii (The Complement Dominion)
  • Probabilities Involving Multiple Events
  • Probability Dominion Four (Improver Rule for Disjoint Events)
  • Finding P(A and B) using Logic
  • Probability Rule 5 (The Full general Improver Rule)
• Rounding Rule of Thumb for Probability
• Let's Summarize

CO-vi: Employ bones concepts of probability, random variation, and commonly used statistical probability distributions.

LO 6.four: Relate the probability of an event to the likelihood of this event occurring.

LO six.5: Apply the relative frequency arroyo to guess the probability of an upshot.

LO 6.half dozen: Apply basic logic and probability rules in lodge to find the empirical probability of an event.

In the previous section, nosotros introduced probability as a way to quantify the uncertainty that arises from conducting experiments using a random sample from the population of interest.

We saw that the probability of an event (for example, the outcome that a randomly chosen person has blood type O) can be estimated past the relative frequency with which the event occurs in a long series of trials. And so we would collect data from lots of individuals to estimate the probability of someone having blood blazon O.

In this section, we will plant the basic methods and principles for finding probabilities of events.

We will as well cover some of the basic rules of probability which can be used to calculate probabilities.

Introduction

Nosotros will begin with a classical probability example of tossing a fair coin iii times.

Since heads and tails are equally probable for each toss in this scenario, each of the possibilities which can upshot from three tosses will also be equally probable so that we tin list all possible values and utilise this list to calculate probabilities.

Since our focus in this grade is on data and statistics (not theoretical probability), in well-nigh of our hereafter problems we will use a summarized dataset, usually a frequency table or two-way table, to calculate probabilities.

EXAMPLE: Toss a off-white coin three times

Let'due south listing each possible consequence (or possible result):

{HHH, THH, HTH, HHT, HTT, THT, TTH, TTT}

Now let's ascertain the following events:

Outcome A: "Getting no H"

Event B: "Getting exactly ane H"

Event C: "Getting at least 1 H"

Note that each event is indeed a statement nearly the effect that the experiment is going to produce. In do, each effect corresponds to some collection (subset) of the possible outcomes.

Event A: "Getting no H" → TTT

Event B: "Getting exactly one H" → HTT, THT, TTH

Event C: "Getting at least i H" → HTT, THT, TTH, THH, HTH, HHT, HHH

Here is a visual representation of events A, B and C.

From this visual representation of the events, it is easy to see that event B is totally included in event C, in the sense that every issue in result B is also an result in event C. As well, note that event A stands apart from events B and C, in the sense that they accept no outcome in mutual, or no overlap. At this indicate these are only noteworthy observations, but as you'll discover later, they are very important ones.

What if we added the new issue:

Effect D: "Getting a T on the commencement toss" → THH, THT, TTH, TTT

How would it look if nosotros added event D to the diagram above? (Link to the answer)

Recollect, since H and T are equally probable on each toss, and since at that place are 8 possible outcomes, the probability of each outcome is 1/8.

See if you lot can answer the following questions using the diagrams and/or the list of outcomes for each event forth with what you have learned and so far most probability.

If yous were able to answer those questions correctly, y'all probable have a adept instinct for calculating probability! Read on to learn how nosotros volition apply this knowledge.

If not, we will try to help y'all develop this skill in this section.

Annotate:

• Note that in event C, "Getting at least 1 head" there is only 1 possible result which is missing, "Getting NO heads" = TTT. We will address this once again when we talk about probability rules, in item the complement rule. At this signal, nosotros just want you to think nearly how these two events are "opposites" in this scenario.

It is VERY important to realize that simply because we can listing out the possible outcomes, this does non imply that each outcome is equally likely.

This is the (funny) message in the Daily Bear witness clip we provided on the previous page. But allow's think about this again. In that clip, Walter is claiming that since there are two possible outcomes, the probability is 0.5. The 2 possible outcomes are

• The earth will be destroyed due to use of the big hadron collider
• The globe will Not be destroyed due to apply of the big hadron collider

Hopefully it is clear that these two outcomes are not equally likely!!

Let's consider a more common example.

Case: Birth Defects

Suppose we randomly select three children and we are interested in the probability that none of the children take whatsoever birth defects.

We apply the notation D to represent a child was born with a nascency defect and Northward to represent the child born with NO nascence defect. We can list the possible outcomes just as we did for the money toss, they are:

{DDD, NDD, DND, DDN, DNN, NDN, NND, NNN}

Are the events DDD (all iii children are born with nativity defects) and NNN (none of the children are born with birth defects) equally likely?

It should be reasonable to you that P(NNN) is much larger than P(DDD).

This is because P(N) and P(D) are non equally probable events.

It is rare (certainly not 50%) for a randomly selected child to be built-in with a nascency defect.

Rules of Probability

Now we move on to learning some of the bones rules of probability.

Fortunately, these rules are very intuitive, and as long every bit they are practical systematically, they will let us solve more than complicated problems; in particular, those problems for which our intuition might be inadequate.

Since most of the probabilities you will be asked to find tin can be calculated using both

• logic and counting

and

• the rules we volition be learning,

we give the following communication as a principle.

PRINCIPLE:

If you lot can calculate a probability using logic and counting you lot do non Demand a probability rule (although the correct rule tin always be practical)

Probability Rule One

Our beginning rule merely reminds the states of the basic property of probability that nosotros've already learned.

The probability of an event, which informs united states of the likelihood of it occurring, can range anywhere from 0 (indicating that the event will never occur) to 1 (indicating that the event is certain).

Probability Rule I:

• For whatever event A, 0 ≤ P(A) ≤ 1.

Notation: One practical utilize of this dominion is that it tin can exist used to identify whatsoever probability calculation that comes out to be more than one (or less than 0) as incorrect.

Before moving on to the other rules, let's commencement look at an example that will provide a context for illustrating the side by side several rules.

EXAMPLE: Claret Types

As previously discussed, all human being claret can be typed as O, A, B or AB.

In addition, the frequency of the occurrence of these blood types varies by ethnic and racial groups.

Co-ordinate to Stanford University'due south Blood Center (bloodcenter.stanford.edu), these are the probabilities of homo blood types in the United States (the probability for blazon A has been omitted on purpose):

Motivating question for dominion 2: A person in the United States is chosen at random. What is the probability of the person having blood type A?

Answer: Our intuition tells us that since the four blood types O, A, B, and AB frazzle all the possibilities, their probabilities together must sum to 1, which is the probability of a "sure" event (a person has one of these four blood types for certain).

Since the probabilities of O, B, and AB together sum to 0.44 + 0.ane + 0.04 = 0.58, the probability of blazon A must be the remaining 0.42 (1 – 0.58 = 0.42):

Probability Dominion Two

This example illustrates our second rule, which tells us that the probability of all possible outcomes together must be 1.

Probability Rule 2:

The sum of the probabilities of all possible outcomes is 1.

This is a proficient identify to compare and contrast what nosotros're doing here with what we learned in the Exploratory Data Assay (EDA) section.

• Discover that in this problem we are essentially focusing on a single categorical variable: blood type.
• We summarized this variable higher up, as nosotros summarized single categorical variables in the EDA section, by listing what values the variable takes and how often information technology takes them.
• In EDA we used percentages, and here we're using probabilities, but the ii convey the same information.
• In the EDA section, we learned that a pie nautical chart provides an advisable brandish when a single chiselled variable is involved, and similarly we can use it hither (using percentages instead of probabilities):

Fifty-fifty though what nosotros're doing hither is indeed similar to what we've done in the EDA section, there is a subtle but important deviation between the underlying situations

• In EDA, we summarized data that were obtained from asampleof individuals for whom values of the variable of interest were recorded.
• Here, when we present the probability of each blood type, nosotros have in listen the entirepopulationof people in the United States, for which we are presuming to know the overall frequency of values taken past the variable of interest.

Probability Rule Three

In probability and in its applications, we are often interested in finding out the probability that a certain issue willnotoccur.

An of import signal to understand here is that "issue A does not occur" isa split up event that consists of all the possible outcomes that are non in A and is called "the complement event of A."

Notation: we will write"not A" to denote the event that A doesnot occur. Here is a visual representation of how outcome A and its complement event "not A" together stand for all possible outcomes.

Comment:

• Such a visual brandish is called a "Venn diagram." A Venn diagram is a unproblematic way to visualize events and the relationships between them using rectangles and circles.

Dominion 3 deals with the human relationship betwixt the probability of an event and the probability of its complement effect.

Given that issue A and result "non A" together brand up all possible outcomes, and since rule ii tells u.s. that the sum of the probabilities of all possible outcomes is 1, the post-obit dominion should be quite intuitive:

Probability Dominion 3 (The Complement Rule):

• P(not A) = 1 – P(A)
• that is, the probability that an event does not occur is 1 minus the probability that it does occur.

EXAMPLE: Blood Types

Dorsum to the claret blazon example:

Here is some additional information:

• A person with typeA can donate blood to a person with typeA orAB.
• A person with typeB can donate claret to a person with typeB orAB.
• A person with typeAB tin can donate claret to a person with typeAB only.
• A person with typeO blood can donate to anyone.

What is the probability that a randomly chosen person cannot donate blood to anybody? In other words, what is the probability that a randomly chosen person does not take claret type O? Nosotros need to find P(not O). Using the Complement Rule, P(not O) = 1 – P(O) = one – 0.44 = 0.56. In other words, 56% of the U.S. population does not take blood blazon O:

Clearly, we could also find P(not O) directly by adding the probabilities of B, AB, and A.

Annotate:

• Notation that the Complement Rule,P(not A) = 1 – P(A)can exist re-formulated asP(A) = i – P(not A).
  • P(not A) = 1 – P(A)
  • can exist re-formulated asP(A) = 1 – P(not A).
  • This seemingly trivial algebraic manipulation has an important awarding, and really captures the strength of the complement rule.
  • In some cases, when finding P(A) directly is very complicated, information technology might be much easier to find P(not A) and then simply subtract it from 1 to get the desired P(A).
  • We will come up back to this annotate soon and provide additional examples.

Comments:

• The complement rule can be useful whenever it is easier to calculate the probability of the complement of the consequence rather than the event itself.

• Notice, we again used the phrase "at least one."
• Now nosotros have seen that the complement of "at least one …" is "none … " or "no …." (as we mentioned previously in terms of the events being "opposites").
• In the to a higher place activity we see that
  • P(NONE of these two side furnishings) = 1 – P(at to the lowest degree one of these two side furnishings )
• This is a common application of the complement dominion which you can frequently recognize by the phrase "at least one" in the problem.

Probabilities Involving Multiple Events

We will often exist interested in finding probabilities involving multiple events such as

• P(A or B) = P(event A occurs or event B occurs or both occur)
• P(A and B)= P(both event A occurs and event B occurs)

A mutual outcome with terminology relates to how we unremarkably think of "or" in our daily life. For example, when a parent says to his or her child in a toy store "Do yous want toy A or toy B?", this means that the child is going to get but one toy and he or she has to choose between them. Getting both toys is usually non an choice.

In contrast:

In probability, "OR" means either 1 or the other or both.

and soP(A or B) = P(event A occurs or event B occurs or BOTH occur)

Having said that, information technology should be noted that there are some cases where information technology is simply incommunicable for the two events to both occur at the same fourth dimension.

Probability Dominion Four

The distinction between events that can happen together and those that cannot is an important one.

Disjoint: Two events that cannot occur at the same fourth dimension are called disjoint or mutually exclusive. (Nosotros will use disjoint.)

Information technology should be clear from the moving-picture show that

• in the commencement case, where the events areNOT disjoint, P(A and B) ≠ 0
• in the second case, where the eventsARE disjoint, P(A and B) = 0.

Hither are ii examples:

Instance:

Consider the post-obit 2 events:

A — a randomly chosen person has blood type A, and

B — a randomly chosen person has blood blazon B.

In rare cases, information technology is possible for a person to have more than one blazon of blood flowing through his or her veins, but for our purposes, we are going to presume that each person can accept only one blood type. Therefore, it is incommunicable for the events A and B to occur together.

• Events A and B are DISJOINT

On the other manus …

EXAMPLE:

Consider the post-obit two events:

A — a randomly chosen person has blood type A

B — a randomly called person is a adult female.

In this instance, itis possible for events A and B to occur together.

• Events A and B are Not DISJOINT.

The Venn diagrams suggest that some other way to call back almost disjoint versus not disjoint events is that disjoint eventsdo not overlap. They exercise not share any of the possible outcomes, and therefore cannot happen together.

On the other hand, events that are non disjoint are overlapping in the sense that they share some of the possible outcomes and therefore can occur at the aforementioned time.

We now begin with a simple rule for finding P(A or B) for disjoint events.

Probability Dominion Four (The Addition Dominion for Disjoint Events):

• If A and B are disjoint events, and so P(A or B) = P(A) + P(B).

Comment:

• When dealing with probabilities, the give-and-take "or" will always be associated with the operation of addition; hence the name of this rule, "The Addition Rule."

EXAMPLE: Blood Types

Recall the claret blazon example:

Hither is some additional data

• A person with typeAcan donate blood to a person with typeA orAB.
• A person with blazonBtin donate blood to a person with typeB orAB.
• A person with typeABtin can donate blood to a person with typeAB
• A person with typeOblood can donate to anyone.

What is the probability that a randomly called person is a potential donor for a person with blood type A?

From the information given, we know that being a potential donor for a person with blood type A means having blood type A or O.

We therefore need to find P(A or O). Since the events A and O are disjoint, we can use the add-on rule for disjoint events to get:

• P(A or O) = P(A) + P(O) = 0.42 + 0.44 = 0.86.

Information technology is easy to see why adding the probability actually makes sense.

If 42% of the population has claret type A and 44% of the population has blood type O,

• then 42% + 44% = 86% of the population has either claret blazon A or O, and thus are potential donors to a person with blood type A.

This reasoning well-nigh why the addition rule makes sense tin exist visualized using the pie nautical chart below:

Comment:

• The Add-on Dominion for Disjoint Events can naturally exist extended to more than two disjoint events. Let's take three, for case. If A, B and C are three disjoint events

so P(A or B or C) = P(A) + P(B) + P(C). The rule is the same for any number of disjoint events.

We are now finished with the first version of the Improver Rule (Rule four) which is the version restricted to disjoint events. Before covering the second version, we must commencement discuss P(A and B).

Finding P(A and B) using Logic

Nosotros at present turn to computing

• P(A and B)= P(both event A occurs and result B occurs)

Later, nosotros volition discuss the rules for computing P(A and B).

First, we desire to illustrate that a rule is not needed whenever you tin determine the answer through logic and counting.

Special Case:

In that location is i special example for which we know what P(A and B) equals without applying any rule.

So, if eventsA and B are disjoint, and then (by definition)P(A and B)= 0. Merely what if the events are not disjoint?

Remember that rule 4, the Add-on Dominion, has two versions. 1 is restricted to disjoint events, which we've already covered, and we'll bargain with the more full general version afterwards in this module. The same volition be true of probabilities involving AND

However, except in special cases, we will rely on LOGIC to find P(A and B) in this course.

Before covering any formal rules, let's look at an example where the events are not disjoint.

EXAMPLE: Periodontal Condition and Gender

Nosotros like to ask probability questions similar to the previous example (using a ii-way table based upon information) as this allows you to make connections betwixt these topics and helps you lot keep some of what yous take learned about data fresh in your mind.

Remember, our master goal in this course is to clarify real-life information!

Probability Dominion 5

We are at present ready to move on to the extended version of the Addition Rule.

In this department, we will larn how to discover P(A or B) when A and B are not necessarily disjoint.

• We'll call this extended version the "General Improver Rule" and country information technology as Probability Rule Five.

We will begin past stating the rule and providing an instance similar to the types of problems we by and large enquire in this course. Then nosotros will present a more another case where we do not have the raw data from a sample to work from.

Probability Rule Five:

• The General Improver Rule: P(A or B) = P(A) + P(B) – P(A and B).
  

Notation: It is best to use logic to find P(A and B), not another formula.

A VERY common error is incorrectly applying the multiplication rule for independent events covered on the adjacent page. This volition simply exist right if A and B are contained (run into definitions to follow) which is rarely the instance in data presented in 2-way tables.

As we witnessed in previous examples, when the two events are not disjoint, there is some overlap betwixt the events.

• If we simply add the two probabilities together, nosotros will get the wrong answer because we take counted some "probability" twice!
• Thus, we must subtract out this "extra" probability to arrive at the correct respond. The Venn diagram and the 2-style tables are helpful in visualizing this thought.

This dominion is more than general since it works for any pair of events (even disjoint events). Our advice is still to endeavor to respond the question using logic and counting whenever possible, otherwise, nosotros must exist extremely careful to choose the right rule for the problem.

PRINCIPLE:

If you can calculate a probability using logic and counting yous do non Demand a probability rule (although the correct rule tin can always be applied)

Discover that, if A and B are disjoint, and so P(A and B) = 0 and dominion five reduces to rule 4 for this special case.

Let'southward revisit the terminal example:

Example: Periodontal Status and Gender

Consider randomly selecting one individual from those represented in the following table regarding the periodontal condition of individuals and their gender. Periodontal status refers to gum affliction where individuals are classified as either healthy, have gingivitis, or have periodontal disease.

Let'south review what we have learned then far. We can calculate any probability in this scenario if we can determine how many individuals satisfy the event or combination of events.

• P(Male) = 3009/8027 = 0.3749
• P(Female person) = 5018/8027 = 0.6251
• P(Salubrious) = 3750/8027 = 0.4672
• P(Not Healthy) = P(Gingivitis or Perio) = (2419 + 1858)/8027 = 4277/8027 = 0.5328
  We could also, calculate this using the complement rule: 1 – P(Good for you)

We also previously establish that

• P(Male person AND Good for you) = 1143/8027 = 0.1424

Recall dominion 5, P(A or B) = P(A) + P(B) – P(A and B). We now apply this dominion to summate P(Male person OR Healthy)

• P(Male person or Good for you) = P(Male) + P(Good for you) – P(Male and Healthy) = 0.3749 + 0.4672 – 0.1424 = 0.6997 or near 70%

We solved this question earlier past only counting how many individuals are either Male or Healthy or both. The flick beneath illustrates the values we need to combine. We need to count

• All males
• All healthy individuals
• BUT, not count anyone twice!!

Using this logical approach we would notice

• P(Male person or Good for you) = (1143 + 929 + 937 + 2607)/8027 = 5616/8027 = 0.6996

We have a minor divergence in our answers in the final decimal place due the rounding that occurred when we calculated P(Male), P(Good for you), and P(Male and Salubrious) and then practical dominion 5.

Clearly the answer is finer the same, about 70%. If we carried our answers to more decimal places or if we used the original fractions, we could eliminate this small discrepancy entirely.

Let's await at one final example to illustrate Probability Rule v when the rule is needed – i.east. when we don't have actual data.

EXAMPLE: Important Commitment!

It is vital that a sure document reach its destination within one day. To maximize the chances of on-time commitment, two copies of the certificate are sent using ii services, service A and service B. Information technology is known that the probabilities of on-time delivery are:

• 0.90 for service A (P(A) = 0.90)
• 0.80 for service B (P(B) = 0.eighty)
• 0.75 for both services being on fourth dimension (P(A and B) = 0.75)
  (Note that A and B are due northot disjoint. They can happen together with probability 0.75.)

The Venn diagrams beneath illustrate the probabilities P(A), P(B), and P(A and B) [not fatigued to scale]:

In the context of this problem, the obvious question of interest is:

• What is the probability of on-time delivery of the document using this strategy (of sending it via both services)?

The document will attain its destination on time as long as it is delivered on fourth dimension by service A or by service B or by both services. In other words, when event A occurs or event B occurs or both occur. so….

P(on fourth dimension delivery using this strategy)=P(A or B), which is represented the by the shaded region in the diagram below:

We can now

• use the iii Venn diagrams representing P(A), P(B) and P(A and B)
• to meet that we tin discover P(A or B) by calculation P(A) (represented by the left circle) and P(B) (represented by the correct circle),
• and so subtracting P(A and B) (represented by the overlap), since we included information technology twice, once equally function of P(A) and once as part of P(B).

This is shown in the following epitome:

If we employ this to our example, we notice that:

• P(A or B)= P(on-time delivery using this strategy)= 0.xc + 0.80 – 0.75 = 0.95.

And then our strategy of using ii delivery services increases our probability of on-time delivery to 0.95.

While the Venn diagrams were great to visualize the General Addition Dominion, in cases similar these it is much easier to display the information in and work with a two-way table of probabilities, much as we examined the relationship between two categorical variables in the Exploratory Data Assay section.

We will simply show you the table, not how we derive information technology equally you won't be asked to practice this for us. Yous should be able to see that some logic and unproblematic addition/subtraction is all nosotros used to fill in the table below.

When using a 2-way table, we must remember to look at the entire row or cavalcade to observe overall probabilities involving only A or only B.

• P(A) = 0.ninety means that in 90% of the cases when service A is used, information technology delivers the document on time. To find this we look at the total probability for the row containing A. In finding P(A), we do not know whether B happens or non.

• P(B) = 0.lxxx means that in 80% of the cases when service B is used, it delivers the document on fourth dimension. To find this we look at the total probability for the column containing B. In finding P(B), nosotros exercise not know whether A happens or not.

Comment

• When we used two-style tables in the Exploratory Data Analysis (EDA) department, information technology was to record values of ii categorical variables for a concrete sample of individuals.
• In contrast, the information in a probability two-fashion tabular array is for an unabridged population, and the values are rather abstruse.
• If nosotros had treated something like the delivery example in the EDA section, nosotros would have recorded the actual numbers of on-fourth dimension (and not-on-fourth dimension) deliveries for samples of documents mailed with service A or B.
• In this section, the long-term probabilities are presented as being known.
• Presumably, the reported probabilities in this delivery instance were based on relative frequencies recorded over many repetitions.

Rounding Rule of Thumb for Probability:

Follow the following general guidelines in this form. If in doubt comport more than decimal places. If we specify give exactly what is requested.

• In general you should carry probabilities to at least 4 decimal places for intermediate steps.
• We often round our terminal answer to two or three decimal places.
• For extremely small probabilities, it is important to accept ane or 2 pregnant digits (non-zero digits), such as 0.000001 or 0.000034, etc.

Many calculator packages might display extremely small-scale values using scientific annotation such as

• 58×ten^-five or 1.58 E^-5 to represent 0.0000158

Permit'south Summarize

So far in our report of probability, you accept been introduced to the sometimes counter-intuitive nature of probability and the fundamentals that underlie probability, such as a relative frequency.

We too gave you some tools to aid you find the probabilities of events — namely the probability rules.

You probably noticed that the probability section was significantly different from the 2 previous sections; it has a much larger technical/mathematical component, and then the results tend to be more than of the "right or wrong" nature.

In the Exploratory Data Analysis section, for the about role, the computer took care of the technical aspect of things, and our tasks were to tell it to exercise the right thing and and then interpret the results.

In probability, nosotros do the work from beginning to cease, from choosing the right tool (dominion) to use, to using it correctly, to interpreting the results.

Hither is a summary of the rules we have presented and then far.

1. Probability Dominion #1 states:

• For any event A, 0 ≤ P(A) ≤ 1

2. Probability Rule #ii states:

• The sum of the probabilities of all possible outcomes is 1

three. The Complement Rule (#3) states that

• P(non A) = 1 – P(A)

or when rearranged

• P(A) = 1 – P(not A)

The latter representation of the Complement Rule is especially useful when nosotros need to notice probabilities of events of the sort "at least one of …"

4. The General Addition Dominion (#v) states that for whatsoever two events,

• P(A or B) = P(A) + P(B) – P(A and B),

where, by P(A or B) nosotros mean P(A occurs or B occurs or both).

In the special case ofdisjoint events, events that cannot occur together, the Full general Addition Dominion tin can exist reduced to the Add-on Rule for Disjoint Events (#4), which is

• P(A or B) = P(A) + P(B). *

*Only use when you are CONVINCED the events are disjoint (they do NOT overlap)

5. Therestricted version of the addition rule (for disjoint events)can be easily extended to more than two events.

half dozen. And so far, we have only constituteP(A and B) using logic and counting in uncomplicated examples

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Source: https://bolt.mph.ufl.edu/6050-6052/unit-3/module-6/

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