probability assignment rule

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7.1 - the rules of probability, example 7.1 section  .

The astragalus (ankle or heel bone) of animals were used in ancient times as a forerunner of modern dice. In fact, Egyptian tomb paintings show that sheep astragali were used in board games as early as 3500 B.C. (see Figure 7.2 ). When a sheep astragalus is thrown into the air it can land on one of four sides, which were associated with the numbers 1, 3, 4, and 6 (see Table 7.1 ). Two sides (the 3 and the 4) are wider and each come up about 40% of the time, while the narrower sides (the 1 and the 6) each come up about 10% of the time. Astragali were used for gambling, games, and divination purposes by the ancients.

  Toss an astragalus once. What's the chance you get a "1"?

Since probabilities can often be viewed as the proportion of times something happens we see our first rule of probability.

  Toss an astragalus once. What's the chance you get at least a 3?

Notice that there is another way to solve the previous problem. The opposite of "at least 3" is "getting a 1" (i.e. the only other possibility) so you can also figure the answer as 100% - 10% = 90% or 0.90. This rule of the opposites is our third rule of probability.

  Suppose you toss an astralgus twice. What's the chance that you get "4s" on both tosses?

  In ancient Rome, the lowest score in tossing four astragali (getting all four 1s) was called the dog throw. What is the probability of getting a dog throw?

Example 7.2 Independent or not? Section  

For the next two single births at Hershey Medical Center: whether the first baby is a boy and whether the second baby is a boy.

For the next two single births at Hershey Medical Center: whether the first baby is a girl and whether both babies are girls.

For the end of the month in February next year: whether there will be snow on the ground at the State College airport on February 27th and whether there will be snow on the ground at the airport on February 28th

Example 7.3 Section  

The highest paid employee has randomly selected from the list of Fortune 500 companies. Which of these probabilities is the largest?

• The chance this person is a college graduate
• The chance this person is a college graduate with a Business degree.
• The chance this person is a college graduate with an Engineering degree.

Other Interpretations Section  

While we can often think of how the process leading to data might be repeated, some events arise in situations that are not easily seen as being repeatable. In such situations, the relative frequency interpretation of probability may seem inappropriate. For example, answering a question like "What is the chance that our next President will be a woman?" would seem to require a different interpretation of the meaning of probability. Luckily, it is perfectly reasonable to assign probabilities to events outside of the relative frequency interpretation as long as they satisfy the above rules of probability. Personal probabilities that satisfy these rules give a coherent interpretation even if they might differ from one person's assignment to another's.

3.3 Two Basic Rules of Probability

In calculating probability, there are two rules to consider when you are determining if two events are independent or dependent and if they are mutually exclusive or not.

The Multiplication Rule

If A and B are two events defined on a sample space , then P ( A AND B ) = P ( B ) P ( A | B ).

This equation can be rewritten as P ( A AND B ) = P ( B ) P ( A | B ), the multiplication rule.

If A and B are independent , then P ( A | B ) = P ( A ). In this special case, P ( A AND B ) = P ( A | B ) P ( B ) becomes P ( A AND B ) = P ( A ) P ( B ).

A bag contains four green marbles, three red marbles, and two yellow marbles. Mark draws two marbles from the bag without replacement. The probability that he draws a yellow marble and then a green marble is

Notice that P ( green | yellow ) = 4 8 P ( green | yellow ) = 4 8 . After the yellow marble is drawn, there are four green marbles in the bag and eight marbles in all.

The Addition Rule

If A and B are defined on a sample space, then P ( A OR B ) = P ( A ) + P ( B ) − P ( A AND B ).

Draw one card from a standard deck of playing cards. Let H = the card is a heart, and let J = the card is a jack. These events are not mutually exclusive because a card can be both a heart and a jack.

If A and B are mutually exclusive , then P ( A AND B ) = 0. Then P ( A OR B ) = P ( A ) + P ( B ) − P ( A AND B ) becomes P ( A OR B ) = P ( A ) + P ( B ).

Draw one card from a standard deck of playing cards. Let H = the card is a heart and S = the card is a spade. These events are mutually exclusive because a card cannot be a heart and a spade at the same time. The probability that the card is a heart or a spade is

Example 3.14

Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska.

• Klaus can only afford one vacation. The probability that he chooses A is P ( A ) = .6 and the probability that he chooses B is P ( B ) = .35.
• P ( A AND B ) = 0 because Klaus can only afford to take one vacation.
• Therefore, the probability that he chooses either New Zealand or Alaska is P ( A OR B ) = P ( A ) + P ( B ) = .6 + .35 = .95. Note that the probability that he does not choose to go anywhere on vacation must be .05.

Example 3.15

Carlos plays college soccer. He makes a goal 65 percent of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P ( A ) = .65. B = the event Carlos is successful on his second attempt. P ( B ) = .65. Carlos tends to shoot in streaks. The probability that he makes the second goal given that he made the first goal is .90.

a. What is the probability that he makes both goals?

a. The problem is asking you to find P ( A AND B ) = P ( B AND A ). Since P ( B | A ) = .90: P ( B AND A ) = P ( B | A ) P ( A ) = (.90)(.65) = .585.

Carlos makes the first and second goals with probability .585.

b. What is the probability that Carlos makes either the first goal or the second goal?

b. The problem is asking you to find P ( A OR B ).

P ( A OR B ) = P ( A ) + P ( B ) − P ( A AND B ) = .65 + .65 − .585 = .715

Carlos makes either the first goal or the second goal with probability .715.

c. Are A and B independent?

c. No, they are not, because P ( B AND A ) = .585.

P ( B ) P ( A ) = (.65)(.65) = .423

.423 ≠ .585 = P ( B AND A )

So, P ( B AND A ) is not equal to P ( B ) P ( A ).

d. Are A and B mutually exclusive?

d. No, they are not because P ( A and B ) = .585.

To be mutually exclusive, P ( A AND B ) must equal zero.

Try It 3.15

Helen plays basketball. For free throws, she makes the shot 75 percent of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P ( C ) = .75. D = the event Helen makes the second shot. P ( D ) = .75. The probability that Helen makes the second free throw given that she made the first is .85. What is the probability that Helen makes both free throws?

Example 3.16

A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly.

a. What is the probability that the member is a novice swimmer?

a. There are 150 members; 75 of these are advanced, and 47 of these are intermediate swimmers. So there are 150 − 75 − 47 = 28 novice swimmers. The probability that a randomly selected swimmer is a novice is 28 150 . 28 150 .

b. What is the probability that the member practices four times a week?

b. 40 + 30 + 10 150 = 80 150 40 + 30 + 10 150 = 80 150

c. What is the probability that the member is an advanced swimmer and practices four times a week?

c. There are 40 advanced swimmers who practice four times per week, so the probability is 40 150 . 40 150 .

d. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and being an intermediate swimmer mutually exclusive? Why or why not?

d. P (advanced AND intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time.

e. Are being a novice swimmer and practicing four times a week independent events? Why or why not?

e. No, these are not independent events. P (novice AND practices four times per week) = .0667 P (novice) P (practices four times per week) = .0996 .0667 ≠ .0996

Try It 3.16

A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college are on their school's sports teams. Thirty of the seniors going directly to work are on their school's sports teams. Five of the seniors taking a gap year are on their schools sports teams. What is the probability that a senior is taking a gap year?

Example 3.17

Felicity attends a school in Modesto, CA. The probability that Felicity enrolls in a math class is .2 and the probability that she enrolls in a speech class is .65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is .25.

Let M = math class, S = speech class, and M | S = math given speech.

• What is the probability that Felicity enrolls in math and speech? Find P ( M AND S ) = P ( M | S ) P ( S ).
• What is the probability that Felicity enrolls in math or speech classes? Find P ( M OR S ) = P ( M ) + P ( S ) − P ( M AND S ).
• Are M and S independent? Is P ( M | S ) = P ( M )?
• Are M and S mutually exclusive? Is P ( M AND S ) = 0?

a. P ( M AND S ) = P ( M | S ) P ( S ) = .25(.65) = .1625

b. P ( M OR S ) = P ( M ) + P ( S ) − P ( M AND S ) = .2 + .65 − .1625 = .6875

c. No, P ( M | S ) = .25 and P ( M ) = .2.

d. No, P ( M AND S ) = .1625.

Try It 3.17

A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P ( B ) = .40, P ( D ) = .30, and P ( D | B ) = .5.

• Find P ( B AND D ).
• Find P ( B OR D ).

Example 3.18

Researchers are studying one particular type of disease that affects women more often than men. Studies show that about one woman in seven (approximately 14.3 percent) who live to be 90 will develop the disease. Suppose that of those women who develop this disease, a test is negative 2 percent of the time. Also suppose that in the general population of women, the test for the disease is negative about 85 percent of the time. Let B = woman develops the disease and let N = tests negative. Suppose one woman is selected at random.

a. What is the probability that the woman develops the disease? What is the probability that woman tests negative?

a. P ( B ) = .143; P ( N ) = .85

b. Given that the woman develops the disease, what is the probability that she tests negative?

b. Among women who develop the disease, the test is negative 2 percent of the time, so P ( N | B ) = .02

c. What is the probability that the woman has the disease AND tests negative?

c. P ( B AND N ) = P ( B ) P ( N | B ) = (.143)(.02) = .0029

d. What is the probability that the woman has the disease OR tests negative?

d. P ( B OR N ) = P ( B ) + P ( N ) − P ( B AND N ) = .143 + .85 − .0029 = .9901

e. Are having the disease and testing negative independent events?

e. No. P ( N ) = .85; P ( N | B ) = .02. So, P ( N | B ) does not equal P ( N ).

f. Are having the disease and testing negative mutually exclusive?

f. No. P ( B AND N ) = .0029. For B and N to be mutually exclusive, P ( B AND N ) must be zero.

Try It 3.18

A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college are on their school's sports teams. Thirty of the seniors going directly to work are on their school's sports teams. Five of the seniors taking a gap year are on their school's sports teams. What is the probability that a senior is going to college and plays sports?

Example 3.19

Refer to the information in Example 3.18 . P = tests positive.

• Given that a woman develops the disease, what is the probability that she tests positive? Find P ( P | B ) = 1 − P ( N | B ).
• What is the probability that a woman develops the disease and tests positive? Find P ( B AND P ) = P ( P | B ) P ( B ).
• What is the probability that a woman does not develop the disease? Find P ( B′ ) = 1 − P ( B ).
• What is the probability that a woman tests positive for the disease? Find P ( P ) = 1 − P ( N ).

a. P ( P | B ) = 1 − P ( N | B ) = 1 − .02 = .98

b. P ( B AND P ) = P ( P | B ) P ( B ) = .98(.143) = .1401

c. P ( B' ) = 1 − P ( B ) = 1 − .143 = .857

d. P ( P ) = 1 − P ( N ) = 1 − .85 = .15

Try It 3.19

• Find P ( B′ ).
• Find P ( D AND B ).
• Find P ( B | D ).
• Find P ( D AND B′ ).
• Find P ( D | B′ ).

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Essential Probability

Part I - Basics

Random Experiments

Sample spaces, populations, rules for combining events, probabilities of events, rules of probability, equally likely outcomes, sampling without replacement, sampling with replacement.

Conditional Probability

Bayes Rule and Total Probability

A random experiment is one whose outcome is uncertain and which can be repeated indefinitely under essentially identical conditions.  Examples are rolling a die, tossing a coin, making a laboratory measurement with a degree of uncontrollable error.

The sample space of a random experiment is the set of all its possible outcomes.  We will denote the sample space by the Greek letter Ω (upper-case Omega).  Individual outcomes of an experiment will be denoted generically by ω (lower-case Omega).  For example, in the experiment of tossing a six-sided die and observing the number of spots, the sample space might be described as Ω={1,2,3,4,5,6}.

A population is a set of individuals or objects that forms the subject of a statistical investigation.  Usually, the population is so large that it is not feasible to examine every member of the population and a smaller subset (called a sample ) is chosen to represent the population.  In such cases, the sample space is not the population.  Rather, it is the collection of all samples (subsets of the population) of a given size.

An event is a subset of the sample space.  Events may be described in mathematical notation or in informal ordinary language.  For example, the event E="The number of spots is even." is also E={2,4,6}.  Events are denoted generically by upper-case Latin letters, such as "E" above.

If ω is the outcome of a random experiment, the event E occurs if we E. 

1.  The complement of E, denoted E c , occurs if and only if E does not occur.  In other words,

2.  The union of a finite or infinite sequence of events is an event.  The union occurs if and only if at least one of the individual events in the sequence occurs.

3.  The entire sample space Ω is an event (the certain event).  The certain event arises naturally when an ordinary language description of an event is satisfied by every outcome of the random experiment.

4.  The empty set φ is an event.  The empty event arises naturally when an ordinary language description cannot be satisfied by any outcome of the random experiment.  Note that

5.  The intersection of a finite or infinite sequence of events in an event.  The intersection occurs if and only if each of the individual events in the sequence occurs.

If E 1 ∩ E 2 = φ, the events E 1 and E 2 are said to be disjoint .  This means that they cannot both occur.

A probability assignment or probability measure is a way of assigning probabilities between 0 and 1 to events.  In other words, it is a function P whose domain is the set of all events associated with a random experiment and whose codomain is the unit interval [0, 1].  A probability assignment is an essential part of modeling a random experiment.  Ideally, it comes from a detailed background knowledge of the phenomena of the experiment.  Probability theory is the study of the mathematical consequences of the basic properties (axioms) of probability assignments given in the next paragraph.  Statistics is the study of methods of using data to assess the correctness of probability models of real life experiments.

Many random experiments have only a finite number of possible outcomes.  For such an experiment, let n denote the number of possible outcomes.  A singleton event is an event that consists of only one outcome ω.  Denote this event by {ω}.  The experiment is said to have equally likely outcomes if P ({ω}) = 1/ n   for all ω.  It then follows from the basic rules of probability that for any event E ,

where #( E ) is the number of outcomes in the event E .  The assumption of equally likely outcomes may or may not be appropriate for a given experiment.  We almost always assume equally likely outcomes for certain simple experiments involving tossing dice, drawing cards, etc.

If a finite population has M members, a sample without replacement of size m is simply a subset of size m of the population.  The number of such subsets is

where, e.g., M ! ( M factorial) is the product of the positive integers from 1 to M , inclusive.  Consider a random experiment whose outcome is a subset of size m from the population.  If all outcomes are equally likely, the probability of each singleton event { ω} is given by

This is what is usually meant by "choosing a random sample of size m ".  Sometimes the order of presentation of the population members selected is important.  In these cases, the outcome of the experiment is a non-repeating sequence (not just a subset) of length m from the population.  These are also called permutations of length m from the population.  The number of permutations of length m is

If all permutations of length m are equally likely, the result of the experiment is called an ordered sample without replacement from the population.

An ordered sample of size m , with replacement , from a population is a possibly repeating sequence of length m from the population.  The number of such sequences is M m .  It is not hard to see that if the sample size m is much less than the population size M , a very large fraction of the ordered samples with replacement will not repeat themselves anyway, so there is little practical difference between ordered sampling with and without replacement.  In such circumstances, samples without replacement are often treated as samples with replacement for mathematical convenience.

Conditional Probability and Independence

Probabilities discussed up to this point have been unconditional probabilities .  If D and E are events and P ( E ) > 0, the conditional probability of D , given that E occurs , is

If the experiment has equally likely outcomes, P ( D | E ) is just the fraction of all the outcomes in E where D also occurs.  Two events D and E are independent if P ( D ∩ E ) = P ( D ) P ( E ).  If P(E) > 0, this is equivalent to P ( D | E ) = P ( D ), i.e., the conditional probability of D given E is the same as the unconditional probability of D .  For example, in the experiment of drawing a single card from a standard deck with equally likely outcomes, the events "Draw a Heart" and "Draw a Queen" are independent.  The events "Draw a Heart" and "Draw a red card" are not independent.  They are dependent.

Bayes' Rule and Total Probability

Let E 1 , E 2 , ..., E k be pairwise disjoint events such that P ( E i ) > 0 for each i and

This means that one of the events E i must occur and that only one can occur.  Let D be another event.  The law of total probability says that

and Bayes' rule says that

For example, let D denote a set of symptoms exhibited by a patient and let E 1 , E 2 , ..., E k be a collection of mutually exclusive disease conditions that might account for the symptoms.   For each disease E i , there is a certain probability P ( D | E i ) that a sufferer of that disease will have the symptoms D and there is a certain probability P ( E i ) that a patient will have that specific disease.  Then Bayes' rule gives the probability that the patient has the specific disease, given that he or she has the symptoms D .

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Lesson 6: Assigning Probabilities to Events; Probability Rules

Hi everyone! Read through the material below, watch the videos, work on the Word lecture and follow up with your instructor if you have questions.

You have, more than likely, used probability. In fact, you probably have an intuitive sense of probability. Probability deals with the chance of an event occurring. Whenever you weigh the odds of whether or not to do your homework or to study for an exam, you are using probability. In this chapter, you will learn how to solve probability problems using a systematic approach.

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• Understand and use the terminology of probability.
• Determine whether two events are mutually exclusive and whether two events are independent.
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Consider three sets A , B , and C . What does $n(A\cup B \cup C)$ look like?

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Physics > Data Analysis, Statistics and Probability

Title: the art of probability assignment.

Abstract: The problem of assigning probabilities when little is known is analized in the case where the quanities of interest are physical observables, i.e. can be measured and their values expressed by numbers. It is pointed out that the assignment of probabilities based on observation is a process of inference, involving the use of Bayes' theorem and the choice of a probability prior. When a lot of data is available, the resulting probability are remarkably insensitive to the form of the prior. In the oposite case of scarse data, it is suggested that the probabilities are assigned such that they are the least sensitive to specific variations of the probability prior. In the continuous case this results in a probability assignment rule wich calls for minimizing the Fisher information subject to constraints reflecting all available information. In the discrete case, the corresponding quantity to be minimized turns out to be a Renyi distance between the original and the shifted distribution.

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A new approach for generation of generalized basic probability assignment in the evidence theory

• Theoretical advances
• Published: 17 February 2021
• Volume 24 , pages 1007–1023, ( 2021 )

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• Yongchuan Tang   ORCID: orcid.org/0000-0003-2568-9628 1 ,
• Dongdong Wu 1 &
• Zijing Liu 1  

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The process of information fusion needs to deal with a large number of uncertain information with multi-source, heterogeneity, inaccuracy, unreliability, and incompleteness. In practical engineering applications, Dempster–Shafer evidence theory is widely used in multi-source information fusion owing to its effectiveness in data fusion. Information sources have an important impact on multi-source information fusion in an environment with the characteristics of complex, unstable, uncertain, and incomplete. To address multi-source information fusion problem, this paper considers the situation of uncertain information modeling from the closed-world to the open-world assumption and studies the generation of basic probability assignment with incomplete information. A new method is proposed to generate the generalized basic probability assignment (GBPA) based on the triangular fuzzy number model under the open-world assumption. First, the maximum, minimum, and mean values for the triangular membership function of each attribute in classification problem can be obtained to construct a triangular fuzzy number representation model. Then, by calculating the length of the intersection points between the sample and the triangular fuzzy number model, a GBPA set with an assignment for the empty set can be determined. The proposed method can not only be used in different complex environments simply and flexibly, but also have less information loss in information processing. Finally, a series of comprehensive experiments basing on the UCI data sets is used to verify the rationality and superiority of the proposed method.

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Yongchuan Tang, Dongdong Wu & Zijing Liu

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Tang, Y., Wu, D. & Liu, Z. A new approach for generation of generalized basic probability assignment in the evidence theory. Pattern Anal Applic 24 , 1007–1023 (2021). https://doi.org/10.1007/s10044-021-00966-0

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Received : 14 May 2020

Accepted : 25 January 2021

Published : 17 February 2021

Issue Date : August 2021

DOI : https://doi.org/10.1007/s10044-021-00966-0

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• Generalized evidence theory
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