Chapter 4 Probability 2013 Statistics for Business 1.

Chapter 4

Probability

2013Statistics for Business

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Probability

4.1 The Concept of Probability4.2 Sample Spaces and Events4.3 Some Elementary Probability Rules4.4 Conditional Probability and Independence

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Probability and statistics

The use of Probability in our life• Life is rife with uncertainty.

– Will it rain tomorrow?– How much oil can be found by drilling here?– Will the economy be better 6 months from now?

• Sometimes we can predict it, many times we cannot.

• Probability quantify the best we can say about the situation.

“The laws of probability, so true in general, so fallacious in particular.”

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Edward Gibbon(English Historian, 1737-1794)

Richard P. FeynmanQED: The Strange Theory of Light and Matter

Philosophers have said that if the same circumstances don't always produce the same results, predictions are impossible and science will collapse. Here is a circumstance—identical photons are always coming down in the same direction to the piece of glass—that produces different results. We cannot predict whether a given photon will arrive at A or B. All we can predict is that out of 100 photons that come down, an average of 4 will be reflected by the front surface. Does this mean that physics, a science of great exactitude, has been reduced to calculating only the probability of an event, and not predicting exactly what will happen? Yes. That's a retreat, but that's the way it is: Nature permits us to calculate only probabilities. Yet science has not collapsed.

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The Concept of Probability

• An experiment is any process of observation with an uncertain (random) outcome.

• A sample space is a collection of all possible outcomes for an experiment .

• A random variable is a function defined on a sample space that characterized an outcome.

• Probability is a measure of the chance that an experimental outcome will occur when an experiment is carried out.

Probability: basic propertiesIf E is an experimental outcome, then P(E) denotes the probability that E will occur with the following basic properties:

1. 0 P(E) 1 such that:• If E can never occur, then P(E) = 0• If E is certain to occur, then P(E) = 1

2. The probabilities of all the experimental outcomes must sum to 1

Classical experiment :Classical experiment : A fair die A fair diePossible Possible outcomesoutcomes: The : The numbers 1, 2, 3, 4, 5, 6numbers 1, 2, 3, 4, 5, 6

One possible One possible eventevent: : The occurrence of an The occurrence of an even number. That is, even number. That is, we collect the we collect the outcomes 2, 4, and 6.outcomes 2, 4, and 6.

An EventEvent is the collection of one or more outcomes of an experiment.

There are six possible outcomes and the sample space consists of six elements: {1, 2, 3, 4, 5, 6}.

Events

• An event is a set (or collection) of experimental outcomes

• The probability of an event is the sum of the probabilities of the experimental outcomes that correspond to the event

outcomestotalofnumberthe

happenedEwhichinoutcomesofnumberthe)E(P

Fair die experiment

P(even number) = (the # of outcomes P(even number) = (the # of outcomes in which an even number appears)/ in which an even number appears)/

(the # of total outcomes)(the # of total outcomes)

= 3/6= 3/6

Discrete Sample Space

• Consider the experiment of flipping two coins. • It is possible to get 0 heads, 1 head, or 2 heads.

Thus, the sample space could be {0, 1, 2}. • Another way to look at it is flip { HH, HT, TH, TT }. • The second way is sometimes more convenient

because each outcome is as equally likely to occur as any other.

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• If the two indistinguishable coins are tossed simultaneously, there are just three possible outcomes, {H, H}, {H, T}, and {T, T}.

• If the coins are different, or if they are thrown one after the other, there are four distinct outcomes: (H, H), (H, T), (T, H), (T, T), which are often presented in a more concise form: HH, HT, TH, TT.

• Thus, depending on the nature of the experiment, there are 3 or 4 outcomes, with the sample spaces.

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Continuous Sample Space

Arrival time. The experimental setting is a metro (underground) station where trains pass (ideally) with equal intervals. A person enters the station. The experiment is to note the time of arrival past the departure time of the last train. If T is the interval between two consecutive trains, then the sample space for the experiment is the interval [0, T], or[0, T] = {t: 0 ≤ y ≤ T}.

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Human height. The experiment is to randomly select a Chinese and measure his or her height. The sample space contains the 1.3 billion humans inhabiting in China. In this case, the height of the selected person becomes a random variable. It is also possible to consider the sample space consisting of all possible values of height measurements of the Chinese people. While the Chinese population is discrete, we may assume that in some height range near the average, all possible heights are realized making up a continuous Sample Space.

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Assigning Probabilities to Experimental Outcomes

• Classical Method (theoretical)– For equally likely outcomes

• Long-run relative frequency (empirical)– Long-run experiment (e.g. throwing a die many times)

– The number of times the event happened over the total number of past data

• Subjective– Assessment based on experience, common

sense, intuition or expertise.

Classical Method

• Frequently used when the experimental outcomes are equally likely to occur

• Example 1: tossing a “fair” coin– Two outcomes: head (H) and tail (T)– If the coin is fair, then H and T are equally likely to

occur any time the coin is tossed– So P(H) = 0.5, P(T) = 0.5

• 0 < P(H) < 1, 0 < P(T) < 1• P(H) + P(T) = 1

Classical Method: Fair die experiment

Let X be the number we get in tossing the die Let X be the number we get in tossing the die once.once.

P(X=2) = 1/6P(X=2) = 1/6

If the die is fair, each number are equally likely to occur any time the die is tossed.

Long-Run Relative Frequency Method

• Let E be an outcome of an experiment• If it is performed many times, P(E) is the

relative frequency of E– P(E) is the percentage of times E occurs in many

repetitions of the experiment• Use sampled or historical data• Example: Of 1,000 randomly selected consumers, 140

preferred brand X• The probability of randomly picking a person who

prefers brand X is 140/1000 = 0.14 or 14%

Example2: Long-Run Relative Frequency Method

1. An accounts receivable manager knows from 1. An accounts receivable manager knows from past data that about 70 of 1000 accounts became past data that about 70 of 1000 accounts became uncollectible. uncollectible.

The manager would estimate the probability of bad The manager would estimate the probability of bad debts as 70/1000 = .07 or 7%. debts as 70/1000 = .07 or 7%.

2. Tossing a fair coin 3000 times, we can see that 2. Tossing a fair coin 3000 times, we can see that although the proportion of heads was far from 0.5 in although the proportion of heads was far from 0.5 in the first 100 tosses, it seemed to stabilize and the first 100 tosses, it seemed to stabilize and approach 0.5 as the number of tosses increased. approach 0.5 as the number of tosses increased.

Long-Run Relative Frequency Method: Example

• Often we determine the probability from a random sample (Long-Run Relative Frequency Method) and apply it to the population.

• Of 5528 Zhuhai residents randomly sampled,445 prefer to watch CCTV-1

• Estimated Share P(CCTV-1) = 445 / 5528 = 0.0805• So the probability that any Zhuhai resident chosen at random

prefers CCTV-1 is 0.0805• Assuming total population in Zhuhai is 1,000,000 :• Size of audience in the city = Population x Share

so 1,000,000 x 0.0805 = 80,500

Long-Run Relative Frequency Method: application

Subjective Probability• Using experience, intuitive judgment, or personal

expertise to assess/derive a probability• May or may not have relative frequency

interpretation (Some events cannot be repeated many times)

• Contains a high degree of personal bias. • What is the probability of your favorite basketball

or football team win the next game? (e.g. sports betting)

Subjective probability & betting

The odds in betting reflect the subjective probability guessed by the mass.Who much are you willing to pay for a ticket which worth $10 if there was life on Mars and nothing if there was not?

Subjective probability usually reflects the mind/opinion more than the reality. It is an area of research in psychologies.

Probabilities: Equally Likely Outcomes

• If the sample space outcomes (or experimental outcomes) are all equally likely, then the probability that an event will occur is equal to the ratio:– The number of ways the event can occur– Over the total number of outcomes

outcomes space sample ofnumber Total

event the tocorrespond that outcomes space sample ofNumber

Watch out!

• Consider, for example, the question of whether or not there is life on Mars. There are only two possible outcomes in the sample space.1. There is life on Mars.2. There is no life on Mars.

• However, you cannot conclude that the probability of life on Mars is p= 1/2.

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Some Elementary Probability Rules

1. Complement2. Union3. Intersection4. Addition5. Conditional probability6. Multiplication

Complement• The complement (Ā) of an event A is the set

of all sample space outcomes not in A• P(Ā) = 1 – P(A)

EventEvent

Sample spaceSample space

“Venn diagram”

Union and Intersection

• The union of A and B are elementary events that belong to either A or B or both– Written as A B

• The intersection of A and B are elementary events that belong to both A and B– Written as A ∩ B

Other rules• Complement of complement

–(E’)’ = E• Complement of intersection/union

–(A∩B)’ = A’ B’–(A B)’ = A’∩B’

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Some Elementary Probability Rules

Entire event spaceEntire event space

Mutually Exclusive

• A and B are mutually exclusive if they have no sample space outcomes in common

• In other words:

P(A∩B) = 0

The Addition Rule (special & general)• If A and B are mutually exclusive, then the

probability that A or B (the union of A and B) will occur is

P(AB) = P(A) + P(B) (special rule)

• If A and B are not mutually exclusive:

P(AB) = P(A) + P(B) – P(A∩B) (general rule) where P(A∩B) is the joint probability of A and B both occurring together

Example: Newspaper Subscribers #1

• Define events:– A = event that a randomly selected household subscribes

to the Atlantic Journal– B = event that a randomly selected household subscribes

to the Beacon News

• Given:– total number in city, N = 1,000,000– number subscribing to A, N(A) = 650,000– number subscribing to B, N(B) = 500,000– number subscribing to both, N(A∩B) = 250,000

Example: Newspaper Subscribers #2

• Use the relative frequency method to assign probabilities

25.0000,000,1

000,250

50.0000,000,1

000,500

65.0000,000,1

000,650

BAP

BP

AP

Contingency table in Table 4.3Example: Newspaper Subscribers (contingency table)

A∩B BA

A contingency table is a tabular representation of categorical data .

Example: Newspaper Subscribers #3• Refer to the contingency table • The chance that a household does not

subscribe to either newspaper

1000000001

000100BAP .

,,

,

Example: Newspaper Subscribers #4

• The chance that a household subscribes to either newspaper:

90.0

25.050.065.0

B)P(AB)B)=P(A)+P(P(A

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1.090.01

)(1

'''1''

)''()''('''

BAP

))BP((A)=BP(A

BABA)B(A

Since:

Therefore:

Other method to find more complex probabilities

Conditional Probability and Independence

• Conditional probability is used to determine how two events are related.

• The probability of an event A, given that the event B has occurred, is called the conditional probability of A given B– Denoted as P(A|B)

• Further, P(A|B) = P(A∩B) / P(B)– assume P(B) ≠ 0

Interpretation

• Restrict sample space to just event B

• The conditional probability P(A|B) is the chance of event A occurring in this new sample space

• In other words, if B occurred, then what is the chance of A occurring

Example: Newspaper Subscribers

• Of the households that subscribe to the Atlantic Journal, what is the chance that they also subscribe to the Beacon News?

• Want P(B|A), where

3846.0

650

250

AP

BAPABP

.

.

|

Example: mutual fund performance • Why are some mutual fund managers more successful than

others? One possible factor is where the manager earned his or her MBA. The following table compares mutual fund performance against the ranking of the school where the fund manager earned their MBA:

Mutual fund outperforms the market

B1

Mutual fund doesn’t outperform the market

B2

A1 - Top 20 MBA program .11 .29

A2 - Not top 20 MBA program

.06 .54

E.g. This is the probability that a mutual fund outperforms AND the manager was in a top-20 MBA program; it’s a joint probability.

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Conditional Probability…

• We want to calculate P(B1 | A1)

Thus, there is a 27.5% chance that that a fund will outperform the market given that the manager graduated from a top-20 MBA program.

B1 B2 P(Ai)

A1.11 .29 .40

A2.06 .54 .60

P(Bj) .17 .83 1.00

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Example: New test for early detection of cancer• Let• C = event that patient has cancer• C’ = event that patient does not have cancer• + = event that the test indicates a patient has cancer - = event that the test indicates that patient does not have cancer

• Clinical trials indicate that the test is accurate 95% of the time in detecting cancer for those patients who actually have cancer: P(+/C) = .95

• but unfortunately will give a “+” 8% of the time for those patients who are known not to have cancer: P(+/ Cc ) = .08

• It has also been estimated that approximately 10% of the population have cancer and don’t know it yet: P(C) = .10

• You take the test and receive a “+” test results. Should you be worried? P(C/+) = ?????

6.45

P(+/C) = .95 P(+/ C’ ) = .08 P(C) = .10

Have Cancer: C Do Not Have Cancer: CC

+

-

True State of Nature

Tes

t R

esul

ts

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False negative

False positive

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6.49

Assignment Problem• The Rapid Test is used to determine whether someone has

HIV. The false positive and false negative rates are 0.05 and 0.09 respectively.

• The doctor just received a positive test results on one of their patients [assumed to be in a low risk group for HIV].

• The low risk group is known to have a 6% probability of having HIV.

• What is the probability that this patient actually has HIV [after the positive test].

• Use a table to work this problem.

Independence of Events

• Two events A and B are said to be independent if and only if:

P(A|B) = P(A)

• This is equivalently to

P(B|A) = P(B)

Example: Newspaper Subscribers

• Of the Atlantic Journal subscribers, what is the chance that they also subscribe to the Beacon News?– If independent, the P(B|A) = P(B)

• Is P(B|A) = P(B)?– Know that P(B) = 0.5– Just calculated that P(B|A) = 0.3846– 0.65 ≠ 0.3846, so P(B|A) ≠ P(B)

• A is not independent of B– A and B are said to be dependent

The Multiplication Rule

• The joint probability that A and B (the intersection of A and B) will occur is

P(A∩B) = P(A) • P(B|A) = P(B) • P(A|B)

• If A and B are independent, then the probability that A and B will occur is:

P(A∩B) = P(A) • P(B) = P(B) • P(A)Special Rule of Multiplication

General Rule of Multiplication

Example: Genders of Two Children

• Let: B be the outcome that child is boy G be the outcome that child is girl

• Sample space S = {BB, BG, GB, GG}• If B and G are equally likely , then

P(B) = P(G) = ½ and• P(BB) = P(BG) = P(GB) = P(GG) = ¼

A Tree Diagram: the Genders of Two Children

Example: Gender of Two Children

• Experimental Outcomes: BB, BG, GB, GG

• All outcomes equally likely:P(BB) = … = P(GG) = ¼

• P(one boy and one girl) =P(BG) + P(GB) = ¼ + ¼ = ½

• P(at least one girl) =P(BG) + P(GB) + P(GG) = ¼+¼+¼ = ¾

Example: Genders of Two Children Continued

• Of two children, what is the probability of having a girl first and then a boy second?

• Want P(G first and B second)?– Want P(G∩B)

• P(G∩B) = P(G) P(B|G)

• But gender of siblings is independent– So P(B|G) = P(B)– Then P(G∩B) = P(G) P(B) = ½ ½ = ¼

• Consistent with the tree diagram

Example: In a bag containing 7 red balls and 5 blue balls, 2 balls are selected at random without replacement. Construct a tree diagram showing this information. Calculate the probability of probability of getting 1 red and 1 bluegetting 1 red and 1 blue ball.

A Tree DiagramTree Diagram is used when you have a list of choices. It clearly shows conditional and joint probabilities.

Using Tree Diagrams

P(R1R2) = (7/12)*(6/11)P(R1R2) = (7/12)*(6/11)

P(B1R2)= (5/12)*(7/11)P(B1R2)= (5/12)*(7/11)

Probability of getting 1 red and 1blue = P(R1B2) + P(B1R2) = Probability of getting 1 red and 1blue = P(R1B2) + P(B1R2) = (5/12)*(7/11) + (7/12)*(5/11)(5/12)*(7/11) + (7/12)*(5/11)

Using Tree Diagrams

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Explaining the solution by conditional probabilityP(R1)=7/12P(B2|R1)=5/11P(R1&B2)=P(B2|R1)*P(R1)=(5/11)*(7/12)P(B1)=5/12P(R2|B1)=7/11P(B1&R2)=P(R2|B1)*P(B1)=(5/12)*(7/11)

P(B1R2 or R1B2)=(2*5*7)/(12*11)

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Another explanation:

All permutations to choose any 2 from 12 balls= 12P2=12*11 (all possible outcomes: sample space)

All permutations to choose 1 R & 1 B = 2* 7C1 * 5C1 = 2*7*5 (outcomes in the event of interest)

P(B1R2 or R1B2)=(2*5*7)/(12*11)

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PROBABILITY AND FREQUENCY DISTRIBUTIONS

Since frequency distribution of a random variable X is constructed from data obtained from a statistical experiment, all possible outcomes of X comprise the sample space of the experiment. In an experiment that accumulated a large number of data, the relative frequency distribution of X naturally reflects the probability distribution of X.

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Consider the following distribution of scores, which has been summarized in a frequency distribution table.For this distribution of scores, what is the probability of selecting a score of X = 8?

Example

p(X = 8) = ƒx=8 / N = 3/10 = 0.30

You can determine a probability from a frequency distribution table by computing the proportion for the X value in question.

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A frequency distribution histogram for a population that consists of N= 10 scores.The shaded part of the figure indicates the portion of the whole population that corresponds to scores greater than X= 4. The shaded portion is two-tenths of the whole distribution. So the probability of X>4 is

P(X>4) = 2/10=0.2

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The normal distribution following a z-score transformation.

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Adult heights form a normal distribution with a mean of 68 inches and a standard deviation of 6 inches. Given this information about the population we can determine the probability associated with specific samples from the normal distribution curve. For example, what is the probability of randomly selecting an individual from this population who is taller than 6 feet 8 inches (X = 80 inches)?

Restating this question in probability notation: p(X > 80) = ?

Example 1:

Solution: Z= (80 – 68)/6 = 12/6 = 2Thus p(X > 80) = p(z>2) = 2.28%

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The Standard Normal Table

• The standard normal table is a table that lists the area under the standard normal curve to the right of the mean (z=0) up to the z value of interest– See Table 6.1, Table A.3 in Appendix A, and the

table on the back of the front cover• This table is so important that it is repeated 3 times in

the textbook!• Always look at the accompanying figure for guidance

on how to use the table

The Standard Normal Table Continued

• The values of z (accurate to the nearest tenth) in the table range from 0.00 to 3.09 in increments of 0.01– z accurate to tenths are listed in the far left

column– The hundredths digit of z is listed across the top of

the table• The areas under the normal curve to the right

of the mean up to any value of z are given in the body of the table

A Standard Normal Table

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Scores on the Scholastic Achievement Test (SAT) form a normal distribution with mean = 500 and sd = 100. What is the probability of selecting an individual from the population who scores above 650? [p(X > 650) = ?]

Example 2:

Solution: z=(650-500)/100=1.5P(X>650) = p(z>1.5) = 0.5 - 0.4332 = 0.0668 = 6.68%

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The height of the young Chinese male population with age between 20 and 35 is normally distributed with mean = 175 cm and sd = 17 cm. What is the probability for finding someone with a height of taller than 199 cm? What is the probability for finding someone taller than Yao Ming in this population ? (His height is 2.29 m)

Exercise:

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The IQ score of the whole population is normally distributed with mean = 100 and sd = 10. What is the percentilerank for someone with IQ = 114?

Example 3: PERCENTILE RANKS

Solution: z=(114-100)/10=1.4P(X>114) = p(z>1.4) = 0.5 - 0.4192 = 0.0808 = 8.08%Percentile rank is 100% - 8.08% =91.92%

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PERCENTILE RANKS: exercise

The height of the young Chinese male population with age between 20 and 35 is normally distributed with mean = 175 cm and sd = 17 cm. What is the percentile rank for someone with a height of 165 cm?

ONEDefine probability in three different way: classical, empirical, and subjective approaches

TWO Understand the terms: event, outcome

THREESpecial addition rule and General addition rule.

Summary

Summary

FOURDefine the terms: conditional probability and joint probability.

FIVE The special multiplication rule and the General multiplication rule.

SIXPROBABILITY AND FREQUENCY DISTRIBUTIONS