chapter 2 probability -...

Chapter 2 Probability

Part 2: Probability Introduction

Section 2.3 Interpretation and Axioms of ProbabilitySection 2.4 Unions of Events and Addition RulesSection 2.5 Conditional ProbabilitySection 2.6 Intersections of Events, Multiplication and

Total Probability Rules

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Introduction to Probability

Probability is used to quantify the likelihood, or chance, that anoutcome of a random experiment will occur

Probabilities for a random experiment are often assigned on the basisof a reasonable model of the system under study.

Definition (Equally Likely Outcomes)

Whenever a sample space consists of N possible outcomes that are equallylikely the probablity of each is 1

N .

Example (Equally Likely Outcomes)

Roll a 6-sided die. Each has probability 16 .

Pick a single card from a deck of 52 cards. Each has probability 152 .

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If you play a lottery where you can choose 3 digits, what is the probabilitythat your number is drawn?

ANS: The sample space is the set of all 3 digits numbers.S = {000, 001, 002, . . . , 999}

All 1000 numbers are equally likely.

Under this assumption of equally likely outcomes, you have a 11000 chance

of having your number drawn.

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Sometimes we’re interested in an event that is composed of severaloutcomes from a sample space.

Definition (Probability of an Event)

For a discrete sample space, the probability of an event E, denoted asP (E), equals the sum of the probabilities of the outcomes in E.


If you roll a fair die, what is the probability of rolling an even?ANS: The sample space S = {1, 2, 3, 4, 5, 6}E is the event of rolling an even and has elements {2, 4, 6}.P (E) = 1

6 + 16 + 1

6 = 36 = 1

2 .

If an experiment can result in any one of N differentequally likely outcomes, and if exactly n of these outcomescorrespond to event A, then the probability of event A is n

N .

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Example (Sampling without replacement)

In a set of 5 bottles, one has a small fracture. If you randomly selected apair of bottles, what is the probability that the fractured bottle is chosen?

ANS:

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Example (Equally likely outcomes)

Samples of emissions from three suppliers are classified for conformance toair-quality specification. The results from 100 samples are summarized asfollows:

ConformsYes No

1 22 8Supplier 2 25 5

3 30 10One sample is drawn at random.

What is the probability that it is from supplier 2?ANS:

What is the probability that it conforms and is NOT from supplier 3?ANS:

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What if we’re interested in an event that is composed of severaloutcomes from a sample space that does not have equally likely outcomes

Example (Not Equally likely outcomes: Loaded die)

Suppose a die is loaded in such away that an even number is twice aslikely to occur as an odd number. If E is the event that a number lessthan 4 occurs on a single toss of the die, find P (E).

The elements in the sample space {1, 2, 3, 4, 5, 6} are NOT equallylikely.

outcome 1 2 3 4 5 6

probability 19

29

19

29

19

29

E = {1, 2, 3}P (E) = 1

9 + 29 + 1

9 = 49

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Example (Printed circuit board defects)

An inspection for defects resulted in the following table:

Number of defects Proportion of boardswith number of defects

0 0.701 0.152 0.053 or more 0.10

If a board is selected at random, what is the probability that...

i) it has no defect? ANS: 0.70

ii) it has 2 or less defects? ANS: 0.05 + 0.15 + 0.70 = 0.90

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Introduction to Probability: Axioms of Probability

Definition (Axioms of Probability)

Probability is a number that is assigned to each member of a collection ofevents from a random experiment that satisfies the following properties:If S is the sample space and E is any event in a random experiment, then

1) P (S) = 1

2) 0 ≤ P (E) ≤ 1

3) For two events E1 and E2 with E1 ∩ E2 = ∅,P (E1 ∪ E2) = P (E1) + P (E2)

If the two events E1 and E2 have nothing in common (i.e. no overlap), theprobability of their union is just the sum of their respective probabilities.

Definition (Probability for complements)

The axioms also imply: P (E) = 1− P (E′)

P (E) + P (E′) = 19 / 30

Addition Rules of Probability

Definition (Probability of a Union, two events)

The event A ∪B is the event that A occurs, B occurs, or both A and Boccur.

P (A ∪B) = P (A) + P (B)− P (A ∩B)

Example (Union)

Roll a 6-sided die.

Let A be the event that an even number turns up and let B be the eventthat a number is divisible by 3 occurs. Find P (A ∪B).

ANS: A = {2, 4, 6}, B = {3, 6}, A ∩B = {6}

P (A ∪B) = P (A) + P (B)− P (A ∩B) = 36 + 2

6 −16 = 4

6 = 23

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Definition (Probability of a Union, three events)

We can extend the union rule to three sets...

P (A ∪B ∪ C) = P (A) + P (B) + P (C)−P (A ∩B)− P (A ∩ C)− P (B ∩ C)+ P (A ∩B ∩ C)

See Montgomery for derivation.

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Definition (Union of Mutually Exclusive Events)

A and B are mutually exclusive events if A ∩B = ∅.

If A and B are mutually exclusive events, thenP (A ∪B) = P (A) + P (B)

A collection of events E1, E2, . . . , Ek is said to be mutually exclusive if forall pairs, Ei ∩ Ej = ∅.

For a collection of mutually exclusive events, E1, E2, . . . , Ek

P (E1 ∪ E2 ∪ . . . ∪ Ek) = P (E1) + P (E2) + . . .+ P (Ek)

←One example of four mutually exclusive events (that alsoform a partition of the sample space).

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Conditional Probability

Sometimes probabilities need to be re-evaluated or adjusted as additionalinformation becomes available that is specific to the situation.

Definition (Conditional Probability)

The probability of event B given an event A, denoted as P (B|A), is

P (B|A) =P (A∩B)P (A) for P (A) > 0

You could also says “The probability of event B occurring given that youknow event A will occur is denoted as P (B|A)”. This includes aconditioning statement of ’given’.

Example (Venn Diagram)

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Example (Sex and Rank of STEM∗ faculty)∗Science, Technology, Engineering, and Mathematics

Randomly choose a STEM faculty member from all 518 faculty members.Let A denote the event that the faculty member is an assistant professor.Let F denote the event that the faculty member is is female.

RANK

Assistant Associate Full

SEX Male 98 87 254

Female 36 25 18

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Example (Sex and Rank of STEM∗ faculty)RANK


SEX Male 98 87 254

Female 36 25 18

Find the unconditional probability of selecting a female.ANS: P (F ) = 79

518 = 0.15

Find the conditional probability of selecting a female given that they arean assistant professor.

ANS: P (F |A) = P (female | assistant professor) = P (A∩F )P (A)

= 36/518134/518 = 36

134 = 0.2687

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Example (Sex and Rank of STEM∗ faculty)RANK


SEX Male 98 87 254

Female 36 25 18

Mosaic Plot for Sex and Rank (conditioning on Rank):

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Example (Manufactured Parts)

Sometimes a visible superficial flaw on a part may signal that there’s afatal error with a product, but sometimes a part with a visible superficialflaw may function just fine.

Can the error that we see on the surface help us know which parts areactually defective?

In a manufacturing process,10% of the parts contain visible flaws90% of the parts do not contain visible flaws

Of the parts containing a visible flaw,25% are defective

Of the parts not containing a visible flaw,5% are defective

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Thus, the chance of being defective depends on whether a ‘visible flaw’ ispresent or not. There is a higher probability that a part is defective if aflaw is observed on it. A mosaic plot...

Bad : 5%

Good : 95%

25%

75%

VF: noVF: noVF: yesVF: yes

0

25

50

75

100

0 25 50 75 100

StatusBad

Good

Visible'Flaw'(1st'sec/oning)'

''10% ' ' ''''''''''90%'with'VF' ' ' ''''''without'VF'

2nd'sec/oning'(condi/onal'on'1st)'

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Define the events and probabilities:

Let D be the event that a part is defective.Let F be the event that a part has a visible flaw.

P (F ) = 0.10P (F ′) = 0.90

Conditional Probabilities:P (D|F ) = 0.25P (D|F ′) = 0.05

Notice that the above implies the following conditional probabilities as well:P (D′|F ) = 0.75P (D′|F ′) = 0.95

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Random sampling and conditional probabilities

Definition (Random Samples)

To ‘select randomly’ implies that at each step of the sample, the itemsthat remain in the batch are equally likely to be selected.

Example (Sequential random sampling without replacement)

Suppose you are to choose two nails from a bin at random. In the bin are50 nails such that 35 are galvanized, and 15 are regular steel.

If you drew a galvanized on the first draw, what is the probability ofchoosing a regular on the second draw?

We assume all nails that are left are equally likely to be drawn.

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If you drew a galvanized on the first draw (and didn’t replace it), what isthe probability of choosing a regular on the second draw?

ANS: Removing a galvanized, there are 49 left, and 15 are regular.

P (regular on second draw given we drew a galvanized on first draw)= P (regular on 2nd|galvanized on 1st)= 15

49 = 0.3061

NOTE 1: Defining events with mathematical notation will allow them to be more concisely

written. If we let G1 denote a galvanized drawn on 1st, and R2 denote a regular drawn on the

2nd then

P (R2|G1) = P (R2|R′1) = 0.3061

NOTE 2: Because this is a dichotomous situation, the event G1 is the same as R′1.

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Multiplication Rule for Probability

Definition (Multiplication Rule for Probability)

The probability of the intersection of two events

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

The joint probability of A and B simultaneously occurring can be writtenas the product of a conditional probability and an unconditional probability.

We’ve actually seen this relationship before in the conditional probabilitydefinition...

P (B|A) = P (A∩B)P (A)

or

P (A|B) =P (A∩B)P (B)

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Continuing the earlier example... Choose two nails without replacement.Bin contains 35 galvanized and 15 regular nails.

What is the probability of choosing a regular first, and then anotherregular nail? (We are stating what happens on both the first and second draw now.)

ANS: We could think of S as an ordered, not equally likely sample spacedenoted by four (x,y) pairs showing the category first and second draws...S = {(g, g), (g, r), (r, g), (r, r)}

Let R1 be the event of drawing a regular on the first draw.R1 = {(r, g), (r, r)}

Let R2 be the event of drawing a regular on the second draw.R2 = {(g, r), (r, r)}

NOTE: R1 ∩R2 = {(r, r)}23 / 30



ANS: Continuing... Using the multiplication rule,

P (R1 ∩R2) = P (R2|R1) · P (R1) =1449 ·

1550 = 0.0857

NOTE: We often find that there is more than one way to ‘perceive’ orset-up a probability question and correctly calculate the probability.Ordered or unordered sample space? Equally or not likely outcomes? Withpractice, we can see which ‘perception’ may lead to the answer with theleast amount of effort.

In the previous example, we could have also calculated the correctprobability by perceiving order to not matter as...(

152

)(350

)(

502

) = 1051225 = 0.0857

And tree diagrams are another way to visualize conditional probabilities.24 / 30


Example (Drawing cards without replacement)

Suppose two cards are randomly drawn from a deck of 52 cards,one-by-one, without replacement. There are 13 of each suit (diamonds,spades, hearts, clubs).

Find the probability that a diamond is drawn on the first AND on thesecond draw.

ANS: Let D1 be the event that a diamond is drawn on the first draw.Let D2 be the event that a diamond is drawn on the second draw.

P (D1 ∩D2) = P (D2|D1) · P (D1)

= 1251 ·

1352

= 1251 ·

14 = 3

51 = 0.0588

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Definition (Extension of Multiplication Rule for probability)

P (A1 ∩A2 ∩A3) = P (A3 ∩ (A2 ∩A1))= P (A3|A2 ∩A1) · P (A2 ∩A1)= P (A3|A2 ∩A1) · P (A2|A1) · P (A1)

The multiplication rule leads into the “Total Probability Rule”...

Consider partitioning the full sample space into an A and A′ (defined byblue line) or a B and B′ (defined by inside of circle and outside of circle).

S

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Total Probability Rule

In the graphic, notice that B = (B ∩A)∪ (B ∩A′), and these two sets aremutually exclusive. Thus, we can write P (B) = P (B ∩A) + P (B ∩A′).

By applying our multiplication rule to this equation, we get the following...

Definition (Total Probability Rule (for two events))

For any events A and B,

P (B) = P (B ∩A) + P (B ∩A′)

= P (B|A) · P (A) + P (B|A′) · P (A′)

In this ‘two-event’ rule, S = A ∪A′. The complements are mutuallyexclusive and comprise the full sample space.

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If we can break-up S into k mutually exclusive sets that form a partition ofS, we can extend the previous rule...

Definition (Total Probability Rule (for multiple events))

Assume E1, E2, . . . , Ek are k mutually exclusive and exhaustive sets. Then

P (B) = P (B ∩ E1) + P (B ∩ E2) + . . .+ P (B ∩ Ek)

= P (B|E1) · P (E1) + P (B|E2) · P (E2) + . . .+ P (B|Ek) · P (Ek)

Example with k = 4:

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Example (Arizona Lung Association)

According to the Arizona Chapter of the American Lung CancerAssociation, 7.0% of the population has lung disease. Of those havinglung disease, 90% are smokers; of those not having lung disease, 25.3%are smokers.

Determine the probability that a person selected randomly is a smoker.

ANS:

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Example (Arizona Lung Association)

Continuing... Consider the tree diagram for this example.LungCancer(L)?

P(L) = 0.07

yes

Smoker(B)?

P(L’)= 0.93

no

yes yes nono

P(B|L)= 0.90 P(B’|L)= 0.10

P(B’|L’)= 0.747 P(B|L’)= 0.253

L∩B L∩B ' L '∩B L '∩B '0.0630 0.0070 0.2353 0.6947

SmokerP(L∩B) = P(L) ⋅P(B | L) = (0.07)(0.90) = 0.0630

P(B) = 0.0630+ 0.2353= 0.2983

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