SS 2 Probability Sp10 Bb

8/8/2019 SS 2 Probability Sp10 Bb

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Probability


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Some definitions.


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Types of Data

Elements of a data set are referred to as data points (= observations ).

Note: the word data is plural (e.g. the data are confusing).

Experiment: a process that generates data.

Qualitative (= categorical ) data: attributes (e.g. gender, occupation).Levels for qualitative variables are attributes or categories .

Quantitative data: numerical valuesLevels for quantitative variables are called variates .

Nominal numbers represent arbitrary codes: 1 = ME, 2 = EE evaluated the same way as qualitative data

Ordinal numbers convey ranking: class rankdifference in numbers is often not meaningful

Interval allows addition and subtraction: temperature100 F is not twice as hot as 50 F.

Ratio allows addition, subtraction, multiplication, division: weightmost statistical work is done with this type


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Sample Space

Sample space, S: all possible outcomes of an experiment

Example What is the sample space for the experiment of tossing adie and observing the number on the top face?{1,2,3,4,5,6}

Example Using a rule, describe the sample space for even faces on adie?{x | x is even}

Event: a subset of S

The vertical bar | is read as such that

www.wikipedia.org


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Multiplication Rule


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Multiplication Rule

A way to count independent eventsApplicable when the probability of any one of them is unaffected by theoccurrence of the other.

n1 x n 2 ... x n k

Example What is the total number of possible outcomes from two tossesof a die?

Example How many odd 4-digit numbers can be formed from the digits 1,2, 3, and 4 if each can be used multiple times ?

6 x 6 = 36

choice for units position = 2 (digits 1 or 3)choice for tens position = 4 (digits 1, 2, 3, or 4 )choice for hundreds position = 4choice for thousands position = 4

4x4x4x2 = 128


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Multiplication Rule

Example How many odd 4-digit numbers can be formed from the digits 1,2, 3, and 4 if each can be used only once ?

choice for units position = 2 (digits 1 or 3)choice for tens position = 3 (digits 2, 4, and 1 or 3)

choice for hundreds position = 2choice for thousands position = 12x3x2x1 = 12


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Probability of an Event


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If a particular outcome must always occur at each trial, then there is aprobability of 1 of that outcome. Similarly, a probability of 0 is equivalent tocertainly will not occur. Probability values between 0 and 1 are reserved for uncertain outcomes.

We can calculate probabilities of outcomes by building up from outcomeswhose probabilities we know.

The two types of events we shall consider are either/or and both/and .

Outcomes are mutually exclusive if only one can occur at any one trial.

Outcomes are independent if the occurrence of either one of them in a

composite trial does not affect the probability of the other occurring.


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The probability of an event A, P(A), if n of N outcomes correspond to A is:

P(A) = n/N , 0 < P(A) < 1, P(S) = 1

ExampleHeads represents one event for the outcome of a coin toss, and heads andtails are the only two possible outcomes of a toss. Thus, P(H) = 1/2.

A is the complement of A with respect to S: P(A)+P(A)=1

Intersection of two events A and B (A B): event containing all elementscommon to A and B

Union of two events A and B (A U B): event containing all elements thatbelong to A or B or both

A and B are mutually exclusive if A B =


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Addition Rule


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S

BA

2

Addition Law

For outcomes of OR type, i.e. A U B (union)

If outcomes A and B are not mutually exclusive , then the probability of theoccurrence in any trial of the event either A or B is equal to

P(A U B) = P(A) + P(B) P(A B)

If outcomes A and B are mutually exclusive , then the probability of theoccurrence in any trial of the event either A or B is equal to P(A) + P(B).

S

A Bexamples: probability of throwingeither 2 or 4 on a die is 1/6 + 1/6;probability of throwing either a heador a tail is + = 1

example: probability of brown hair or brown eyes

intersection


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S

BA 2

S

B

C

A 22

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Addition Law, Extension to 3 Cases

If outcomes A, B, and C are not mutually exclusive , then the probability of

the occurrence in any trial of the event either A or B or C is equal toP(A U B U C) = P(A) + P(B) + P(C)

P(A B) P(A C)

P(B

C)+ P(A B C)

light blueareacovered

twice

red areacovered

thrice

S

B

C

A1

S

B

C

A2

S

B

C

A3

green areasubtracted

once toooften


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If events A 1, A2, ... A n are mutually exclusive , thenP(A 1 U A2 U ... A n) = P(A 1) + P(A 2) + ... P(A n)

Example What is the probability, when tossing a die, of getting aneven number?

The total sample space is:{1,2,3,4,5,6} 6 points

The probability of getting any one of these outcomes is the same, 1/6.Getting one face precludes getting any other, so the events are

mutually exclusive.Three of the faces are even.

The probability is therefore1/6 + 1/6 + 1/6 = 3/6 = 1/2

This example is trivial, but the logic is important.

special case of theaddition law


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Addition Law vs. Multiplication Law

Addition Law

For outcomes of OR type, i.e. A U B

Multiplication Law

For outcomes of AND type, i.e. A B

S

BA

S

BA


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Multiplication Law

For outcomes of AND type, i.e. A B (intersection)

If A and B can occur then:

P(A B)= P(A)P(B|A), if P(A)>0

= P(B)P(A|B), if P(B)>0

Conditional probability:

Probability event B occurring when event A has occurred: P(B|A)(probability of B, given A)

P(B|A)= P(A B) / P(A)

ExampleProbability that a regularly scheduled flight departs on time is 0.83, probabilitythat it arrives on time is 0.82, probability it departs and arrives on time is 0.78.Find the probability that a plane (a) arrives on time given that is departed ontime, (b) departed on time given it has arrived on time.


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Independence


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Independence

Events are independent if and only if P(B|A) = P(B) or P(A|B) = P(A)

OR

P(A B) = P(A)P(B)

This concept is critical, and will be used several times later in the course.

Rearranging:

P(A B) = P(A)P(B) = P(A)P(B|A)= P(B)P(A|B)

Here the vertical bar | is read as given.


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More Complex Multiplication Rule Examples


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Multiplication Rules

P(A

B) = P(A)P(B|A)

example 2.37 in Walpole:

One of the balls from the blue bag is put into the pink one.What is P(black from 2 nd bag) = P(B 2)?


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P(W) = 4/7

P(B) = 3/7

P(W) = 4/9

P(B) = 5/9

P(W) = 3/9

P(B) = 6/9

P(W 2 W1) = 4/7 * 4/9= 0.254

P(B 2 W1) = 4/7 * 5/9= 0.317

P(W 2 B1) = 3/7 * 3/9= 0.143

P(B 2 B1) = 3/7 * 6/9= 0.286

P(B 2) = 0.317 + 0.286 = 0.603

reality check: P(W 2) = 0.254 + 0.143 = 0.397

Since these events aremutually exclusive, canuse the addition rule.


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Example

One box contains 5 roses and 3 daffodilsA second box contains 3 daffodils and 6 carnationsA third box contains 4 roses and 1 daffodil.

A flower is removed from the first box and placed unseen in the second.Then a flower is removed from the 2nd box and placed unseen in the third box.Finally, a flower is removed from the 3rd box.What is the probability that it is a rose?

www.thedailygreen.org

Rose?


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roses daffodils carnations totalBox 1 5R 3D 0C 8FBox 2 0R 3D 6C 9FBox 3 4R 1D 0C 5F

5R 3D1

1R 3D 6C2 4D 6C

5/8R 3/8D

3/10D 6/10C 4/10D 6/10C

4R 2D 4R 1D 1C 4R 2D 4R 1D 1C5R 1D3

R 1/10

R 5/6 D 1/6 R 4/6 D 2/6 R 4/6 C 1/6 R 4/6 D 2/6 R 4/6 C 1/6D 1/6 D 1/6

5/8 x 1/10 x 5/60.052 +

5/8 x 3/10 x 4/60.125 +

5/8 x 6/10 x 4/60.250 +

3/8 x 4/10 x 4/60.100 +

3/8 x 6/10 x 4/60.150 = 0.677

(originally 0.66 from box 3)


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Some card problems

What is the probability of getting, in a hand of 5 cards 1. an ace, a jack, a seven, a three, and a two, in that order?2. an ace, another ace, a jack, another jack, and a third jack, in that order?

Probability of an Event: Counting n and N Outcomes

The probability of an event A, P(A), if n of N outcomes corresponds to A is:P(A) = n/N

www.wikipedia.org


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Example

1. What is the probability of getting, in a hand of 5 cards, A, J, 7, 3, and 2, inthat order?

The probability of getting an A with the first card is 4/52The probability of getting a J with the 2 nd card is 4/51The probability of getting a 7 with the 3 rd card is 4/50The probability of getting a 3 with the 4 th card is 4/49The probability of getting a 2 with the 5 th card is 4/48

Each is independent, so (4) 5/52*51*50*49*48= 1024/311,875,200 = 3.3*10 -6


Note that any other order for these cards occurs with the same probability.

Note that to get the probability of getting this hand with the cards in any order, add the probabilities for all the different possible orders (5!); add

because the cases are mutually exclusive.

b b l f d


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Example

2. What is the probability of getting, in a hand of 5 cards, an ace, another ace, a jack, another jack, and a third jack, in that order?

The probability of getting an ace with the first card is 4/52The probability of getting an ace with the 2 nd card is 3/51

The probability of getting a jack with the 3 rd card is 4/50The probability of getting a jack with the 4 th card is 3/49The probability of getting a jack with the 5 th card is 2/48

Each event is independent, so (4*3*4*3*2)/52*51*50*49*48= 288/ 311,875,200 = 3456 / 311,875,200= 9.2 * 10 -7

Probability of getting the same type of card diminishes.


Conditional probability. Events are not independent.But we can treat them as independent by decrementing.


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An Application of these Rules: System Reliability

S t R li bilit


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System Reliability

An important application of probability is reliability analysis of systems.

Reliability of a component i can be defined in terms of a survival probability :

R i(t) = P(component i survives longer than time t)

The system reliability is a function of the components:R s(t) = f{ R 1(t), R 2(t), R 3(t)}

There are two basic system forms: parallel and series . We can illustrate themsimilarly to electrical circuits.

(adapted from Lapin, Modern Engineering Statistics, Duxbury Press, 1997)

www.wikipedia.org

S t R li bilit


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System Reliability

The system functions as long as ALL the components function.Failure of any component causes the system to fail.(Multiplication rule: 1 survives AND 2 survives AND 3 survives.)

Example: If the toner cartridge of a photocopier is empty, the machine nolonger makes copies. If the paper jams at position C, the machine can nolonger make copies. If the lamp bulb blows, no copies.

From the multiplication law for independent events:R s(t) = R 1(t) R 2(t) R 3(t)R n(t)

(adapted from Lapin, Modern Engineering Statistics, Duxbury Press, 1997)

1 2 3R1(t) = 0.90 R 2(t) = 0.80 R 3(t) = 0.95

Components in Series

R s(t) = 0.90*0.80*0.95 = 0.68

failures areindependent(assumption)

System Reliability


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System Reliability

The system functions as long as ANY ONE the components function.The system fails only when all the components fail.

Example: a photocopier has two printer trays, so the copier can still make

copies when one is empty.

We need to find the failure probabilities, rather than the reliability probabilities:

Components in Parallel

Rp(t) = 1 - 0.10*0.20*0.05 = 0.99

1

2

3

R1(t) = 0.90

R2(t) = 0.80

R3(t) = 0.95

P(component failure) =1 P(component survival)

P(system survival) =1 P( all components failure)

Rp(t) =1 [1 - R 1(t)][1 - R 2(t)][1 -R n(t)]

Increasing System Reliability


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1 2 3R1(t) = 0.90 R 2(t) = 0.80 R 3(t) = 0.95

Original System

Rs

(t) = 0.68

Increasing System Reliability

1. Increase the reliability of individual components

1 2 3R1(t) = 0.90 R 2(t) = 0.95 R 3(t) = 0.95

Upgraded Component

1 3R2(t) = 0.80

2

R1(t) = 0.90 R3(t) = 0.95

Duplicated Component

Rs (t) = 0.81

Rs (t) = 0.82

R2(t) = 0.80

2

Notice that if the first component is upgraded instead of the second, the reliabilitydoes not increase much: the least reliable component most strongly affects theoverall reliability in a series system.

2. Increase redundancy byduplicating components

1 2 3R1(t) = 0.99 R 2(t) = 0.80 R 3(t) = 0.95

Upgraded Component

Rs (t) = 0.75


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