Download - 5A-1. Probability (Part 1) Random Experiments Random Experiments Probability Rules of Probability Rules of Probability Independent Events Independent

Probability (Part 1)Probability (Part 1)Probability (Part 1)Probability (Part 1)

Random Experiments

Probability

Rules of Probability

Independent Events

Chapter5A5A5A5A

McGraw-Hill/Irwin © 2008 The McGraw-Hill Companies, Inc. All rights reserved.

5A-3

• A A random experimentrandom experiment is an observational process is an observational process whose results cannot be known in advance.whose results cannot be known in advance.

• The set of all The set of all outcomesoutcomes ( (SS) is the ) is the sample spacesample space for for the experiment.the experiment.

• A sample space with a countable number of A sample space with a countable number of outcomes is outcomes is discretediscrete..

Sample SpaceSample Space

Random ExperimentsRandom ExperimentsRandom ExperimentsRandom Experiments

5A-4

• For example, when CitiBank makes a consumer For example, when CitiBank makes a consumer loan, the sample space is:loan, the sample space is:

SS = {default, no default} = {default, no default}

• The sample space describing a Wal-Mart The sample space describing a Wal-Mart customer’s payment method is:customer’s payment method is:

SS = {cash, debit card, credit card, check} = {cash, debit card, credit card, check}



5A-5

• For a single roll of a die, the sample space is:For a single roll of a die, the sample space is:

S = {1, 2, 3, 4, 5, 6}

• When two dice are rolled, the sample space is When two dice are rolled, the sample space is the following pairs:the following pairs:



{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

SS = =

5A-6

• Consider the sample space to describe a randomly Consider the sample space to describe a randomly chosen United Airlines employee by chosen United Airlines employee by 2 genders, 2 genders, 21 job classifications, 21 job classifications, 6 home bases (major hubs) and 6 home bases (major hubs) and 4 education levels4 education levels

• It would be impractical to enumerate this sample It would be impractical to enumerate this sample space.space.

There are: 2 x 21 x 6 x 4 = 1008 possible outcomesThere are: 2 x 21 x 6 x 4 = 1008 possible outcomes



5A-7

• If the outcome is a If the outcome is a continuouscontinuous measurement, the measurement, the sample space can be described by a rule. sample space can be described by a rule.

• For example, the sample space for the length of a For example, the sample space for the length of a randomly chosen cell phone call would berandomly chosen cell phone call would be

SS = {all = {all XX such that such that X X >> 0} 0}

• The sample space to describe a randomly chosen The sample space to describe a randomly chosen student’s GPA would bestudent’s GPA would beSS = { = {XX | 0.00 | 0.00 << X X << 4.00} 4.00}

or written as or written as SS = { = {XX | | X X >> 0} 0}



5A-8

• An An eventevent is any subset of outcomes in the sample is any subset of outcomes in the sample space.space.

• A A simple eventsimple event or or elementary eventelementary event, is a single , is a single outcome.outcome.

• A discrete sample space A discrete sample space SS consists of all the consists of all the simple events (simple events (EEii):):

SS = { = {EE11, , EE22, …, , …, EEnn}}

EventsEvents


5A-9

• What are the chances of observing a H or T?What are the chances of observing a H or T?

• These two elementary events are These two elementary events are equally likelyequally likely..

SS = {H, T} = {H, T}

• Consider the random experiment of tossing a Consider the random experiment of tossing a balanced coin. balanced coin. What is the sample space?What is the sample space?

• When you buy a lottery ticket, the sample space When you buy a lottery ticket, the sample space SS = {win, lose} has only two events. = {win, lose} has only two events.

EventsEvents


• Are these two events equally likely to occur?Are these two events equally likely to occur?

5A-10

• For example, in a sample space of 6 simple For example, in a sample space of 6 simple events, we could define the compound eventsevents, we could define the compound events

• A A compound eventcompound event consists of two or more simple consists of two or more simple events. events.

• These are These are displayed in a displayed in a Venn diagramVenn diagram::

AA = { = {EE11, , EE22}}

BB = { = {EE33, , EE55, , EE66}}

EventsEvents


5A-11

• Many different compound events could be defined.

• Compound events can be described by a rule.

S = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

• For example, the compound event A = “rolling a seven” on a roll of two dice consists of 6 simple events:

EventsEvents


5A-12

• The The probabilityprobability of an event is a number that of an event is a number that measures the relative likelihood that the event will measures the relative likelihood that the event will occur.occur.

• The probability of event The probability of event A A [denoted [denoted PP((AA)], must lie )], must lie within the interval from 0 to 1:within the interval from 0 to 1:

0 0 << PP((AA) ) << 1 1

If P(A) = 0, then the event cannot occur.

If P(A) = 1, then the event is certain to occur.

DefinitionsDefinitions

ProbabilityProbabilityProbabilityProbability

5A-13

• In a discrete sample space, the probabilities of all In a discrete sample space, the probabilities of all simple events must sum to unity:simple events must sum to unity:

• For example, if the following number of purchases For example, if the following number of purchases were made bywere made by

PP((SS) = ) = PP((EE11) + ) + PP((EE22) + … + ) + … + PP((EEnn) = 1 ) = 1

credit card: credit card: 32%32%

debit card:debit card: 20%20%

cash:cash: 35%35%

check:check: 18%18%

Sum = Sum = 100%100%

DefinitionsDefinitions


P(credit card) =P(credit card) = .32.32

P(debit card) =P(debit card) = .20.20

P(cash) =P(cash) = .35.35

P(check) =P(check) = .18.18

Sum =Sum = 1.01.0

ProbabilityProbability

5A-14

• Businesses want to be able to quantify the Businesses want to be able to quantify the uncertaintyuncertainty of future events. of future events.

• For example, what are the chances that next For example, what are the chances that next month’s revenue will exceed last year’s average?month’s revenue will exceed last year’s average?

• How can we increase the chance of positive future How can we increase the chance of positive future events and decrease the chance of negative future events and decrease the chance of negative future events?events?

• The study of The study of probabilityprobability helps us understand and helps us understand and quantify the uncertainty surrounding the future.quantify the uncertainty surrounding the future.


5A-15

• Three approaches to probability:Three approaches to probability:

ApproachApproach ExampleExample

EmpiricalEmpirical There is a 2 percent chanceThere is a 2 percent chance of twins in a randomly- of twins in a randomly-chosen birth.chosen birth.

What is Probability?What is Probability?


ClassicalClassical There is a 50 % probability There is a 50 % probability of heads on a coin flip.of heads on a coin flip.

SubjectiveSubjective There is a 75 % chance that England will There is a 75 % chance that England will adopt the Euro currency by 2010.adopt the Euro currency by 2010.

5A-16

• Use the Use the empiricalempirical or or relative frequencyrelative frequency approach to approach to assign probabilities by counting the frequency (assign probabilities by counting the frequency (ffii) )

of observed outcomes defined on the experimental of observed outcomes defined on the experimental sample space.sample space.

• For example, to estimate the default rate on For example, to estimate the default rate on student loans:student loans:

PP(a student defaults) = (a student defaults) = f f //nn

Empirical ApproachEmpirical Approach


number of defaultsnumber of loans

==

5A-17

• Necessary when there is no prior knowledge of Necessary when there is no prior knowledge of events.events.

• As the number of observations (As the number of observations (nn) increases or the ) increases or the number of times the experiment is performed, the number of times the experiment is performed, the estimate will become more accurate.estimate will become more accurate.

Empirical ApproachEmpirical Approach


5A-18

• The The law of large numberslaw of large numbers is an important is an important probability theorem that states that a large sample probability theorem that states that a large sample is preferred to a small one.is preferred to a small one.

• Flip a coin 50 times. We would expect the Flip a coin 50 times. We would expect the proportion of heads to be near .50. proportion of heads to be near .50.

• A large A large nn may be needed to get close to .50. may be needed to get close to .50.

• However, in a small finite sample, any ratio can be However, in a small finite sample, any ratio can be obtained (e.g., 1/3, 7/13, 10/22, 28/50, etc.).obtained (e.g., 1/3, 7/13, 10/22, 28/50, etc.).

Law of Large NumbersLaw of Large Numbers


• Consider the results of 10, 20, 50, and 500 coin Consider the results of 10, 20, 50, and 500 coin flips.flips.

5A-19


5A-20

• Actuarial scienceActuarial science is a high-paying career that is a high-paying career that involves estimating empirical probabilities.involves estimating empirical probabilities.

• For example, actuaries For example, actuaries - calculate payout rates on life insurance,- calculate payout rates on life insurance, pension plans, and health care plans pension plans, and health care plans- create tables that guide IRA withdrawal- create tables that guide IRA withdrawal rates for individuals from age 70 to 99 rates for individuals from age 70 to 99

Practical Issues for ActuariesPractical Issues for Actuaries


5A-21

• In this approach, we envision the entire sample In this approach, we envision the entire sample space as a collection of equally likely outcomes.space as a collection of equally likely outcomes.

• Instead of performing the experiment, we can use Instead of performing the experiment, we can use deduction to determine deduction to determine PP((AA).).

• a prioria priori refers to the process of assigning refers to the process of assigning probabilities probabilities before before the event is observed.the event is observed.

• a priori probabilitiesa priori probabilities are based on logic, not are based on logic, not experience.experience.

Classical ApproachClassical Approach


5A-22

• For example, the two dice experiment has 36 For example, the two dice experiment has 36 equally likely simple events. The equally likely simple events. The PP(7) is(7) is

• The probability is The probability is obtained obtained a prioria priori using the using the classical classical approachapproach as shown as shown in this Venn diagram in this Venn diagram for 2 dice:for 2 dice:

number of outcomes with 7 dots 6( ) 0.1667

number of outcomes in sample space 36P A

Classical ApproachClassical Approach


5A-23

• A A subjective subjective probability reflects someone’s probability reflects someone’s personal belief about the likelihood of an event.personal belief about the likelihood of an event.

• Used when there is no repeatable random Used when there is no repeatable random experiment.experiment.

• For example,For example,- What is the probability that a new truck - What is the probability that a new truck product program will show a return on product program will show a return on investment of at least 10 percent? investment of at least 10 percent?- What is the probability that the price of GM- What is the probability that the price of GM stock will rise within the next 30 days? stock will rise within the next 30 days?

Subjective ApproachSubjective Approach


5A-24

• These probabilities rely on personal judgment or These probabilities rely on personal judgment or expert opinion.expert opinion.

• Judgment is based on experience with similar Judgment is based on experience with similar events and knowledge of the underlying causal events and knowledge of the underlying causal processes.processes.

Subjective ApproachSubjective Approach


5A-25

• The The complementcomplement of an event of an event AA is denoted by is denoted by AA′′ and consists of everything in the sample space and consists of everything in the sample space SS except event except event AA..

Complement of an EventComplement of an Event

Rules of ProbabilityRules of ProbabilityRules of ProbabilityRules of Probability

5A-26

• Since Since AA and and AA′′ together comprise the entire together comprise the entire sample space, sample space, PP((AA) + ) + PP((AA′′ ) = 1 ) = 1

• The probability of The probability of AA′′ is found by is found by PP((AA′′ ) = 1 – ) = 1 – PP((AA) )

• For example, For example, The Wall Street JournalThe Wall Street Journal reports that reports that about 33% of all new small businesses fail within about 33% of all new small businesses fail within the first 2 years. The probability that a new small the first 2 years. The probability that a new small business will survive is:business will survive is:

P(survival) = 1 – P(failure) = 1 – .33 = .67 or 67%

Complement of an EventComplement of an Event


5A-27

• The The oddsodds in favor of event in favor of event AA occurring is occurring is

• Odds are used in sports and games of chance.Odds are used in sports and games of chance.

• For a pair of fair dice, For a pair of fair dice, PP(7) = 6/36 (or 1/6). (7) = 6/36 (or 1/6). What are the odds in favor of rolling a 7?What are the odds in favor of rolling a 7?

( ) ( )Odds =

( ') 1 ( )

P A P A

P A P A

(rolling seven) 1/ 6 1/ 6 1Odds =

1 (rolling seven) 1 1/ 6 5/ 6 5

P

P

Odds of an EventOdds of an Event


5A-28

• On the average, for every time a 7 is rolled, there On the average, for every time a 7 is rolled, there will be 5 times that it is not rolled.will be 5 times that it is not rolled.

• In other words, the odds are 1 to 5 In other words, the odds are 1 to 5 in favorin favor of of rolling a 7.rolling a 7.

• The odds are 5 to 1 The odds are 5 to 1 againstagainst rolling a 7. rolling a 7.



• In horse racing and other sports, odds are usually In horse racing and other sports, odds are usually quoted quoted againstagainst winning. winning.

5A-29

• If the odds against event If the odds against event AA are quoted as are quoted as bb to to aa, , then the implied probability of event then the implied probability of event AA is: is:

• For example, if a race horse has a 4 to 1 odds For example, if a race horse has a 4 to 1 odds againstagainst winning, the winning, the PP(win) is (win) is

PP((AA) =) =a

a b



1 10.20

4 1 5

a

a b

PP(win) = (win) = or 20%or 20%

5A-30

• The union of two events consists of all outcomes in the sample space S that are contained either in event A or in event B or both (denoted A B or “A or B”).

may be read as “or” since one or the other or both events may occur.

Union of Two EventsUnion of Two Events


5A-31

• For example, randomly choose a card from a deck For example, randomly choose a card from a deck of 52 playing cards. of 52 playing cards.

• It is the possibility of drawing It is the possibility of drawing eithereither a queen (4 ways) a queen (4 ways) oror a red card (26 ways) a red card (26 ways) oror both (2 ways). both (2 ways).

• If If QQ is the event that we draw a is the event that we draw a queen and queen and RR is the event that we is the event that we draw a red card, what is draw a red card, what is Q Q RR??

Union of Two EventsUnion of Two Events


5A-32

• The The intersectionintersection of two events of two events AA and and BB (denoted (denoted A A BB or “ or “AA and and BB”) is the event ”) is the event consisting of all outcomes in the sample space consisting of all outcomes in the sample space SS that are contained in that are contained in bothboth event event AA and event and event BB. .

may be read may be read as “and” since as “and” since bothboth events events occur. This is a occur. This is a joint probabilityjoint probability..

Intersection of Two EventsIntersection of Two Events


5A-33

• It is the possibility of getting It is the possibility of getting bothboth a queen a queen andand a red card a red card (2 ways).(2 ways).

• If If QQ is the event that we draw a is the event that we draw a queen and queen and RR is the event that we is the event that we draw a red card, what is draw a red card, what is Q Q RR??

• For example, randomly choose a card from a deck For example, randomly choose a card from a deck of 52 playing cards. of 52 playing cards.

Intersection of Two EventsIntersection of Two Events


5A-34

• The The general law of additiongeneral law of addition states that the states that the probability of the union of two events probability of the union of two events AA and and B B is:is:

PP((A A BB) = ) = PP((AA) + ) + PP((BB) – ) – PP((A A BB))

When you add When you add the the PP((AA) and ) and PP((BB) together, ) together, you count the you count the P(A and B) P(A and B) twice.twice.

So, you have So, you have to subtract to subtract PP((A A BB) to ) to avoid over-avoid over-stating the stating the probability.probability.

A B

A and B

General Law of AdditionGeneral Law of Addition


5A-35

• For the card example:For the card example:

PP((QQ) = 4/52) = 4/52 (4 queens in a deck) (4 queens in a deck)

= 4/52 + 26/52 – 2/52= 4/52 + 26/52 – 2/52

PP((Q Q RR) = ) = PP((QQ) + ) + PP((RR) – ) – PP((Q Q QQ))

Q4/52

R26/52

Q and R = 2/52

General Law of AdditionGeneral Law of Addition


= 28/52 = .5385 or 53.85%= 28/52 = .5385 or 53.85%

PP((RR) = 26/52 (26 red cards in a deck)) = 26/52 (26 red cards in a deck)PP((Q Q RR) = 2/52 (2 red queens in a deck)) = 2/52 (2 red queens in a deck)

5A-36

• Events Events AA and and BB are are mutually exclusivemutually exclusive (or (or disjointdisjoint) ) if their intersection is the null set (if their intersection is the null set () that contains ) that contains no elements.no elements. If If A A BB = = , then , then PP((A A BB) = 0) = 0

• In the case of mutually In the case of mutually exclusive events, the exclusive events, the addition law reduces addition law reduces to:to:

PP((A A BB) = ) = PP((AA) + ) + PP((BB))

Mutually Exclusive EventsMutually Exclusive Events


Special Law of AdditionSpecial Law of Addition

5A-37

• Events are Events are collectively exhaustivecollectively exhaustive if their union is if their union is the entire sample space the entire sample space SS..

• Two mutually exclusive, collectively exhaustive Two mutually exclusive, collectively exhaustive events are events are dichotomousdichotomous (or (or binarybinary) ) eventsevents..

For example, a car repair For example, a car repair is either covered by the is either covered by the warranty (warranty (AA) or not () or not (BB). ).

WarrantyWarrantyNoNo

WarrantyWarranty

Collectively Exhaustive EventsCollectively Exhaustive Events


5A-38

• More than two mutually exclusive, collectively exhaustive events are polytomous events.

For example, a Wal-Mart customer can pay by credit card (A), debit card (B), cash (C) or check (D).

CreditCreditCardCard

DebitDebitCardCard

CashCash

CheckCheck

Collectively Exhaustive EventsCollectively Exhaustive Events


5A-39

• Polytomous events can be made dichotomous Polytomous events can be made dichotomous (binary) by defining the second category as (binary) by defining the second category as everything everything notnot in the first category. in the first category.

Polytomous Events Polytomous Events Binary Binary ((Dichotomous) VariableDichotomous) Variable

Vehicle type (SUV, sedan, truck, Vehicle type (SUV, sedan, truck, motorcycle)motorcycle)

XX = 1 if SUV, 0 otherwise = 1 if SUV, 0 otherwise

Forced DichotomyForced Dichotomy


A randomly-chosen NBA player’s A randomly-chosen NBA player’s heightheight

XX = 1 if height exceeds 7 feet, 0 = 1 if height exceeds 7 feet, 0 otherwiseotherwise

Tax return type (single, married filing Tax return type (single, married filing jointly, married filing separately, head jointly, married filing separately, head of household, qualifying widower) of household, qualifying widower)

XX = 1 if single, 0 otherwise = 1 if single, 0 otherwise

5A-40

• The probability of event The probability of event AA givengiven that event that event BB has has occurred.occurred.

• Denoted Denoted PP((A A | | BB). ). The vertical line “ | ” is read as “given.”The vertical line “ | ” is read as “given.”

( )( | )

( )

P A BP A B

P B

for for PP((BB) > 0 and ) > 0 and

undefined otherwise undefined otherwise

Conditional ProbabilityConditional Probability


5A-41

• Consider the logic of this formula by looking at the Venn diagram.

( )( | )

( )

P A BP A B

P B

The sample space is restricted to B, an event that has occurred.

A B is the part of B that is also in A.

The ratio of the relative size of A B to B is P(A | B).

Conditional ProbabilityConditional Probability


5A-42

• Of the population aged 16 – 21 and not in college:Of the population aged 16 – 21 and not in college:

UnemployedUnemployed 13.5%13.5%

High school dropoutsHigh school dropouts 29.05%29.05%

Unemployed high school dropoutsUnemployed high school dropouts 5.32%5.32%

• What is the conditional probability that a member What is the conditional probability that a member of this population is unemployed, given that the of this population is unemployed, given that the person is a high school dropout?person is a high school dropout?

Example: High School DropoutsExample: High School Dropouts


5A-43

• First defineFirst defineUU = the event that the person is unemployed = the event that the person is unemployedDD = the event that the person is a high school = the event that the person is a high school

dropoutdropoutPP((UU) = .1350) = .1350 PP((DD) = .2905) = .2905 PP((UUDD) = .0532) = .0532

( ) .0532( | ) .1831

( ) .2905

P U DP U D

P D

or 18.31%or 18.31%

• PP((U | DU | D) = .1831 > ) = .1831 > PP((UU) = .1350) = .1350

• Therefore, being a high school dropout is related Therefore, being a high school dropout is related to being unemployed.to being unemployed.

Example: High School DropoutsExample: High School Dropouts


5A-44

• Event Event A A is is independent independent of event of event BB if the conditional if the conditional probability probability PP((AA | | BB) is the same as the marginal ) is the same as the marginal probability probability PP((AA).).

• To check for independence, apply this test:To check for independence, apply this test:

If If PP((A | BA | B) = ) = PP((AA) then event ) then event AA is is independentindependent of of BB. .

• Another way to check for independence:Another way to check for independence:

If If PP((A A B B) = ) = PP((AA))PP((BB) then event ) then event AA is is independentindependent of event of event BB since since

PP((A | BA | B) = ) = PP((A A BB)) = = PP((AA))PP((BB)) = = PP((AA)) PP((BB) ) PP((BB))

Independent EventsIndependent EventsIndependent EventsIndependent Events

5A-45

• Out of a target audience of 2,000,000, ad Out of a target audience of 2,000,000, ad A A reaches 500,000 viewers, reaches 500,000 viewers, BB reaches 300,000 reaches 300,000 viewers and both ads reach 100,000 viewers.viewers and both ads reach 100,000 viewers.

• What is What is PP((AA | | BB)?)?

500,000( ) .25

2,000,000P A

300,000( ) .15

2,000,000P B

100,000( ) .05

2,000,000P A B


Example: Television AdsExample: Television Ads

( ) .05( | ) .30

( ) .15

P A BP A B

P B

.3333 or 33%.3333 or 33%

5A-46

• So, So, PP(ad (ad AA) = .25) = .25 PP(ad (ad BB) = .15) = .15 PP((AA BB) = .05) = .05 PP((AA | | BB) = .3333 ) = .3333

• PP((AA | | BB) = .3333 ≠ ) = .3333 ≠ PP((AA) = .25) = .25

• PP((AA))PP((BB)=(.25)(.15)=.0375 ≠ )=(.25)(.15)=.0375 ≠ PP((AA BB)=.05 )=.05

• Are events Are events AA and and BB independent? independent?


Example: Television AdsExample: Television Ads

5A-47

• When When PP((AA) ≠ ) ≠ PP((AA | | BB), then events ), then events AA and and BB are are dependentdependent..

• For dependent events, knowing that event For dependent events, knowing that event BB has has occurred will affect the occurred will affect the probabilityprobability that event that event AA will will occur.occur.

• For example, knowing a person’s age would affect For example, knowing a person’s age would affect the the probabilityprobability that the individual uses text that the individual uses text messaging but causation would have to be proven messaging but causation would have to be proven in other ways.in other ways.


Dependent EventsDependent Events

• Statistical dependence does Statistical dependence does not not prove causality.prove causality.

5A-48

• An An actuaryactuary studies conditional probabilities studies conditional probabilities empirically, using empirically, using - accident statistics - accident statistics - mortality tables - mortality tables - insurance claims records- insurance claims records

• Many businesses rely on actuarial services, so a Many businesses rely on actuarial services, so a business student needs to understand the business student needs to understand the concepts of conditional probability and statistical concepts of conditional probability and statistical independence.independence.


Actuaries AgainActuaries Again

5A-49

• The probability of n independent events occurring simultaneously is:

• To illustrate system reliability, suppose a Web site has 2 independent file servers. Each server has 99% reliability. What is the total system reliability? Let,

P(A1 A2 ... An) = P(A1) P(A2) ... P(An)if the events are independent

F1 be the event that server 1 failsF2 be the event that server 2 fails


Multiplication Law for Independent EventsMultiplication Law for Independent Events

5A-50

• Applying the rule of independence:

• The probability that at least one server is “up” is: The probability that at least one server is “up” is:

PP((FF11 FF22 ) ) == PP((FF11) ) PP((FF22)) == (.01)(.01) = .0001(.01)(.01) = .0001

1 - .0001 = .9999 or 99.99%1 - .0001 = .9999 or 99.99%

• So, the probability that both servers are down So, the probability that both servers are down is .0001.is .0001.


Multiplication Law for Independent EventsMultiplication Law for Independent Events

5A-51

• RedundancyRedundancy can increase system reliability even can increase system reliability even when individual component reliability is low.when individual component reliability is low.

• NASA space shuttle has three independent flight NASA space shuttle has three independent flight computers (triple redundancy). computers (triple redundancy).

• Each has an unacceptable .03 chance of failure Each has an unacceptable .03 chance of failure (3 failures in 100 missions).(3 failures in 100 missions).

• Let Let FFjj = event that computer = event that computer jj fails. fails.


Example: Space ShuttleExample: Space Shuttle

5A-52

• What is the probability that all three flight What is the probability that all three flight computers will fail?computers will fail?PP(all 3 fail)(all 3 fail) = = PP((FF11 FF22 FF33))

= 0.000027 = 0.000027 or 27 in 1,000,000 missions.or 27 in 1,000,000 missions.

= = PP((FF11) ) PP((FF22) ) PP((FF33)) presuming presuming that failures that failures are are independentindependent

= (0.03)(0.03)(0.03)= (0.03)(0.03)(0.03)


Example: Space ShuttleExample: Space Shuttle

5A-53

• How high must reliability be?How high must reliability be?

• Public carrier-class telecommunications data links Public carrier-class telecommunications data links are expected to be available 99.999% of the time.are expected to be available 99.999% of the time.

• The The five nines rulefive nines rule implies only 5 minutes of implies only 5 minutes of downtime per year.downtime per year.

• This type of reliability is needed in many business This type of reliability is needed in many business situations.situations.


The Five Nines RuleThe Five Nines Rule

5A-54

• For example,For example,


The Five Nines RuleThe Five Nines Rule

5A-55

• Suppose a certain network Web server is up only 94% of the Suppose a certain network Web server is up only 94% of the time. What is the probability of it being down?time. What is the probability of it being down?

• How many independent servers are needed to ensure that How many independent servers are needed to ensure that the system is up at least 99.99% of the time (or down only the system is up at least 99.99% of the time (or down only 1 - .9999 = .0001 or .01% of the time)?1 - .9999 = .0001 or .01% of the time)?

PP(down) = 1 – (down) = 1 – PP(up) = 1 – .94 = .06(up) = 1 – .94 = .06


How Much Redundancy is Needed?How Much Redundancy is Needed?

5A-56

• So, to achieve a 99.99% up time, 4 redundant So, to achieve a 99.99% up time, 4 redundant servers will be needed.servers will be needed.

2 servers: 2 servers: PP((FF11 FF22) = (0.06)(0.06) = 0.0036) = (0.06)(0.06) = 0.0036

3 servers: 3 servers: PP((FF11 FF22 FF33) )

= (0.06)(0.06)(0.06) = 0.000216 = (0.06)(0.06)(0.06) = 0.0002164 servers: 4 servers: PP((FF11 FF22 FF33 FF44) )

= (0.06)(0.06)(0.06)(0.06) = (0.06)(0.06)(0.06)(0.06) =0.00001296 =0.00001296


How Much Redundancy is Needed?How Much Redundancy is Needed?

Download - 5A-1. Probability (Part 1) Random Experiments Random Experiments Probability Rules of Probability Rules of Probability Independent Events Independent

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