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8/13/2019 Conditional Probability and Probability Trees Statistics

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4.5 General Probability Rules

Rules of Probability

Rule 1. 1)(0 AP for any event A

Rule 2. 1)( =SP

Rule 3. Complement rule: For any event A,

)(1)( APAP c =

Rule 4. Addition rule: If A and are dis!ointevents, t"en

AP( or )()() BPAPB +=

Rule #. $ultipli%ation rule: If A and areindependent events, t"en

AP( and )()() BPAPB =

&nion

'"e union of any %olle%tion of events is t"e

event t"at at least one of t"e %olle%tion o%%urs.

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Figure4.1# '"e addition rule for dis!oint events:P(A or or C)*P(A)+P()+P(C) "en events A,, and C are dis!oint.

Addition Rule for -is!oint vents

If events A, , and C are dis!oint in t"e senset"at no to "ave any out%omes in %ommon, t"en

(P one or more of )()()(),, CPBPAPCBA ++=

'"is rule e/tends to any number of dis!ointevents.

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Figure 4.10 '"e eneral addition rule:

P(A or )*P(A)+P()P(A and ) for any eventsA and .

eneral Addition Rule for &nions of 'o events

For any to events A and ,

AP( or

APBPAPB ()()() += and

)B

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Figure 4.1 5enn diaram and probabilities for/ample 4.36.

/ample 4.3 6 -ebora" and $att"e arean/iously aaitin ord on "et"er t"ey "avebeen made partners of t"eir la firm. -ebora"uesses t"at "er probability of ma7in partner is6.0 and t"at $att"e8s is 6.#.

P(at least one is promoted)*6.0+6.#6.3*6.9

P(neit"er is promoted)

* 1 P(at least one is promoted)

* 1 6.9 * 6.1.

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Conditional Probability and ProbabilityTrees

'"e ne notation )|( BAP is a %onditionalprobability. '"at is, it ives t"e probability of oneevent under t"e %ondition t"at e 7no anot"erevent. ;ou %an read t"e bar < as =iven t"einformation t"at.>

-efinition of Conditional Probability

?"en 0)( >AP , t"e %onditional probability of iven A is

)(

)()|(

AP

BandAPABP =

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/ample @et8s define to events:

A* t"e oman %"osen is youn, aes 1 to 29B * t"e oman %"osen is married'"e probability of %"oosin a youn oman is

217.0870,103

512,22)( ==AP .

'"e probability t"at e %"oose a oman "o is

bot" youn and married is075.0

870,103

842,7)( ==BandAP .

'"e %onditional probability t"at a oman ismarried "en e 7no s"e is under ae 36 is

348.0512,22

842,7

)(

)()|( ===

AP

BandAPABP .

$ultipli%ation Rule

'"e probability t"at bot" of to events A and "appen toet"er %an be found by

)|()()( ABPAPBandAP = .

/ample 4.3 4 lim is still at t"e po7er table. Att"e moment, "e ants very mu%" to dra todiamonds in a ro. As "e sits at t"e table

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loo7in at "is "and and at t"e upturned %ards ont"e table, lim sees 11 %ards. Bf t"ese, 4 arediamonds. '"e full de%7 %ontains 13 diamonds

amon its #2 %ards, so 9 of t"e 41 unseen %ardsare diamonds.'o find lim8s probability ofdrain to diamonds, first %al%ulate

(P first %ard diamond 419

) =

(P se%ond %ard diamond < first %ard diamond 408

) =

$ultipli%ation rule )|()()( ABPAPBandAP =

no says t"at

(P bot" %ards diamonds 044.0408

41

9) == .

lim ill need lu%7 to dra "is diamonds.

Probability Trees

$any probability and de%ision ma7in problems%an be %on%eptualied as "appenin in staes,and probability trees are a reat ay to e/presssu%" a pro%ess or problem.

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/ample. %onsider t"e problem of flippin a fair%oin ti%e. Bn t"e first flip or stae t"e out%ome%an eit"er be a "eads or a tails. '"is is

e/pressed in t"e tree diaram belo by movinfrom left to ri"t it" one bran%" oin up ordon to represent t"e to out%omes. Doti%e t"att"e probabilities of t"e to out%omes are rittenne/t to t"eir respe%tive bran%"es, bot" bein .#for t"is e/ample. '"e ne/t set of bran%"esrepresent t"e ne/t toss of t"e %oin. '"e ne/ttoss %an also be eit"er a "eads or a tails. '"esebran%"es e/tend from eit"er of t"e firstout%omes bran%"es be%ause t"e ne/t toss %ouldbe a "eads or a tails no matter "at t"eout%ome of t"e first toss as. Doti%e also t"att"e probabilties for t"e se%ond set of bran%"es is

also .#, t"is means t"at t"e probability of "eadsor tails on t"e se%ond toss does not depend on"at "appened on t"e first toss. '"is is ane/ample of t"e first and se%ond tosses beinindependent. Bne event does not %"ane t"eprobability of t"e ot"er event "appenin.

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'"e probability of any final out%ome of t"ee/periment li7e EE, or "eads on bot" tosses isfound by multiplyin t"e bran%" probabilities ittoo7 to rea%" t"e final tree out%ome on t"e ri"t.

Alays remember: multiply t"e bran%"

probabilities.'"e probability of any ot"er finalout%ome ill be found in t"e same fas"ion, li7e'E, tails folloed by "eads ill also be 12 times12 ivin 14 also.

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Fiure. 'ree diaram for /ample. '"eprobability P() is t"e sum of t"e probabilities oft"e to bran%"es mar7ed it" asteris7s (G).

/ample

'"ere are to dis!oint pat"s to (professionalplay). y t"e addition rule, P() is t"e sum oft"eir probabilities. '"e probability of rea%"in t"rou" %ollee (top "alf of t"e tree) is

)|()()( ABPAPAandBP =

00085.0017.005.0 == .'"e probability of rea%"in it"out %ollee(bottom "alf of t"e tree) is

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)|()()( ccc ABPAPAandBP =

00095.0001.095.0 == .

About 9 "i" s%"ool at"eletes out of 16,666 illplay professional sports.

Doti%e.

In t"e last tree t"e tree as dran it" t"e iven

information in t"e first set of bran%"es. In aproblem des%ription at%" for "at is t"e Hivent"atH information. '"ese items ill be t"e firstbran%"es of t"e tree, and t"en t"e ot"er eventsill be t"e ne/t set. If you follo t"is uidelineyou "ave a better %"an%e of %onstru%tin t"etree %orre%tly.

ayes8s Rule

If A and are any events "ose probabilitiesare not 6 or 1,

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)()|()()|(

)()|()|(

ccAPABPAPABP

APABPBAP

+

= .

Independent vents

'o events A and t"at bot" "ave positiveprobability are independent if

)()|( BPABP = .