Probablty s commo sese reduced to calculato. Laplace Our ma objectve ths lesso s to develop the art of descrbg ucertaty terms of probablstc models, as well as the skll of probablstc reasog (ferece). The models we cosder assg probabltes to collectos (sets) of possble outcomes. For ths reaso, we beg wth a short revew of set theory.
SET THEORY A set s collecto of objects called elemets. C A x : x A Complemet A B x : x A or x B Uo A B AB x : x A ad x B Itersecto A B AB A C C A A B B A ad A B B A A B C A B C A B C A B C A B C A B A C A B C A B AC C C C C A B A B C C C A B A B S A A Mutually Exclusve or Dsjot Double Complemet Commutatve Propertes Assocatve Propertes Dstrbutve Propertes De Morga's Law
SET THEORY Cosder uversal set S ad ts subsets A1, A2,, A. f A A, k j for all k ad j ad k k1 A j k S the t s sad that A1, A2,, A are mutually exclusve ad collectvely exhaustve. S A B S A B S A B C C C Mutually Exclusve Collectvely Exhaustve A1, A2,, A also form a partto of the uversal set S. Mutually Exclusve ad Collectvely Exhaustve
Fudametals of Probablty: Probablstc Models Probablty Law Radom Expermet Evet A Evet B P A P B Sample Sapce or S (Set of Outcomes) Radom Expermet: Expermet whose result s ucerta before t s performed. Tral: Sgle performace of a radom expermet. Outcome: Result of a tral. Sample Space S: The set of all possble outcomes of a radom expermet. Evet: Subset of the sample space S (to whch a probablty ca be assged). A collecto of possble outcomes. Sure Evet: Sample space S (a evet for sure to occur) Impossble Evet: Empty set (a evet mpossble to occur). A B Evets
Fudametals of Probablty: Probablstc Models Example: Dscrete Sample Space Sgle roll of a de. Dscrete Sample Space: S 1,2,3,4,5,6 Some Evets: A a odd umber shows up 1,3,5 A A A A A A A 1 2 3 4 5 6 7 8 a umber less tha 3 shows up 1,2 a eve umber shows up 2,4,6 2 shows up 2, (sgle outcome) 2 or 4 show up 2,4 2 ad 4 show up a umber greater tha 6 shows up, (o outcome) a umber from 1 to 6 shows up 1,2,3,4,5,6 S, (all outcomes)
Fudametals of Probablty: Probablstc Models Example: Cotuous Sample Space Rado Recever 5V 5V st Output of the rado recever s measured at tme t t. 1 S s : 5 s 5 Cotuous sample space has fte umber of outcomes. 1 2 3 A s: 2.5 s 2.5 A s: 1 s1 A s: s 2.3
Fudametals of Probablty: Probablstc Models Example: Cotuous to Dscrete Sample Space Rado Recever 5V 5V st 8-bt ADC s There wll be oly 8 2 256 outcomes at the output of the 8-Bt ADC. S 5, 4,9609375,, 0,0390625,0,0,0390625,, 4,9609375 or S= 128, 127,, 1,0,1,,127 Dscrete sample space ca have ftely or ftely may elemets
Fudametals of Probablty: Probablstc Models Example: 2D Sample Space Rollg a de two tmes. S, j : 1,1, 1,2, 1,3,, 4,3,, 6,4, 6,5, 6,6 6 5 4 3 2 1 1 2 3 4 5 6
Secod Roll Fudametals of Probablty: Probablstc Models Tree Based Sequetal Descrpto: Sample Space Par of Rolls Sequetal Tree Descrpto 6 5 4 3 Root 1 2 1,1 1,2 1,3 1,4 1,5 1,6 Leaves 2 6 1 1 2 3 4 5 6 Frst Roll
Fudametals of Probablty: Probablstc Models Probablty Laws: Itutvely, probablty law specfes the lkelhood of ay outcome, or of ay set of possble outcomes (a evet). The probablty law assgs to every evet A, a umber P A, called probablty of A, satsfyg the followg axoms. Axoms of Probablty: 1) Noegatvty: P A 0 for every evet A. 2) Addtvty: If A ad B are two dsjot evets, the the probablty of ther uo satsfes P A B P A PB Furthermore, f A1, A 2, s a sequece of dsjot evets, the the probablty of uo satsfes, P A A P A P A 1 2 1 2 3) Normalzato: The probablty of the etre sample space S s equal to 1. PS 1
Fudametals of Probablty: Probablstc Models Example: Sgle Co Toss. H, T, possble evets: S H, T, H, T, If the co s far,.e. we beleve that heads ad tals are equally lkely, the we assg equal probabltes to the two possble outcomes. P H P T 0.5 Addtvty axom mples that P H, T P H P T 1 whch s cosstet wth the ormalzato axom. Noegatvty axom s also satsfes P "Ay Evet" 0
Fudametals of Probablty: Probablstc Models Dscrete Probablty Law: If the sample space cossts of a fte umber of possble outcomes, the the probablty law s specfed by the probabltes of evets that cossts of a sgle elemet. I partcular the probablty of ay evet s1,s 2,, s s the sum of the probabltes of ts elemets P s1,s 2,, s P s1 P s2 P s Uform Probablty Law: If the sample space cossts of possble outcomes whch are equally lkely (.e. all sgle-elemet evets have the same probablty), the the probablty of ay evet A s gve by Number of elemets of A P A
Secod Roll Fudametals of Probablty: Probablstc Models Example: Cosder the expermet of rollg a par of far dce. Far dce assumpto meas that each of the 36 possble outcomes has the same probablty 1/36. S, j : 1,1, 1,2, 1,3,, 4,3,, 6,4, 6,5, 6,6 6 5 4 3 2 1 Sample Space Par of Rolls 1 2 3 4 5 6 Frst Roll P P P P the sum of the rolls s eve the sum of the rolls s odd 18 1 36 2 18 1 36 2 the frst roll s larger tha the secod roll at least oe roll s equal to 4 11 36 15 36
Fudametals of Probablty: Probablstc Models Example: A wheel of fortue s cotuously calbrated from 0 to 1, so the possble outcomes of a expermet cosstg of a sgle sp are the umbers the terval 0,1. Assumg a far wheel, t s approprate to cosder all outcomes equally lkely, but what s the probablty of the evet cosstg of a sgle elemet? Aswer: It caot be postve, because the, usg the addtvty axom, t would follow that evets wth a suffcetly large umber of elemets would have probablty larger tha 1. Therefore, the probablty of ay evet that cossts of a sgle elemet must be 0. What s the probablty of ay subterval ab?, t makes sese to assg probablty b a to ay subterval abof, 0,1, ad to calculate the probablty of a more complcated set by evaluatg ts legth. Ths assgmet satsfes the three probablty axoms ad qualfes as a legtmate probablty law. P Legth of Iterested Iterval b a a, b Cotuous Uform Law Legth of Sample Space 1
Fudametals of Probablty: Probablstc Models Example: Romeo ad Julet have a date at a gve tme, ad each wll arrve at the meetg place wth a delay betwee 0 ad 1 hour, wth all pars of delays beg equally lkely. The frst to arrve wll wat for 15 mutes ad wll leave f the other has ot yet arrved. What s the probablty that they wll meet? 0,1 0,1 S whose elemets are possble pars of delays. y 1 14 M 14 1 x 7 M x, y x y 1 4,0 x 1,0 y 1, PM 16
Fudametals of Probablty: Probablstc Models Propertes of Probablty Laws: 1) If A B, the P A P B 2) P A B P A P B P A B 3) P A B P A P B C C C 4) P A B C P A P A B P A B C Proof of these propertes ca be obtaed aalytcally or by usg Ve dagrams. Proof s also gve the text book.
Codtoal Probablty Codtoal probablty provdes us wth a way to reaso about outcome of a expermet, based o partal formato. o I a expermet volvg two successve rolls of a de, you are told that the sum of the two roll s 9. How lkely s that the frst roll was 6? o I a word guessg game, the frst letter of the word s a t. What s the lkelhood that the secod letter s a h? o How lkely s t that a perso has a dsease gve that a medcal test was egatve? o A spot shows up o a radar scree. How lkely s t correspodg to a arcraft? Defto: The probablty of a evet A uder the codto that evet B has occurred s called codtoal probablty of A gve B. def P A B P A B, PB 0 P B
Codtoal Probablty Iterpretato of codtoal probablty: Assume that A s a evet defed uder the sample space S. We kow that A S A ad PS 1, the P A S P A P A S P A P S 1 Therefore, ths result dcates that codtoal probablty of A gve B s smply the probablty of A assumg the etre sample space s B, whch s reasoable sce the evet B has occurred (whch meas that occurrece of ay outcome outsde of B s mpossble). S A B A B
Codtoal Probablty Example: We toss a far co three successve tmes. The evets A ad B are defed as follow: A= {more heads tha tals come up}, B= {frst toss s a head} P A B. Fd the codtoal probablty S HHH,HHT,HTH,THH,TTH,THT,HTT,TTT A HHH, HHT, HTH, THH P A 4 8 1 2 B HHH, HHT, HTH, HTT P B 4 8 1 2 AB HHH, HHT, HTH P AB 3 8 P A B P A B 3 8 3 P B 4 8 4
Codtoal Probablty Example: Cosder rollg a de. Let A= {umber 1 shows up}, B= {a odd umber shows up}, C= {umber 1 or 2 shows up} S 1,2,3,4,5,6 P A B P A B P A 1 6 1 P B P B 3 6 3 P C B A P C B P A 1 6 1 P B P B 3 6 3 P B C P B C 1 6 1 PC 2 6 2
Codtoal Probablty Example: Let t be the age of a perso whe he/she des. The probablty that he/she des at a age ot older tha t0 0 0 P t t t dt t 0 s gve by where t s a fucto determed from mortalty records ad gve by t 2 0, otherwse 9 2 3 10 t 100 t, 0 t 100 years
Codtoal Probablty Example: (Cotug) a) What s the probablty that a perso wll de betwee the ages 60 ad 70? P 60 t 70 # of people who de betwee 60 ad 70 total populato 70 60 70 t dt t dt t dt 0.154 0 0 60 b) What s the probablty that a perso wll de betwee the ages 60 ad 70 assumg that hs/her curret age s 60? # of people who de betwee 60 ad 70 P60 t 70 t 60 total populatoof age older tha 60 60 t 70 70 P60 t 70t 60 tdt 60 0.486 100 Pt 60 t dt 60
Codtoal Probablty Example: (Cotug) c) What s the probablty that a perso wll de betwee the ages 20 ad 50 assumg that hs/her curret age s 60? P 20 t 50 t 60 # of people who de betwee 20 ad 50 total populatoof age older tha 60 20 50 60 Pt P t t P 60 P t 60 0
Codtoal Probablty Example: Radar Detecto If a arcraft s preset a certa area, a radar detects t ad geerate ad alarm sgal wth probablty 0.99. If a arcraft s ot preset, the radar geerates a (false) alarm, wth probablty 0.10. We assume that a arcraft s preset wth probablty 0.05. What s the probablty of o arcraft presece ad a false alarm? What s the probablty of arcraft presece ad o detecto? Let A ad B be two evets defed as A A C a arcraft s preset, Bthe radar geerates a alarm C a arcraft s ot preset, B the radar does ot geerate a alarm C C C C C P Not Preset, False Alarm P A B P A P B A 0.95 0.10 0.095 P Preset, No Detecto P A B P A P B A 0.05 0.01 0.005
Codtoal Probablty Example: Radar Detecto Arcraft Preset P A 0.05 C 0.95 P A Arcraft ot Preset PB A0.99 C P B A 0.01 P B C A 0.10 C C P B A 0.90 Mssed Detecto False Alarm
Codtoal Probablty Multplcato Rule 1 1 1 2 1 3 1 2 1 P A P A P A A P A A A P A A Verfcato of the multplcato rule: Multplcato rule ca be verfed by usg the defto of the codtoal probablty as follow: 1 P A P A P A2 A1 P A1 A2 A3 P 1 1 1 1 2 1 1 P A P A A P A A 1 1 P A P A P A A 2 1 P A 1 P A A A 1 2 3 P A A 2 1 P P 1 1 1 A A
Codtoal Probablty Example: Three cards are draw from a ordary 52-card deck wthout replacemet (draw cards are ot placed back the deck). We wsh to fd the probablty that oe of the three cards s a heart. We assume that at each step, each oe of the remag cards s equally lkely to be pcked. By symmetry, ths mples that every trplet of cards s equally lkely to be draw. Not a Heart 39 52 Not a Heart 38 51 13 51 Not a Heart 37 50 Heart 13 50 Heart the th card s ot a heart, 1,2,3 P A1 A2 A3 A We wat to calculate P A A A P A P A A P A A A 1 2 3 1 2 1 3 1 2 39 38 37 P A1, P A2 A1, P A3 A1 A2 52 51 50 39 38 37 P A1 A2 A3 P A1 P A2 A1 P A3 A1 A2 52 51 50 Heart 13 52
Total Probablty Theorem ad Bayes Rule Total Probablty Theorem: Let A1, A2,, A be dsjot evets that forms a partto of the sample space, ad assume P A for all 1,,. The for ay evet B, we have that 0 1 2 P A PB A P A PB A P A PB A P B P A B P A B P A B 1 1 2 2 A 1 A AB 1 AB 2 AB A 2 B
Total Probablty Theorem ad Bayes Rule Example: You eter a chess touramet where your probablty of wg a game s 0.3 agast half the players (call them type 1), 0.4 agast a quarter of the players (call them type 2), ad 0.5 agast the remag quarter of the players (call them type 3). You play a game agast a radomly chose oppoet. What s the probablty of wg? Let A be the evet of playg wth a oppoet of type. We have P A 0.5, P A 0.25, P A 0.25 1 2 3 Let also B be the evet of wg. We have P B A 0.3, P B A 0.4, P B A 0.5 1 2 3 P B P A P B A P A P B A P A P B A 1 1 2 2 3 3 0.5 0.3 0.25 0.4 0.25 0.5 0.375
Total Probablty Theorem ad Bayes Rule Bayes Rule: Let A1, A2,, A be dsjot evets that forms a partto of the sample space, ad assume P A for all 1,,. The for ay evet B such that PB 0, we have that 0 P A P A B PB P A PB A P A1 PB A1 P A2 PB A2 P A PB A P A PB A P A P B A 1 P A P B A : a pror probablty of evet Probablty of evet A wthout kowg evet B has occurred. P A B : a posteror probablty of A Probablty of evet A kowg evet B has occurred. A
Total Probablty Theorem ad Bayes Rule Example: Radar Detecto Let us retur to radar detecto problem. Let A a arcraft s preset, B the radar geerates a alarm C a arcraft s ot preset, B the radar does ot geerate a alarm C A We are gve that C P A 0.05, P B A 0.99, P B A 0.10 Applyg Bayes rule wth A1 A ad A2 A P arcraft preset alarm P A B C PB P A PB A P A P B A C C P A P B A P A P B A 0.050.99 0.3426 0.050.99 0.950.1
Total Probablty Theorem ad Bayes Rule R from Example: A box s cotag 11 resstors. The probablty of selectg th resstor the box s gve the table below. A resstor s selected from the box ad measured ts value by usg a multmeter whch has a measuremet characterstcs gve Fg 1. R 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 P R 0.028 0.056 0.083 0.111 0.139 0.166 0.139 0.111 0.083 0.056 0.028 0.35 0.3 0.25 Fgure 1: Measuremet Characterstcs of the Multmeter 0.2237 0.2503 What s the probablty of the selected resstor beg 1300 gve that the measured value of the resstor s 1600? P(B R) 0.2 0.15 0.1 0.1312 0.1920 0.1023 What s the probablty of the selected resstor beg 1300 gve that the measured value of the resstor s 1300? 0.05 0.0456 0.0374 0.0071 0.0090 0.0013 0.0001 0 R-150 R-100 R-50 R R+50 R+100 R+150 R+200 R+250 R+300 R+350 B (Ohm) Fgure 1
Total Probablty Theorem ad Bayes Rule Example: (Cotug) R 8 :'Selected resstor s 1300 ', B :'Measured value of the resstor s 1600 ' By usg Bayes rule, PR8 PB R8 PR8 PB R8 PR8 B 11 PB P R P B R 8 11 P B P R P B R P R 1 8 PB R8 PB 1 0.028 0 0.056 0 0.083 0 0.111 0 0.139 0 0.166 0 0.139 0 0.111 0.0013 0.083 0.0374 0.056 0.1920 0.028 0.2237 0.0202641 0.0203 P R 0.111 0.0013 B 0.0071 0.0203
Total Probablty Theorem ad Bayes Rule Example: (Cotug) R :'Selected resstor s 1300 ', B :'Measured value of the resstor s 1300 ' 8 8 11 P B P R P B R P R 1 0.028 0 0.056 0 0.083 0 0.111 0 0.139 0.0013 0.166 0.0374 0.139 0.1920 0.111 0.2237 0.083 0.0456 0.056 0 0.028 0 0.0616926 0.0617 8 PB R8 PB P R 0.111 0.2237 B 0.4024 0.0617 Moreover, assume that the measured value of the resstor s 1600, ad let us fd the P R B, 1,,11. probabltes
7 PB PB PB 7 PB R7 PB 8 PB R8 PB 9 PB R9 PB 10 PB R10 PB 11 PB R11 PB PB PB PB P R1 P B R1 0.0280 P R2 P B R2 0.0560 PR1 B 0, PR2 B 0 0.0203 0.0203 P R3 P B R3 0.0830 P R4 P B R4 0.1110 PR3 B 0, PR4 B 0 0.0203 0.0203 P R5 P B R5 0.1390 P R6 P B R6 0.1660 PR5 B 0, PR6 B 0 0.0203 0.0203 P R P R 8 10 11 B P R 0.139 0 0 0.0203 P R 0.111 0.0013 B 0.0071 0.0203 11 P R 11 0.083 0.0374 PR9 B 0.1529 P R B 1 0.0203 P R P R P R 0.056 0.1920 B 0.5297 0.0203 B P R 0.028 0.2237 0.0203 0.3086 1 1
Idepedece of Evets If kowledge of occurrece of a evet B does ot alter the probablty of some other evet A, the t would be atural to say that evet A s depedet of evet B. I terms of probabltes ths stuato occurs whe P A B P A P A B, PB 0 PB We wll defe two evets A ad B to be depedet f P AB P APB Ths result mples that P A B P A ad PB A PB The evets A1, A2,, A are sad to be depedet f 1 1 P A P A
Idepedece of Evets Example: Cosder a expermet volvg two successve rolls of a 4-sded de whch all 16 possble outcomes are equally lkely ad have probablty 1/16. a) Are the evets A 1st roll s a 1, B sum of the two rolls s a 5 depedet? 1 P A B Pthe result of the two roll s 1, 4 16 # of elemets of A 4 P A total umber of possble outcomes 16 # of elemets of B 4 PB total umber of possble outcomes 16 P AB P APB ad the evet A ad B are depedet.
Idepedece of Evets Example: (Cotug) b) Are the evets maxmum of the two roll s 2, B mmum of the two rolls s a 2 A 1 P A B Pthe result of the two roll s 2,2 16 # of elemets of A 3 P A A 1,2, 2,1, 2,2 total umber of possble outcomes 16 # of elemets of B 5 P B B 2,2, 3,2, 2,3, 4,2, 2,4 total umber of possble outcomes 16 P AB P A P B ad the evet A ad B are ot depedet.
Idepedece of Evets Example: Two umbers x ad y are selected at radom betwee zero ad oe. Let the evets B, D ad F are defed as follows: B y 1 2, D x 1 2, F x 1 2 ad y 1 2 x 1 2 ad y 1 2 y y y 1 1 B 12 D 12 1 F F 1 x 12 1 x 12 1 x The fgure gve above shows that these three evets are parwse depedet. 1 1 PB D PB PD, PB F PB PF 4 4 1 PD F PD PF 4
Idepedece of Evets Example: (Cotug) However, the three evets are ot depedet sce B D F, so 1 PB D F P 0 PBPDPF 8
Idepedece of Evets Codtoal Idepedece: Gve a evet C, the evets A ad B are called codtoally depedet f P A B C P A C P B C If A ad B are codtoally depedet, the the defto of the codtoal probablty ad the multplcato rule yeld P A B C P A B C PC Assumg that PC PB C P A B C PC PB C P A B C P B C s ozero, P A B C P A C. I words, ths relato states that f C s kow to have occurred, the addtoal kowledge that B has also occurred does ot chage the probablty of A.
Idepedece of Evets Example: Cosder two depedet far co tosses, whch all four possble outcomes are equally lkely. Let H D H the two tosses have dfferet results 1st toss s a head, 2d toss s a head, 1 2 The evets H 1 ad H 2 are ucodtoally depedet. But So that 1 1 PH1 D, PH2 D, PH1 H2 D 0 2 2 P H H D P H D P H D 1 2 1 2 ad H 1, H 2 are ot codtoally depedet.
Idepedet Trals ad Bomal Probabltes Idepedet Trals: If a expermet volves a sequece of depedet but detcal stages, the each stage s called depedet tral. Beroull Trals: I partcular, f each stage of a depedet trals has oly two possble results, the the sequece of these depedet trals s called depedet Beroull trals. Example: Cosder a expermet that cossts of depedet tosses of a based co, whch probablty of heads s p ad probablty of tals s 1 p. I ths expermet, depedece meas that the evets A1, A2,, A are depedet where A th toss s a head Ths expermet ca be vsualzed for 3 by usg tree-based sequetal descrpto as follows:
Idepedet Trals ad Bomal Probabltes Example: (Cotug) p H p 1 p HH HT p 1 p p 1 p 3 HHH P HHH p 2 1 HHT P HHT p p 2 1 HTH P HTH p p 1 2 HTT P HTT p p 1 p T p 1 p TH TT p 1 p p 1 p 2 1 THH P THH p p 1 2 THT P THT p p 1 2 TTH P TTH p p 1 3 TTT P TTT p
Idepedet Trals ad Bomal Probabltes Bomal Probabltes: I ths example, ay partcular outcome (3-log sequece of heads ad tals) that volves k heads ad 3 k tals has probablty: k " heads a partcular 3 log sequece" 1 3 P k p p Ths formula ca be exteded to the case of geeral tosses. I ths case, the probablty of ay partcular log sequece that cotas k heads ad k tals s Cosder the probablty k " heads a partcular log sequece" 1 P k p p " heads a toss sequece" p k P k k k
Idepedet Trals ad Bomal Probabltes Bomal Probabltes: " heads a toss sequece" p k P k The probablty gve above cosders the all possble log sequeces that cotas k heads. Therefore, the probablty pk ca be calculated by summg the probabltes of all dstct toss sequeces that cotas k heads, sce the dstct toss sequeces are mutually exclusve. k k pk p 1 p Bomal Probablty Law k : umber of dstct toss sequeces cota k heads. k k k k k a b a b, a b 1 a b 1 k0 k k0 k Bomal Theorem Normalzato
Idepedet Trals ad Bomal Probabltes Bomal Probabltes: The umbers k are called choose k. The umbers k are kow as the bomal coeffcets, ad the probabltes kow as the bomal probabltes. pk are!, k k! k! k 0,1,2,, where for ay postve tegers we have! 1 23, 0! 1
Idepedet Trals ad Bomal Probabltes Example: A teret servce provder has stalled c modems to serve the eeds of a populato of customers. It s estmated that at a gve tme, each customer wll eed a coecto wth probablty p, depedetly of the others. What s the probablty that there are more customers eedg a coecto tha the umber of modems? It s asked that that the probablty of more tha c customers smultaeously eed a coecto. It s equal to where kc1 p k p k p p k k 1 k
Idepedet Trals ad Bomal Probabltes Example: A commucato system trasmts bary formato over a chael that 3 troduces radom bt errors wth probablty 10. The trasmtter sed each formato bt three tmes ad a decoder takes a majorty vote of the receved bts to decde o what the trasmtted bt was. Fd the probablty that the recever wll make a correct decso. The recever ca correct the sgle error, but t wll make the wrog decso f the chael troduces two or more errors. If we vew each trasmsso as a Beroull tral whch a success correspods to the troducto of a error, the the probablty of two or more errors three Beroull trals s 3 3 Pk 2 0.001 0.999 0.001 310 2 3 2 3 6
Coutg Coutg Prcple: The coutg prcple s based o a dvde-ad-coquer approach, whereby the coutg s broke dow to stages through the use of a tree. Cosder a expermet that cossts of two cosecutve stages: The possble results of the frst stage are a1, a2, a m The possble results of the secod stage are b1, b2, b The, the possble results of ths expermet would be all possble ordered pars a, b, 1,, m, j 1,, j Ad the umber of such ordered pars s m. leaves 1 Choces 2 Choces 3 Choces 4 Choces Stage 1 Stage 2 Stage 3 Stage 4
Coutg Coutg Prcple: Cosder a process that cossts of r stages. Suppose that: 1 possble results for the frst stage. o There are o For every possble result of the frst stage, there are stage. o More geerally, for all possble results of the frst 1 stages, there are results at the th stage. o The, the total umber of possble results of the r stage process s 1 2 r 2 possble results at the secod possble Example: A telephoe umber s a 7-dgt sequece, but the frst dgt has to be dfferet from 0 or 1. How may dstct telephoe umbers are there?
Coutg Example: (Cotug) Ths problem ca be vsualzed as a sequetal process where oe dgt s selected at a tme. Thus, there wll be 7 stages, ad choce of oe out of 10 elemets at each stage. There s a excepto for the frst dgt where there are oly 8 elemet s avalable. 6 810 10 8 10 Example: Cosder a -elemet set s1, s2,, s. How may subsets does t have (cludg tself ad the empty set)? The problem ca be terpreted as examato of oe elemet at a tme ad decde whether to clude t the set or ot. There wll be total of stages, ad a bary choce at each stage. 2 2 2 2 tmes
Coutg Permutato: Selectg k objects out of the objects wth payg atteto to order of the selecto s called a permutato. k Permutato: Assume that there s dstct object ad we wat to select k of them. How may dfferet ways ca be foud where the order of the selecto matters? The aswer s smple: o Ay of the objects ca be chose to be the frst oe. o After choosg frst oe, there are oly 1 possble choce for the secod oe. o Gve the choce of frst two, there oly remas 2 avalable objects for the thrd selecto, etc. o At the kth selecto, k 1 objects have already bee selected ad there would be k 1 objects avalable for the last selecto. By the Coutg Prcple, the umber of possble sequeces s called k permutatos s 1 k 1 k 21! 1 k 1 k 21 k!
Coutg Example: Let us cout the umber of words that cosst of four dstct letters. Ths problem ca be terpreted as the coutg the umber of 4 permutatos of the 26 letters Eglsh alphabet.! 26! 26 25 24 23 358,800 k! 22! 1 classcal musc CDs, 2 rock musc CDs, ad 3 coutry musc Example: You have CDs. I how may dfferet ways ca you arrage them so that the CDs of the same type are cotguous? The problem ca be broke dow two stages: 1) Selectg order of CD types 3! 2) There are! (or!, or 1 2! 3 ) permutatos of the classcal (or rock, or coutry, respectvely) CDs. The desred total umber s 3! 1! 2! 3!.
Coutg Combato: Selectg k objects out of the objects wthout payg atteto to order of the selecto s called a combato. 2 permutatos of the letters A, B, C, ad D are AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, DC However, combato of two out four of these letters are AB, AC, AD, BC, BD, CD The umber of possble combatos s gve by! k k! k!
Coutg Example: Fd the umber of subsets of a -elemet set by usg combato. Sce k s the umber of k-elemet subsets of the -elemet set, oe ca fd the total umber of subsets by summg k over k. k0 k 2 Parttos ad Multomal Coeffcets: We have dstct objects ad the equalty gve below s hold for sum tegers 1, 2,, r 1 2 r Whch meas that objects are dvded to r dsjot groups where umber of elemets of th group s. How may ways are there to dvde objects to r dsjot groups?
Coutg Parttos ad Multomal Coeffcets: Each group ca be form oe at a tme. o There are ways of formg frst group. 1 1 o There are ways of formg secod group, ad so o. 2 o Use the Coutg Prcple 1 1 2 1 2 r 1 1 1 3 r whch s equal to! 1! 1 r 1!! 1! 1! 2! 1 2! r! 1 r! 1! 2! r! 1, 2,, r Multomal Coeffcets
Coutg Example: A class cosstg of 4 graduate ad 12 udergraduate studets s radomly dvded to four groups of 4. What s the probablty that each group cludes a graduate studet? We frst determe the ature of the sample space. A typcal outcome s a partcular way of parttog the 16 studets to four groups of 4. 16 16! N S 4,4,4,4 4!4!4!4! o Number of ways of dstrbutg four graduate studets to the four group s 4! o Number of ways of dstrbutg remag 12 udergraduate studets to the four groups where each group wll have three udergrads. 12 12! 3,3,3,3 3!3!3!3! Total umber of elemets of the terested evet ca be obtaed by usg Coutg Prcple 12! N A 4! 3!3!3!3!
Coutg Example: (Cotug) The probablty of ths evet s P A N N A S 12! 4! 3!3!3!3! 16! 4!4!4!4!