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TnumberXXI TofXXqToutcomesXX5TinXX-TsampleXX÷TspaceXXmä@XlDXXƒ’3XXƒT8XXmäXX].XX375XXUäXXE37XX9.XXk5XXå% ŸP`(`0~€heads,~€3~tails)€~=~€{\#€~of€~favorable~€outcomes~€\in€~sample~€space}€over€Ð  Ð{total€~\number€~of~€outcomes~€\in€~sample€~space}€~=~€XX!PXXŸ(XXÿ0XXÑheadsXXñ,XX3XXStailsXXç)XX‰äQVXˆDXXÈ’#XXÈ ’ofXXð ’ favorableXXÂ’outcomesXX†’inXX~’sampleXXH’spaceXXŸTtotalXX¹ TnumberXXà TofXXëToutcomesXX¯TinXX§TsampleXXqTspaceXXçä žP`(`1~€head,€~2~tails)€~=~€{\#€~of€~favorable~€outcomes~€\in€~sample~€space}€over€Ð  Ð{total€~\number€~of~€outcomes~€\in€~sample€~space}€~=~€XX!PXXŸ(XXÿ1XX±headXXo,XXÿ2XXÑtailsXXe)XXäQVXDXXF’#XXF ’ofXXn ’ favorableXX@’outcomesXX’inXXü’sampleXXÆ’spaceXXTtotalXX7 TnumberXXA TofXXiToutcomesXX-TinXX%TsampleXXïTspaceXXeä ŸP`(`3~€heads,€~0~tails)€~=~€{\#€~of€~favorable~€outcomes~€\in€~sample~€space}€over€Ð  Ð{total€~\number€~of~€outcomes~€\in€~sample€~space}€~=~€ KÝ ƒi½Ö"ÝÔUSUS.,ÔÝ  ÝÔ_Ôà@vv*àßi€e=@A1 !b ` ‹@ÿEh œh œ%h wi߈Ð  ÐÌÌà@rr*àßi€eBCD1!b ` ‹@ÿEsq œq œsq œißXX!PXXŸ(XXÿ3XXÑheadsXXñ,XX0XXStailsXXç)XX‰äQVXˆDXXÈ’#XXÈ ’ofXXð ’ favorableXXÂ’outcomesXX†’inXX~’sampleXXH’spaceXXŸTtotalXX¹ TnumberXXà TofXXëToutcomesXX¯TinXX§TsampleXXqTspaceXXçä 1P`(B€~€given€~€A)€~=~€{P`(A€~\and~B)}€over€{P(A)} _P`(A)€~=~€{\#€~of€~\times~€event~€A€~€occurred}€over€Ð  Ð{\#€~of€~repetitions}XX!PXXŸ(XXéAXX})XXäy ?/b `€@ÿE°Dqgqg°Dqgg߈Ð DV  ÐÌÓ*°,X…X<À”X°Ü ,*Óà0  àòòExampleóóÐØê(#(# ÐÓ- °,X›X°,X…X<À”X-Óà0  àrandom€phenomenon:€à0h(#(#àa€Ô_ÔstyrofoamÔ_Ô€cup€is€tossed€and€its€landingÐ  Ðposition€is€notedÐ4Fh(#h(# Ðà0  àsample€space:€€{side,€top,€bottom}Ðbt(#(# ÐÌà0  àThe€outcomes€are€not€equally€likely€so€we€must€repeat€the€randomÐ ¾Ð Ðphenomenon€many€times€and€count€the€number€of€times€it€lands€on€itsÐ ìþ Ðtop.€€Suppose€in€50€repetitions,€it€lands€on€its€top€5€times,€side€44Ð , Ðtimes,€and€bottom€1€time.ÐHZ(#(# ÐÌà@¡¡*ìàßg€eE F/b `€@ÿE°¤!—g—g°¤!—g g߈Ð ¤!¶ ÐÌà0  àBased€on€50€tosses€of€a€Ô_ÔstyrofoamÔ_Ô€cup,€we€estimate€that€theÐ 8%J" Ðprobability€of€a€Ô_ÔstyrofoamÔ_Ô€cup€landing€on€its€top€is€.10.€€That€is,€weÐ f&x# Ðestimate€that€a€Ô_ÔstyrofoamÔ_Ô€cup€lands€on€its€top€10%€of€the€time,€in€theÐ ”'¦$ Ðlong€run.ÐÂ(Ô%(#(# ÐÌà0  àAs€the€random€phenomenon€is€repeated€again€and€again,€the€relativeÐ +0(  Ðfrequency€probability€of€an€event€tends€to€approach€the€trueÐ L,^)! Ðprobability.€€(This€is€called€the€Law€of€Large€Numbers.)Ðz-Œ*"(#(# ÐÐ ¨.º+# Їà@(ìà„€6€„ˆÐ î ÐÌà0  àòòExampleóóÐJ\(#(# ÐÌà0  àIn€a€certain€state,€37,052€boys€and€35,192€girls€were€born€last€year.€Ð ¦¸ ÐUse€the€information€to€estimate€the€probability€of€a€birth€resulting€inÐ Ôæ Ða€baby€boy.Ð (#(# ÐÌà@õõ*ìàßg€eG H/ b `€@ÿE°^ ïgïg°^ ïg g߈Ð ^ p  ÐÌà0  àWhich€probability€estimate€do€you€feel€more€comfortable?€€Ðò (#(# Ðà0  àP(Ô_ÔstyrofoamÔ_Ô€cup€lands€top)€=€10%€€or€€P(baby€boy)€=€51.3%?Ð 2 (#(# ÐÌÌòòINDEPENDENT€AND€DEPENDENT€EVENTSóóÐ ª¼ ÐÌIn€the€study€of€probability,€it€is€important€to€know€whether€the€outcome€ofÐ  Ðone€event€affects€the€outcome€of€another.Ð 4F ÐÌConsider€the€following€events.Ð ¢ ÐÌà  àà ` àà ¸ àR€=€rain€tomorrowÐ ìþ Ðà  àà ` àà ¸ àU€=€you€carry€an€umbrella€tomorrowÐ , Ðà  àà ` àà ¸ àH€=€coin€flipped€tomorrow€lands€on€headsÐ HZ ÐÌDoes€the€probability€of€R€affect€the€probability€of€U?€€€Ð ¤!¶ Ðà0  àYes.€€R€and€U€are€dependent€events.ÐÒ"ä(#(# ÐÌDoes€the€outcome€of€the€coin€flip€affect€whether€or€not€it€will€rainÐ .%@" Ðtomorrow?€€€Ð \&n# Ðà0  àNo.€€R€and€H€are€independent€events.Њ'œ$(#(# ÐÌTwo€events€are€ò òòòindependentó óóó€if€the€probability€of€one€remains€the€sameÐ æ)ø&  Ðregardless€of€how€the€other€turns€out.€€Events€that€are€not€independent€areÐ +&(! Ðò òòòdependentó óóó.Ð B,T)" ÐÐ  p-‚*# Ðà@(ìà„€7€„ˆÐ î ÐÌòòExampleóóÐ J\ ÐÌYou€roll€a€regular€red€die€and€a€regular€green€die.€€Consider€the€followingÐ ¦¸ Ðevents.Ð Ôæ ÐÌà  àà ` àà ¸ àà  àA€=€a€4€on€the€red€dieÐ 0 B Ðà  àà ` àà ¸ àà  àB€=€a€3€on€the€green€dieÐ ^ p  Ðà  àà ` àà ¸ àà  àC€=€a€sum€of€9€on€the€two€diceÐ Œ ž  ÐÌTell€whether€each€pair€of€events€is€independent€or€dependent.Ð èú  ÐÌ(a)à0  àA€and€BÐDV (#(# Ðà0  àindependent€€€ð!ð€€à0(#(#àThe€outcome€of€the€red€die€is€not€affected€by€theÐ r„ Ðoutcome€of€the€green€die.€€The€probability€of€BÐ  ² Ðoccurring€is€the€same€no€matter€how€A€turns€out€(andÐ Îà Ðvise€versa).Ðü(#(# ÐÌ(b)à0  àB€and€CÐXj(#(# Ðà  àdependent€€€€€ð!ðà0  àIf€the€green€die€is€a€3€(event€B€occurs),€then€theÐ †˜ Ðprobability€that€the€sum€of€the€two€dice€is€9€is€equalÐ ´Æ Ðto€1/6.€€This€probability€does€not€necessarily€stayÐ âô Ðthe€same€if€the€green€die€is€not€a€3€(i.e.,€event€B€doesÐ " Ðnot€occur).€€€For€example,€if€the€green€die€is€a€2,€thenÐ >P Ðevent€C€is€impossible.€€€Ðl ~(#(# ÐÌTwo€events€are€ò òòòindependentó óóó€if€knowing€whether€one€event€occurs€does€notÐ È"Ú Ðalter€the€probability€that€the€other€event€occurs.Ð  ö#! Ðà@(ìà„€8€„ˆÐ î ÐÌò òòòRuleóó:€€à@33 ìàIf€A€and€B€are€independent€events,€P(A€and€B)€=€P(A)€ðð€P(B).òòóóˆÐ J\ Ðó óÌòòExampleóóÐ ¦¸ ÐÌà0  à(a)à0` (#(#àA€drawer€contains€3€red€paper€clips,€4€green€paper€clips,€and€5Ð   Ðblue€paper€clips.€€One€paper€clip€is€taken€from€the€drawer€andÐ 0 B Ðthen€replaced.€€(Sampling€with€replacement.)€€Another€paper€clipÐ ^ p  Ðis€taken€from€the€drawer.€€What€is€the€probability€that€the€firstÐ Œ ž  Ðpaper€clip€is€red€and€the€second€paper€clip€is€blue?ÐºÌ ` (#` (# ÐÌòòà  àà0 ` àSolutionóó:à0` (#` (#àÐ( (#(# ÐÌà0  àà0` (#(#àBecause€the€first€paper€clip€is€replaced,€the€sample€space€of€12Ð r„ Ðpaper€clips€does€not€change€from€the€first€event€to€the€secondÐ  ² Ðevent.€€The€events€are€independent.ÐÎà` (#` (# ÐÌà0  àà0` (#(#àP(red€then€blue)€=€P(red)€ðð€P(blue)€=€3/12€ðð€5/12€òòò òóóð ð€ó ó.1042€=€10.42%Ð*<` (#` (# ÐÌà0  à(b)à0` (#(#àA€drawer€contains€3€red€paper€clips,€4€green€paper€clips,€and€5Ð †˜ Ðblue€paper€clips.€€One€paper€clip€is€taken€from€the€drawer€and€isÐ ´Æ ÐNOT€replaced.€€(Sampling€without€replacement.)€€Another€paperÐ âô Ðclip€is€taken€from€the€drawer.€€What€is€the€probability€that€theÐ " Ðfirst€paper€clip€is€red€and€the€second€paper€clip€is€blue?Ð>P` (#` (# ÐÌà0  àà0` (#(#àòòSolutionóó:К!¬` (#` (# ÐÌà0  àà0` (#(#àBecause€the€first€paper€clip€is€NOT€replaced,€the€sample€spaceÐ ö#! Ðof€the€second€event€depends€on€what€color€the€first€paper€clipÐ $%6" Ðwas.€€Note€that€the€sample€space€for€the€second€event€is€11Ð R&d# Ðpaper€clips.€€Ð€'’$` (#` (# ÐÌà0  àà0` (#(#àThen,€for€example,€P(second€clip€is€blue)€=€5/11€if€the€first€clipÐ Ü)î&! Ðwas€not€blue.€€P(second€clip€is€blue)€=€4/11€if€the€first€clip€wasÐ  +(" Ðblue.€€Ð8,J)#` (#` (# Ðà0  àà0` (#(#àÐ ` (#` (# Ðà0  àà0` (#(#àThe€two€events€are€dependent.Ð ”.¦+%` (#` (# Ðà@(ìà„€9€„ˆÐ î ÐÌThe€notation€P(B€given€A)€means€the€probability€of€B,€given€that€A€hasÐ J\ Ðoccurred.€Ð xŠ ÐÌò òòòMultiplication€Rule€for€Probabilitiesó óóóÐ Ôæ ÐÌWhen€two€events,€A€and€B,€are€dependent,€then€the€probability€of€bothÐ 0 B Ðoccurring€is:Ð ^ p  ÐÌò òà@G G ìàP(A€and€B)€=€P(A)€ðð€P(B€given€A)ˆÐ ºÌ  Ðó óÌActually,€the€multiplication€rule€applies€to€òòanyóó€two€events,€A€and€B.€€ForÐ (  Ðòòindependent€eventsóó,€P(B€given€A)€=€P(B),€since€the€probability€of€B€occurringÐ DV  Ðis€the€same€no€matter€how€A€turns€out.€€Therefore,€when€A€and€B€areÐ r„ Ðindependent,€the€multiplication€rule€agrees€with€the€independent„eventsÐ  ² Ðformula,€P(A€and€B)€=€P(A)€ð ð€P(B).Ð Îà ÐÌLetððs€apply€the€multiplication€rule€to€the€last€example.€€Ð *< ÐÌLet€A€=€first€paper€clip€is€redÐ †˜ ÐLet€B€=€second€paper€clip€is€blueÐ ´Æ ÐÌA€drawer€contains€3€red€paper€clips,€4€green€paper€clips,€and€5€blue€paperÐ " Ðclips.€€One€paper€clip€is€taken€from€the€drawer€and€is€NOT€replaced.€Ð >P Ð(Sampling€without€replacement.)€€Another€paper€clip€is€taken€from€theÐ l ~ Ðdrawer.€€€Ð š!¬ ÐÌWe€wish€to€find€P(A€and€B)€=€P(first€clip€is€red€and€second€clip€is€blue).Ð ö#! ÐÌP(A€and€B)€à0 ` à=€P(A)€ðð€P(B€given€A)€ÐR&d#` (#` (# ÐÒ9°Òà  àà ` à=€P(first€clip€is€red)€ðð€P(second€clip€is€blue€given€that€first€clip€is€red)Ð €'’$ Ðà  àà ` à=€3/12€ðð€5/11Ð ®(À%  Ðà  àà ` àð ð€.1136€=€11.36%Ð Ü)î&! ÐÌÌÐ  f-x*$ ÐÔ_ÔÔ_Ôà@šš)§à„€10€„ˆÐ î ÐÌòòCONDITIONAL€PROBABILITYóóÐ J\ ÐÌòòMotivating€Exampleóó:à0  àA€math€teacher€gave€her€class€two€tests.€€25%€of€theÐ ¦¸ Ðclass€passed€both€tests€and€42%€of€the€class€passed€theÐ Ôæ Ðfirst€test.€€What€percent€of€those€who€passed€the€firstÐ   Ðtest€also€passed€the€second€test?€€That€is,€what€is€theÐ 0 B Ðprobability€of€a€student€passing€the€second€test,€givenÐ ^ p  Ðthat€the€student€passed€the€first€test?ÐŒ ž Ÿ$Ÿ$ ÐÌà0  àThis€problem€describes€a€òòò òconditional€Ô_ÔprobablityÔ_Ôó óóó€since€it€asks€us€to€find€theÐ èú  Ðprobability€that€the€second€test€was€passed€given€that€the€first€test€wasÐ (  Ðpassed.ÐDV Ÿ$Ÿ$ ÐÌThe€formula€for€the€conditional€probability€of€an€event€can€be€derived€from€theÐ  ² Ðmultiplication€rule.Ð Îà ÐÌò òà@’’§àó óP(A€and€B)€=€P(A)€ðð€P(B€given€A)ˆÐ *< Ðßp€k;< ,(bvM÷ᢠ`€@ÿE° 9<¢ttTè qÌ2 pßâ ´Ÿ$´Ÿ$°Ÿ$°Ÿ$âà@© © E© àˆÌà@© © E© àò òó óˆÌÌÌÌÌÌÌÌâ °Ÿ$°Ÿ$´Ÿ$´Ÿ$âLetððs€solve€the€problem€at€the€top€of€the€page.Ð ö#! ÐÌP(pass€2òòndóó€test€|€pass€1òòstóó€test)€€€=€€€òòP(pass€1òòstóó€test€and€pass€2òòndóó€test)óó€Ð R&d# Ðà  àà ` àà ¸ àà  àà h à€€€€€€€€à0 À àà0ÀŸ$ÀŸ$àP(pass€1òòstóó€test)Ѐ'’$Ÿ$Ÿ$ Ðà  àà ` àà ¸ àÌà  àà ` àà ¸ àà  àà h à€€€=€€€.25/.42€ð ð€.5952€=€59.52%Ð  Ü)î&! Ðà@««)§à„€11€„ˆÐ î ÐÌòòExercisesóóÐ J\ ÐÌAt€Kennedy€Middle€School,€the€probability€that€a€student€takes€Technology€andÐ ¦¸ ÐSpanish€is€.087.€€The€probability€that€a€student€takes€Technology€is€.68.€€What€isÐ Ôæ Ðthe€probability€that€a€student€takes€Spanish€given€that€the€student€is€takingÐ   ÐTechnology?€€(Ans:€12.79€%)Ð 0 B Ðà  àà ` àà ¸ àà  àà h à€€Ð ^ p  Ðà  àà ` àà ¸ àà  àà h àà À àà  àà p àà È àà  àà x àÌÌÌÌÌÌÌÌÌÌAt€a€middle€school,€18%€of€all€students€play€football€and€basketball€and€32€%€ofÐ Xj Ðall€students€play€football.€€What€is€the€probability€that€a€student€playsÐ †˜ Ðbasketball€given€that€the€student€plays€football?€€(Ans:€56.25%)Ð ´Æ ÐÌÌ