PROBABILITYChapter 6CONCEPT OF PROBABILITYIn the study of probability, we shall consider activitiesforwhichtheoutcomecannotbe predicted with certainty.EXPERIMENT- Study of probability which shall be consideractivitiesforwhichtheoutcome cannot be predicted with certainty. - Set of all possible outcomes - Individual element or outcome in a sample space.Example 1Consider the activity of rolling a die. Anyofthenumber1to6isasamplepointofS.we cansaythattherearesixsamplepoints.Ifwelet A betheeventofgettinganoddnumberandBan event of getting a perfect suare! then"hisactivityhassixpossibleoutcomes!thatis 1!#!$!%!&! and 6. thus!S ' (1!#!$!%!&!6)A' *1!$!&+andB' *1!%+,otethattheelementsofAareelementsofthe samplespaceS. "henumberofsamplepointsina samplespaceS!aneventsAandBareusually written as n*S+! n*A+! and n*B+ respectively. "hus!n*S+ ' 6!n*A+ ' $!n*B+ ' #-xample #If a pair of dice is rolled! then determine thenumber of sample points of the following.a. Sample spaceb. -vent of getting a sum of &c. -vent of getting a sum of at most %Solution to -xample #a. /etn1bethenumberofpossibleoutcomesfor the first die/et n# be the number of possible outcomes for the second dien*S+ ' n1 0 n# ' 6 0 6' $6Solution to -xample #b./etAbetheeventofgettingasumof&."he number of sample points in A can be listed. "hus!A ' *1!%+! *%!1+! *#!$+!*$!#+ n*A+ ' % Solution to -xample #c. /et B be the event of getting a sum of at most %. "hen! B ' *1!1+! *1!#+! *1!$+!*#!1+! *#!#+! *$!1+ n*B+ ' 6 -xample $A box contains 6 red and % green balls. If three balls are drawn from the box! then determine the number of sample points of the following.a. "he sample spaceb. "he event of getting all red ballsc. "he event of getting 1 red and # green balls Solution to -xample $/et S be the event of drawing $ balls from the box/et A be the event of getting all red balls/et B be the event of getting 1 red and # green ballsa. n*S+ ' 11C$ ' 112 ' 1#1

*11 3 $+2 $2 b. n*A+ ' 6C$ 0 %C1 '620%2' #1*6 3 $+2$2 *% 3 1 +212 c. n*B+ ' 6C1 0 %C# '620%2' $6*6 3 1+212*% 3 #+2#2 PROBABILITYIsthechancethataneventwillhappen.The probabilityofanevent AdenotedbyP(A)refers toanumberbetweenand!includin"the valuesand!.Thiscanbee#pressedasa fraction,asadecimalorasapercent.$henwe assi"n a probability ofto event A, it means that it is impossible for event A to occur. $hen event Ais assi"ned a probability of !, then we say that event A will really occur.Theprobabilityofoccurrenceplusthe probabilityofnon%occurrenceofaneventis alwayse&ualto!.Therefore,ina"iven observationsore#periment,aneventmustoccur ornot.IfweletA'betheeventthatAwillnot occur, then we shall say that 4*A+ 5 4*A6+ ' 1 -7, 18rom the above formula! we can have4*A+ ' 1 3 4*A6+-7, #9r4*A6+ ' 1 3 4*A+-7, $ -xample %Astudentinastatisticsclasswasableto compute the probability of passin" the sub(ect to be e&ual to .)*. +ased on this information, what is the probability that he is not "oin" to pass the sub(ect,-et A be the event of passin" the sub(ect-et A' be the event of not passin" the sub(ect.sin" /01 2 we have4*A6+ ' 1 3 4*A+ ' 1 3 1.%64*A6+' 1.&% THREEAPPROACHES OF PROBABILTY1. Subject!e app"#ac$-Theprobabilityofaneventis determinedbasedonanindividual'se#perience orperception.Thisapproachdoesnotre&uire e#tensivedatatosupportone's(ud"ementbut simplye#pressesthestren"thofone'sbelief with re"ard to uncertainties involved. THREEAPPROACHES OF PROBABILTY%.P"#bablt't$eRelat!e F"e(ue)c&-thesecondapproachofinterpretin"the probabilityofaneventisthrou"hthe determinationoftherelativefre&uencyof occurrence. /#ample 34ecordsshowthat!outof!3students who entered in a certain colle"e leave the school due to financial problem. $hat can we say about theprobabilitythatafreshmenenterin"this colle"ewillleavetheschoolduetofinancial reason,/et A be the event that a freshmen will leave the colle"e due to financial reason. Then based on the past records, the relative fre&uency shall be4*A+ ' 111:1!&11' 1.166; or 6.6;< /#ample *4ecords show that in a certain university, 23 outof!,53"raduateswhotoo6the7PA e#amination were able to pass. $hat can we say aboutthepossibleperformanceofthefuture "raduatesofthisuniversitywhowillta6ethe 7PA e#amination,-etAbetheeventthata"raduatesofthis universitywhowillta6ethee#aminationwill pass.Then,byusin"therelativefre&uency occurrence, we have4*A+ ' $&1:1!;&1' 1.# or #1< THREEAPPROACHES OF PROBABILTY *. Cla++cal P"#bablt&-Inthisapproach,ane#perimentshallbe performed.Thepossibleoutcomecanbe predictedevenbeforetheperformanceofthe e#periment.8neoftheassumptionsinthe classicalprobabilityisthattheprobabilityof each sample point must be e&ual. The computin" formula for the classical probability of an event A is "iven byP(A) 9 n(A) /01 )n(S)where: n(A) represents the number of samplepointsin event An(S) represents the number of sample points in sample space S /#ample 5Ifacoinistossed,whatistheprobabilityof "ettin" a head,S ' (head! tail)=ence! we may say that n*S+ ' #If we let A be the event of getting head!then A ' (head) and n*A+ ' 1=ence! by applying the classical probability! we have4*A+ ' n*A+:n*S+ ' 1:# /#ample ;If two coins are tossed, what is the probability of "ettin" both heads, S ' (==! ="! "=! "")=ence! we may say that n*S+ ' %If we let A be the event of getting two heads! then A ' (==) and n*A+ ' 1=ence! by applying the classical probability! we have4*A+ ' n*A+:n*S+ ' 1:% /#ample +. The probability of A . + denoted by P(A.+) is "iven by.4*ACB+ ' 4*A+ 5 4*B+ 3 4*ADB+-7, &Using classical probability, EQN 5 shall becomeP(AUB) = n(A) n(B) !n(A"B) EQN 6n*S+ n*S+n*S+ A,,ITION R-LETwoeventsAand+aresaidtobemutually e#clusive if they can not occur both at the same time. This implies that the occurrence of event A e#cludestheoccurrenceofevent+andvice versa.?ence,n(A>+)98.Thisfollowsthat P(A>+) 9 8 then /01 3 will be reduced to4*ACB+ ' 4*A+ 5 4*B+ -7, ;/#ample !27onsider the activity of rollin" a die and let A, +, and7betheeventsof"ettin"anoddnumber, anevennumber,andaperfects&uare respectively. @etermine the probability of "ettin" a.) an odd or an even number. b.) an even number and a perfect s&uareSolution to /#ample !2"he sample space is S ' (1!#!$!.........6)! thus n*S+ ' 6"heelementsofthethreeeventsaregiven below.A' (1!$!&)! n*A+ ' $B' (#!%!6)n*B+ ' $C' (1!%)n*C+ ' #Solution to /#ample !2a.)1oticethateventsAand+donothavea commonsamplepoint.?ence,theseevents cannotoccurbothatthesametime.So,wecan saythatthetwoeventsaremutuallye#clusive events. .sin" /01 5, we have4*ACB+ ' 4*A+ 5 4*B+' n*A+:n*S+ 5 n*B+:n*S+' $:6 5 $:6' 6:6' 1Solution to /#ample !2b.) events + and 7 have a common sample point. Thisimpliesthatthetwoeventscanoccurboth atthesametime.?ence,wemaysaythattwo eventsarenon%mutuallye#clusiveevents. .sin" /01 *, we have4*BCC+ ' 4*B+ 5 4*C+ 3 4*BDC+' $:6 5 $:6 3 1:6' %:6' #:$/#ample !)Acardisdrawnfromanordinarydec6of3= playin" cards. Aind the probability of "ettin"a. an ace or a 6in"b. a 6in" or a face cardc. a blac6 or a &ueenSolution to /#ample !)S ' a. /et A be the event of getting an ace/et B be the event of getting a Eing4*ACB+ ' 4*A+ 5 4*B+' n*A+:n*S+ 5 n*B+:n*S+' %: 5 %:' B:' #:1$Solution to /#ample !)S ' b. /et C be the event of getting a face card/et B be the event of getting a Eing4*BCC+ ' 4*B+ 5 4*C+ 3 4*BDC+' n*B+:n*S+ 5 n*C+:n*S+ 3 n*BDC+:n*S+' %: 5 1#: 3 %:' 1#:' $:1$Solution to /#ample !)S ' c. /et A be the event of getting blacE card/et - be the event of getting a ueen4*AC-+ ' 4*A+ 5 4*-+ 3 4*AD-+' n*A+:n*S+ 5 n*-+:n*S+ 3 n*AD-+:n*S+' #6: 5 %: 3 #:' #B:' ;:1$CON,ITIONAL PROBABILITY7onditionalprobabilityistheprobabilitythat asecondeventwilloccurifthefirstevent alreadyhappened.Symbolically,conditional probabilityiswrittenasP(Al+)andisreadas the probability of event A "iven that event + has occurred.Thecomputin"formulaforthe conditional probability of A "iven + is "iven by4*A l B+ '4*ADB+:4*B+!-7, Bprovided 4*B+ is not eual to 1CON,ITIONAL PROBABILITYInconditionalprobability,theoccurrenceof event+specifiesthenewsamplespacefor which we want to calculate the probability of that part of A that is contained in +.By using classical probability!4*ADB+' n*ADB+:n*S+and4*B+ ' n*B+:n*S+"hus! -7, B shall become4*A l B+ ' n*ADB+:n*S+n*B+:n*S+4*A l B+ 'n*ADB+:n*B+-7, F/#ample !3/et 4*A+ ' 1.%!4*B+ ' 1.&!4*ADB+ ' 1.$8ind the value of the following.a.+ 4*A l B+b.+ 4*B l A+a.+ 4*A l B+ ' 4*ADB+:4*B+ '1.$:1.& ' 1.6 b.+ 4*A l B+ ' 4*ADB+:4*A+ ' 1.$:1.% ' 1.;&/#ample !*Adieisrolled.Iftheresultisanevennumber, what is the probability that is a perfect s&uare,Solution to /#ample !*/et A be the event of getting a perfect suare./et B be the event of getting an even number./et ADB be the event of getting a perfect suare and an even number"hus! 4*A l B+ ' n*ADB+:n*B+ ' 1:$ /#ample !5SupposethatP(A)9.5,P(+)9.2,and P(AB+)9.=.Civenthattheoutcomeofthe e#perimentbelon"sto+,whatisthenthe probability of A, Solution to /#ample !5See dia"ram below. 1ote that we are definin" anewsamplespacefromSto+.8fthe probabilityP(+)e&uals.2,thevalue.= corresponds to P(AB+). ?ence, 1.11.#*A+*B+ *AGB+4*A l B+ ' 4*ADB+:4*B+ ' 1.#:1.$' 1.; /#ample !;A card is drawn from a dec6 of 3= playin" cards. Civenthatthecarddrawnisafacecard,then what is the probability of "ettin"a. a 6in",b. a spade,c. a red card,Solution to /#ample !;/et A be the event of getting a face card/et B be the event of getting a Eing/et C be the event of getting a spade/et A be the event of getting a red carda. 4*B l A+ ' n*BDA+:n*A+ ' %:1# ' 1:$ Solution to /#ample !;b. 4*C l A+ ' n*CDA+:n*A+' $:1#' Hc. 4*A l A+ ' n*ADA+:n*A+' 6:1#' 1:#/#ample !