AP Statistics Notes Unit Five: Randomness and Probability

Transcription

1 AP Statistics Ntes Unit Five: Randmness and Prbability Syllabus Objectives: 3.1 The student will interpret prbability, including the lng-term relative frequency distributin. 3.2 The student will discuss the Law f Large Numbers cncept. This unit intrduces yu t the cncept f chance behavir. Prbability is the branch f mathematics that describes the pattern f chance utcmes. When we prduce data by randm sampling r randmized cmparative experiments, prbability helps us answer the questin, What wuld happen if we did this many times? Prbability calculatins and an understanding f randm behavir are the basis fr inference, which will be studied in units 8 thrugh 12. The Idea f Prbability Randm Phenmenn individual utcmes are uncertain, but there is a regular distributin f utcmes in a large number f repetitins. Prbability prprtin f times an event ccurs in many repeated events f a randm phenmenn. Prbability can als be thught f as lng-term relative frequency. Independent events (trials) utcme f an event (trial) des nt influence the utcme f any ther event (trial). Prbability Mdels Chance experiment any activity r situatin in which there is uncertainty abut which f tw r mre pssible utcmes will result. Sample Space the cllectin f all pssible utcmes f a chance experiment. Sample spaces can be stated in different ways. List Ex: A sample space can be a list f all the pssible utcmes. Cnsider flipping a cin. There are tw utcmes. The sample space wuld be: S = {H, T}. Cnsider rlling ne die. The sample space can be represented as S= {1, 2, 3, 4, 5, 6}. List Ex: An experiment is t be perfrmed t study student preferences in the fd line in the cafeteria. Specifically, the staff wants t analyze the effect f the student s gender n the preferred fd line (burger, salad r main entrée). The sample space cnsists f the fllwing pssible utcmes: 1) A male chsing the burger line. 2) A female chsing the burger line. 3) A male chsing the salad line. 4) A female chsing the salad line. 5) A male chsing the main entrée line. 6) A female chsing the main entrée line. The sample culd be represented by using set ntatin and rdered pairs. S = {(male, burger), (female, burger), (male, salad), (female, salad), (male, main entrée), (female, main entrée)}. This can be simplified t: S = {MB, FB, MS, FS, ME, FE}. 1

2 Tree diagram anther way f illustrating a sample space using a picture. Burger Outcme (Male, Salad) Male Salad Main Entree Outcme (Female, Burger) Burger Female Salad Main Entree This tree has tw sets f branches crrespnding t the tw bits f infrmatin gathered. T identify any particular utcme f the sample space, yu traverse the tree by first selecting a bracket crrespnding t gender and then a branch crrespnding t the chice f fd line. Event any cllectin f utcmes frm the sample space f a chance experiment. Simple event an event cnsisting f exactly ne utcme. If we lk at the lunch line example, the event that the student selected is male is given by male = {MB, MS, ME}. The event that the preferred line is the burger line is given by burger = {MB, FB}. The event that the persn selected is a female that prefers the salad line is {FS}. This is an example f a simple event. Anther tree diagram example: In this experiment, first the persn will flip a cin and then rll a die. The tree illustrates the sample space. There are 12 different utcmes. This leads t the fllwing rule. Multiplicatin Principle: If yu can d ne task in n1 number f ways and a secnd task in n2 number f ways, then bth tasks can be dne in n1 n2number f ways. In this example, there are 2 pssible results fr the cin tss and 6 pssible results fr the die, giving us 2 6r 12 different utcmes. Venn Diagram an infrmal picture that is used t identify relatinships. The cllectin f all pssible utcmes f a chance experiment are represented as the interir f a rectangle. The rectangle represents the sample space and shaded area represents the event A. 2

3 Cmplement Let A and B dente tw events. The event nt A cnsists f all experimental utcmes that are nt in event A. Nt A is called the cmplement f A and is c dented by A. The shaded area represents nt A. Unin (r) - Frming new events: Let A and B dente tw events. The event A r B cnsists f all experimental utcmes that are in at least ne f the tw events that is in A r in B r in BOTH f these. A r B is called the unin f the tw events and is dented A B. The shaded area represents A B. Intersectin (and) Let A and B dente tw events. The event A and B cnsists f all experimental utcmes that are in bth f the events A and B. A and B is called the intersectin f the tw events and is dented A B. The shaded area represents A B. Disjint r Mutually Exclusive Tw events that have n cmmn utcmes. The tw events cannt happen simultaneusly. Belw, A and B are disjint events. Mre Venn Diagrams illustratins shwing mre than tw events. A B A B C A B C C A, B & C are Disjint A B A B C A B C C c A B C 3

4 Prbability number f times A ccurs Fr any event A, the prbability is PA= ( ). ttal number f trials Empirical Apprach: The prbability f any utcme f a randm phenmenn is the prprtin f times the utcme wuld ccur in a very lng series f repetitins. That is, prbability is lng-term relative frequency. Cnsider the chance experiment f rlling a fair die. We wuld like t investigate the prbability f getting a 1 fr the up face f the die. The die was rlled and after each rll the up face was recrded and then the prprtin f times a 1 turned up was calculated and pltted. Repeated Rlls f a Fair Die Prprtin f 1's / Relative Frequency # f Rlls This prcess culd be repeated ver and ver again and the results wuld be similar. Ntice that the prprtin f 1 s seems t stabilize and in the lng run gets clser t the theretical value f 1/6. In many real-life prcesses and chance experiments, the prbability f a certain utcme r event is unknwn, but nevertheless, this prbability can be estimated reasnably well frm bservatin. The justificatin is the Law f Large Numbers. Law f Large Numbers: As the number f repetitins f a chance experiment increases, the chance that the relative frequency f ccurrence fr an event will differ frm the true prbability f the event by mre than any very small number appraches zer. 4

5 Syllabus Objective: 3.7 The student will create prbability distributins by using simulatin techniques. Simulatin The imitatin f chance behavir, based n a mdel that accurately reflects the phenmenn under cnsideratin, is called a simulatin. Simulatin steps: 1. State the prblem r describe the randm phenmenn. 2. State the assumptins. 3. Assign digits t represent utcmes. 4. Simulate many repetitins (trials). 5. State yur cnclusins. Calculatrs and cmputers are particularly useful fr cnducting simulatins because they can perfrm many repetitins quickly. Since this cannt be tested n the AP exam, we will nt cncentrate n this technique. We will simulate using the randm digit table, as this can be tested n the exam. Of curse cins, dice, cards and ther bjects can be used fr simulatin, but again, these cannt be tested. Ex: A cuple plans t have children until they have a girl r until they have fur children, whichever cmes first. Use a randm digit table t simulate this phenmenn. Step 1: Lking fr a girl r fur children. Step 2: Assume each child has prbability f 0.5 f being a girl r by and the sexes f successive children are independent. Step 3: Assign digits. One digit simulates the sex f ne child. 0, 1, 2, 3, 4 = girl and 5, 6, 7, 8, 9 = by. This gives each event (by r girl) a 5/10 prbability. Step 4: T simulate ne repetitin, read digits frm the randm digit table until the cuple has either a girl r fur children. The number f digits needed t simulate ne repetitin can vary frm 1 t 4. We wuld want t simulate this many, many times. Belw is 14 simulatins using line 130 f the randm digit table BBG BG BG BG BBBG BG B BBG BBG BG G G BG BBBB In these 14 repetitins, a girl was brn 13 times. Step 5: The estimate f the prbability that this strategy will prduce a girl is therefre, 13 estimated prbability = = This is actually very clse t the actual prbability which 14 is

6 Syllabus Objectives: 3.3 The student will slve prblems using the additin rule, multiplicatin rule, and cnditinal prbability. 3.4 The student will determine if tw events are independent. Frmal Prbability The Basic Prbability Rules The prbability P(A) f an event A satisfies 0 PA ( ) 1. If S is the sample space fr an experiment, PS ( ) = 1. If tw events A and B are disjint meaning they can never ccur simultaneusly, then PA ( Br) = PA ( ) + PB ( ). PA ( B) = PA ( ) + PB ( ) This is the additin rule fr disjint events. This rule can be extended fr mre than tw events. If events A, B and C are disjint in the sense that n tw events have any utcmes in cmmn, then PA ( r Br C) = PA ( ) + PB ( ) + PC ( ). PA ( B C) = PA ( ) + PB ( ) + PC ( ) This rule extends t any number f disjint events. Fr any event A, ( ) ( c c PA+ PA) = 1s, PA ( ) = 1 PA ( ). Equally likely utcmes If a randm phenmenn has k utcmes that are all equally likely, then each individual utcme has prbability, 1 k. The prbability f any event A is cunt f utcmes in A cunt f utcmes in A PA ( ) = = cunt f utcmes in S k Ex: Rll a fair die. Let A be the event f rlling a ne. Let B be the event f rlling an dd number. S = {1, 2, 3, 4, 5, 6}. PA= ( ) and PB ( ) = + + =. Since btaining a 1, 3 r 5 (dd numbers) are disjint, use the additin rule t find the prbability. Ex: Cnsider the experiment cnsisting f randmly picking a card frm an rdinary deck f playing cards (52 card deck). Let A stand fr the event that the card chsen is a King. The sample space is given by S = {A, K,,2, A, K,, 2, A,, 2, A,, 2 } and cnsists f 52 equally likely utcmes. The event is given by A={K, K, K, K } and cnsists f 4 utcmes, s 4 1 P(A) = = = Ex: Cnsider the experiment cnsisting f rlling tw fair dice and bserving the sum f the up faces. Let E stand fr the event that the sum is 11. The sample space is given by S = {(1, 1), (1, 2), (1, 3), (6, 6)} and cnsists f 36 equally likely utcmes. The event E is given by E = {(5, 6), (6, 5)} and cnsists f 2 utcmes, s 2 1 PE ( ) = =

7 Ex: Cnsider the experiment cnsisting f rlling tw fair dice and bserving the sum f the up faces. Let E stand fr the event that the sum is 11 and Let F stand fr the event that the sum is 7. F={(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}. It has 6 6 utcmes s PF ( ) =. Since E and F are disjint events, PE ( F) = PE ( ) + PF ( ) = + = = Mre Cmplex Prbability Rules Multiplicatin Rule fr Independent Events Tw events are independent if knwing that ne ccurs des nt change the prbability that the ther ccurs. Example: Flipping a cin tw times. The cin des nt have memry and each time yu flip the cin, getting heads r tails is equally likely. Getting head n the first flip and head n the secnd flip are independent events. Example: If yu rll tw dice and btain a sum f 7, the result f that rll has n effect n the next rll, s the tw rlls are 4 independent. But if yu draw an ace frm a deck f cards P (ace) = and 52 withut replacing it draw a secnd card, the prbability that the secnd card 3 is als an ace is P (ace) =. These events are NOT independent. 51 PA ( an d B) = PAPB ( ) ( ) If A and B are independent, then PA ( B) = PAPB ( ) ( ) This rule ONLY applies t independent events; yu cannt use it if events are nt independent. Typically independence is assumed fr the purpses f calculating prbabilities if the samples are being drawn frm a large ppulatin. Disjint events cannt be independent. If A and B are disjint, then the fact that A ccurs tells us that B cannt ccur (see the Venn Diagram n p. 3). Unlike disjintness f cmplements, independence cannt be pictured by a Venn diagram, because it invlves the prbability f the events, rather than just the utcmes that make up the events. Ex: We ve determined that the prbability that we encunter a green light at the crner f Cllege and Main is 0.35, a yellw light is 0.04, and a red light Let s think abut yur mrning cmmute in the week ahead. Questin: What is the prbability yu find the light red bth Mnday and Tuesday? Because the clr f the light n Mnday desn t influence the clr n Tuesday, these are independent events. P(red Mnday red Tuesday)=P(red) P(red) =(0.61)(0.61)= Questin: What s the prbability yu dn t encunter a red light until Wednesday? P(nt red) = 1 P(red) = = 0.39 P(nt red Mn. nt red Tues. red Wed.) = (0.39)(0.39)(0.61) = Questin: What s the prbability that yu ll have t stp at least nce during the week? Having t stp at least nce means that I have t stp fr the light 7

8 either 1, 2, 3, 4, r 5 times next week. It s easier t think abut the cmplement: never having t stp at a red light. Having t stp at least nce means that I didn t make it thrugh the week with NO red lights. P(having t stp at the light at least nce in 5 days) = 1 P(n red lights fr 5 days) = 1 P(nt red nt red nt red nt red nt red) = 1 (0.39)(0.39)(0.39)(0.39)(0.39) = = There s ver a 99% chance I ll hit at least ne red light smetime this week! General Additin Rule If events E and F are nt disjint, they can ccur simultaneusly. The prbability f their unin is than less than the sum f their prbabilities. The utcmes cmmn t bth (shwn as the shaded regin belw) are cunted twice when we add the prbabilities, s we must subtract this prbability nce. PA ( r B) = PA ( ) + PB ( ) PA ( and B) Fr any tw events A and B, PA ( B) = PA ( ) + PB ( ) PA ( B) If A and B are disjint, the event A Bthat bth ccur has n utcmes in it. The empty event is just zer, s the general additin rule just becmes the simple additin rule fr disjint events. Ex: Debrah and Matthew are anxiusly awaiting wrd n whether they have been made partners f their law firm. Debrah guess that her prbability f making partner is 0.7 and that Matthew s is 0.5. Debrah als guesses that the prbability that bth she and Matthew are made partners is 0.3, then by the general additin rule, P(at least ne is prmted) = PD ( r M) PD ( r M) = PD ( ) + PM ( ) PD ( and M) = = 0.9 This can als be illustrated as a Venn diagram, Nte the intersectin f the tw events is 0.3 and placed in the center f the Venn diagram. We can nw btain the answer f 0.9 by adding the three regins 0.4, 0.3 and 0.2 tgether. Nte als that 0.1 lies utside thse events. The prbability that neither is prmted is

9 Ex: A survey f jb satisfactin f teachers was taken, giving the fllwing results. Jb Satisfactin Satisfied Unsatisfied Ttal Cllege High Schl L E V E L Elementary Ttal If we pick a teacher at randm, find the prbability that they teach cllege (C) r they are satisfied with their jb (S). We want t find PC ( S). Since these events are NOT disjint, use the general additin rule. PC ( ) = 0.150, PS ( ) = and PC ( S) = frm the table abve, s PC ( S) = PC ( ) + PS ( ) PC ( S) = = Cnditinal Prbability Cnditinal prbability gives the prbability f ne event under the cnditin that we knw anther event has ccurred. The new ntatin: PA ( B ). The bar is read given the infrmatin that Let A and B be tw events. The cnditinal prbability f the event A given PA ( B) that the event B has ccurred is PA ( B) = PB ( ) A cnditin has the effect f reducing the size f the sample space, and therefre, the value f the denminatr in a prbability fractin. This can be seen in the frmula abve. Ex: A study was perfrmed t lk at the relatinship between mtin sickness and seat psitining in a bus. The fllwing table summarizes the data: Seat Psitin in Bus Frnt Middle Back Ttal Nausea N Nausea Ttal Let s use the symbls, N, N c, F, M, B t stand fr the events Nausea, N Nausea, Frnt, Middle and Back respectively. Cmputing the prbability that 417 an individual in the study gets nausea, we have, PN ( ) = = Other prbabilities are easily calculated by dividing the numbers in the cells by 3256 t get 9

10 Seat Psitin in Bus Frnt Middle Back Ttal Nausea N Nausea Ttal P(N and F) P(N) P(F) P(M and N C ) The event a persn gt nausea given he/she sat in the frnt seat is an example f what is called a cnditinal prbability. Of the 928 peple wh sat in the frnt, 58 gt nausea s the prbability that a persn gt nausea given he/she sat in the frnt seat is 58 = This can als be fund 928 PN ( and F) using the values n the table PN ( F) = = It is PF ( ) nt exactly due t rundff errr. Ex: Back t the survey f jb satisfactin f teachers. Find the prbability that a teacher chsen at randm is a cllege teacher, given that he r she is satisfied. L E V E L Jb Satisfactin Satisfied Unsatisfied Ttal Cllege High Schl Elementary Ttal PS ( ) =, PC ( and S) = PC ( and S) PC ( S) = = = = PS ( ) is the prprtin f satisfied that are cllege teachers. This is different than the prprtin f teachers wh are satisfied, given that 74 PS ( C) 74 they are cllege teachers. PS ( C) = = 778 = = PC ( ) is the prprtin f cllege teachers that are satisfied. **Nte: PC ( S) PS ( C). This is always true. 10

11 Ex: One survey fund that 56% f cllege students live n campus, 62% have a campus meal prgram, and 42% d bth. Questin: While dining in a campus facility pen nly t students with meal plans, yu meet smene interesting. What is the prbability that yur new acquaintance lives n campus? P(student lives n campus given that the student has a meal plan) PC ( M) 0.42 P(n campus meal plan) = = = PM ( ) 0.62 There is a prbability f abut that a student with a meal plan lives n campus. Ex: A simple randm sample f adults living in a suburb f a large city was selected. The age and annual incme f each adult in the sample were recrded. The resulting data are summarized in the table belw. Annual Incme Age Categry $25,000-$35,000 $35,001-$50,000 Over $50,000 Ttal Over Ttal (a) What is the prbability that a persn chsen at randm frm thse in this sample will be in the age categry? 89 P(31 45) = (b) What is the prbability that a persn chsen at randm frm thse in this sample whse incmes are ver $50,000 will be in the age categry? Shw yur wrk. 35 P(31 45 ver$50, 000) P(31 45 ver$50, 000) = = P( ver $50, 000) (c) Based n yur answers t parts (a) and (b), is annual incme independent f age categry fr thse in this sample? Explain. N, these events are dependent. If these tw events were independent, then P(31 45) = P(31 45 ver$50, 000), but

12 (d) What is the prbability that a persn chsen at randm frm thse in this sample will be in the age categry r have an incme ver $50,000? Need t find the unin f these tw events. Since these events are NOT disjint, we need t use the general additin rule. P(46 60 ver$50, 000) = P(46 60) + P( ver $50, 000) P( bth) = + = General Multiplicatin Rule fr any tw events The jint prbability that events A and B bth happen can be fund by PA ( and B) = PAPB ( ) ( A). Here, PB ( Ais ) the cnditinal prbability that B ccurs, given the infrmatin that A ccurs. This general rule wrks fr independent AND dependent events. If the tw events are independent, then PB ( A) = PB ( ). If A and B are nt independent, then they are said t be dependent events. With replacement If we select an bject and place it back in the sample space, it is said t be sampling with replacement. This is an example f independent events. Withut replacement If we select an bject and DO NOT replace it in the sample space and select anther bject, this is sampling withut replacement. This is an example f dependent events. Ex: Suppse we are ging t select three cards frm an rdinary deck f cards. Cnsider the events: E = {event that the first card is a king}, F = {event that the secnd card is a king}, G = {event that the third card is a king}. If we select the first card and then place it back in the deck befre we select the secnd, and s n, the sampling will be with replacement. 4 PE ( ) = PF ( ) = PG ( ) = 52 PE ( F G) = PEPFPG ( ) ( ) ( ) = = If we select the cards in the usual manner withut replacing them in the deck, the sampling will be withut replacement PE ( ) =, PF ( ) =, PG ( ) = PE ( F G) = PEPFPG ( ) ( ) ( ) = = *Nte: We are still applying the multiplicatin rule as we did with independent events t find ur prbability, it s just that with dependent events, we have t think abut the 2 nd and 3 rd fractin that we are multiplying, nw that the previus events have ccurred. 12

13 Ex: Plice reprt that 78% f drivers are given a breath test, 36% a bld test, and 22% bth tests. Define events: A = {suspect is given a breath test} B = {suspect is given a bld test}. It is knwn: PA ( ) = 0.78, PB ( ) = 0.36, PA ( B) = 0.22 Questin: Are giving a DWI suspect a bld test and a breath test mutually exclusive? N. Disjint events cannt BOTH happen at the same time, s check t see if PA ( B) =. Since sme suspects are given bth tests and the prbability is equal t 0.22, they are nt mutually exclusive. Questin: Are the tw tests independent? Let s make a table. Breath Test Bld Test Yes N Ttal Yes N Ttal Des getting a breath test change the prbability f getting a bld test? That is, des PB ( A) = PB ( )? PA ( B) 0.22 PB ( A) = = 0.28 PA ( ) 0.78 PB ( ) = 0.36 since PB ( A) PB ( ) Overall, 36% f the drivers get bld tests, but nly 28% f thse wh get a breath test d. Since suspects wh get a breath test are less likely t have a bld test, the tw events are nt independent. Extended Multiplicatin Rule Extend the multiplicatin rule t the prbability that all f several events ccur. The key is t cnditin each event n the ccurrence f ALL f the preceding events. PA ( and Band C) = PAPB ( ) ( APC ) ( Aand B) Best way t figure these types f prblems is t use a tree diagram. The tree diagram keeps straight what has ccurred and the prbabilities f each ccurring. Tree diagrams cmbine the additin and multiplicatin rules. The multiplicatin rule says that the prbability f reaching the end f any cmplete branch is the prduct f the prbabilities written n its segments. Ex: A labratry test fr the detectin f a certain disease gives a psitive result 5 percent f the time fr peple wh d nt have the disease. The test gives a negative result 0.3 percent f the time fr peple wh have the disease. Large-scale studies have shwn that the disease ccurs in abut 2 percent f the ppulatin. (a) What is the prbability that a persn selected at randm wuld test psitive fr this disease? Shw yur wrk. 13

14 .02 Have Disease Psitive Negative.98 N Disease.05 Psitive.95 Negative P( psitive ) = (.02)(.997) + (.98)(.05) = (b) What is the prbability that a persn selected at randm wh tests psitive fr the disease des nt have the disease? Shw yur wrk. Find P(N disease tests psitive) P( NDisease test+ ) (.98)(.05).049 = = = P( + ) Ex: Only 5% f male high schl basketball, baseball and ftball players g n t play at the cllege level. Of these, nly 1.7% enter majr league prfessinal sprts. Of thse athletes that d nt cmpete in cllege, nly 0.01% enter prfessinal sprts. Belw is a tree diagram illustrating these percents. Questin: What is the prbability that a male high schl athlete will g n t prfessinal sprts? T find P(B), we multiply the branches alng the tw that reach B and add them tgether. PB ( ) = (0.05)(0.017) + (0.95)(0.0001) = = Abut 9 high schl athletes ut f 10,000 will play prfessinal sprts. Questin: What prprtin f prfessinal athletes cmpeted in cllege? This is a cnditinal prbability. It is the cnditinal prbability PA ( B) where B = {cmpetes prfessinally} and A = {cmpetes in cllege}. PA ( B) PA ( B) = = = PB ( ) Almst 90% f prfessinal athletes cmpeted in cllege. 14