II. PROBABILITY OF AN EVENT

Transcription

1 II. PROBABILITY OF AN EVENT As ndcated above, probablty s a quantfcaton, or a mathematcal model, of a random experment. Ths quantfcaton s a measure of the lkelhood that a gven event wll occur when the experment s performed. A probablty functon s a functon P defned on the subsets of a sample space S. To defne ths functon we proceed as follows: of a Smple Event: Let S { e e,..., } =, 2 e n be the sample space of an experment. To each smple event P ( e assgn a real number ) two condtons are satsfed: e we, called the probablty of the event e, such that the followng () The probablty of each smple event s a number between 0 and, nclusve. That s, 0 P( e ), =,2,..., n () The sum of the probabltes of the smple events space s. That s, e,...,, e2 en n the sample P ( e ) + P( e2) P( en ) = Any assgnment that satsfes condtons () and () s an acceptable probablty assgnment. Examples 2.:. { e, e, e, e e } S = 2 3 4, 5 s the sample space of an experment. Whch of the followng are acceptable probablty assgnments? If the assgnment s not acceptable, explan why. (a) P e ) = 0.3, P( e ) = 0.2, P( e ) = 0.5, P( e ) = 0., P( e ) ( = (b) P e ) = 0.3, P( e ) = 0.2, P( e ) = 0.5, P( e ) = 0.3, P( e ) ( = (c) P e ) = 0.3, P( e ) = 0.2, P( e ) = 0.5, P( e ) = 0., P( e ) ( = 2. Roll a de. Fnd the sample space and determne two dfferent acceptable probablty assgnments. 37

2 Solutons:. (a) Condton () s satsfed; each value s a number between 0 and. Checkng condton (): = The sum s not ; the assgnment s not acceptable. (b) Pe ( 3) = 0.5 ; condton () s not satsfed; the assgnment s not acceptable. (c) Condton () s satsfed; each value s a number between 0 and. Checkng condton (): =. Ths s an acceptable probablty assgnment. 2. The sample space : S = {,2,3,4,5,6}. There are nfntely many solutons to ths problem. We ll gve 3: Assgnment #: P() =, P(2) =, P(3) =, P(4) =, P(5) =, P(6) =. Ths s the assgnment we would make f we know that the de s far; each number s as lkely to turn up as any other number. For another assgnment, we smply have to nsure that condtons () and () are satsfed; there are nfntely many choces. Here s one: Assgnment #2: P() =, P(2) =, P(3) =, P(4) =, P(5) =, P(6) =. Ths assgnment would ndcate that the de s loaded n such a way that an even number s twce as lkely to come up as an odd number. Assgnment #3: P() = P(2) = P(4) = P(5) = P(6) = 0, P(3) =. Ths s the assgnment we would make f we knew that the de was loaded and that a 3 always comes up. In addton to beng acceptable, we want the probablty assgnment to accurately reflect the experment; we want the assgnment of probabltes to be reasonable. For example, suppose we flp a far con. The sample space s S = { H, T}, and the smple events are H the con comes up heads and T the con comes up tals. Whle the assgnment 38

3 PH ( ) = 3/ 4, PT ( ) = / 4 s acceptable, t s not reasonable. Snce the con s far, we reasonably assume that a head s just as lkely to turn up as a tal. Ths mples that we should set P(H) = P(T) from whch t follows that s the assgnment we should make. PH ( ) = / 2, PT ( ) = / 2 Examples 2.2:. A box contans marbles: 6 red, 3 blue, and 2 whte. After shakng the box thoroughly, you draw a marble out of the box. What s the sample space for ths experment and what probablty would you assgn to the smple events. 2. A con s based n such a way that a head s twce as lkely to come up as a tals. Determne a probablty assgnment for the smple events of the sample space S = { H, T} that reflects ths bas. 3. A de s rolled 000 tmes wth the results gven n the table. Outcome Number of tmes Gve two probablty assgnments for the smple events. Solutons:. The marble wll ether be red (R), blue (B), or whte (W). Therefore, S = { R, B, W}. Snce 6 of the marbles are red, 3 are blue and 2 are whte, we assgn 6 3 2,, PR ( ) = PB ( ) = PW ( ) =. 39

4 2. We ll let PT ( ) = x, 0 x, be the probablty that a tal comes up. Then the probablty that a head comes up s PH ( ) = 2 x. Now, by condton (), x + 2x = 3x = x =. 3 Thus, 2, PH 3 3 PT ( ) = ( ) =. 3. For the frst assgnment we ll use the results gven n the table: P() = = 0.66, P(2) = = 0.70, P(3) = = 0.69 P(4) = = 0.70, P(5) = = 0.65, P(6) = = These numbers are close to 6 assumpton, then we set so t appears that the de s far. If we make that P() =, P(2) =, P(3) =, P(4) =, P(5) =, P(6) =. Examples 2. and 2.2 llustrate two typcal ways n whch acceptable and reasonable probablty assgnments are made for the smple events n a sample space S.. Theoretcal. We use assumptons and analytcal reasonng (the de s far so the outcomes have equal probablty; the con s based n a certan way), and mpose the two condtons that the probablty assgnment must satsfy. No experments are actually performed. 2. Emprcal. For one of our assgnments n Examples we used the results of an actual experment. Emprcal probablty can be defned more precsely as follows: Suppose we perform an experment n tmes and the smple event e occurs wth frequency f ( e ). That s, the outcome e occurs f ( e ) tmes. Then the rato f ( e ) / n s called the relatve frequency or emprcal probablty of e. 40

5 frequency of occurence of e f( e) Pe ( ) = =. number of trals n We leave t as an exercse to verfy that ths assgnment s acceptable. The connecton between theoretcal and emprcal probablty assgnments for a gven random experment s contaned n a theorem whch states that the emprcal probabltes approach the theoretcal probabltes as n gets larger and larger, provded the theoretcal probablty assgnment s correct. Look at Example agan. The emprcal probablty assgnments to the outcomes are all close to /6. Based on ths evdence, f we rolled the de 0,000 tmes or a 00,000 tmes, we would expect that a would occur essentally /6 of the tme, a 2 would occur essentally /6 of the tme, and so on. So, we assume that the de s far and assgn (theoretcally) the probabltes P()=/6, P(2)=/6,, P(6)=/6. The Equally Lkely Assumpton: If we toss an unbased con, we would expect that a heads s as lkely to turn up as a tals. There are two possble outcomes, they are equally lkely to occur, and so we assgn the value, /2, to each outcome; P ( H ) = P( T ) = / 2. To say ths another way, f we were to toss the con a large number of tmes, we would expect that half (or at least very close to half) the tosses would result n heads, and half n tals. If we roll a far de, there are sx possble outcomes and any one number s as lkely to turn up as any other. Thus, we assgn the values P ( ) = P() =... = P(6) = / 6. If we draw a card from a well-shuffled, standard deck, then any one card s a lkely to be drawn as any other and so we assgn the value P ( e) = / 52 to each of the 52 smple events. In general, f a sample spaces S has n elements, { e e } S =, 2,..., e n and f we determne that each of the outcomes e s as lkely to occur as any other, then we assgn the probablty value / n to each. That s, Examples 2.3: P( e ) =, =,2,3,..., n n. A red de and a green de are rolled. The dce are far. Let e j denote the outcome on the red de, j on the green de. What s P e )? ( j 4

6 2. Two dentcal de are rolled and the sum of the numbers that turn up s noted. What s the sample space? Are the outcomes equally lkely? 3. A combnaton lock has 3 dals, each labeled wth the 9 dgts from 0 to 9. An openng combnaton s a sequence of 3 dgts wth no dgt repeated. How many openng combnatons are there? What s the probablty of a person guessng the rght openng combnaton? 4. A 5-card hand s dealt from a standard, well shuffled, 52-card deck. How many smple events are n the sample space? Are they equally lkely? If so, what are the probablty assgnments for the smple events? Solutons:. The dce are far; each outcome s as lkely to occur as any other; there are 36 possble outcomes (6 possbltes on the red de, 6 possbltes on the green de, 6 x 6 =36). Therefore, Pe ( ) =., j The sample space s S = {2, 3, 4, 5,,, 2}. There are outcomes. If we assume that the outcomes are equally lkely, then we would assgn to each outcome. Whle ths assgnment s acceptable, s t reasonable? Note that there s only one way to get a 2, both dce have to come up. Smlarly, there s only one way to get a 2, both dce have to come up 6. On the other hand, there are several ways to get a 7 a and a 6, a 2 and a 5, etc. Therefore a 7 s more lkely to occur than a 2 or a 2. We conclude that the outcomes are not equally lkely and so the equally lkely assgnment s not approprate. We wll fnsh ths example below. 3. The frst dgt can be any one of the 0 numbers. Snce no dgt can be repeated, the second dgt can be any one of the remanng 9 and the thrd any one of the 8 that reman after the frst 2 numbers have been chosen. Thus, by the Multplcaton Prncple, the number of openng combnatons s 0 x 9 x 8 = 720 Note that the number of openng combnatons s the number of permutatons of 0 thngs taken 3 at a tme, P 0, 3. The sample space s the set of possble openng combnatons. The probablty of guessng the correct one s / The number of 5-card hands that can be dealt from a standard deck s the number of combnatons of 52 thngs taken 5 at a tme; C 52, 5. Each hand s a lkely to occur as any other so each hand has probablty /C 52, 5. 42

7 of an Event: Now suppose that we have been gven a random experment, determned the sample space S, and devsed an acceptable probablty assgnment for the smple events. How do we defne the probablty of an arbtrary event E? =, 2 3 e n s the sample space for an experment, and suppose E s an arbtrary subset of S. We defne the probablty of E as follows: Suppose S { e e, e,..., } () If E s the empty set, then P ( E) = 0. () If E s a smple event, then P(E) has already been assgned. () E = { e, e,..., e } s a compound event, then 2 k P ( E) = P( e ) + P( e ) P( e ). 2 k That s, P(E) s the sum of the probabltes of the smple events that comprse E. (v) If E = S, then P ( E) =.Ths follows from () and the defnton of the probablty of smple event. Note: An event E such that P(E) = 0 s called an mpossble event. An event E such that P(E) = s called a certan event. Examples 2.4:. Toss a far con twce. What s the probablty that at least one tal appears? 2. Roll a red de and a green de. (a) What s the probablty that the sum of the numbers that turn up s 7? (b) What s the probablty that the sum of the numbers that turn up s less than 4? (c) What s the probablty that the sum of the numbers s 3? 43

8 3. A company receves orders from women n the followng age groups: Age Number of Orders Under and over 3 00 A customer s selected at random from the set of 00 customers. Solutons: (a) Gve the sample space for the experment and determne the probabltes of the smple events. (b) What s the probablty that the customer selected s under 40?. The sample space s S = { HH, HT, TH, TT}. The outcomes are equally lkely and so we assgn probablty to each. Let E = at least one tal. Then E = { HT, TH, TT} and 4 3 PE ( ) + + = =. 2. The sample space s llustrated n the fgure. GREEN DIE (,) (,2) (,3) (,4) (,5) (,6) RED DIE (2,) (2,2) (2,3) (2,4) (2,5) (2,6) (3,) (3,2) (3,3) (3,4) (3,5) (3,6) (4,) (4,2) (4,3) (4,4) (4,5) (4,6) (5,) (5,2) (5,3) (5,4) (5,5) (5,6) (6,) (6,2) (6,3) (6,4) (6,5) (6,6) 44

9 As we dscussed prevously, the outcomes are equally lkely to occur and so we assgn the value to each. 36 (a) Let A = the sum of the numbers s 7. Then and A = {(,6),(2,5),(3,4),(4,3),(5,2),(6,)} 6 PA= ( ) = = (b) Let B = the sum of the numbers s less than 4. Then and B = {(,), (, 2), (2,)} 3 PB ( ) = + + = = (c) Let C = the sum s 3. Snce the sum of the numbers on the dce s a number between 2 and 2. C = and PC ( ) = 0; C s an mpossble event. 3. (a) Based on ths data, P( < 20) = 22 /00, P(20-29) = 22 /00, P(30-39) = 47 /00, P(40-49) = 8/00, P( > 49) = 3/00 (b) The probablty that she s under 40 s: P ( < 20) + P(20 29) + P(30 39) = = It s clear that the crtcal step n determnng the probabltes of events assocated wth a gven experment s the assgnment of acceptable, reasonable probabltes to the smple events n the sample space S. Once that has been done, we can fnd the probablty of any event connected wth the experment. 45

10 The equally lkely case revsted: As llustrated above, probablty experments that result n equally lkely smple events s the easest case; each of the smple events s assgned probablty / n where n s the total number of possble outcomes. Whle not all probablty experments have equally lkely outcomes, many can be restated or refned so that the outcomes are equally lkely and ths s an effectve approach to solvng probablty problems. So now we consder the specal case of the probablty of an event E when the outcomes of the experment are equally lkely. Suppose that { e e, e } S =,...,, 2 3 s the sample space of an experment and that the outcomes are equally lkely. Then e n P( e ) =, =,2,3,..., n. n If E = { e, e,..., e }, (E conssts of k smple events), then 2 k P( E) = P( e ) + P( e 2 ) P( e That s, the probablty of the event E s smply k ) = = n n n k terms k n. the number of elementsn E n( E) P ( E) = =. the number of elementsn S n( S) Examples 2.5:. Toss two dentcal cons. (a) What s the probablty that the cons match (both heads or both tals)? (b) What s the probablty that at least one head turns up? 2. Roll two dentcal dce (the completon of Example 2.3.2). 46

11 (a) What s the probablty that the sum of the numbers that turn up s 7? (b) What s the probablty that the sum of the numbers that turn up s less than 4? 3. A commttee has 6 female and 5 male members. A 5-member subcommttee s to be selected at random from the commttee members. (a) What s the probablty that the subcommttee wll consst of 3 females and 2 males? (b) What s the probablty that the subcommttee wll have at least 4 females? Solutons:. If we toss two dentcal cons the set of possble outcomes s S = { HH, HT, TT}. Stated n terms of the number of heads that appear, S = {2,,0}. However, these smple events are not equally lkely; there s only one way to get 2 heads and only one way to get 0 heads, but exactly head can occur n two ways. If the cons were dstngushable, then the sample space would be R = { HH, HT, TH, TT} and these events are equally lkely wth each havng the probablty. Note that the 4 compound event { HT, TH } n R corresponds to the smple event HT (or exactly head) n S. Snce the cons don t know whether or not they are dentcal, we wll assume that they are and then go on to determne the probablty assgnment for S:,, PHH ( ) = PHT ( ) = PTT ( ) =. (a) Let A = the cons match. Then A= { HH, TT} and 2 PA= ( ). (b) Let B = at least head. Then B = { HH, HT, TH} and 3 4 PB ( ) =. 2. The sample space s: S = {2, 3, 4,..., 2}. As we have seen, these outcomes are not equally lkely; the assgnment 47

12 s acceptable, but t s not reasonable. P(2) = P(3) = P(4) =... = P(2) = As n part, we handle ths stuaton by defnng a sample space for the experment n whch the outcomes are equally lkely. Assume that one of the dce s red and the other s green. There s no loss n generalty here; the dce don t know what color they are. Now see Example (a) If A = the sum of the numbers s 7, then PA= ( ). 6 (b) If B = the sum of the numbers s less than 4, then PB ( ) = The sample space s the set of 5-member subcommttees and, snce the subcommttee wll be selected at random, any partcular 5-member subcommttee s as lkely to be chosen as any other. (a) Let E = the subcommttee has 3 female and 2 male members. Then no. of subcommttees wth 3 females and 2 males PE ( ) =. no. of 5-member subcommttees The number of 5-member subcommttees s the combnatons of thngs taken 5 at a tme, C,5. Vew the selecton of a subcommttee wth 3 female members and 2 male members as a 2-step process: Step: Select the 3 female members. Ths can be done n C 6,3 ways. Step2: Select the 2 male member. Ths can be done n C 5,2 ways. Therefore, 6! 5! C6,3C 5,2 3!3! 2!3! PE ( ) = = = = =. C!, !6! (b) Let F = at least 4 females. The number of subcommttees wth at least 4 female members equals the number wth exactly 4 female members plus the number wth exactly 5 female members. Proceedng as above, the number of subcommttees wth 4 female and male member s gven by C6,4C 5, = 5 5 = 75. The number of subcommttees wth 5 female members s gven by C 6,5 = 6. Thus, 48

13 PF ( ) = = = Exercses 3.2:. Let E be an event n a sample space S. (a) How would you nterpret P(E)=0? (b) How would you nterpret P(E)=? 2. A probablty experment s conducted. Whch of the followng cannot be consdered a probablty of an outcome. (a) 2/3 (b) (c) 75% (d) 0 (e) 2.5 (f) -/2 (g).6 (h) 3. A probablty experment has 4 possble outcomes A, B, C, and D. Whch of the followng s an acceptable probablty assgnment? Gve reasons for your answer. (a) P(A) = 0.5, P(B) = -0.35, P(C) = 0.50, P(D) = 0.70 (b) P(A) = 0.32, P(B) = 0.28, P(C) = 0.24, P(D) = 0.30 (c) P(A) = 0.26, P(B) = 0.4, P(C) = 0.42, P(D) = A de s loaded n such a way that an even number s twce as lkely to turn up as an odd number. Determne the probablty assgnments for the smple events. 5. In a famly wth 3 chldren, excludng multple brths and assumng that a boy s as lkely as a grl at each brth, what s the probablty of havng: (a) 2 boys and grl n that order? (b) 2 boys and grl n any order? (c) At least one chld of each sex? 6. An experment conssts of tossng 3 far (.e., not weghted) cons, except one of the cons s 2-headed. Fnd the probablty of gettng: (a) head. (b) 2 heads. (c) 3 heads. (d) 0 heads. (e) At least one head. (f) More than tal. 7. An experment conssts of rollng 2 far dce and addng the numbers on the faces that turn up. Fnd the probablty that the sum s: (a) 2 (b) 5 (c) Greater than 8. (d) Less than 7. (e) An odd number. (f) A prme number. (g) Dvsble by 3. 49

14 8. An experment conssts of rollng 2 far dce and addng the numbers on the faces that turn up. However, each de has the number on two opposte faces, the number 2 on two opposte faces and the number 3 on two opposte faces. Fnd the probablty that the sum s: (a) 6 (b) 3 (c) An odd number. (d) 7 (e) Greater than Among 00 students n a college fraternty, 50 are majorng n the humantes majors, 30 are majorng n a scence and 20 are engneerng majors. If a student s selected at random, what s the probablty that he s: (a) An engneerng major? (b) Nether a scence nor an engneerng major? 0. A roulette wheel has 38 spaces numbered, 2,, 36, 0 and 00. Fnd the probablty of gettng: (a) An odd number. (b) A number greater than 24. (c) A number less than 5. (d) A prme number.. An experment conssts of dealng 5 cards from a standard deck of 52 cards. What s the probablty of beng dealt: (a) 5 hearts? (b) 5 black cards? (c) 4 of a knd (.e., 4 aces, or 4 kngs, or 4 queens, and so on)? (d) 3 aces and 2 queens? (e) 3 of one knd and two of another (a full house )? 2. A 4-person commttee s to be composed of employees from two departments, department A wth 5 employees and department B wth 20 employees. If the 4 people are selected at random from the pool of 35 people, what s the probablty that: (a) All are from department B? (b) 3 are from B and s from A? (c) 2 are from A and 2 are from B? (d) At least 3 are from department A? 50