not to be republishe NCERT PROBABILITY Chapter 16.1 Introduction

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1 Chapter 6 Where a mathematcal reasonng can be had, t s as great a folly to make use of any other, as to grope for a thng n the dark, when you have a candle n your hand. JOHN ARBUTHNOT 6. Introducton PROBABILITY In earler classes, we studed about the concept of probablty as a measure of uncertanty of varous phenomenon. We have obtaned the probablty of gettng an even number n throwng a de as 3 6.e.,. Here the 2 total possble outcomes are,2,3,4,5 and 6 (sx n number). The outcomes n favour of the event of gettng an even number are 2,4,6 (.e., three n number). In general, to obtan the probablty of an event, we fnd the rato of the number of outcomes favourable to the event, to the total number of equally lkely outcomes. Ths theory of probablty Kolmogorov s known as classcal theory of probablty. ( ) In Class IX, we learnt to fnd the probablty on the bass of observatons and collected data. Ths s called statstcal approach of probablty. Both the theores have some serous dffcultes. For nstance, these theores can not be appled to the actvtes/experments whch have nfnte number of outcomes. In classcal theory we assume all the outcomes to be equally lkely. Recall that the outcomes are called equally lkely when we have no reason to beleve that one s more lkely to occur than the other. In other words, we assume that all outcomes have equal chance (probablty) to occur. Thus, to defne probablty, we used equally lkely or equally probable outcomes. Ths s logcally not a correct defnton. Thus, another theory of probablty was developed by A.N. Kolmogorov, a Russan mathematcan, n 933. He

2 384 MATHEMATICS lad down some axoms to nterpret probablty, n hs book Foundaton of Probablty publshed n 933. In ths Chapter, we wll study about ths approach called axomatc approach of probablty. To understand ths approach we must know about few basc terms vz. random experment, sample space, events, etc. Let us learn about these all, n what follows next. 6.2 Random Experments In our day to day lfe, we perform many actvtes whch have a fxed result no matter any number of tmes they are repeated. For example gven any trangle, wthout knowng the three angles, we can defntely say that the sum of measure of angles s 80. We also perform many expermental actvtes, where the result may not be same, when they are repeated under dentcal condtons. For example, when a con s tossed t may turn up a head or a tal, but we are not sure whch one of these results wll actually be obtaned. Such experments are called random experments. An experment s called random experment f t satsfes the followng two condtons: () It has more than one possble outcome. () It s not possble to predct the outcome n advance. Check whether the experment of tossng a de s random or not? In ths chapter, we shall refer the random experment by experment only unless stated otherwse Outcomes and sample space A possble result of a random experment s called ts outcome. Consder the experment of rollng a de. The outcomes of ths experment are, 2, 3, 4, 5, or 6, f we are nterested n the number of dots on the upper face of the de. The set of outcomes {, 2, 3, 4, 5, 6} s called the sample space of the experment. Thus, the set of all possble outcomes of a random experment s called the sample space assocated wth the experment. Sample space s denoted by the symbol S. Each element of the sample space s called a sample pont. In other words, each outcome of the random experment s also called sample pont. Let us now consder some examples. Example Two cons (a one rupee con and a two rupee con) are tossed once. Fnd a sample space. Soluton Clearly the cons are dstngushable n the sense that we can speak of the frst con and the second con. Snce ether con can turn up Head (H) or Tal(T), the possble outcomes may be

3 PROBABILITY 385 Heads on both cons = (H,H) = HH Head on frst con and Tal on the other = (H,T) = HT Tal on frst con and Head on the other = (T,H) = TH Tal on both cons = (T,T) = TT Thus, the sample space s S = {HH, HT, TH, TT} Note The outcomes of ths experment are ordered pars of H and T. For the sake of smplcty the commas are omtted from the ordered pars. Example 2 Fnd the sample space assocated wth the experment of rollng a par of dce (one s blue and the other red) once. Also, fnd the number of elements of ths sample space. Soluton Suppose appears on blue de and 2 on the red de. We denote ths outcome by an ordered par (,2). Smlarly, f 3 appears on blue de and 5 on red, the outcome s denoted by the ordered par (3,5). In general each outcome can be denoted by the ordered par (x, y), where x s the number appeared on the blue de and y s the number appeared on the red de. Therefore, ths sample space s gven by S = {(x, y): x s the number on the blue de and y s the number on the red de}. The number of elements of ths sample space s 6 6 = 36 and the sample space s gven below: {(,), (,2), (,3), (,4), (,5), (,6), (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)} Example 3 In each of the followng experments specfy approprate sample space () A boy has a rupee con, a 2 rupee con and a 5 rupee con n hs pocket. He takes out two cons out of hs pocket, one after the other. () A person s notng down the number of accdents along a busy hghway durng a year. Soluton () Let Q denote a rupee con, H denotes a 2 rupee con and R denotes a 5 rupee con. The frst con he takes out of hs pocket may be any one of the three cons Q, H or R. Correspondng to Q, the second draw may be H or R. So the result of two draws may be QH or QR. Smlarly, correspondng to H, the second draw may be Q or R. Therefore, the outcomes may be HQ or HR. Lastly, correspondng to R, the second draw may be H or Q. So, the outcomes may be RH or RQ.

4 386 MATHEMATICS Thus, the sample space s S={QH, QR, HQ, HR, RH, RQ} () The number of accdents along a busy hghway durng the year of observaton can be ether 0 (for no accdent ) or or 2, or some other postve nteger. Thus, a sample space assocated wth ths experment s S= {0,,2,...} Example 4 A con s tossed. If t shows head, we draw a ball from a bag consstng of 3 blue and 4 whte balls; f t shows tal we throw a de. Descrbe the sample space of ths experment. Soluton Let us denote blue balls by B, B 2, B 3 and the whte balls by W, W 2, W 3, W 4. Then a sample space of the experment s S = { HB, HB 2, HB 3, HW, HW 2, HW 3, HW 4, T, T2, T3, T4, T5, T6}. Here HB means head on the con and ball B s drawn, HW means head on the con and ball W s drawn. Smlarly, T means tal on the con and the number on the de. Example 5 Consder the experment n whch a con s tossed repeatedly untl a head comes up. Descrbe the sample space. Soluton In the experment head may come up on the frst toss, or the 2nd toss, or the 3rd toss and so on tll head s obtaned. Hence, the desred sample space s S= {H, TH, TTH, TTTH, TTTTH,...} EXERCISE 6. In each of the followng Exercses to 7, descrbe the sample space for the ndcated experment.. A con s tossed three tmes. 2. A de s thrown two tmes. 3. A con s tossed four tmes. 4. A con s tossed and a de s thrown. 5. A con s tossed and then a de s rolled only n case a head s shown on the con boys and 2 grls are n Room X, and boy and 3 grls n Room Y. Specfy the sample space for the experment n whch a room s selected and then a person. 7. One de of red colour, one of whte colour and one of blue colour are placed n a bag. One de s selected at random and rolled, ts colour and the number on ts uppermost face s noted. Descrbe the sample space. 8. An experment conssts of recordng boy grl composton of famles wth 2 chldren. () What s the sample space f we are nterested n knowng whether t s a boy or grl n the order of ther brths?

5 PROBABILITY 387 () What s the sample space f we are nterested n the number of grls n the famly? 9. A box contans red and 3 dentcal whte balls. Two balls are drawn at random n successon wthout replacement. Wrte the sample space for ths experment. 0. An experment conssts of tossng a con and then throwng t second tme f a head occurs. If a tal occurs on the frst toss, then a de s rolled once. Fnd the sample space.. Suppose 3 bulbs are selected at random from a lot. Each bulb s tested and classfed as defectve (D) or non defectve(n). Wrte the sample space of ths experment. 2. A con s tossed. If the out come s a head, a de s thrown. If the de shows up an even number, the de s thrown agan. What s the sample space for the experment? 3. The numbers, 2, 3 and 4 are wrtten separatly on four slps of paper. The slps are put n a box and mxed thoroughly. A person draws two slps from the box, one after the other, wthout replacement. Descrbe the sample space for the experment. 4. An experment conssts of rollng a de and then tossng a con once f the number on the de s even. If the number on the de s odd, the con s tossed twce. Wrte the sample space for ths experment. 5. A con s tossed. If t shows a tal, we draw a ball from a box whch contans 2 red and 3 black balls. If t shows head, we throw a de. Fnd the sample space for ths experment. 6. A de s thrown repeatedly untll a sx comes up. What s the sample space for ths experment? 6.3 Event We have studed about random experment and sample space assocated wth an experment. The sample space serves as an unversal set for all questons concerned wth the experment. Consder the experment of tossng a con two tmes. An assocated sample space s S = {HH, HT, TH, TT}. Now suppose that we are nterested n those outcomes whch correspond to the occurrence of exactly one head. We fnd that HT and TH are the only elements of S correspondng to the occurrence of ths happenng (event). These two elements form the set E = { HT, TH} We know that the set E s a subset of the sample space S. Smlarly, we fnd the followng correspondence between events and subsets of S.

6 388 MATHEMATICS Descrpton of events Correspondng subset of S Number of tals s exactly 2 A = {TT} Number of tals s atleast one B = {HT, TH, TT} Number of heads s atmost one C = {HT, TH, TT} Second toss s not head D = { HT, TT} Number of tals s atmost two S = {HH, HT, TH, TT} Number of tals s more than two φ The above dscusson suggests that a subset of sample space s assocated wth an event and an event s assocated wth a subset of sample space. In the lght of ths we defne an event as follows. Defnton Any subset E of a sample space S s called an event Occurrence of an event Consder the experment of throwng a de. Let E denotes the event a number less than 4 appears. If actually had appeared on the de then we say that event E has occurred. As a matter of fact f outcomes are 2 or 3, we say that event E has occurred Thus, the event E of a sample space S s sad to have occurred f the outcome ω of the experment s such that ω E. If the outcome ω s such that ω E, we say that the event E has not occurred Types of events Events can be classfed nto varous types on the bass of the elements they have.. Impossble and Sure Events The empty set φ and the sample space S descrbe events. In fact φ s called an mpossble event and S,.e., the whole sample space s called the sure event. To understand these let us consder the experment of rollng a de. The assocated sample space s S = {, 2, 3, 4, 5, 6} Let E be the event the number appears on the de s a multple of 7. Can you wrte the subset assocated wth the event E? Clearly no outcome satsfes the condton gven n the event,.e., no element of the sample space ensures the occurrence of the event E. Thus, we say that the empty set only correspond to the event E. In other words we can say that t s mpossble to have a multple of 7 on the upper face of the de. Thus, the event E = φ s an mpossble event. Now let us take up another event F the number turns up s odd or even. Clearly

7 PROBABILITY 389 F = {, 2, 3, 4, 5, 6,} = S,.e., all outcomes of the experment ensure the occurrence of the event F. Thus, the event F = S s a sure event. 2. Smple Event If an event E has only one sample pont of a sample space, t s called a smple (or elementary) event. In a sample space contanng n dstnct elements, there are exactly n smple events. For example n the experment of tossng two cons, a sample space s S={HH, HT, TH, TT} There are four smple events correspondng to ths sample space. These are E = {HH}, E 2 ={HT}, E 3 = { TH} and E 4 ={TT}. 3. Compound Event If an event has more than one sample pont, t s called a Compound event. For example, n the experment of tossng a con thrce the events E: Exactly one head appeared F: Atleast one head appeared G: Atmost one head appeared etc. are all compound events. The subsets of S assocated wth these events are E={HTT,THT,TTH} F={HTT,THT, TTH, HHT, HTH, THH, HHH} G= {TTT, THT, HTT, TTH} Each of the above subsets contan more than one sample pont, hence they are all compound events Algebra of events In the Chapter on Sets, we have studed about dfferent ways of combnng two or more sets, vz, unon, ntersecton, dfference, complement of a set etc. Lke-wse we can combne two or more events by usng the analogous set notatons. Let A, B, C be events assocated wth an experment whose sample space s S.. Complementary Event For every event A, there corresponds another event A called the complementary event to A. It s also called the event not A. For example, take the experment of tossng three cons. An assocated sample space s S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} Let A={HTH, HHT, THH} be the event only one tal appears Clearly for the outcome HTT, the event A has not occurred. But we may say that the event not A has occurred. Thus, wth every outcome whch s not n A, we say that not A occurs.

8 390 MATHEMATICS Thus the complementary event not A to the event A s A = {HHH, HTT, THT, TTH, TTT} or A = {ω : ω S and ω A} = S A. 2. The Event A or B Recall that unon of two sets A and B denoted by A B contans all those elements whch are ether n A or n B or n both. When the sets A and B are two events assocated wth a sample space, then A B s the event ether A or B or both. Ths event A B s also called A or B. Therefore Event A or B = A B ={ω : ω A or ω B} 3. The Event A and B We know that ntersecton of two sets A B s the set of those elements whch are common to both A and B..e., whch belong to both A and B. If A and B are two events, then the set A B denotes the event A and B. Thus, A B = {ω : ω A and ω B} For example, n the experment of throwng a de twce Let A be the event score on the frst throw s sx and B s the event sum of two scores s atleast then A = {(6,), (6,2), (6,3), (6,4), (6,5), (6,6)}, and B = {(5,6), (6,5), (6,6)} so A B = {(6,5), (6,6)} Note that the set A B = {(6,5), (6,6)} may represent the event the score on the frst throw s sx and the sum of the scores s atleast. 4. The Event A but not B We know that A B s the set of all those elements whch are n A but not n B. Therefore, the set A B may denote the event A but not B.We know that A B = A B Example 6 Consder the experment of rollng a de. Let A be the event gettng a prme number, B be the event gettng an odd number. Wrte the sets representng the events () Aor B () A and B () A but not B (v) not A. Soluton Here S = {, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {, 3, 5} Obvously () A or B = A B = {, 2, 3, 5} () A and B = A B = {3,5} () A but not B = A B = {2} (v) not A = A = {,4,6}

9 PROBABILITY Mutually exclusve events In the experment of rollng a de, a sample space s S = {, 2, 3, 4, 5, 6}. Consder events, A an odd number appears and B an even number appears Clearly the event A excludes the event B and vce versa. In other words, there s no outcome whch ensures the occurrence of events A and B smultaneously. Here A = {, 3, 5} and B = {2, 4, 6} Clearly A B = φ,.e., A and B are dsjont sets. In general, two events A and B are called mutually exclusve events f the occurrence of any one of them excludes the occurrence of the other event,.e., f they can not occur smultaneously. In ths case the sets A and B are dsjont. Agan n the experment of rollng a de, consder the events A an odd number appears and event B a number less than 4 appears Obvously A = {, 3, 5} and B = {, 2, 3} Now 3 A as well as 3 B Therefore, A and B are not mutually exclusve events. Remark Smple events of a sample space are always mutually exclusve Exhaustve events Consder the experment of throwng a de. We have S = {, 2, 3, 4, 5, 6}. Let us defne the followng events A: a number less than 4 appears, B: a number greater than 2 but less than 5 appears and C: a number greater than 4 appears. Then A = {, 2, 3}, B = {3,4} and C = {5, 6}. We observe that A B C = {, 2, 3} {3, 4} {5, 6} = S. Such events A, B and C are called exhaustve events. In general, f E, E 2,..., E n are n events of a sample space S and f E E E... E = E = S 2 3 then E, E 2,..., E n are called exhaustve events.in other words, events E, E 2,..., E n are sad to be exhaustve f atleast one of them necessarly occurs whenever the experment s performed. Further, f E E j = φ for j.e., events E and E j are parwse dsjont and n = E events. = S, then events E, E 2,..., E n are called mutually exclusve and exhaustve n n =

10 392 MATHEMATICS We now consder some examples. Example 7 Two dce are thrown and the sum of the numbers whch come up on the dce s noted. Let us consder the followng events assocated wth ths experment A: the sum s even. B: the sum s a multple of 3. C: the sum s less than 4. D: the sum s greater than. Whch pars of these events are mutually exclusve? Soluton There are 36 elements n the sample space S = {(x, y): x, y =, 2, 3, 4, 5, 6}. Then A = {(, ), (, 3), (, 5), (2, 2), (2, 4), (2, 6), (3, ), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, ), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)} B = {(, 2), (2, ), (, 5), (5, ), (3, 3), (2, 4), (4, 2), (3, 6), (6, 3), (4, 5), (5, 4), (6, 6)} C = {(, ), (2, ), (, 2)} and D = {(6, 6)} We fnd that A B = {(, 5), (2, 4), (3, 3), (4, 2), (5, ), (6, 6)} φ Therefore, A and B are not mutually exclusve events. Smlarly A C φ, A D φ, B C φ and B D φ. Thus, the pars, (A, C), (A, D), (B, C), (B, D) are not mutually exclusve events. Also C D = φ and so C and D are mutually exclusve events. Example 8 A con s tossed three tmes, consder the followng events. A: No head appears, B: Exactly one head appears and C: Atleast two heads appear. Do they form a set of mutually exclusve and exhaustve events? Soluton The sample space of the experment s S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} and A = {TTT}, B = {HTT, THT, TTH}, C = {HHT, HTH, THH, HHH} Now A B C = {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH} = S Therefore, A, B and C are exhaustve events. Also, A B = φ, A C = φ and B C = φ Therefore, the events are par-wse dsjont,.e., they are mutually exclusve. Hence, A, B and C form a set of mutually exclusve and exhaustve events.

11 PROBABILITY 393 EXERCISE 6.2. A de s rolled. Let E be the event de shows 4 and F be the event de shows even number. Are E and F mutually exclusve? 2. A de s thrown. Descrbe the followng events: () A: a number less than 7 () B: a number greater than 7 () C: a multple of 3 (v) D: a number less than 4 (v) E: an even number greater than 4 (v) F: a number not less than 3 Also fnd A B, A B, B C, E F, D E, A C, D E, E F, F 3. An experment nvolves rollng a par of dce and recordng the numbers that come up. Descrbe the followng events: A: the sum s greater than 8, B: 2 occurs on ether de C: the sum s at least 7 and a multple of 3. Whch pars of these events are mutually exclusve? 4. Three cons are tossed once. Let A denote the event three heads show, B denote the event two heads and one tal show, C denote the event three tals show and D denote the event a head shows on the frst con. Whch events are () mutually exclusve? () smple? () Compound? 5. Three cons are tossed. Descrbe () Two events whch are mutually exclusve. () Three events whch are mutually exclusve and exhaustve. () Two events, whch are not mutually exclusve. (v) Two events whch are mutually exclusve but not exhaustve. (v) Three events whch are mutually exclusve but not exhaustve. 6. Two dce are thrown. The events A, B and C are as follows: A: gettng an even number on the frst de. B: gettng an odd number on the frst de. C: gettng the sum of the numbers on the dce 5. Descrbe the events () A () not B () A or B (v) A and B (v) A but not C (v) B or C (v) B and C (v) A B C 7. Refer to queston 6 above, state true or false: (gve reason for your answer) () A and B are mutually exclusve () A and B are mutually exclusve and exhaustve () A = B

12 394 MATHEMATICS (v) A and C are mutually exclusve (v) A and B are mutually exclusve. (v) A, B, C are mutually exclusve and exhaustve. 6.4 Axomatc Approach to Probablty In earler sectons, we have consdered random experments, sample space and events assocated wth these experments. In our day to day lfe we use many words about the chances of occurrence of events. Probablty theory attempts to quantfy these chances of occurrence or non occurrence of events. In earler classes, we have studed some methods of assgnng probablty to an event assocated wth an experment havng known the number of total outcomes. Axomatc approach s another way of descrbng probablty of an event. In ths approach some axoms or rules are depcted to assgn probabltes. Let S be the sample space of a random experment. The probablty P s a real valued functon whose doman s the power set of S and range s the nterval [0,] satsfyng the followng axoms () For any event E, P (E) 0 () P (S) = () If E and F are mutually exclusve events, then P(E F) = P(E) + P(F). It follows from () that P(φ) = 0. To prove ths, we take F = φ and note that E and φ are dsjont events. Therefore, from axom (), we get P (E φ) = P (E) + P (φ) or P(E) = P(E) + P (φ).e. P (φ) = 0. Let S be a sample space contanng outcomes ω, ω2,..., ω n,.e., S = {ω, ω 2,..., ω n } It follows from the axomatc defnton of probablty that () 0 P (ω ) for each ω S () P (ω ) + P (ω 2 ) P (ω n ) = () For any event A, P(A) = P(ω ), ω A. Note It may be noted that the sngleton {ω } s called elementary event and for notatonal convenence, we wrte P(ω ) for P({ω }). For example, n a con tossng experment we can assgn the number to each 2 of the outcomes H and T..e. P(H) = 2 and P(T) = () 2 Clearly ths assgnment satsfes both the condtons.e., each number s nether less than zero nor greater than and

13 PROBABILITY 395 P(H) + P(T) = = Therefore, n ths case we can say that probablty of H = 2, and probablty of T = 2 If we take P(H) = 4 and P(T) = 4 3 Does ths assgnment satsfy the condtons of axomatc approach?... (2) Yes, n ths case, probablty of H = 4 and probablty of T = 3. 4 We fnd that both the assgnments () and (2) are vald for probablty of H and T. In fact, we can assgn the numbers p and ( p) to both the outcomes such that 0 p and P(H) + P(T) = p + ( p) = Ths assgnment, too, satsfes both condtons of the axomatc approach of probablty. Hence, we can say that there are many ways (rather nfnte) to assgn probabltes to outcomes of an experment. We now consder some examples. Example 9 Let a sample space be S = {ω, ω 2,..., ω 6 }.Whch of the followng assgnments of probabltes to each outcome are vald? Outcomes ω ω 2 ω 3 ω 4 ω 5 ω 6 (a) (b) (c) (d) (e) Soluton (a) Condton (): Each of the number p(ω ) s postve and less than one. Condton (): Sum of probabltes = =

14 396 MATHEMATICS Therefore, the assgnment s vald (b) Condton (): Each of the number p(ω ) s ether 0 or. Condton () Sum of the probabltes = = Therefore, the assgnment s vald (c) Condton () Two of the probabltes p(ω 5 ) and p(ω 6 ) are negatve, the assgnment s not vald (d) Snce p(ω 6 ) = 3 >, the assgnment s not vald 2 (e) Snce, sum of probabltes = = 2., the assgnment s not vald Probablty of an event Let S be a sample space assocated wth the experment examnng three consecutve pens produced by a machne and classfed as Good (non-defectve) and bad (defectve). We may get 0,, 2 or 3 defectve pens as result of ths examnaton. A sample space assocated wth ths experment s S = {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG}, where B stands for a defectve or bad pen and G for a non defectve or good pen. Let the probabltes assgned to the outcomes be as follows Sample pont: BBB BBG BGB GBB BGG GBG GGB GGG Probablty: Let event A: there s exactly one defectve pen and event B: there are atleast two defectve pens. Hence A = {BGG, GBG, GGB} and B = {BBG, BGB, GBB, BBB} Now P(A) = P(ω ), ω A = P(BGG) + P(GBG) + P(GGB) = and P(B) = P(ω ), ω B = = P(BBG) + P(BGB) + P(GBB) + P(BBB) = = = Let us consder another experment of tossng a con twce The sample space of ths experment s S = {HH, HT, TH, TT} Let the followng probabltes be assgned to the outcomes

15 PROBABILITY P(HH) =, P(HT) =, P(TH) =, P(TT) = Clearly ths assgnment satsfes the condtons of axomatc approach. Now, let us fnd the probablty of the event E: Both the tosses yeld the same result. Here E = {HH, TT} Now P(E) = Σ P(w ), for all w E 9 4 = P(HH) + P(TT) = + = For the event F: exactly two heads, we have F = {HH} and P(F) = P(HH) = Probabltes of equally lkely outcomes Let a sample space of an experment be S = {ω, ω 2,..., ω n }. Let all the outcomes are equally lkely to occur,.e., the chance of occurrence of each smple event must be same..e. P(ω ) = p, for all ω S where 0 p Snce n P(ω ) =.e., p + p p (n tmes) = = or np =.e., p = n Let S be a sample space and E be an event, such that n(s) = n and n(e) = m. If each out come s equally lkely, then t follows that m Number of outcomes favourable to E P(E) = = n Total possbleoutcomes Probablty of the event A or B Let us now fnd the probablty of event A or B,.e., P (A B) Let A = {HHT, HTH, THH} and B = {HTH, THH, HHH} be two events assocated wth tossng of a con thrce Clearly A B = {HHT, HTH, THH, HHH} Now P (A B) = P(HHT) + P(HTH) + P(THH) + P(HHH)

16 398 MATHEMATICS If all the outcomes are equally lkely, then 4 P( A B) = = = Also P(A) = P(HHT) + P(HTH) + P(THH) = 3 8 and P(B) = P(HTH) + P(THH) + P(HHH) = 3 8 Therefore P(A) + P(B) = = It s clear that P(A B) P(A) + P(B) The ponts HTH and THH are common to both A and B. In the computaton of P(A) + P(B) the probabltes of ponts HTH and THH,.e., the elements of A B are ncluded twce. Thus to get the probablty P(A B) we have to subtract the probabltes of the sample ponts n A B from P(A) + P(B).e. P(A B) = P(A) + P(B) P(ω ), ω A B = P(A) + P(B) P(A B) Thus we observe that, P(A B) = P(A) + P(B) P(A B) In general, f A and B are any two events assocated wth a random experment, then by the defnton of probablty of an event, we have ( ) ( ) P A B = p ω, ω A B. Snce A B = (A B) (A B) (B A), we have P(A B) = [ P(ω ) ω (A B)] + [ P(ω ) ω A B ] + [ P(ω ) ω B A] (because A B, A B and B A are mutually exclusve)... () Also P(A) + P(B) = [ p(ω ) ω A ] + [ p(ω ) ω B] = [ P(ω ) ω (A B) (A B) ] + [ P(ω ) ω (B A) (A B) ] = [ P(ω ) ω (A B) ] + [ P(ω ) ω (A B) ] + [ P(ω ) ω (B A)] [ P(ω ) ω (A B) ] = P(A B) [ P(ω ) ω A B] + [usng ()] = P(A B)+P(A B). +

17 PROBABILITY 399 Hence P(A B) = P (A)+P(B) P(A B). Alternatvely, t can also be proved as follows: A B = A (B A), where A and B A are mutually exclusve, and B = (A B) (B A), where A B and B A are mutually exclusve. Usng Axom () of probablty, we get P (A B) = P (A) + P (B A)... (2) and P(B) = P ( A B) + P (B A)... (3) Subtractng (3) from (2) gves P (A B) P(B) = P(A) P (A B) or P(A B) = P(A) + P (B) P (A B) The above result can further be verfed by observng the Venn Dagram (Fg 6.) Fg 6. If A and B are dsjont sets,.e., they are mutually exclusve events, then A B = φ Therefore P(A B)=P( φ)=0 Thus, for mutually exclusve events A and B, we have P (A B) = P(A) + P(B), whch s Axom () of probablty Probablty of event not A Consder the event A = {2, 4, 6, 8} assocated wth the experment of drawng a card from a deck of ten cards numbered from to 0. Clearly the sample space s S = {, 2, 3,...,0} If all the outcomes, 2,...,0 are consdered to be equally lkely, then the probablty of each outcome s 0

18 400 MATHEMATICS Now P(A) = P(2) + P(4) + P(6) + P(8) = = 4 = Also event not A = A = {, 3, 5, 7, 9, 0} Now P(A ) = P() + P(3) + P(5) + P(7) + P(9) + P(0) = 6 = Thus, P(A ) = 3 5 = 2 = P(A) 5 Also, we know that A and A are mutually exclusve and exhaustve events.e., A A = φ and A A = S or P(A A ) = P(S) Now P(A) + P(A ) =, by usng axoms () and (). or P( A ) = P(not A) = P(A) We now consder some examples and exercses havng equally lkely outcomes unless stated otherwse. Example 0 One card s drawn from a well shuffled deck of 52 cards. If each outcome s equally lkely, calculate the probablty that the card wll be () a damond () not an ace () a black card (.e., a club or, a spade) (v) not a damond (v) not a black card. Soluton When a card s drawn from a well shuffled deck of 52 cards, the number of possble outcomes s 52. () Let A be the event 'the card drawn s a damond' Clearly the number of elements n set A s 3. Therefore, P(A) = 3 = 52 4.e. Probablty of a damond card = 4 () We assume that the event Card drawn s an ace s B Therefore Card drawn s not an ace should be B. 4 2 We know that P(B ) = P(B) = = =

19 () Let C denote the event card drawn s black card Therefore, number of elements n the set C = e. P(C) = = 52 2 Thus, Probablty of a black card = 2. PROBABILITY 40 (v) We assumed n () above that A s the event card drawn s a damond, so the event card drawn s not a damond may be denoted as A' or not A 3 Now P(not A) = P(A) = = 4 4 (v) The event card drawn s not a black card may be denoted as C or not C. We know that P(not C) = P(C) = = 2 2 Therefore, Probablty of not a black card = 2 Example A bag contans 9 dscs of whch 4 are red, 3 are blue and 2 are yellow. The dscs are smlar n shape and sze. A dsc s drawn at random from the bag. Calculate the probablty that t wll be () red, () yellow, () blue, (v) not blue, (v) ether red or blue. Soluton There are 9 dscs n all so the total number of possble outcomes s 9. Let the events A, B, C be defned as A: the dsc drawn s red B: the dsc drawn s yellow C: the dsc drawn s blue. () The number of red dscs = 4,.e., n (A) = 4 4 Hence P(A) = 9 () The number of yellow dscs = 2,.e., n (B) = 2 2 Therefore, P(B) = 9 () The number of blue dscs = 3,.e., n(c) = 3

20 402 MATHEMATICS 3 Therefore, P(C) = = 9 3 (v) Clearly the event not blue s not C. We know that P(not C) = P(C) 2 Therefore P(not C) = = 3 3 (v) The event ether red or blue may be descrbed by the set A or C Snce, A and C are mutually exclusve events, we have 4 7 P(A or C) = P (A C) = P(A) + P(C) = + = Example 2 Two students Anl and Ashma appeared n an examnaton. The probablty that Anl wll qualfy the examnaton s 0.05 and that Ashma wll qualfy the examnaton s 0.0. The probablty that both wll qualfy the examnaton s Fnd the probablty that (a) Both Anl and Ashma wll not qualfy the examnaton. (b) Atleast one of them wll not qualfy the examnaton and (c) Only one of them wll qualfy the examnaton. Soluton Let E and F denote the events that Anl and Ashma wll qualfy the examnaton, respectvely. Gven that P(E) = 0.05, P(F) = 0.0 and P(E F) = Then (a) The event both Anl and Ashma wll not qualfy the examnaton may be expressed as E F. Snce, E s not E,.e., Anl wll not qualfy the examnaton and F s not F,.e., Ashma wll not qualfy the examnaton. Also E F = (E F) (by Demorgan's Law) Now P(E F) = P(E) + P(F) P(E F) or P(E F) = = 0.3 Therefore P(E F ) = P(E F) = P(E F) = 0.3 = 0.87 (b) P (atleast one of them wll not qualfy) = P(both of them wll qualfy) = 0.02 = 0.98

21 PROBABILITY 403 (c) The event only one of them wll qualfy the examnaton s same as the event ether (Anl wll qualfy, and Ashma wll not qualfy) or (Anl wll not qualfy and Ashma wll qualfy).e., E F or E F, where E F and E F are mutually exclusve. Therefore, P(only one of them wll qualfy) = P(E F or E F) = P(E F ) + P(E F) = P (E) P(E F) + P(F) P (E F) = = 0. Example 3 A commttee of two persons s selected from two men and two women. What s the probablty that the commttee wll have (a) no man? (b) one man? (c) two men? Soluton The total number of persons = = 4. Out of these four person, two can be selected n 4 C ways. 2 (a) No men n the commttee of two means there wll be two women n the commttee. Out of two women, two can be selected n 2 C2 = way. Therefore P( no man) 2 2 = = = 4 C2 C (b) One man n the commttee means that there s one woman. One man out of 2 can be selected n 2 C ways and one woman out of 2 can be selected n 2 C ways. Together they can be selected n 2 C 2 C ways. Therefore POneman ( ) C2 C C = = = (c) Two men can be selected n 2 C way. 2 Hence P( Two men) C2 C2 C = = = 6 EXERCISE 6.3. Whch of the followng can not be vald assgnment of probabltes for outcomes of sample Space S = { ω, ω, ω, ω, ω, ω, ω }

22 404 MATHEMATICS Assgnment ω ω 2 ω 3 ω 4 ω 5 ω 6 ω 7 (a) (b) (c) (d) (e) A con s tossed twce, what s the probablty that atleast one tal occurs? 3. A de s thrown, fnd the probablty of followng events: () A prme number wll appear, () A number greater than or equal to 3 wll appear, () A number less than or equal to one wll appear, (v) A number more than 6 wll appear, (v) A number less than 6 wll appear. 4. A card s selected from a pack of 52 cards. (a) How many ponts are there n the sample space? (b) Calculate the probablty that the card s an ace of spades. (c) Calculate the probablty that the card s () an ace () black card. 5. A far con wth marked on one face and 6 on the other and a far de are both tossed. fnd the probablty that the sum of numbers that turn up s () 3 () 2 6. There are four men and sx women on the cty councl. If one councl member s selected for a commttee at random, how lkely s t that t s a woman? 7. A far con s tossed four tmes, and a person wn Re for each head and lose Rs.50 for each tal that turns up. From the sample space calculate how many dfferent amounts of money you can have after four tosses and the probablty of havng each of these amounts. 8. Three cons are tossed once. Fnd the probablty of gettng () 3 heads () 2 heads () atleast 2 heads (v) atmost 2 heads (v) no head (v) 3 tals (v) exactly two tals (v) no tal (x) atmost two tals 9. If 2 s the probablty of an event, what s the probablty of the event not A. 0. A letter s chosen at random from the word ASSASSINATION. Fnd the probablty that letter s () a vowel () a consonant

23 PROBABILITY 405. In a lottery, a person choses sx dfferent natural numbers at random from to 20, and f these sx numbers match wth the sx numbers already fxed by the lottery commttee, he wns the prze. What s the probablty of wnnng the prze n the game? [Hnt order of the numbers s not mportant.] 2. Check whether the followng probabltes P(A) and P(B) are consstently defned () P(A) = 0.5, P(B) = 0.7, P(A B) = 0.6 () P(A) = 0.5, P(B) = 0.4, P(A B) = Fll n the blanks n followng table: P(A) P(B) P(A B) P(A B) () () () Gven P(A) = 5 3 and P(B) = 5. Fnd P(A or B), f A and B are mutually exclusve events. 5. If E and F are events such that P(E) = 4, P(F) = 2 and P(E and F) = 8, fnd () P(E or F), () P(not E and not F). 6. Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually exclusve. 7. A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.6. Determne () P(not A), () P(not B) and () P(A or B) 8. In Class XI of a school 40% of the students study Mathematcs and 30% study Bology. 0% of the class study both Mathematcs and Bology. If a student s selected at random from the class, fnd the probablty that he wll be studyng Mathematcs or Bology. 9. In an entrance test that s graded on the bass of two examnatons, the probablty of a randomly chosen student passng the frst examnaton s 0.8 and the probablty of passng the second examnaton s 0.7. The probablty of passng atleast one of them s What s the probablty of passng both? 20. The probablty that a student wll pass the fnal examnaton n both Englsh and Hnd s 0.5 and the probablty of passng nether s 0.. If the probablty of passng the Englsh examnaton s 0.75, what s the probablty of passng the Hnd examnaton?

24 406 MATHEMATICS 2. In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students s selected at random, fnd the probablty that () The student opted for NCC or NSS. () The student has opted nether NCC nor NSS. () The student has opted NSS but not NCC. Mscellaneous Examples Example 4 On her vacatons Veena vsts four ctes (A, B, C and D) n a random order. What s the probablty that she vsts () A before B? () A before B and B before C? () A frst and B last? (v) A ether frst or second? (v) A just before B? Soluton The number of arrangements (orders) n whch Veena can vst four ctes A, B, C, or D s 4!.e., 24.Therefore, n (S) = 24. Snce the number of elements n the sample space of the experment s 24 all of these outcomes are consdered to be equally lkely. A sample space for the experment s S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB BACD, BADC, BDAC, BDCA, BCAD, BCDA CABD, CADB, CBDA, CBAD, CDAB, CDBA DABC, DACB, DBCA, DBAC, DCAB, DCBA} () Let the event she vsts A before B be denoted by E Therefore, E = {ABCD, CABD, DABC, ABDC, CADB, DACB ACBD, ACDB, ADBC, CDAB, DCAB, ADCB} Thus P( E) ( ) ( ) n E 2 = = = n S 24 2 () Let the event Veena vsts A before B and B before C be denoted by F. Here F = {ABCD, DABC, ABDC, ADBC} Therefore, P( F) ( ) ( ) n F 4 = = = n S 24 6 Students are advsed to fnd the probablty n case of (), (v) and (v).

25 PROBABILITY 407 Example 5 Fnd the probablty that when a hand of 7 cards s drawn from a well shuffled deck of 52 cards, t contans () all Kngs () 3 Kngs () atleast 3 Kngs. Soluton Total number of possble hands = 52 C7 () Number of hands wth 4 Kngs = 4 C 48 C (other 3 cards must be chosen from the rest 48 cards) 4 3 Hence P (a hand wll have 4 Kngs) = C7 C C = 7735 () Number of hands wth 3 Kngs and 4 non-kng cards = 4 C 48 C Therefore P (3 Kngs) = C7 C C 9 = 547 () P(atleast 3 Kng) = P(3 Kngs or 4 Kngs) = P(3 Kngs) + P(4 Kngs) 9 46 = + = Example 6 If A, B, C are three events assocated wth a random experment, prove that P( A B C) = P( A) + P( B ) +P( C) P( A B) P( A C) P ( B C) + P ( A B C) Soluton Consder E = B C so that P (A B C ) = P (A E ) = P( A) + P( E) P( A E)... () Now P E = P B C ( ) ( ) P( B) P( C) P( B C) Also A E= A ( B C) = ( A B) ( A C) = +... (2) [usng dstrbuton property of ntersecton of sets over the unon]. Thus P( A E) = P( A B) + P( A C) P ( A B) ( A C)

26 408 MATHEMATICS Usng (2) and (3) n (), we get = P( A B) + P( A C) PA [ B C] [ ] = ( ) + ( ) + ( ) ( ) P A B C P A P B P C P B C P( A B) P( A C) + P( A B C) Example 7 In a relay race there are fve teams A, B, C, D and E. (a) (b)... (3) What s the probablty that A, B and C fnsh frst, second and thrd, respectvely. What s the probablty that A, B and C are frst three to fnsh (n any order) (Assume that all fnshng orders are equally lkely) Soluton If we consder the sample space consstng of all fnshng orders n the frst three places, we wll have 5 3 P,.e., ( ) 5! = = 60 sample ponts, each wth 5 3! a probablty of 60. (a) A, B and C fnsh frst, second and thrd, respectvely. There s only one fnshng order for ths,.e., ABC. Thus P(A, B and C fnsh frst, second and thrd respectvely) = 60. (b) A, B and C are the frst three fnshers. There wll be 3! arrangements for A, B and C. Therefore, the sample ponts correspondng to ths event wll be 3! n number. So P (A, B and C are frst three to fnsh) 3! 6 = = = Mscellaneous Exercse on Chapter 6. A box contans 0 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what s the probablty that () all wll be blue? () atleast one wll be green? 2. 4 cards are drawn from a well shuffled deck of 52 cards. What s the probablty of obtanng 3 damonds and one spade?

27 PROBABILITY A de has two faces each wth number, three faces each wth number 2 and one face wth number 3. If de s rolled once, determne () P(2) () P( or 3) () P(not 3) 4. In a certan lottery 0,000 tckets are sold and ten equal przes are awarded. What s the probablty of not gettng a prze f you buy (a) one tcket (b) two tckets (c) 0 tckets. 5. Out of 00 students, two sectons of 40 and 60 are formed. If you and your frend are among the 00 students, what s the probablty that (a) you both enter the same secton? (b) you both enter the dfferent sectons? 6. Three letters are dctated to three persons and an envelope s addressed to each of them, the letters are nserted nto the envelopes at random so that each envelope contans exactly one letter. Fnd the probablty that at least one letter s n ts proper envelope. 7. A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A B) = Fnd () P(A B) () P(A B ) () P(A B ) (v) P(B A ) 8. From the employees of a company, 5 persons are selected to represent them n the managng commttee of the company. Partculars of fve persons are as follows: S. No. Name Sex Age n years. Harsh M Rohan M Sheetal F Als F Salm M 4 A person s selected at random from ths group to act as a spokesperson. What s the probablty that the spokesperson wll be ether male or over 35 years? 9. If 4-dgt numbers greater than 5,000 are randomly formed from the dgts 0,, 3, 5, and 7, what s the probablty of formng a number dvsble by 5 when, () the dgts are repeated? () the repetton of dgts s not allowed? 0. The number lock of a sutcase has 4 wheels, each labelled wth ten dgts.e., from 0 to 9. The lock opens wth a sequence of four dgts wth no repeats. What s the probablty of a person gettng the rght sequence to open the sutcase?

28 40 MATHEMATICS Summary In ths Chapter, we studed about the axomatc approach of probablty. The man features of ths Chapter are as follows: Sample space: The set of all possble outcomes Sample ponts: Elements of sample space Event: A subset of the sample space Impossble event : The empty set Sure event: The whole sample space Complementary event or not event : The set A or S A Event A or B: The set A B Event A and B: The set A B Event A and not B: The set A B Mutually exclusve event: A and B are mutually exclusve f A B = φ Exhaustve and mutually exclusve events: Events E, E 2,..., E n are mutually exclusve and exhaustve f E E 2... E n = S and E E j = φ V j Probablty: Number P (ω ) assocated wth sample pont ω such that () 0 P (ω ) () P( ω ) for all ω S = P ω for all ω A. The number P (ω ) s called probablty of the outcome ω. Equally lkely outcomes: All outcomes wth equal probablty Probablty of an event: For a fnte sample space wth equally lkely outcomes n(a) Probablty of an event P(A) =, where n(a) = number of elements n n(s) the set A, n(s) = number of elements n the set S. If A and B are any two events, then P(A or B) = P(A) + P(B) P(A and B) equvalently, P(A B) = P(A) + P(B) P(A B) If A and B are mutually exclusve, then P(A or B) = P(A) + P(B) If A s any event, then P(not A) = P(A) () P(A) = ( )

29 PROBABILITY 4 Hstorcal Note Probablty theory lke many other branches of mathematcs, evolved out of practcal consderaton. It had ts orgn n the 6th century when an Italan physcan and mathematcan Jerome Cardan (50 576) wrote the frst book on the subject Book on Games of Chance (Bber de Ludo Aleae). It was publshed n 663 after hs death. In 654, a gambler Chevaler de Metre approached the well known French Phlosopher and Mathematcan Blase Pascal ( ) for certan dce problem. Pascal became nterested n these problems and dscussed wth famous French Mathematcan Perre de Fermat (60 665). Both Pascal and Fermat solved the problem ndependently. Besdes, Pascal and Fermat, outstandng contrbutons to probablty theory were also made by Chrstan Huygenes ( ), a Dutchman, J. Bernoull ( ), De Movre ( ), a Frenchman Perre Laplace ( ), the Russan P.L Chebyshev (82 897), A. A Markov ( ) and A. N Kolmogorove ( ). Kolmogorov s credted wth the axomatc theory of probablty. Hs book Foundatons of Probablty publshed n 933, ntroduces probablty as a set functon and s consdered a classc.