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mF, Peantioat relinmey Aewetrepent 4 the Patelor anes Phe limbennene « eh pe A plab a Rareeay we CROUEE Gene he AHERN) ey Vike anvehveee, tee, Heed tie sbtlity ANALY OLE @ OLEAT CE) mene ble aot ee whether the ain vielted « . BEOOIAION, Aaa, SVEAL 9?) epee: alsa be aa ab Benevated ty PM toal damortitiin vata Aah SANS AYE @atsaN, WIEN Baul Senora ly WAU the feeb hae ewan ACODT IO Ad phReee Breas shemung (te operation hy SOM pOKe ; APPrOXlnation of x Aigner — Qu the . : Yr he other hand. 7 hb . if the came at = outoo ee i ependent events. Me of One event te atfhoted by other, then tt ‘ . Jone evente aes called 5.3 PROBABILITY Probability may be defined aa the ¢ rT m CRE ARETE Poe EewapatansGncerceiniaranet Sere oe random experiments, Tr any random experinent, (here te Sree ee carr ee ak ae SPOUSES IE TOU COR Tn aaalaneds Tf (t ta eure that an event will oss, y¥ that its probability ia 0% or 0, If it HAT UES SUCCES y will ogowy, then We oan eeeraiitais batcemLa nace ver the event will ooour of hot, then te Det us consider an example, The probability of ocounrence of wall February In & year tet alee i" is certain to occur every year, On the other hand, the probability of eeou Hot Pebraaey bag year is O since it never comes, Again, the probability: Y ooourvence of vy in a year ta neither O not 1. Itis always between 0 and 1, Actually, it ia 1/4 singe it oeoure every leap year bey once in four year: “Therefore, from above dis Probability of an event A, t Number of poasible favourable outcomes P(A) = aT : reece i > umber of outcomes rota er in tossing of a die the probability of getting an even number 6 equally likely way® sal expreanion for probability as athematic cussion, we can write ar let us consider For example, vin an eve \ In tossing of a die, an even umber can occur in § ways out 3 1 4 a tAy ee we 2 0.5 ON Therefore, P|WY 6 5.4 PROPERTIES fe] PROBABILITY — — — epee asible outcomes © © : = en ace S contains all poss? 1 Oa ockiTan . As discussed earlier, the sample space © oventa are anid to De rnareurntly clase Me ga am Pal vent is ne ne. accurrence of ¢ : m per 4) W 9 orate of one of them Pry ludeatty se eidie, tha ocourrencs | num ne outcomes then tt ls < ied : qutually exclusiv’ eas 1 6, If an event contains all th ace =“ “ ¥ " 2 3 5 ane 3, ps sbers 1 2. 9 ahe ae oe “4 f eobability ye this event is unity: © : ren, ve ae ; PCA) PLO) Rg listed aa unde: fp ebability may be , ten ‘ The properties of pr weeny oka certain event is unity Se eee ptr _ “ —— 908 f@ Communioalion Systane f Pye) wh) Property % The probabiling of any event 16 always lows than or equal ta 1 and nonnegative, thematically, a Oe PAs) MD Property & If A and are Wwo mutually exdudiye events then PA 4 B) = MALS rt) wi Property 4 If A is any event, then the probability of not happening of A is P(A) #1 = PIA) wl) where A represents the complement of event A, Property :1fA and B are any two events (not mutually exclusive events), then 55) P(A+B) = POA) + PUB) ~ PUAB) de ca jaa P (AB) \e called the probability of events A.and B both occurring smu taneously, an bi ed B) ie called the joint probability, event is called joint event of A nnd B, ond the probability 2 (A, Now, ifevents A and # are mutually exclusive, then the joint probability, PAB) = 0, 5.5 CONDITIONAL PROBABILITY t of conditional probability is used in conditional occurrences of the events, Let us consider ich involves two events A and B, Now the probability of event B, given that event A has occurred, 1s represented by P (BIA) Similarly, P (AIP) represents probability of event A given that event B has already occurred, Therefore, P (BIA) and P(A/B) are called conditional probabilities, These conditional probabilities may be defined in terme of their independent and Joint probabilities ae under: wid given that event A has already happened, The concep! an experiment whi The conditional probability of ave! A) = paw = Fan) where P (AB) is the joint probability of A and B, Similarly, the conditional probability of event A given that event B has already happened. (5.6) P(AB) P(AIB) = PB) w(5,7) At this point, it may be noted that the joint probability has commutative property which states that P(AB) = P (BA) wB.8) 5.6 PROBABILITY OF STATISTICALLY INDEPENDENT EVENTS IA and B are two events in an experiment, and possibility of occurrence of event B does not depend upon occurrence of event A, then these two events A and B are known as statistically independent events, The probability of event B, given that event A has already happened is expressed as P(AB) P(BIA) = Dc Ay (5.9) Again, since occurrence of event B does not depend on the occurrence of event A, then the probability of event B will be same as conditional probability P (BIA), Mathematically, (BIA) = P(B) v5.10) Now, putting the value of P(BIA) from equation (5.10) in equation (6.9), we have \PAB) = P(A) P(B) wB,11) aA E Tine atter the secon t,t ay = A. drawing, there O16 black balls out of w total of 6 balla, fore the required robabilivy jy 42 hoe Par, AP) HAA) Bhp dnb nd a 8 Ans, PLE 5.5, Ab foams: Pind Pipe Containg 5 white, 4 red and 2 black balls, Three balls are drawn in © Probability that the balls Will be of different colours, Solution; Let W, he 1 Ht, B66 the oventy of drawing white, red and black balls, respectively. It may be served that there arg 1% powsible Way of govting the balle of different colours a6 follow: WHE, WERK, h WH, RBW, BWH, BRW Now, the Probability of firg Combination WRB wi) be PHB) = £8 2 10’ 9'g Here, it May be noted that = ¢ ie the probability of wetting first ball white, 2 '4 the probability of getting second bal red, P= (Ta*h5)+( nbd) (2 cded) *(i*5*8) m Probability, Random Signals and Random Process m ny MPLE 5.6. Three students A, B and Care given 4 problem in Maths, The 4g 1 Probabilities of ewe wel wine bi a’ g °nd > respectively, Determine the Probability that the problem is solved if all of them try to solve the problem. (PTU, 1999) Solution: The probabilities that A, B and C will n jot be able to solve the Pp i :) ( " 4 4 3 Toblem are 43 Therefore, the probability that no one will solve the problem = Sy cee 4. 5. deauda Hence the probability that the problem will be solved hag. cia te Ans. EXAMPLE 5.7. A box contains three white balls W,, W,, W, and two red balls R,, R. are drawn in succession at random. Find the probability that the first drawn ball hae the second ball is red. and Solution: Probability of event W, i.e. first ball is white will be No. of white balls 3 ~ Total no. of balls 5 Now, if a white ball is drawn, there remain two white and two red balls. Now, the conditional probability of the event of second ball red with first white R)\_2 aia Hence, the probability of the event that the second ball is red with first white, will be = P[W,, Ry] = P (first ball white, second red) =p). PRE Aa «sp -12 PUR) = 5 on ae ne EXAMPLE 5.8. In a coin-tossing experiment, if the coin has head, one die is thrown and the result is recorded. But, if the coin has tail, two dice are thrown and their sum is recorded. Find the probability that the recorded number is 2. Solution: We know that in a coin-tossing experiment, the probability of getting head = Similarly, the probability of getting tail = 1- ; = ; Now, the probability of getting head with recorded number 2. iy £4 1 = (3\3)=a9 «td Similarly, the probability of getting tail with recorded number 2, will be rao Ge 4 ¢ Gy) er aa) Therefore, the required probability of getting recorded number 2 will be the sum of the equations @O&@ bole a Probability: Random Sianale and Random Process @ anid ? — , ewer nn vanorntmPle: the con Oe asda gcd at random variabl copt of random variabl From this example, crete random variable ot! X takes on anly a fnite number of values Le. 6. a 5.7.2. Continuous Random Varlabies . A random vari ‘ teu. “ped email that takes on an infinite number of values ie called a conti or outcomes, Such s: ie are several physical ayal periments) generate continuous outputs Continuous [oe tbotiied saben number of outputs.or outcomes within the finite period — RiPhsanexaraple, the noise voltae 4 used to define the outputs of such systems. This means that sam noise volta 6 generated by an electronic ampli has a continuous amplitude. ple space S of the noise voltage amplitude is conti z é, in case, the random variable X has a continuous range of values. 5.8 PROBABILITY FUNCTION OR PROBABILITY DISTRIBUTION OF A __DISCRETE RANDOM VARIABLE tia te discrete random variable and also let 4; xg X «- be the values that Xcan take. P (X=) = (x) 45.45) bs ade where j=1,2,3,.0 ’ will be the probability of x,. This f(x) or Simply f(z) is called the probability function or probab ility distribution of the discrete 5.10. In an experiment, three coins are tossed simultaneously. If the number of is ig the random variable, find the probability function for this random variable. Golution: In the three tosses of coins, there are eight possible outcomes. ‘This is the sample space Ss. Let the random variable, number of heads, be X. Since, in the given experiment, the random variable X, number of heads, takes only finite values, this is a discrete random variable. Writing the sample space S$ and values of discrete random variable Xas S={HHH HHT HTH THH HTT THT TTH TTT) Kew Fig, 5.1. = Fy ()-0= Fz) Hence,wehave yx) = Jfxtrdz Hence Proved. DenPerty 4: Probability of the : fx, < X< x) is sim i the area under ~~ Sttnemcta er he range 2, < Kee a ra PG, 0. x x Therefore, F(x) = fect aes foctars f ao 3 ae Kea x=0 ot This factor for x <0 musta fata ee Hence, for x < 0, we have 3 Jac ax SAM =p dae ees Similary, for x2 0, we have 3 foe dx+ foe™ dx x=0 a o* ei-s[e"f. ; [e*= &) oi ola x ~ i) " ~ 2 jets Gh ee tr: 2-e=55 R-e (Q-e*)=1- 3 eo DF will be given by wleola cla ole " Therefore, the expression for Cc Fyx) = ; a for x <0 1-; ee forx20 lative distribution function (CDF) with respect to Plot of cumu We know from previous discussion that for 8 CDF Fx) Fy(x) atx =- 9 =0 ————————) CaP ENT treed anton RR =zmaey te . ea pee Sa w WP semaadias Re=2 e ARs = Tamim fes=3 whee sa RRsa: See 7” R= xe. Theis, RFR<> —— Ss AX<> >We aa : @@ Tie POP ic ewesiias a< §e=—F See = bd ~wexsd * = 2) fr O xd 10 Q 070 [= colse xan | |f = oxi x0 ts 4\2 #' G>xld-1= GX)d qo oF 0 4 += [Sl-24] = (L>X)d ) u 1! x! ‘ vias on eT yi Ozh s0F a = (MY pus | = Pepnp(t ») a) 02x30 5a = (2)X) ee ‘so[quitea Wopues uepuedaput axe 4 pus x ‘soUaEE (Of: OH = 02-29 = 4x9 = (C97) se WOAts st oHOUNS AyIsuap AyyIGeqord quzol yey uaa :wOYM[OS (1>4 +X) & 
