Special Discrete Possibility

8/14/2019 Special Discrete Possibility

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ln dris chapter, studnts will:(a) understand the concepts ofdiscrete random variable, binomial variablc, aDd poissonvariable;(b) understand that B(X) and Var(X) of a probability distribution are related to the mean andvariance respectively ol a frequency distribution;(c) understand that the binomial distribution B(r,p) and poisson dislribution po(,tr) arcxamples ofprobabilily distributions; and know the menn and variance ofeach iistribution;(d) recognise conditions unde. which the binomial or poisson distdbution can be use.l iomodel a random variable and use them to model practical situations;(e) use a graphic calculator to calculate probabilities;(0 commcnt on thc appropajate usc of a binomial or poisson nrodel and the assumphonsmade;(g) us(r the fact thal fhc surn oftwo or rnore indcpendent poisson variables is a porssonvariablc;(h) use the Poisson distribution as an apprcximation to the binomial disr!:ibution whcrcappropriatc (n > 50 atul ry) < 5, approximately).

1. l)iscrete Random VariablcA discrcte randon variable (r.v.) -f has the following properties:(i) it assumes a discrcte (countable) sel ofreal values (e_g. the nunber ofpeople waiting at a

bus-stop)(ii)itassumesvaluesr/,_r-r,_rr,._,r,r,_._withcorrcspondingprobabiliticspT,p2,pr,..-,

p", -.. such thatllrU-uS--t

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When referritrg to mndom variables, capital letters such as X { Z etc. are used. Whenrefening to the values these random variables assume, lower case letters such as .r, _jr, z etc.are used. For instalrce, let X be the variable "ihe number obtaind when a die is thrown".Hence, X takes values { 1 , 2, 3, 4, 5, 6} .

IP(x: 1): P(x:2): P(x:3): P(x:4): P(x:5):P(x:6)-7)r1x=x;: P(X: l) + P(X: 2) + P(,r: 3) + P(X= 4) + P(x= 5) + P(x= 6)-l

. . X is a discrele random variable.Example 1LetXbcthevariable'thenumberoffoursobtainedwhentwofairdicearethrown.'Showthat X is a discrete random variable.Solution:The outcome could be one ofthe following: no fou$, one fiour or two fours-Hence. X assumes the '"alues 0, I and 2 only. (P(.{ 0, rt t'' nor q. z"o no, +t - I 5 I 5 ) 'z 1Q)1 |\ " /ln.J - 'dP(X- 1) : P(l"t 4, 2"d nor 4) + P(l"t not 4, 2'd 4)(t\(s\ (s(1\ El;llal-loiol- lrP(X= 2) - P(1.' 4, 2.d 4):(:)t:) ?,

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)r1x=";: P(x: 0) + P(x: t) ). P(x = 2)255r,+-= I16 18 lt'

Jnu'lLHencc. X is a?andom r ariahle.

2- Probabilitv DistributionWe can w.ite the results in Example I in a table form which is known as the probabilitvdistribution of .{

X 0 2P(X:,r) 25/36 t0/36 t/36general, a probability distribution tabl shows all the possible values ofa random variableand their corresponding probabilities_

XP(X- ir)ttote: f r,1x=r1= | .2LA/x' "'/x')'"'r,*e,+r,+'7^=I

3. The Binomial DistributionConsider an expe.iment with two possible outcomes, one ofwhich may be tenned as"success" and the other "failure,,. This is known as "Bemoulli Trial,,_ When this experimentis repeated a number oftimes, it gives rise to a ',binomial distribution,,.For example,. toss a coin 6 times, consider obtaining a head on a single toss as a success, and obtaininga tail as a failure;. throw a die 10 times, consider obtaining a score of'6' on a single throw as a suciess andnot obtaining a '6' as a failure.

InX

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Conditions for a Bino(i) The experiment coNists of, independent rcpcatcd trials-(ii) Each tial has two possible outcomes: either a 'success' or a 'fiilure'-(iii) The probability of a success is the same in each trial-lfthe above coDditiols are satisfied, thcn the random variblc -tr representing the number ofsuccesses in dre n trials is called a rrromi.rl random varieble-

3.1 Definition:Ifthe binomial trial results in a success with probabilityp, theD we haveP(success) :t,Thc probability distribution ofXis given by

& P(failure) - 4

a hinomial distribution

whereq=l-p.rr,L-,1

This random variable X hasrepresented sFnbolically as with parameters r and 17 thatphtl i?,4 ot- .lacl

ls

[xampl 2A box contains a largc number ofballs in the proporlion ol1 white ball to 4 black balls.Onc hundrcd and twenty balls are randomly drawrr, one by one and \rvilhout replacementlrom lhe box. The r:rndom variablc X is thc nurDber of black balls drawn. State a suitablemodel for the distribution ofx. Justily your answer.Solution:The binomial model is suitable because:. there is a frxed munber ofindependent triats. (r : 120)' each trial has two possible outcomcs: cither a 'success' (drawing a black ball) or a'failure' (drawing a white ball).. sioce the box contains a large nurnber of balls (r > 30), thc probability of drawing a

black ball is approximately a constant although the balls are drawn without rcplacement.Note: The last point is dn dssumption made to ensure thal wc confonn to the conditionfor abinomial motlelwhich rcquires lhdt the probdbility of 'succe*' is the sane in each trial.

h,l*u,1,2,...,n whcrc 0


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Example 3Find the probability ofgettirg 5 heads in 12 flips ofa balanced coin.Solution: Note the chamcteiistics of the binomial distribution here:. The experimcnt consists of 12 independant trials (there is a fixed number of trials).. Each trial has two oossiblc ourcomes. either getting a head (a .success') or a tail (a.fairure') li)()u)6)el(il .. The probability olgen'ng ,r\r6rh {'succcs" ; is rhc scme in each flrp.

hrJ-, f)oaiJ"Exanple 4 (Do it yourself)An autoinobile salety engineer claims thai I in l0fatigue. What is the probability that at most 2 of 5l'atiglrc?

automobile accidents is due to driverautomobile accideDts are due to driver

,rrutu - ,-6 rlr >''

l-et X bc lhe r.v- "the no. of heads olrtaincd jn t 2 0ips,...-State thc distribufion ofthe random variable x-B( t)._ +\

P, 'tka,4z ll- ynb,L,l,,Write dowi lteiequiredprobability

P(,Y =5) -0.1934=0.193 (correctro 3 s.f.)i Key in the parameters for binompdf (n, p, ;)to compute P(X = 5) : binompdf(l2, l/2, 5)

Tasks:Lc1X be the t.v. "lhe

": 5 ' p:6 :no. ofaccidenis due to driver l'atigue".

lx-B( , . ro )1. Define thc mndomvariablc2. State the distribution ofthe random variable,1,r4/l. Write down the required

probability

?(xt2)" P(x --o)l flx=l ydv --71" i-ffiv--o tat Q',ru!4i,r)gg!)dA KN A >.ur.,lL](t < :.)/ t> (/>''X " pyaryto ki o''*,,111",r,') f- ,,*f'fu fq'-4 |I/ k'',)


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Example 5In a large populaiion, it is found that on average 4 out of 5 people will rccover from a certaindisease. Find the probability that in a random sanlple of l0 people, at least 7 will recoverfrom the disease

Example 6A social scientist claifirs that 507u ofall high school seniors capablc ofdoing colloge workactually go to college. Find the probabilities (correct to 6 decimal places) that arnong 18high school seniors capable oldoing college work, (a) cxactly l0 will go to college, (b) atleast 1 will go to college, (c) at lcast 2 but rot more than 6 will go to college.

Tasks: I-elJl be the r.v-disease".

"thc no. of pcople thai \\'ill reco\,er fiom the.. - 1:-7.,\"D,ttLt (/4/t/ ( t t / i41. Define the randornVanablcl. Statc the distribution ofthe random variable 10. f :o8 .r-- B(r0, 0.8) .,*r'Tt,-;1"VG /)

l- Write down the requircdProbability

p(x >'t):t ?(Xtb) *o.s7gl=0.879 (orrect to 3 s.f.)-ElnOmC,6) .87912611A4

Note that the GC is r.ulable to compute P(X > x).

l-el I be thc r.\. "tbc no. ofhigh sclrool senioN that wili go tocollcgc". ,i . fc, trr r.r, n,:.r)l iJ1. DeLine the randomVariablel. State the distribution ofthe random variahlc //: 18, 1i = o.5 ; "'>1)

Write down the requiredI'robability (a) P(x = r 0):0.166e24(b) P(x>1)=1- = 0.9eee96G) P(2


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ExamplcTln order to dctenniDe whether to accept a large batch ofmaDulactured articles, a sample of20 is inspccted. Thc balch is ac.epted ifthere are no delective articles in the sarnple, and itis rejected ifthere are two or more- Ifthcre is one defcclive article, a second sampl o[20 isexamined and the batch is acceptcd ilthis sccond samplc contains no delective afticlcs.Previous experiencc has shown lolo ofthe articlcs ar: defeclive.(D Calculate the probability that the batch is acceptcd after thc lirst sample is taken.(iD Calculate the probability that the batch is rejected_Solution:Lct,Y be lhc r-v. "the numbcr of dclcctjve articlcs in a samplc of 20 alticlcs_.'hr ihe lst sanple. if Xr : 0. Lhe batch js acccpred.If ,(1 ) 2. the barch is rejccled.11' )( : I, a seboud san4rlc of20 is exalnincd.In lhe 2nd sample, if/r - 0, rhc batch is accepted.Il Xr -: 0, thc batch is rcjccledGiven that p : 0,01, Xt - B(20, 0.01) for i :1, 2.(D P(batch is accepted aftcr the lsl samplc is taken)

:0.818P(batch is reiectcd) = p(xr > 2)+p(Xr = t)p(x, > 0)= 0.0469

J.2 Fxpcclation {nd \ arianccExpcctation oI a discrete random variablcIfwc have a statistical expcriment (lh.owing an unbiased dic 120 timcs and rccord thcresuhs):(a) A practical approach rcsults in atz4rrel c! dktributhh .ntl a fiesn v^l]oe

F reque n cy dis tribu I io n tableFrequenc 120'fhc nrean score is t 1x 15+2x 22 +,..+ 6x l8

(i0

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(b) A tlreoretical apprcachrcsults in a probabiliE distribution a1d an e$pecte.l value (orPxoecled ,xean\Supposc we toss an unbiased die and -Y is the 'number otl the die'.Then thc litv distribution ofX is as shown:

Score, x I ) 3 4 5 6P(X : x) 1/6 116 r/6 1t6 1/6 t/6The expectation ofX(or expected value) is

Notc:(i) E(-\l is a constant, not a function ol-,11:(ii) ECf) rcpreselts the average score that is expected to be obtained ifa suflicieltly trargenumbcr olthe same cxperirnent wcrc repeated under the same set ofconditions.Variance and Standard l)eviationWrile expcctation is a measurernent of ccntral tendeDcy (i.e. it attcmpts to locatc a tlricalvalue about which the distribution clusters), variance is a measure oldispersion_Variance measurcs lhe "spread" ofa sel ofobservations, i.c. how scattercd a distributioois. 11 is defined as the averagc ofthe squared dcviatioDs liom the mcan-Standard dviation is sirnply the positive squdre root ofthc variance That is,

Srrrr,laril Deviarion ../V a ricnc.Expcctation and Varialce ofa binomial random variablcIftblJ random varidbleXis such thatX- B(,,p), then

n(x)= np and Vat(X) = np(1 p) - npqNote: The fonnula list MF15 rcfeN to expectation as the mean_ Byrunning a sirnulation ola binomial experimcnt, we can see that the expectation is thc average in the long run.

Examole 8If on averagc 2 out of5 days are sunny, find the expected number ofsunny days in a week,and the standard deviation.

(,,.:)"(,":). .(.,.*)+l+.-.+6) =1 (orl.5)l'"ll\ 6lI= (l+

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Solotion:Let X be the r.v. "lhe no. ofsunny days in a r.veek.'.n=7; p:0.4; q:0.6. So I-B(7,0.4)E(X)=np= : Va(.{) =np(1- -"tondu.,l d"r.iotion = n&-@ =

Examplc 9 (I)o it yourselo'Ihe randorn variable X is binomially distributed with paramterst /] and p_ Civcn thatE(x)': rr.l v-(,Y) :1 .findrhcralucsornandp.llSolutiolr]

Example l0A pottery produces royal souvenir mugs. It is knowt thal 6% are defcctive.selectcd at random, find the probability that ihe san4)le contains less than 5and the most likely (most probablc) number ofdefectivc mugs.Solutiorr]Let l. be thc r.v. "the no_ ofdclcctivc mugs in the sample of 20,,.n:20, p: I-06:. X- B(20,0.06)

P(r


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3-3 Fittins a Binomial DistributionIt is sometimes useful to compare experimental results with a theoretical distribution.(a) Using the results obtained in an experiment, we can find the values of the parametersa andp by using i a F,(X) - np(b) Generate the binomial probability distribution forl- Bfn ,p)(c) To hnd the thcoretical expected frequencies, multiply each probability found by thetotal frequency ofthe given distribution.Examplc 1lGroups ofsix people are chosen at random and the number, -!, ofpcople in each group whonormally wear glasses is recorded. The results obtained from 200 groups of six are shownin thc table-No. in group wearing glasses (.x) 0 1 2 3 4 5 6No- ofoccurrenccs l7 53 65 45 1B 2 0Calculate, from the abovc datr, the mean value of]r.Assuming that the situation can be modellcd by a bitromial dislibution having the samenean as the one calculated above, state the appropfiate values for the binomialpaaameters /, and p.Calculate thc theoretical fiequencies corrcsponding to thosc in the table-Solution:LetX be the r.v. "thc no- ofpeople who wear glassesFrom the frequency dislribution,lr"" \-.L10(17)+1(53)+2(65)+3(45)+4(18)+5(2)+6(0)

200

in a g.oup".

Assuming that the situation follows a binomial distribution, i.e. ,f-' 8(6, p), thenIi \ E{l.) = ",a = 2 - p = -

.. I - Bro. 1rlUsirg GC, from tho lablc. the theoretical frequencies can be obtained by multiplying theprobabilitics by 200 (round offto the nearest integer).No- in grouo wearins slasscs (n) 0 1 2 3 1 5 6No. of occurrences l8 53 44 1'1 l 0

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Plot1 PlrtZ flct]rV r Ebi nonedf (6, 1rVzE26B*Vr\Vl=l\V6=

VrUII .oEtre I 1?jss1 l.26iitlsz-6?s2 l.le922 l6s-E'r'r3 1.219\E I'il.S96Ir 1.0!Zl lt6.16L5 1.016'rE11.2922E l.{0117 l-u7q35H=B

Note: Due io lound-up errorsthe lolal theolciical liequcncvis 201- flowcver. ihis docsnot iilvalidatc the rcsults.

particularp.ulicular;ntcrvalany other disjr)int

4. the Poisson DistributionThe l'oisson distribution is uscl'ur forcoosidcring the distribution ofmndom events. l,oissonprobabilities are used for data that arise from lhe number of occurences (X.) ofan cvent perunit timc or space e_g. 1he number ot yeasl cells per cm2 on I rnicroscopc slide.Namcd alier the French mathematician Silreon poisson, it is also used as an approximdtionto thc binorrial distriburion

I has a Poisson dishibution ifXis the r.v. "thc no- of occurrcnces of:ln event irr a givcninterval of timc or space" such that:(i) The event occurs in a specificd continuous intetwal of timc or space. At arypoint of iimc or spacc, an event either occuni or does no1 occur_{ rr r l llc r'\ cnr nc, urs independently. i.e. th" uccunence ol l]ny evclt ill aof tirne or spacc does not influencc the occunlcc of thc cvent ininteNal o1-liiDc or space.(iii)'lhe cvent occurs unifonnly, i.e. the expected number ofevellls (or mean) in a grven. jntcNal ol'tin1eorspaccis proportional to thc size ofthc intelv loftimcorspacc_(iv) Ihc event is rare, i.e. the probability ofsucccss ofan event is small (or unlikcly).Erarnplcs ofcvcnts which follow poisson distriblrtiol:The nr-rrnber.,f(a) flaws ir a given lcngth ofmatcrial,(b) car accidents oD a particular stretch of road in one day,(c) tclephone calls made to a switchboard in a given minute.(d) pafticlcs emittcd by a ftdioaclive source in a givcn time,(c) sandstorms occurring in a given pedo


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Example l2The number of times a radio set breaks down in a yeiu (52 weeks) may be taken to be a Poissonvariable with mean l Calculate(a) the probability that a radio set will break down morc than twice in a year,(b) rhe probability that a mdio set will go for at least 3 years without brcaking dowrl(c) the probability that a radio set will break down twice in one pa(icular week.

In short, X -Po(l)The probability distribution of.r is

Solution:(a) Let X be the r.v. '1he no. of tim{rs a radiosct breaLs down in a yeai'.So I- Po( )P(X>2)=

- 0.0803E 1 < C:Poissoncdf(>

-POrSSOnCdt(t,. 68436139

P(X:x)=e 2{ where r>tt, r=0,1,2,3...

(b) Let f be the r.v. "the no. of times a mdiosct brcaks down in I yeals".So r- Po( ) i.c. r-Po( )

otEP(I =0) =0 04q8

P(ttl=2)=0.0001,31

(c) Lct tf bc thc r.v. "thc no. of timcs a radioset breaks down iD I week".S,' l,' n,f )tl

Example 13An insurance company receives on average 2 claims per week from a certain factory. Find theprobability that(a) it receives more than 3 claims iD a given weelq(b) it receives more than 2 claims in a giver fortnighl,(c) it receives no claims on a given day, assuming that the factory operates on a 5-day week,(d) the company receives no claim on cxactly 3 days in a 5-day wcck.

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Solution:(a) LctX bc the r.v. "the no. ofclaims perweek"- X-Po( )P(X >3)= ,0141(b) Let I bc the r,v. "the no. ofclaims in 2

r- Po( )(d) Lel1) bc the r.v. "the no. ol days in aweek that the company receives no claim".,-B( )

given day''.w- P,,( )r.j

P( t4/ = 0) = 0.6 70

P(r>2)= = 0.'762 P(D = 3):0.328(c) Let lil be dre r.v. "thc no. of claims on aNOTE: A binomful distribttion has a fixed maxirnum number of surcesses whereas aPor'ssar distribution does not.Exarnple 14A student js studing the distdbution ofbullfrogs in a large rural field wherc thcrc is an a,"eragcof 500 bullfrogs. per 400 kr1r2. Onc part of the field i; identificd and dividcd ioto 50 equ,rlsquares' cacir with sidcs measuring 2 km- The student wishes to modcl tbc distribulion ol.bullfrogs in each squarc by usirrg (he poisson distdbutiqr with mean )-_(i) State the value of7. and one assurnption made in using the l,oisson clistribuhon.(ii) Four squarcs arc sclected al mndom. Fincl thc probability that two ofthese squares containno bullfrogs and each ofthe other two squarcs contains ai lcast 2 bullliogs-(iii) The student suggesls using ihc samc tnodel on anothcr nual ficld ii anorhcr courtry.Conrnrent on the suilability ofthc modcl in this siluation.lofUlqq(D Number ofsquares in an arca of400 k1[, :400/4: 100Hence, 7, ..

One assumption made is lhat bullfrogs are randomly scattered ii each square.(ii) LctX bc the r.v- '1he no. of trullfrogs in each squarc,. Then {- l,o( ).The required probability =

=[t'(x=0)]rx[l p(X < t)1, x6= 0.000251(iii) 'lhc rnodel may not bc suitable bccause ofthe diffcrent distributio ofbullfrogs in thc othercountry duc to cnvironmcnlar / climatc r'ectors- As a rcsult, thc co.ditio^s for a poissondistribution may no longcr apply_

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Altemative aeasons: (l) The environment / climate in another couDtry may not allow foran uniform occurence of bullfrogs throughout the ficld. (2) The environment in the freldin another country may be very colducive in the brceding of bullfrogs and thus it is nolonger considered a rare event.4.2 Expectation And Variance

IUa Po (1"), then E(x)- ),-Yar(X)Example 15 (Do it yourselfl(a) If x'' Po(2), find (i) P(x-4), (iDPtx 2): 1-P(),


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Siucer- 500(> 50) and ry = I (< 5 ),I-po( ) approximately_P(X = 2) = 0. t84

Itxample l7Each week a sccurity fiIm transports a largc sum of money belween two placcs. The day onwhich thcjounley is made is varied at raDdom, and in any weck each oftire fivc days lionMonday to Fdday is equally likely to bc chosen.(i) Calculate thc prlbability that, in a period of l0 wccks, Friclay wiil bc chosen arl(ast t linL(\.(ii) The evcnt that' in a 4-wcek pcriod, the samc day is crrosen on all rour occasio,s is denotedby S_- Show that rhe probability of s occun ing is 0.00g. rsrt dle the probability that, inone hundrcd 4 wcck periods, the evenl S will occur at lcast 3 times.

Friday is chosen ir a pcriod oftcn weeks,'.)= o.322

(ii) P(S occurs) : P(all Mondays) + f(all Tucsdays) + . .. + l,(all F.idays): = 0.008Let f be drc r-v. "the no. ol timcs S occurs in olle hundr.ed 4 week periods,,)'- B( )

SolutiolL(D LelX bc the r.v- "the no. oftimes.{-' B(P(,].>l)=

So )' - Po(P(v>3)-

) approximately.

trxamplc 18IfX- Po(3.4), find the mosr likcly value of,y.SolutioniNote: A I'oisson disrribution is bimodal if )" is an integcrA Poisson distribution is unimodal if L is rot an integcr.Using GC, f(nu thc krblc. thc most likely valuc ol'X is 3.

Plntl FIrtZ gi{tlrVr EroiEsonpdf(3-(11117.111q7.is19.zlr62.1EgEZ.11616.'!716

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4-,1 Fittins a Poisson DistributiopSteps:(a) Fronr the liequency distdbution, find the mean ;.(b) Use the mean i as the parameter of Poisson {:lislribution, gcncmle the poissondistribution forf - Po(;J.(c) To find the thcorctical expected frequencies, inultiply cach probability found by thctotal frequency ofthc given distribution.

Exarnplc I9The number olphone calls rr:ccived over a pcriod of 150 days was rccorded.Number ofcalls 0 2 l 4

5l 54 6 3(a) Find the avcrage nrmber of calls per day.(b) Calculatc the frequencies of the compamblc Poisson distdbution.Solution:

Oarl, tli14)_2(lot . t{o).t 4{l) 1.04rr, I50(b) Let X be the r.v. "the no. of phonc calls received per day"- So .Y- Po( 1.04),Multiplying the probabilities by 150 (rounding offto thc nearest inteSer) gives the theoreticalPoisson frequencies.

Pl{t1 PlotZ Pl+t3rVt Epoissonpdf( I.64, H\Vz E 158{,V I\V] =T\V9=

41z3sl-tls59.13S:8.6723.9197Z-58r{f.5]75q.09317

.lSlrrs.3A7Sg.19115.06626.01723.0035tE-2E -r'l

L'1umber ofcalls 0 2 3 4Number of days 53 55 29 t0 3

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4.5 T_hc Distribution ofTwo or More Indcpendcnt poisson VariablesIfXand Y arc two independent Poisson random variables with parameters p and Lrespectivcly, that is, X- Po(p) and I- po(,,), thenX+)'-Po(p+?,)

Thc abovc result can be extnded to 3 or more independcnt poisson random variablcs.If XL Xr, -.--.. X. are ll independent obsewations of a poisson r.v. L with pammeters )",thcn.

X,+ X,+'..+ X,- Po(nL)Examplc 20During a certail morning, the nurnbcr ofpeople entcring a ccriailr LRT station pcr minute has aPoisson dislibution with mean J and the nurnber- of people Icaving thc station ier minutc has anindependent Poissoo distribulion with mcan 2_(a) Find, to 3 siSnilicant figurcs, the p.obabilily that during a particular minule in the monring.(i) not more than one pcrson willettcr thc statiol,fii) n,,,'ne will ..lrcr nr lcrre rhc.r.rrion,(iii) exacily onc will enter thc shtior given th:rt thc number ol.people entering alld leavingthe statioo during thc particular rinute is cxactlv 4_(b) Detennine thc least numbc., such thar rhe probrhiliiy of nur more than , people leavingthe station in a:10 second inteNal is srea1cr lhat 0.9i.$ol!l!!rn:Lct,{ be the r.v. "the no_ ofpeople cnlering lccrtain LRl station pcr lrrinute,, and I bc ther.v. "the no. ofpeople leaving a certain Ll{Tstation pcr minute".

(iii) x+ Y- Po( )P(exactly one will entcr the stalion lno_ol'pple entcring & ,cavtng the stdtiun 4t

P(X =l ft X +Y =4:)P(X+Y =4)P(X:l),P(1.-3)

SoI'-Po( )and l/-Po(irrdependent. );Xand f are(a) (i) l'(not more than I will enter the station)= = 0.199(ii) P(no onc willenter or lcavethestation)

= 0.00674

P(X +Y =4):0 t54

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O) Let Wbe the r,v- "the no. ofpeopleleaving the station in a 30 second intewal".So I/- Po( 1)P(W < n)> 0.9sUsing GC, from rhc table.the least value ofn is 3.

Example 21 (Do it yourself)The number of flaws in a roll ofa paflicular design ofwallpaper is known to follow a poissolrdisldbution with mean 0.2. Calculate the p.obability that(i) in each oftwo rolls selected at Dndom thcre are no flaws,(ii) in a random sample of5 rolls ther.e will be a total ofexactly 2 flaws.Solution:Let -tr be the r-v. "the no, offlaws in a roll(i)X+Xr-po( )..P{,rl+x,=0)=

of a parlicular design ofwallpapei'. SoX- t o((iD{+Xr+...+Xj-Po( )..P(;Y r,Y. ,...- X,- 2) -

).

Example 22An advcrtising display contains a large nurnber of light bLrlbs which are continually beingswitched on and off- Individual lights fail at ra0dom timcs, and each day the display is inspcr:tcdand any failcd lights are rcplaced- Thc number of lights that lail in irny one day period has aPoisson distriblLtion with mcalr 2-2-(a) Find the probability that at least four lights will need to bc replaced on a particular day_(b) Find the probability that in a period ofseven days, at lcast four lights need to be replacgd onat least two days.(c) Caiculate the least number ofconsccutive days after which the probability of at least onelight having to be rcplaced excecds 0.9999.Solution:LetX be the r-v. "the no- oflights that fail in a o[e-day period". SoX^,po( ).(a) P(-r > 4) = =0.1806!0.181(b) Let y be the r.v.'1he no. oldays out o[seven where at least lour lights need to be replaccd,,So f- B( ). * 0.169

Pl+tl fl+tz Plot3\Vr EFoissoncdf(, Hlr\v5- 0IL

E

.36?EE.7]976.9197.9S101.9963rt.999q1.99992

P( )' > 2) =

c25&26 l8


19/19

(c) Let Itl be the r.v. "dre no. of lights that fail and are replaced in z consecutive days"-So W- Po(2-2 ).> 0.9999t p(n = 0) > 0.9999

P(n/=0)

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Special Discrete Possibility