gk math 2.pdf

PROBABILITY & STATISTIC S

PROBABILITYIt is a concept ofmathematics which measures the degreeofcertainty or uncertainty ofthe occurrence ofevents.. If any event can happen in rn ways and.fails in n

ways and each of the (m + n) ways are equallylikelyto occur, then probability ofthe happening ofthe

events is defrned as the ,utio, L and that of itsm+n

ntallrng as

-

.m+nIfprobability ofthe happening is denoted byp andnot happening by q, then p + q = l.

o If event is certain to happen, its probability is unity.. Ifhappening is impossible, then its probability is zero.SAMPLINGA small section selected from the population is calleda sample and the process of drawing a sample is calledsamplirug.

Random sampling.It is essent e must be a randomselection so ofthe population hasthe same chance of beingincludes in the sample. Thusthe fundamental assumption underlying theory ofsampling is random sampling

Simple sampling.A special case of random sampling in which each eventhas the same probability p of success and the chanceofsuccess ofdifferent events are independent whetherprevious trials have been made or not, is called simplesampling.

Parameters.The statistical constants of the population such as mean( p), standard deviati on (o) etc. are called th e p arameters.

Statistics.Constants for the sample drawn from the givenpopulation i.e. mean (x ), standard deviation (S) etc.are called statistic. The population parameters are ingeneral, not known and their estimates given by thecorresponding sample statistic are used. We use theGreek letters to denoted the population parametersand Roman letters for sample statistic.

Objectives of Sampling.Sampling aims at gathering the maximum informationabout the population with the minimum effort, costand time. The object of sampling studies is to obtainthe best possible value of the parameters under specificconditions. Sampling determinesthe reliability of theseestimates. The iogic of the samplingtheory is the logicof induction in which we pass from a particular(sample) to general (population). Such a generalizationfrom sample to population is called statisti.cal inference.

Sampling Distribution.Consider all possible sample of size n which can bedrawn from a given population at random. For eachsample, we can compute the mean. The mean of thesamples will not be identical. If we group thesedifferent means according of their frequencies, thefrequency distribution so formed is known as samplingdistribution of the uleqn. Similarly we can havesampling distribution ofthe standard divination etc.When drawing each sample, we put back the previoussample so that the parent population remains the same.This is called sampling with reploremenf and all subsequentformulae will perbain to sampling with replacements.

STANDARD ERROR.The standard error is used to assess the differencebetween the expected and observed values. Thestandard deviation ofthe sarnpling distribution is calleclstandard error (S.E.). Thus the standard error of thesampling distribution ofmeans is called standard errorof means. The reciprocai of the standard error is calledprecision..

If n > 30, a sample is called large otherwise small.The sampling distribution of large samples is assumedto be normal.

POPUI,ATION.Collection of all possible samples is population. It isimpractical to collect data on each sample of apopulation. Statistics helps to determine the bestestimate of the population parameter from randomlyselected samples. Because random erros are involvedin determination of parameters, the estimate willrepresent a parameter with a given probability only.

3.2

}IEASIJRES OF CEI{TRAL TENDENCYThese'are the Mean (Arithmetic Mean A.M or theaverage), Median (positional average), Mode (value ofthe variable which occurs most, frequently) Geometric\Iean (G.M. used extensively in frnding the rate ofpopulation growth) and the Harmonic Mean (H. M.)Let rr, xr, .............. r., be the observed values and letJ- be the frequency of x, (t,) value of r repeated f,times).i.e.. if the data is 3, 1, 3, 3, 2, I, 3, 2,2, then

x, = 1, xz= 2 and xs = 3,and /, = fz= 3 andf,, = 4.

Arithmetic mean(.)._ 1 1+, = ; (tr' +x2 + .... + r^, = i\tt

and for frequcncy distributionIi= *-{11x1

+ f2x2+....+fnxr)

PROBABILITY & STATISTICS

<nn

"-1 Y4",,whereN=In* N ?=t' r.=r

Median.Median divides the collection of data into two equalparts. Hence this is a positional average.

Mode.Mode is the value which occurs most frequently.

Mode =o+c(fi- ft-)

2fi-fi-t-fr*rwhere (a - b) = modal class

4 = rnaximum frequency

C = constant difference for each class.

Example. Find median of the following data.

Cost : r0-20 20-30 30-40 40-50 50-60Items in a group: 4 5 3 6 3

Solution.

Here. lN

Hencc 2

Frcm fon-n

= 10.5. The median class is B0 - 40..t0

ula. Median=go+i(10.5-9)=30+1.25 = 51.25I2Example. Find mode for the following distribution:

Solution.Maximum frequency = 2g.'. Modal class = 40 _ b0

Mode= o*-9!t-tr)-=40+2fi - [i-t - fi -t

10(28 - 12)= 4O + 6.666 = 46.G7(2 x 28) -12-20

Er" he following frequency table

Cost Numberof iterhs in thegroup

Cumulative frequency

10-20 4 4

20-30 o 9

30- 40 3 L2

40-50 6 18

50-60 aL2

N=21.

Class interval

Nlonthly Expenditure(in hundreds of rurreesr

0-10 t0 -20

,

20 -30 30-40 40-50Number of families I4 27 2 15If mode for the distribution is 24, calculate the missing f."q""*t""


Solution.Let missing frequencies for the classes 10 - 20 and 30 - 40 be f, and I respectively.

N = 100 =86+fr+fzfr+f" = 1gg-5$=44

is given tobe24, which lies in the class 20 - 30. So 20 - 30 is the modal class.

C(fi - fi:,)

(')

Mode

Mode = o+

24 =20 +

2fi-fi;-fi*t10(27 -ft)

(2x27)-ft-fz

"' fr=23Substituting in (r) , we get fz = 21.

Relation between Mean, Mode and Median.For a symmetrical distribution

Mean =Median=Modeand for a moderately asymmetrical distribution

Mode=3Median-02Mean.

= Mean-Mode = 3 (Mean-Medain)

Geometric mean (G).

G = n.,l x1x2......xn = (xrxr...........*n)''"

270 -r0f1 2t0 -r0f1a'ri _()A =

-

54-tr-li 54-44('.'fr+fr= 44)

Harmonicmean.It is the reciprocal of mean of reciprocals, i.e.

H = -----l---- .Ir n I

I 'F(tlr,) |[";i )and for a frequency distribution,

n

where N = tfi

Skewness.Skewness is defined as lack of symmetry.

Coeffrcient of skewness =Mean-Mode

Standard deviation

Negatively skewed Positively skewed

),*r))

and for a frequency distribution,

G = (rf r*....... "# )t'* , where N =

For both cases, taking log,1n

IogG= a)log(r;)n-

and for a frequency distribution,1+"

logG= *ltloe(ri)If n' and n2ate sizes and G, and Grare geometricmeans of two series, then geometric mean of thecombined series is given by

logG= ntlogGt+nzlogGzn1+n2

Geometric mean is used to find the rate ofpopulation growth and the rate ofinterest and it isalso used in the construction of index numbers.

ir,i=1

Expenditure Number of families Cumulative frequency

0-10 L4 l4

10-20 Il t4 7 f

20-30 27 4I r E

30-40 f, 4l+f,+f,40-50 15 56+t+f,

5.4

MEASURES OF DISPERSION.The set of constant which would in a concise way explainthe "variability" or "spread" in a data is known as"measures of dispersion or variability".The average for two groups of the same number ofmeasurements may be equal, but one group may bemore variable than the other.e.g. set of frve values 5, 6,7,8, t has the mean as 7;while another set of frvevalues 1,6,4, 10, 14 alsohasthe same mean 7. The second set has, obviously morevariability than the frrst.UsuaIIy four rneasures of dispersion are defined:

1. Range.This is the difference between two extremeobservations in the data given.In a frequency distribution ,

fi, = (largest r value) - (smailest r value).Obviously, this is not a very satisfactory measureexcept in cases where the number of observationsare small and a quick calculation for the scatter isneeded. It is used in statistical quality control studiesrather widely.

2. Quartile deviation.Median bisects the distribution. If we divide thedistribution, into four parts, we get what are calledquartiles, Q' Q, = median and Q.. The frrst quartileQ,, would have 25 percent of the values below itand the rest above it;the third quartile would have75 percent ofvalues below it and the rest above it.The method of calculation of quartiles is similar tothat of the median with slight variations.

O=.9

-xl/1 .,^-\i't*'


for discrete data

N!:n - lQt

Q,=/+ I . rC;I

where we locate Q, - class and Q" - class properlyI = lower limit of the quartile classC = cofirrnon factor

Quartile deuiation is defined as

Q.D.= ] re,_Q,l

3. Average deviatiorJ(l).Ifaverage chosen is A (say), then average deviationabout A is A.D.'t-A. D. (A) = : > .l(x; - A) | for discrete datan.-

= 1; ftlxi - A I for a frequency disiribution.n /_/.,

Usually, we take either median or the mean as theaYerage.

4. Standard deviation (Root mean squaredeviation).Standard deviation.

il q

= ./- I f, (" i - i)' for frequency distribution.li1\

Square ofthe standard deviation , o2 is defrned asthe variance (V).

v = oz = fr>f,tr, -t)'Out of these measures, the last , o is widely usedas a companion to r on which is based, whendealing with dispersion or scatter.

Calculation of V or o.f y rsz t\ rs2Pl

v _ | Ltivl _\z-tiul t lrg2LN N ]

Some important resrrlts.. Average d,eviation is least when taken from the

median.. Standarsd deviation is not less than the average

deviation from the mean.. For raw data xr, x2, ......x,,,

!-2 /l-.\2l4*lv=--t-l N \N/

5"-2or V= 4 - @)'.n

If n r,t?2are sizes oftwo fJroups, x17 x2 theirmeansor, 6, their standard deviations, then standarddeviation of the combined group is determined fron

(n, + nr) o2= ,r G? - a?) + ,,, ("1+ al)where, dr= i -11 and dr= x-x2,r being the combined mean.. This result can beextended to more than two groups.

Coefficients to Dispersion.When two series of measurements have to becompared, then averages nray be different and theunits which the measurements are recorded mayalso be different. Hence, coefficients of dispersionare provided as

Ranse. n;, wltere A is largest qnd B is smallest ofA+ ljthe ualues.

Qs-Qr' Qs *Qr

. Average deviation aboutAA

. Eaclt of these is free from units of measurementand is a pure number.

Coefficient of variation (C.V) = 9 , 100

Thus coeffrcient of variation is r Our""rrf"*".


ELENIENTS OF PROBABILITYExperiment.An experiment is a process in which a certain work isrepeated under the same conditions. the outcomes (orresults) of which need not be the same.e.g. tossing a coin or rolling a die are experiments

Sample space.The set S of all possible outcomes of a given experimentis called sample spoce for the experiment. An outcomean element of S, is called a sample point.e.g. for experiment of tossing a fair (or unbiased) coin

sample space S = (H, T)where H, T refer to head. tail respectively.

Event.A subser of the sample space S is ca\led an euellt. The

A, B. We can form new events usingons of union, intersectio., urrJ

(l) A u B is the event that occurs if and only if Aoccurs or B occurs (or both).(ii) Aa B is the event that occurs if and only if both Aand B occurs.(iii) 6c or 4^,the co.mqlement of A, is the event thatoccurs ifand only ifA does notoccur.

5.5

Sinite sannp\e space.A sample space S ia said to be a f,rnite sample space, ifS is a frnite set.

Finite probability space.Let S = (apar......., o,) be afrnite sample space. S is. said to be a finite piobability space or probabiiitymodel, if each sample point a, in S we can assign a realnumber p,, called probability of a such that(t) p,2 , for each i(ii) pr+ Pz+........ + p,,= |If A is an event, then probability of A, denoted by P(A)is sum of the probabilities of the sample points in A.For the elementary event [o,] we write P(o.) instead ofP(o).

\qu\1ro\r\\eslace.A finite probability space S in which each sample pointhas the same probability is called an equfprobablespctce.

IfE is an event, then

P(E) = _ lEllsl

number of outcomesw favourable to Ebotal number of possible o"t.or*.

Mutually exclusive events.Thro events A and B are said to beif and only if they cannot occurifAnB=0.Three or more events are called mutually exclusiue, ifevery two of'them are mutuall5, exclusive.

Example. Find total number of possible events thatcan occur for an experiment.Solution.Let sample space S consist of n sample points. Thentotal number of possible events

= total number of subsets of S= lp(S) |

wherep(S) is power set of S = 2,,.Example. In_ an experiment, a fair coin is tossed 4timcs. Describe the sampleSolution.The sample space S consists of 16 (= 2a) sample points.S = IHHHH, HHHT,-HHTH, HTHH, THHH, HHTT,

HTTH, TTHH, THTH, HTHT, THHT, HTTT,

THTT, TTHT TTTH, TTTTI

. We use the word.,,at random,, only when dealingwith a equiprobable space. By the sLtement, ,,a baIIis drawn at random from a bag condisining 10 balls,,we mean that each ball in the bag has the sameprobability of being chosen.

. If p = probability of happening of an event E = p(E)

and q = probability of not happening of E = e (E)

then s=P(E) = #=11#=r-l# =t_p= p+e=I

Conditional probability.. P(AB) = P(A) . P(B/A). n P(AB) = P(B). p(A/B)

. P(A/B) = P(AB)

P(B)

. P(A/B) - P(A)'-|{B/A), this is Baye,s retation

P(B)

where, P(B/A) is probability to happening B, whenA has already happened.

mutually exclusivesimultaneously, i.e.

Fig. Conditional Event

P(A nE)P (A/E) = ,(E

P(A/E) represents the probability of A with respect tothe reduced sample space E.

Wheg S is an equipobable space,

p(An.E)= !+f,, p (E) = j|_i

P(wE)= "'3,:3' ='o':P'PIEI IEInumber of elements inAnE

5.6

o Let E be an event in a sample space S with P(E) > 0.

Let an event A occurs after the occurence ofE. Thenco,n.ditional probability of a given E,


Mutual independence.Let (E, : A is a positive integer) be a collection ofevents.These events are said to be mutually independent iffor each frnite non-empty subset {E' E2, .......E,}

P (E, n E, n....n E,,) = P (Er). P(Er) ....P (E,)

Pairwise independence.Let {E* :ft is a positive integer) be a collection of events.These events are said to be pairwise independent if

P (E, n E) = P (E,). P (E) for alli + j.

Possible event.An event E is said to be a possible event, if P (E) > 0.

. Possible independent events are not mutuallyexclusive.

. Possible mutuallv exclusive events are notindependent.

Example. Let p be the probability that a man agedyyears will meeet with an accident in a year. What isthe probability that a man among n men all aged yyears u.ill meet with an accident frrst ?

Solution.Probabilitythat aman agedy years willnotmeet withan accident =l-p

P (none meets with an accident)

(t-p)(t-p) . ..(1- pl

n times

= (1-p)''P (atleast one man meets with an accident)

= 1- (1 -p),,So, P (atleast one man meets with an accident

a person is chosen)

1

=;"t1-(1-P)"1

Example. A box contains 4 white, 3 blue and 5 greenballs. Four balls are chosen. What is the probabilitythat aII three colours are represented ?

Solution.Total number of balls in the box is 12.

Hence total number of wavs in which 4 balls can bechosen

l2xl-l-x 10 x 9-12rl -v4- = 495

E

S

number of elements in E

Note : P (A/E). P (E) = P (An E) is calledMrzltiplicationTheorem for conditional probability.

Example. Suppose a pafu of fair dice is rolled. If sum is6, then what is the probabiJity that one die shows a 2 ?

Solution.Sample space S consists of 6 x 6 = 36 sample points(1, 1), (1, 2), .... (6,5). (6, 6)).

E = sum is 6 = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}

A= 2 appears on atleast one die

= I(2, l), (2,2), (2,3), (2, 4), (2, 5), (2,6), (I,2),(3, 2), (4, 2), (5, 2), (6, 2))

Also, An E = {(2, 4),(4,2)l5 11 2Thus, P(E)= ;,P(A)= ;,P(AnE)= ;

P(A n E)Requiredprobability= P (AlE) = + /- P(E)

2

= 36- 2ob

36

DEPENDEIVT E\MNTS.Events A and B in the space S are said to beindependent if occurence of one of them does notinfluence the occurence ofthe other

i.e. B is dependent of A if P(B/A) = P(B).

Now P (A n B) = P(A). P(B) (B lA) = p141 p13;

Thus events A and B are independent

ifP(AnB)=P(A).P(B)

4 x3 x2xl


Each colour will be represented in the followingmutually exclusive ways :

White Blue Green(t)2r1(iilL21uii\112.'. Number of ways of drawing four balls in the above

fashion

- nC"" 3C, x 5C, + aC, x;lC, x 5C, + aC. x 3C, x 5C,

=90+60+120=270270.'. Required probability =

- .

LAW OF TOTAL PROBABILITY.Let B, Br, ....... B^ be mutually exclusive and let anevent A occur only ifanyone ofB. occurs. Then

P(A) = ; nto / B; )P(B; )

i=L

Bay's theorm.Let E,, Er, ...... E" be mutualJy exclusive events suchthat P(E,) > 0 for each i.

Then for any event A - tt n such that P(A) > 0, we have1

P(A/E,)P(E,)P(E,/A) =fr,i=1,2,.,n

F pra / E, )p(E, )/-,/i=r

Example. Three urns A, B and C have 1 white , 2black, 3 red balls, 2 white I black, 1 red balls and 4white, 5 black, 3 red balls respectively. One urn ischosen at random and two balls are drawn. Theyhappen to be white and red balls. What is the probabilitythat they came from urn B ?

Solution.Let E' E, and E" be the events of choosing urns A, Band C respectively.

IThen P(Er) P(Er) = P(E.) = E ... Q)

Let X be the event of choosing two balls, white andred. To frnd P(EolX).By Baye's theorem,(E/x)

_ P(X|E2)P(E2)P(X / El)P(Er) + P(X I E2)P (E2) + P(X /EB)P(Es)

P(X/E2){from (l)}

P(X /E1) + P(X /Ez) + P(X /Es)

5;7

4r- x3o ,.and P(x/E3) = -ju- =;1

From (ii), P(E/X) = T+ 2 = #5-5-ii

BERNOUILLI TRIALS.Independent repeated trials of an experiment with twooutcomes only are called Bernouilli trials, call One ofthe outcomes is called success and the other outcomeis called failure.Let p = probability of success in a Bernouilli trial

q = probability of failure = 1 -p.A binomial experiment consisting of a frxed number noftrials is denoted hy B (n, p).Probability of r success in the experiment B (rz, p) isgiven by

p (r) = ,,C, p, q"-,

Thc function P(r) for r = 0 ,I,2........., n for B (n, p) iscalled binomial distribution.Example. If four coins are tossed , find the charrcethat there should be two heads and two tails.Solution.

p = probability ofa head j

q = probability of a tait = i -p = +2

.'. Probability of getting 2 heads (and 2 tails) whenfour coins are tossed

=P (2

=6x

RANDONVARIABLE.Many variables of interest in a system are notexpressible on explicit function of time, all such variableon explicit function of time, all such variable areclassifred as random variable. These are implyinguncertaintv about their exact nature.

Discrete Continuons

) = nCrp' qn-t

[l')' " (r)'- q

\2) [z/ - s

lC, "

3C' IP(X{E.)= -#=-'-'---r' 6Cz b'2n ln

P()vE)-"ct^'cr-1,. qC2 B

l)iscrctc ProbabilityTtrcory

Continu<.rus Plobabi) ityTheory

Random Variable

5.8

Let X be a random variable.. Function F(r) = P (X < .r) is called distributiott

function of X.

. MeanorExpectation of X= U =E (X)

=X(o,) P (o,) + +X(o,,) p (o,)n

- Yvr- rp- L'-\-t t-'(Qi)'l,=1

where S is probability space (a,., a", .....ct,)o Variance of X = Var (X) = 02 = E (Xr) - tE(X)],

Standard deviation ofX = o

Example. Find expectation of the number of pointswhen a fair die is rolled.Solution.Let X be the random variable showing number of points.

Then X= 1,2,3,4,5,6

Qi P(X = a,) =P (a,') Product

1l-t)

E(X) =

' Expectation = i.Example. What is the-expectation of the number offailure preceding the frrst success in an infrnite series

of independent trials with constant probability ofsuccess p in each trial ?

Solution.Let p = probability of succes

.'. q = probability of failure=l_ p

Let X be the random variable representing the numberoffailure preceding the frrst success.

Since an infinite series of independent trials areconducted, X=0 ,1,2


(X = r) = probability that there are r failurespreceding the frrst success.

= qxqx.......xqxp=q.pr times

,)

E()0 = )xp(X -- r).r=0

€,.= Lxq'- P

r=0

= L qp + 2q'p + Sq3p + ......

= qp (1+ 2q + 3q2 + ......)

= qp (L -Sf' ['.'(l-gf'zbyBilromialtherre.-.

=I+2q+3q3+....qpqpq

-9-Z-(t- q)' P- P

Example. The Bernouilli probability lau'rnriparameter p in which 0 < P < l is defrned by a ranC,_=variable X taking the value 1 with probability P adthe value 0 with probability q. Find f,hs msan :'a-variance of X.

Solution.P (o,) Productppq0

E(X) =p.'. Mean = f, (X) =p.Now to find Var (X) :

ei t? P(o,)11p

Product [a! xPtp0

EOG) =p

Var (X) = E()f) - [E(X)]'?

=p-p2=(t_p)=pq

Absolute frequency.Let a value x, is repeated { times, then ffrequency or absolute frequency. f.Let N = sum of all frequencies f then frrelative frequency, i = 1, 2 .........n.

Sum ofall relative frequencies

1

6

2;o.J

6

4

6

5

6

6;o

1;o1

6

1;t)

I;oI;t)

4,.I

1

0

21 7

62

fi fz fn- N N '""""""' N

fi+fz fn

N=-N-1


PROBABILITY DISTRIBUTION.Suppose that a random experiment is repeated n times.If an event A occurs n,, times, then its probability p(A)is defined as

P(A) = P(xu) = lT, "o= Prob"ilility of outcome A.

where, ne = times occur An = total time experiment

. This limit may not exist.

o This definition presumes that all outcomes are

equaily likely to occur'

p(A + B) = P(A) + P(B) - P(AB)

here, p(AB) = joint ProbabilitY.

1. DISCRETE DISTRIBUTION(i) Binomial Distribution.It is concerned with trials of a repetitive nature inwhich only the occurrence or non-occurrence'success or failure, acceptance or erejection, yes or

no ofa particular event is ofinterest'

If we perform a series of independent trials such

that for each trial p is the probability ofsuccess and

q that ofa failure, then the probability ofr successes

in a series of n trials is given by .C.P'Q"-', where rtakes any integral value from 0 to n.

The probabilities of 0, 1,2,......t,....-,fl successes are

therefore, given bY

e^, "C, pen - t, nCz ptqn - 2,...."C.p'qn -',.'.'..,P^

The probability of the number of successes so

obtained is called the binomial distribution for

the simple reason that the probability are the

successive terms in the expansion of thebinominal(q + p)".

Hence sum of the Probabilities

= q" + "C, pqn-t + "Crprgn-, +....+ p.

=(q+p)"=1

Constants of the binomial distribution'

Moment generating function about the origin is

M^(0 = Bis*)

= I,,C, p"q,, * c,*

= In C, (pet)* q. *

- (q + Pet)"

Diff'erentiating with respect to t and puttingI = 0. we get the mean given b;/

M,,,ttl = e' nPr iq + Pc')"

= (qe-rL + Peqi\n

5.9

. t2 , z ,tJ= l+pq,+pqtq'-n"o

+pq(q3 *p'i**...t2 +3 t4LOV

= 1+ g,t*1r"D.*frn*Un n! *.t2 +3

=1+npqh*"oo,o-p)3t*

+npq [1 + 3(n - 2) Pq] e +....

Equating coefficients of like power of t on eitherside, we have

F' = nPq' Pt'

= npq(q-P)

fr+ = npe [1+ 3(n - 2)pq]

v? (q-p)' .:r-2P)2=-.f\lSO, P,=..3tt2 nPq nPq

p, ^ 1-6pqO

-=.)T-rr P; nPq

.'. Mean = np

Standard deviation = {(ttpq)7-2p

Skwenessr/nPq

Kurtosis = Fz

Obs. Skewess is Positive for P <

/r)and negative for p t

[;J

r1)\t)

s i'e. ProbabilitY

I be sYmmetrical

a efrnitelY, 0, -+ 0

Binomial fo equencY distribution.If n independent trials constitute one experimentand this experiment be repeated N-times. then

frequency of r successes = Nn C. P'Q"-'.The possible number of successes together withthese expected. frequencies constitute binomialfrequency distribution.

Applications of binomial distribution.This distribution is applied to problems concerning(o) Number of defectives in a sample frorn

production line.(b) Estimation of reliability of sysiem.(c) Number of rounds frred from a gun hitting a

target.(d Rader detection.

5-10

(ii) Poisson Distribution.It is a distribution related to the probabilities ofevents which are extremely rare, but which have alarge number of independent opportunities foroccurrence.e.g. number of persons born blind per year in alarge city is the phenomena, in which this law isfollowed.This distribution can be derived as limiting case ofthe binomial distribution by making n very large andp very small keeping np frxed (=m, say)

The probability of r successes in a binomialdistributionP(r) = nC.P'q"-'

= n(n - t) (n - 2)...(n - r + 1) p"qn -'r!

_ np(np - p)(np - 2p)..(np - r - 1p)(t - p)n-'r!

Asn-+@,p+0(np=6;,m' Lr (1- m / n)2 m'

p(r)= -

il,.

---

e-m- rl " (l- m/n)' rlso that probabi\i\ies of 0, \, 2, . ..'.', r,...'.. success inpoisson distributio is gvenbY

n'. ,,'c.r TnZ gm-tt ,....

*t "-* ,..,..2\ r\

Sum of thcse probabilities is unity as it should be.

Z CONTINUOUS PROBABILITY DISTRIBI.]TION.When a variate X takes e'i'ery value in an interval,it gives rise to continuous distribution of X.e.g. distributions defined by the variates like heightsor weights are continuous distributions.A major conceptual difference, however, existsbetween discrete and continuous probabilities. Whenthinking in discreate terms, the probability associatedwith an eventis mcaningfuI. With continuous event,however, where number of events is infrnitely large,the probability that a specifrc event will occur ispractically zero. For this reason, continuousprobability statements must be worded somewhat,differently from discrete ones. Instead offinding theprobability that z equals some value, we frnd theprobability ofx falling in a small interval.

Thus probability distribution of a continuous variatex is defrned by a function 1(x) such that probabilityof the variate x falling in the small interval

11[x - ^ dxJto [x + ; dxlis f(x) dx.'2.2

Let X be a continuous random variable. If /(r) isprobabilitv density functions of X, then

.:

. plX.o)= Jf?)dx


bF

. p(a<X<b) = lf(*)d,;

aIo Expectation ofX = E (X) = Mean = | xf(x)dx

j-br^

. variance ofX=y(X) = J tt -E(x)"f(x)dxl

f ^r.,. f I\x)ax=L:-

(i) Uniform Distribution.A continuous random variable X is said to followuniform distribution or rectangular distribution ifits probability density function is given by

1

flr) = L, n(x3bo-a

= O, otherwise

The uniform distrihution geneta\\y arises in thestudy of round ott errors wtrete measutements ane

recorded upto a certain level of accuracy.

Curmtl,atiue distribution function of X is giuen by

f (x) = p (X< r).

= Jft'u'_d)

lol" if x<alx-a ;^I . ,tf a< )c<b

= l-o-q if )c>bt1

Mean = o:b, and Varian

"u = 9--t-

2t2Example. The melting point X of a certainspecimen be assumed to be a continuous randomvariable which is uniformly distributed over theinterval [110, 120]. Find the density function of X,mean of X, variance of X and P (lI2 < r < 115.).

Solution.Here.a=100.b=120

(t| ^ 110<r<120Itx)= i120_110,I o, otherwise(rl - 110<r<120

= 110,lo, otherwise

a+b 110+120Meanof X= ------= - =115=2=2


q

VarianceofX = 9:-9-L2

(120 - 110)2

1225-T1 1-5

P[L2<r<115 = lf{xldxtiz

lrar I,= | ----4XJ10

rt2a

10

(ii) Exponential Distribution.A continuous random variable X is said to have theexponential distribution if its probability densityfunction is given by

f k)= )ae-"*'x>0[0, otherwise

a being some positive number, called its parameter

This distribution arises in the study of life-length ofelectronic components.Also the inter-arrival time and service time inqueuing theory have exponential distributions.

llMean - i, Variance = *a a"-_-b^

Example. The sales tax retuYn oYg salesman is'l

exponentially distributed^Mith paramdter i . What.4

is the probability that his sale will exceed Rs, 10,000

assuming that sales tax is charged at the ratepf 57o

on the sales ?

Solution.

It -: r>o)-e =,Here, lU) =)*

lu , otherwiseSales tax for the sale of Rs. 10,000

= 10000 * 9100

= Rs. 500Now, P(x>500)

500

=IP(x<500)=1- JfG)dx\0

-x/4 6,

= 1+ ie-rl4)500

=e 125

5.11

(iii) Normal Distribution.A continuous random variable X is said to follorvnormal distribution if its probability density functionis

(t-p,I - o,:-

nx1 = __7e ."o"l2r

-@<r( 6,o)0,- co <ft( o6

We write in such a situation X - N (u, o')

The distribution involves hvo parameters pt and o.

Properties.(o) The distribution is symmetrical.(b) Mean - p, Variance = o2.

Area = 0.5 !Y 4.." = O.S

\ /-T-V

Nor-!l curve

(c) For this distribution, mean, median and modecoincide.

(d) fk) > 0 for all r.€

(e) Jffrla*=7, i.e. total area under the curve

-@y = l\x) bounded by the axis of r is 1.

Some important points.. The curve y = f(x), called normal curve is a

bell-shaped curue.It is symmetrical about r = rti

the two tails on the left and right sides of themean extend to infrnity'.

c PuL z= r - l-t . Then z is called a stanclard norrnal

ouariate and its probability density function is given

bY z,1 -;

FQ)= r c o ,-(,/i 1z1a-

^l2rMean of the standard normal distnbution is 0 and

Variance is 1. We \mtEZ- N (0. 1).

. Area under the standard normal curve.(o) Between z = - 1 and z = L is 0.6827

(since total area under standard normalcurvg is 1)

i.e.' p (-l <z <I) =0.6827(b) Betweerr z = - 2 and z = 2 is 0.9545

i.e., p (-2 <z <2) = 0.9545

(c) Betwee\ z = - 3 and z = 3 is 0.9973i.e., p(-3 <z<3) =0.9973

500

= 1-1 l"4un

5.12

In other words,

p(p-o<x<V+o)={.6827p0r2c <r < p +2o)= 0.9545p(r3c<r<V+3o)=0.9973p(p- 1.96 o <x <p + 1.96o) = 0.95pqtlZ.Sgo <x < rr + 2.58 o) = 0.99

Since distribution is symmetric, we consideronly positive values of z.

Then 0Q) = dz glves the area

under the nor'z=0antdz=2.

For different values ofz, Q(z) can be calculated.These values of 0(z) have been tabulated.

There is another table c,glled table of

ord,inates.Weread pQ) =" ; .o.r"roonding

to a value of z from this table.

Many distributions tend to a normal distributionin the limit.When a variable is not normal, it can be madenormal using some suitable transformation.

When the sample size is large, distributions ofthe sample mean, sample variance etc. approachnormality. Thus the distribution forms a basisfor tests of significance.

Normal distribution is also known asdistributionof errors.

NORMALTZED GAUSSIAN RANDOM VARIABLE.It mean of the standard normal distribution is 0 andvariance is 1, we write Z - N (0, 1).

The distribution function of Xare plotted respectively.

Mean and variance of X are

p =ElXl=[o*t =E[(X-F*)2J=o2

We use the notation N(p ;o2) to denote that X is normal(or Gaussian) with mean m and variance o2.

In particular, X = N (0 ; 1), i.e. X with zero mean andunit variance is defrned as a normalized Gaussianre,ndom uariable.


CEI',{TRAL UNIT THEOREM.

The normal (or Gaussian) distribution has played asignificant role in the study of random phenomena innature. Many naturally occurring random phenomenaare approximately normal . Central-limit theore statesthat sum of a la,rge number of independent randomuarinbles, undcr certainconditions can be approximatedby a normal distribution.

__2

| "-TJ0mal

1 (x-u)2lxt = offi exP - -r;r-

If mean = 0, i.e. p = 0,

curve between then (x) = -L "*o - (*)l

a^l2r - 2o"

GAUSSIAN PROCESS.

fly) = ----l--u"oLo-!rf"l(2xor')

- 2or"' where, !tr= mean

o 2= variance

A Gausslaut Pulse

Properties of Gaussian process.(t) If a Gaussian process x(t) is applied to a stable

linear filter, then random process V(t) developed.The output of the frlter is also Gaussian.

(ii) Consider the set of random variables or samplesx(t,), x(tr), x(t,) ---- obtained by observing a randomprocess x(t) at time t' t", -----.

Second moment or Mean square value.Consider random variable X which assume the possiblevalue x,, x2, xB ----- x,,. fn a sequence of n experiment,let event xr occur n, time, x2 occur n, times etc.Arithmetic average or mean of xu is

.!a _[lXt +I]2x2 +----+xm nm

1m=atn,H.

K=l

xt Dt

'ROBABILITY & STATISTICS

t7If n -+ co, then -lL --> p(x,)

n4", = Pr"t * Pzxz* ---- P- X-

m

-\-= L Ptxr'k=1

andIJ=avg

CORREI"ATION.In a divariate distribution, ifthe classes in one variableare associated by clases in the other, then variablesare called correlated.If ratio of two variable deviationsis constant, then correlation is called,perfect.

Co-effr cient of correlation.Numerical measure of correlation is called co-effrcientofcorrelation and is defined as

'XYf =

-

n6rcoy

where, X = deviation from mean, ,=)C_,C

Y = deviation from mean, y

=y- y

o, = standard deviation ofr series

oy = standard ofy series

n = number of values of the two variables

Method of calculation :

Direct ntethod.' Substituting the value of o, and o, inthe above formula, we get

' J>r':v''"' ^vL r. - J)

Rank correlation.A group of n individuals may be arranged in order ofmerit respect to some characteristic. The same groupwould give different order for different characteristics.Consider order corresponding to two characteristic Aand B. The correlation between these n pairs of ranksiscalled rank corcelation in the charq,cteristics A and,B for that group of individuals.Let x,, y, be the ranks of the i,/,individuals in A and Brespectively. Assuming that no two individuals arebracketed equal in either case, each ofthe variablestaking the values L,2,3,....., n, we have

l+2+3+...+nY=V=

If X, Y be the deviation of x, y fromtX,'=I(r,- i),

5.13

n(n+1)_n+12n2

their mean, then

nt"r'* ',' zxz2 +----N

-Ixi2+n(x)2-2;Ix,

=In2+q{-,'*t.I'n(n+1)(2n+1) n(n + 1)2

4

= f t,r,-r,)

[ , l,.z , xu,-l...i':o; ou=LlI n n)

Similarly

Now let

tY''

d,

d,

Id,,

14Y'

t,"

xi-Y'(",- i)-(y,- y)X,-Y,I4' 2IXrY,

1= ,(2X,2 * IY,, - Idi2)

11= ftn3-n)- r>a,,

Hence correlation coefficient between these variables-is

=1_+{n- -nThis is called ranh correlation cofficierul and is denotedbv p.

Line of regression.If dots of scatter diagram generally, tend to clusteralong a well defrned direction which suggests a linearrelationship between variabies x and y, such a line ofbest - frt for the given distribution of dots is called theline of regression.

In fact there are two such lines, one giving the bestpossible mean values of y for each specifred value of xand the other giving the best possible mean values forx for given values of y. The former is called the line forregression ofy on r and the later is called the line ofregression of x on y.

5.1 4

First consider the line of regression of y on x. Letstraight line satisfying the general trend ofn dots in ascatter diagram be

J=u+bx ...(r)

We have to determine the constants a and b so thatequation (l) gives for each valuc ofx, the best estimatefor the average value of y in accordance with theprinciple of least squares. Thus normal equations fora and b are

Iy = ,ra + blx ...fti)

IxY=alx+blx2Equation (ii) gives

1-_ r

-IY=a+b':J-xnny = a+ bx

This shows that ( x , y ), i.e. means of x and y lie on (l)

Shifting the origin to ( x , y ), equation (fii) takes theform

I(x- i ) (V- V ) = al(x-x) + bl(x- x ),

But I(x-x)=0.l.(x-x,(y-y,b-

PROBABILITY & STAT]STICS

ov [ >xv't_r_ 1...I'=-' o* \ no*o, J

Thus the line of best frt becomes

(y-yl= r3(*-i)which is equation ofthe line ofregression ofy on x. Itsslope is called regressiort co-efficient ofy on xInterchanging x and y, we find that the line ofregression ofx on y is

(x-x) = ro.. (v-y)o"

Thus regression co-efficient ofy on x6y

=f-"xand regression co-efficient ofx on y6

=rj 6v

con Correlation co-effrcient is the geometric meanbetween the two regression cofficients

ov o-y-:Lyy--!- = y2or o,

Regression co-efEcient.It is defrned as the hope oflines ofregression,i,.e. regression coeffrcient of line x on y

o-o^,

X(x - x)'

'XYtx'IXYtro*2

5:15PROBABILITY & STATISTICS

OBJECTTVE QUESTIONS

1. A card is taken out of a pack of 52 cards numbered

2 to 53. The probability that the numbers on the

card is a prime number less than 20 is

12to) ,, {6) t',

3 .4tc) ,,

(d) 19

2. The probability of getting a number between 1 and

100. Which is divisible by one and itself only is

8. If A and B are any two events then which one of

the following pairs of events are mutuallyexclusive ?

(o) AuBand AnB (b) AuBand AnB(c) AuBand At-''g (d) AuBandAnB

9. There are three events A, B, C one of which must'

and only one can haPPen, the odds are

8 to 3 against A, 5 to 2 against B' The odds against

C are

27(n\

-18525(rl -98

23{b) *(d) none of these

(o) 43 to 34

(c) 34 to 13

10. There are two bags one of which contains3 black and 4 white balls while other contains 4

black and 4 white balls. A dies is cast, if the face 1

or 3 turns uP, a and ifany other face from

second bag. Th black

ball is11(a) U12rc)d

(b) 32to23(d) none of these

2ltb) tt

2l{d) u

129

440

none ofthese

(b) 16:7G) 14:6

If A and B are two elements such that

p(A) = ?. "r", =

q, P(A n B) = 1, then

3' 4' 2P(AuB)=

qr([r i (b)

6

t7(ct I (d)18

If A and B are two events such thatP (A u B) = 0.65, P(A n B) = 0'15, then

P(A)+ P(E) =

(a) 0.6 (b) 0.8

k) 1.2 @) r.4

5. If A, B, C are threc evcnts, then PIA n (B v C)] =

(o) P(A) + P(B) + P(C) - P(A n B) - P(A n C)

(b) P(A) + P(B) + P(C) - P(B) P(C)

(c) P(An B) + P(An C)- P(An B n C)

(;) P(B n C) + P(A n B) - P(An B n C)

6. If A and B are two arbitrary events, thenP(A n B) cannot be

(o) Iess than P(A) + P(B) - 1

(b) gre.rter than P(A) + P(B)

(c) equal to P(A) + P(B) - P(A u ts)

(d) equal to P(A) + P(B) + P(A u B)

7. I1M and N are two events. then probability thatexactly one ofthem occurs is not equal to

(ot P(M)+P(N)-2P(MnN)(b) P(M) + P(N) - P(M n N)

r"t P(M)+ P(N)- 2P(M n N)

tal P(M. x) * e(M n N)

11

nn

n

11. An urn contains 5 red and 10 black balls' Eight of

them are placed in another urn' The chance thatthe latter then contains 2 red and 6 black balls is

140 /, \kll -

(o,' 429139@ffi (d)

12. A has one share in a lottery in rvhich there is1 prize and 2 blanks; B has three shares in a lottery

inwhich there are 3 prizes and 6 blanks;comparethe probability of lt's success to that of B's success

is

@) 7 :16(c) 6:14

13. IfA and B are independent, then A and B are

also(o) independent (b) dePendent

(c) both (o) and (b) (d) none of these

14. The chance that a leap year sel'ected at random

will contain 53 SundaYs is7@)t3@7

2(b) ;I

nI

{d) ,

II

I

5.1 6

This is= P (E,) + P (Ez) + ..... + P (E,.).

(a) Law of total probability(6) Law o1'probability(c) Both.4effi (6)

(r{) None of thesc

18. Srhat is the probability of correctly choosing anunknown integer between 0 to 9 within thrce


21. The probability of dr wing an ace or a spade orboth from a deck ofcards is

,^'34,o) 18 (b) tS

,^t 4 ,t\ 5

'") 17 tcl) *22.P(AruAzuAs)

1 -P (A,) P (A"/A,) p (A,/A, n A").(o) Tlue (6) False(c) Both (o) and (6) (d) None of these

23. IfP(A/C)> p(B/C)andp(A/C, ) >p(B/Cr ), rlrcn(6) p gr )< P (B)(d) P(A)>P(B)

24, If Aand B are independent and p (C) = 0, then A,B and C are independent.(o) Tlue (6) False(c) Both (a) and (b) (d) None of fl-rese

25. Aman altcrnately tosses acoin and throws a clice.beginningwith thc coin. Then the probability thathe will get a head before he gets a 5 or 6 on dice is

,^tl 3\at i (bt i4+

1c) F ktt 126. Six dice are thrown siorultancously. The

probability that all will show different faces is5l

(o) ab"

(c) *6'

(o) P(A)>P(B)(c) P(A)<P(B)

,'\ 5!\o) ,

b'(d) nrine of'these

chances ?

963(0.1

---1000

t. t 983

1000

19. A five figure number is fbrmed by the digits0,7,2,3, 4 without repetition. Then the probabilitythat thc number fbrmed is divisible by 4 is

3 .,. 5(clr -

(b) -

,'-\ ! ,', s(cr 16 dl

16

20. A bag contains 8 white and 6 red balls. The- probability of drawing two balls of the same colourl"s

27. Given two events A and B and

P(A)= 1,p1s7a;= 142

Iand P(A/B)= i/+State that following is true'/(a) A is sub-event of B

o

(6) P (A,/B)= "4

(c) P(A/B)+P(A/B)=1(d) None of these

28. AboxAcontains 2 whitc and 4 black balls. Anotherbox B contains 5 white and Z black balls. A ball istransferred from the box A to thc box B. Then aball is drawn from the box Ii. The probabilitv thatit is white is

@)973

1000

None of these

AA(a)

-6U

43(c) 93

43ror _(d) none of these

16'*' 39

I2(c,,

39

T4(o) 39q

tdt -39

PROBABIL|TY & SriTigTfc*sl' .'

29. IfP (A) = 0, then P (An B) = 0.

5.17

35. In an experiment a coin is tossed 4 times. What isthe size of the sample space?

(a) 12

(c) 16

(b) 14

(d) 20

36. For any two events A and B

(o) P(A nB) =P(B)-P(AnB)(6) P (A u B) = P (A) + P (B) - P (An B)

(c) P(A/B)<P(A).(d) All of these

37. Which of the following is correct?

(o) P (Aw B/C) = P (A/C) + P (B/C) - P (AnB/C)

(6) P (A n B/C) + P (A n B/C) = P (A/C)

(c) IfA and B are independent, then A and B arealso independent.

(d All of these

38. Which of the following is correct?

(o) If A, B and C are mutually independent, thenA u B and C are also independent.

(6) The events Er, Er, ......., E. are mutuallyn

exclusive.LetE=U E,.i:1

If P (A/Ei) = P (B/Ei), i = L, 2,...., n, then

P(A/E)=P(B/E).

(c) Both (o) and (b)

(d None of these

39. Let p be the probability that a man aged y will getinto an accident in a year. What is the probabilitythat a man among n men of all aged y will getinto an accident frrst ?

1, r 1, r/r\{cr :(1-(1*p)") (b) :(t-ft* p)" )n'

(c) "(1- (1* p)") (d None of'these

40. What would be the expectation of the numbel offailures preceding the first success in an infiniteseries of independent trials with the constantprobability ofsuccess p ?

1\a) -p

@!p

(b) -lq

(d) Nonc of thesc

(o) o

(a) *(6) 1

(b) none of these

31.

Which of the following is correct?

(o) IfP(A) > 0, P (B) > 0 and P(A/B) = P (B/A), then

P(A)=P(B).

(b) If A and B mutually exclusive, then

P (A/B) = P (A)/(1 - P (B))'

k) If Aand B are independent, then

. P(AuB)=1-Pfel.Ptel(d) All of these

A problem in mechanics is given to three students

A, B and C whose chances of sloggingit r"u 1. 1'2'31

and ; respectively. Then probability that the4-problem will be solved is

1\d) -4

a(c)

4

(o) 0.155

(c) 0.355

(b) 0.255

(d) 0.455

1(D) -2

(d) none of these

32. A and B throw alternately with a pair of diCe. Awins if he throws 6 before B throws 7 and B winsif he throws 7 before Athrows 6. If Abegins, thenhis chance of winning is

31 30(o) 60 tar 61

31( ^\ _ (d) none of these

61.1 i)

33. If P (A) = ; and P (B) = ;, then4A(a) P(AuB)>+ (6) :<P(AnB)<:'488

1-3(c) * <Pfen B )<; (d) All of thesed6

34. Three machines M' M2 and M, produce identicalitems. Of their respective output 57o,4o/o and3%of items are faulty. On a certain day, M, hasproduced 25Vo of the total output, M, has produced307o and M, the remainder. An item selected atrandom is found to be faulty. What are the chancesthat it was produced by the machine with thehighest output?

5.18

41. A rcstaurant serves two special dishes A and B toits customers consisting of 60%, men and 40Vo

women. SOVo of men order dish A and the rest BandTOa/c' women order dish B and the rest A. Inwlrat ratio of Ato B should the rcstaurant preparethe dishes?

12. If A and B are mutually exclusive andp (A,._, B) = p (A) + p (B), then


If cov (X, Y) = 0, thep u'hic[ of thc fbllqrving isnegative?(o) 6,, (b) b,.',

(c) p (X, Y) (td) nonc of these

Rcgression equation ofY on X is8r-10y+66=0ando"=3

Flence cov (X, Y) is cqual is

50.

51.

(b) 6:9@) 7 :16

(o) 19:6(c) 16:7

P(A/AuB)=

P(A)P(A) + P(B)

P(B)P(A) + P(B)

43. IfP(A)=P(B),then

(<z) A= B

(c) A* B

52. The mean of a set of observations is - . If cachobservation is divided by cx., a + 0 and then isincreased by 10, then the nrean ofnew set is

(a) 11.25

(c) 2.4

x(cr) -0,

r+ 10cr(cl --

C[

(b) 7.2

(d) none of thcse

r+10(b)

-0

(d) cx F +10

(6) -io(d) o

of thesc

(6) -1 and 2

@) -2and-1

M. If P (A) = 0, then(a) A=O (6) A=0(c) A=.o (d) none of these

45. The mcan of a set of number is r . If each numberis increased by )., then mean of the new set is

h)i (6) i+Itc) 1, (d) none of these

46. If each observation of raw data whose variance iso2, is increased by 1", then variance ofthe new setIS(a) &

D\.c) t +o'

47. If mode of amedian is(o) 18

(c)22

(6) B=0(d) None of these

(b) )"2o2

(d) ),2 + o2

data is 18 and mean is 24, then

(b) 24

@) 2L

53. Coeffrcient of correlation betwecn two variates Xand Y is

(o) 0 (6) -1(c) 1 (d) none of these

54. If mean = (3 median - mode) r, then value of.r is

(a) | (b) 213

r") 2 {d) ,55. S.D. of n observation n,, a2t e^,........ an is o, then

If S.D. of'the observations tro,, l.{zr, ),o.,,..... tro,, is(o) to(c) ll.lo

56. If median - (mode + 2 mean) p, then pL is equal to

(o) 31

(6) ;.)

(c) 2 (d) none of thesc

48. If pt is mean of distribution, thenlf,{1,-p) is

equal to

(o) M.D(c) o

49. b,u x b,,, is equal to(a) p(X,Y1s,t lp (x, y)12

57. If two lines of regressioY = 2x - 4, then p(X, Y) is

(n) 1 and -2(c) 2 and -1

Y=3x-5andto

6,, il;E

", l;If two lines regression are 3x - y - 5 = 0 and

2x-y-4=0,then r and t arerespectively

(6)

(d)

n areequal

IIt-1o

none(b) stamdard deviation(d) none of these

(b) cov k, y)(d) None of these

X 1 2 J 4 o

Y 5 4 ,) 2 I

58.


59. If regressiolt equation ofY on X is Y = Lt + 4 and

that ofX on Ybc 4x =Y - 5, then

5.19

If two variables X and Yhave a pcrfect correlation(direct or indirect), then they may bc colrncctedby a relation ofthe typc

(q) xy = a2 (b)

(c) {+ { =r (d) none of thesc

25o/o of the items of dat,a are lcss than 35 and25%

of the items are more than 75' Q'D of thc data is

(o) 0<4I<1(c) 0<)'<4

(b) 0<r<1(d) nonc of these

(6) -1(d) 0

67.

68.

69.

o,b_, _ _txy

60. If cov (X.Y) = 0. then two lines of the rcgrcssion are

(o) parallel (b) concident

(c) at right angles (d) none of these

61. Iftwo litres ofregression are atright angles, thenp (X, Yt is eqiral to(cr) 1

(c) 1 or -1

62. If in a rcgression analysis problem, 6,,. = -0'9 ancl

b,,, = 0.4, then P(r, Y) is

(a) o'6 (b) -0'6(c) 0 (d) none of thcse

63. It b,,, and b.,. bc the two rcgression coeffrcients,

then coef'flcient of correlation p is given by

F-.--(a) P = Vb..ub.".,

.. F-,lb) p= -{oy*o.",

k) p=(-s*,,6r,)J-u*u*(d) none of these

64. Given ', n = 10, by = 4 )j = 3, )-.t' = 8, b" =

and Iry = 3, then coeffrcient ofcorrelation is

65. It standard deviation for two variables X and Yare 3 and 4 respectively and thejr covariance is 8,

then corrclation coefficient between them is

2@)t

I\") ETz

Mean deviation of the dataa, a + d,a +2d,...a + 2nd from the mean is equal to

(a) 55

(c) 10

(a) m>p(c) m<P

ct(a) ;

o(c) -1

72. If X and

that iifu--3x(o) 0

(b) 50

(d) nonc of these

b.,,. + b,...If p>0andrn =

-t-,then

70. If two lines of rcgression are x + 2y = 7 and

2x + y = 7, lhetr regression equation ofY artd Xis@)2r+y=7 (6)r+2Y='i(c)r+2y=0' @)2x+Y=0

71. TWo variables X and Y are contlocttrd by therelation ar + by * c = 0 where a6 < 0. thenp r, y is equal to

(b) m=P(d.) none of these

b(6) -a(d)1

Y are two independent variables such

-5,y=10o?=+of;=9and+ 4y,u - 3*-y,then p (u, u) is eaual to

(b) 1

(d) none of these

1(b)

-72

t4(d) J

1(a)

A

15lc)

-4 2(c) g

n(n +I\d(b) 2n +L'-

(d) none of these

1,2,3,......n rvithncrr....,..r'tcn ia

ZtL+L

n*L

none of these

ball mav be brown is13

30

(b)8

------3.12

73. N{ean of the numbers 0,

respective weight'Co,'C'gn

b) : (6),z+1n(c) ^ (d)22(ils

66.

Common Data Q. 74-76

Box A contains three balls with colours red, greenand blue and Box B contains a balls with coloursred, yellow, blue, white and brown. A box is chosenand a ball is picked.

74. The probability that the1(dl

-108

(?\ -15

(b)

(d) none of these

5.20

lD. The probability that the ball may be green orblue is

I(') to8

(c) G

blue is red or blue is1

{o) ,8

(c) G

13tor

30

(d) none of these


82. The chance that two bel<lng to the same class andthird to the different class. is

DDld. ) -:--

8456

d/

54tbt *dl)57(r/) *6I

5(o) 845(c) tt

Common Data Q.84-85Ttvo cards are drawn in succession from a pack of 52cards. First cards should be a king and the second a qucen

84. The chance when frrst card is rcplaced, is1

ta) -169

2tcl u*

85. The chance when first card is not replaced, is

1 .-. 1(O)

-

/L\ .'"' 669 '"' 6632_4trt _ (dl ^_663 ob.t

Common Data Q.86-87'n'Ietters are placed-at random in n con-ectly addressedenvelopes.

86. The probabilities that, at lcast onc lcttel goes tothe correct envelooe is

5to) _

6b

(d) none of thcsc

I(h\

-J169

4(n\ -

'"' 169

(a)L-p1

\c) -n87. The probabilities that all lettcrs go to thc correct

envelope is

@) r-P

I\c) -n

Common Data Q.88-89

Box t has 10 light bulbs of which 3 are defective. Box IIhas 6 light bulbs which 2 are defective; Box III has 8light bulbs of which 5 are defective. A box is chosen atrandom and a bulb is drawn at random.88. What is the probability that the bulb drarvn is

defective?

720

32032

(c) -720

13/A\ -30

(d) none of these

76. The probability that the ball may be green or 83. The chance that three belons to the same clerss is


Given : P(A, (A, n Arl = l1and Pf At) = i.

77. P(Ar) and P(Ar) respectively are2l(^\ -\u/ B' 2

11\L/ 2'2

78. Ar and A, are independent


Consider an ordinary six faced die.

79. The chances of throwine four isI

\a) -21

\c) ;e)

80. The chanccs of throwing an even number is

(o) True(c) Uncertain

I@),

,'1(c, 6

I2(b) -. -2'322(d) ;, ;t).)

(b) False(d) Can not be said

1(Dl ;.l

1(ds d

1(b) ;1

tat ;d

1(Dl ,n!

(cl) none of thesc

1(6) ;n:

(d) none of these

302

i2072

302

Common Data Q.81-83A committee consists of 9 students two of which arefronr lstyear, three from 2nd year and four from 3rdy'car. Three studcnts are to be removed at random.81. The chance that three students belong to different

classes, is3

Io.) ; \b)

,+

ic/ = \d)I

2n

5;

PROBABILTTY & STATISTICS

89. What is the probability that the defective bulb was

from box I ?

720(ct\

-' 320

32tct- d)-

720 302

90. Four fair coins are tossed simultaneously' The

probabilitv that at lcast one head and one tail turnup rs

IInl

-167k):6

91. How many four4 digits distinct ?

(a) 2240

k) 2620

92. Seven (distinct) car accidents occurred in a week'

What is thc probability that they all occurred on

the same day ?

1I

{a\ * (b\I

I(c) "2'

The nrinimum number of cards to be dcalt from an

arbitrarily shuflled deck of'52 cards to guarantee

that three cards are frr-rm same suit is

1b):615(d\ -16

digit EVEN numbers have all

(bt 2296

@) 4536

1

nGI

7

27

@)3(c) 9

A dic is rollecl three times. The probability thatexactly one ODD number turns up among the three

outcomes is1(a) ;(]1(c) ;6

E, and E, are events in a probability space satisfying

the following constraints :

o P,(Er) = P,(Ez)

o P,.(E, u Er) = 1

o E. and Eo are indePendentThe value of P, tEr), the probability of event Er is

1(b) n

d)t

,^E"tt

96. Consider two events Er and E, such th'

probability of'E,, P.(Er) = |,

probability of E' P.(Er.1 = ! , u.td

1probability of Er and E2, P, (Er and Er) = S '

Which of the following statement(s) is/are TRU'

(a) P, (E, or Erl is l(b) Events Er and E, and E" are independent

(c) Events Er and E, are not independent

.,."[Et) 4\clt _r

lprJ = s

97. TWo girls have picked 10 Roses, 15 Sunflowers

and 14 Daffodils. What is the number of ways

thev can divide the flowers amongst themselves ?

(a) 1638

k) 2640

(b) 2100

(d) None of these

98. The probability that it will rain today is 0'5' The

probability that it will rain tomorrow is 0'6' The

probability that it will rain either today or.

tomorrow is 0.7. What is the probability that itwill rain todav and tomorrow ?

302(h)

-72072

(d)

(6) 8

@) 12

1(b) ;D

1\u/2

99. The probability that top and bottom cards of a

randomly shuffled deck are both aces is

(o) 0.3

(c) 0.35

44(n\ -

X -'*' 52 52

43r'n)

-x-\L/ s2 51

(o.) 0.2048

(c) 0.4096

(b) 0.25 : a

(d) 0.4

43(6) -x-52 52

44(.:l\ -

x -'*' 52 51

(b) o.4ooo

(d) 0.942r

10O. The probability that a number selected at random

between 100 and 999 (both inclusive) will not

10f. If 20 per cent managers are technocrats, theprobability that a random committee of 5

managers consists of exactly 2 technocrats is

contain the digit 7 is

16 re\'o) n (D) [loJ

27 18@i @*

(cr) 0

I(r'l

2

5.22

102. The probability that two frineds share the samebirth-month is

fO3. A regression moodel is used to express a variableY as a function of another variable X' This impliesthat(a) there is a causal relationship between

YandX(b) a value of X may be used to estimate a value

ofY(c) values of X exactly determine values of Y(d) there is no causal relationship between Y and

X

lO4. Arrivals at a telophone booth are considered tobe Poisson, with an average time of 10 minutesbetween successive arrivals. The length of a phone

call is distributed exponentially with mean 3minutes. The probability that an awival does nothave to wait before service is

(b) 0.5

(d) o.e

lO5. Abox contains 5 black and 5 red balls. Tlvo ballsare randomly picked one after another from thebox, without replacement. The probability for bothballs being red is(a) 1/90

(c) 19190

106. Thc following data about the flow of liquid wasobserved in a continuous chemical processplantFlow rate (litres/sec) Frequency

7.5to7.7 1

i.7 to 7.9 5

7.9 to 8.1 35

8.1 to 8.3 17

8.3 to 8.5 12

8.5 to 8.7 10

Mcan flow rate of the liquid is(o) 8.00 litres/sec (6) 8.06 litres/sec(c) 8.16 litres/sec @) 8.26 litres/sec


10?. From a pack ofregular playing cards, tu'o cardsare drawn at random. What is the probability thatboth cards will be Kings, if the frrst card is NOTreplaced?

108. Abox contains 10 screws, 3 of which are defective.

Two screws are drawn at random withreplacement. The probability that none of thc twoscrews is defective will be

1\b) t2

1td)

24

1\a)

6

1(t'l

-t44

1(d\

-'*' 26

1(nl

-169

1(b) _5Z

1(/'l\

-221

109. A fair coin is tossed three times in succession. Ifthe frrst toss poduces a head, then the probabilityof getting exactly two heads in thlee tosscs is

.J(c) :d

ll0. In a frequencey distribution, mid value of a class

is 15 and class interval is 4. The lower limit ofthe class is

(o) 14 (b) 13

G) 12 (d\ L0

lll. Folloiving marks were obtained by the studentsin a test :

81,72,90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85,

79,62. Rangc of the marks is

(b) 17

(d) 33

Il2. Standard deviation for7,9,11. 1,11, 15 is

ra.) 2.4 6) 2.5

ft) 2.7 @) 2.8

113. If probabilities that A and B will clie rvith ina yearare p and q respectively, then probability thatonly one of them will be alive at thc end of theyear is

1(h) -'"' 2

a(CL) -.A

1tu);6

@) 1007o

(c) 497o

(b) 50ck

(d) None of these

(b't p1t- q1

(d) p+7-2pt7

(o) 0.3

(c) 0.7

(b) r/5(d) 2t9

@)e(c) 27

@) Prl

k) q(7- p)


1-l4. In a binomial distribution, mean is 4 and variance

is 3. Then, its mode is

(b) 6

(d) none of these

115. Ifsum and product ofthe mean and variancc ofabinomial distribution are24 and 18 respectively',

thcn distriution is

t1 t\' (t 3\'o

^, lr. ,) 6) [;.aj

5.23

116. Shen correlation coefficient | =*1 , Lhen two

regcssion lines

(d are perpendicular to each othct'

(b) coincide

k) are parallel to each other

(d) do not exist

lL7. If1=Q,then(o) there is a perfec[ corrclation between :r and y

(U x andy are not correlated

(c) there is a positive correlation bctween x and y

(cl) there is a negative correlation between r andy

(a) 5

(c) 4

kt) (:.!r)"., (;.;)"

ANSWERS

1. (6)

r1. (o)

2r. (b)

31. (c)

4r. @)

51. (b)

61. (d)

71. @)

81. (b)

el. (b)

101. (o)

1r1. (d)

2. (c)

12. (a)

22. @)

32. (b)

42. (a)

52. (c)

62. (d)

72. (a)

82. (a)

92. (b)

102. (b)

rr2. (d)

3. (b)

13. (o)

23. (.q,)

33. (d)

43. (a)

53. (6)

63. (c)

73. (c)

83. (o)

93. (c)

103. (b)

113. (d)

4. (c)

14. (b)

24. (a)

34. (c)

44. (a)

54. (c)

64. (a)

74. (a)

84. (a.)

94. (c)

LO4. (a)

rr4. k)

5. (c)

r5. (a)

25. (b)

35. (c)

45.ft)aa. (c/

65. (o)

75.@)

85. (d)

e5. @)

105. (d)

rr5. (d)

6. (c)

16. (o)

26.6)36. (d)

46. (a)

56. (b)

66. (o)

76. (c)

86. (o)

96. (c)

106. (c)

116. (b)

7.ft)17. (a)

27. (d)

37. (a)

47. (c)

D t. \a)

67. (c)

77. (b)

87. (b)

97. (c)

ro7. (d)

rr7. (b)

8. (c)

18. (6)

28. @)

38. (o)

48. (c)

58. (o)

68. (b)

78. (a)

88. (b)

98. (d)

108. (d)

9. (o)

19. (b)

29. (u)

39. (o)

49. k)59. (c)

69. (o)

7e. k)8e. (d)

99. (c)

LOe. (d)

10. (o)

20. ft)30. (a)

4O. (c)

50. (d)

60. (c)

70. (b)

80. (cr)

90. (c)

100. (d)

110. (b)

5.24 PROBABILITY & STATISTICS

EXPLANATIONS

13.9. Since odds are 8 to 3 against A^\)

.'. P(A) = -:- ; P(B) = :ll 7

and P(A) + P(B) + P(C) = 1

.'. P(C)= I 3 2 34

I ll 7 77Hence odds against C are 43 to 34.

Let E. be the event that a baII is drawn fromfirst bag, E" the event that a ball is drawnfrom second bag and E the event a black ballis chosen, therefore

(e\ (n\P(E) = P(E )Pl-al--P(8")Pl-:-l

IE' ) -'(8,/

Number of ways in which 8 balls can be drawnout of 15 is tsCr.

Number of ways of drawing 2 r:ed balls is sC,

and corresponding to each ofthese 5C, waysof drawing a red ball, there are r0C. ways ofdrawing 6 black balls.

Hence total number of ways in which 2 redand 6 black balls can be drawn = 5C^ x roC-.

Required probability = tu c,L40

429

A can draw a ticket in 3C, = 3 ways.Number of cases in which A can get a prizcis clearly 1.

1.'. Probability of A's success = i.9.8.7

Again B can draw a ticket in eC, = g .2 .l= g4way.

Number of rvays in which B gets all blanks

= 6c^ = 6.5.4

=zo." 3.2.r

Number of ways of getting a prize=84-20=64

Thus, probability of B's success = 9+ =1684 2rA's probability of success : B's probabilityof success

P(Ane) = P(A)-P(AnB)= P(A) - P(A) P(B)

= P(A) (l - P(B)) : P(A) P(B)Hence, A and B are independent.

14. A leap year consists of366 days, so that thereare 52 full weeks (and hence 52 Sundays) andtwo extra days. These two days can be

(i) Monday, Thesday(ii) T\r.esday, Wednesday,

(iil) Wednesday, Thursday(iu) Thursday, Friday(u) Friday, Saturday(ul) Saturday, Sunday(uii) Sunday, Monday

Of these 7 cases, the last two are favourable

Required probability = 1.

r rrd P(AnBnC) PrAnBnC)f

P(C) P(C)

P(AnBnC)+P(AnBnC)P(C)

P(AnBnC)vP(Ar-,BnC)lP(C)

_ P(AnC)P(C)

AnB=0.'. P(AnB) = 0

s = f0, 1,2, ...,91

Suppose we want a particular integer "3" to bechosen.

10.

lltI

J

2

o

3 44-f-.- =7 67

3 24_L__-t)t

11.

15.

_ .rA 'l

- rt-l\.c/

12.

16.

18.

Probability of not choosing

chance)Probability of not choosingchances

999=-Xl0 10 l0

Hence probability of choosing it in at least oneofthe three chances

=1- 27 - 973

1000 1000

19. The five digits can be arranged in 5! ways, outof which 4! will begin with zero.

.'. Total number of 5-frgure numbers formed

=5!-4!=96.

"3" is fr (in one

3 in all the three

11632L =7:16


Those nrtmbers forntcd u'ill be divisible by 4which will l-rave t',vo extrcme right digitsdivisible b.t'4,

r.c , numbet's ending in 04. 12, 20.24,32,40.

Now, Itumbers ending in 04 = 3! = 6,

numbcrs ending in 12 = 3t - 2t = 4,

numbers erlding in 20 = 3l = 6,

numJrers ending in 24 = 3t, - 2t = 4.

ttunbers ending in 32 = 31, -2t = 4,

atrdrturnbers ending in '40 = 3l = 6.

INumbe rs having 12,24,32 in thc extreme tightarc (3! - 2!), since thc numbers har'ing zero on

the cxtrernc lc{t are to excludcd.l

.'. Total nunrber of favourablc ways

= 6 + 4 + 6 + 4 + 4 + ti = 30

Hetrce; rcquiredProbabilitY= + = 1196 16

20. T\'o balls out 14 catr be drawn in 1'C., ways whichis thc total nrtmbe r of'outcomcs.

Two rvhite balls out of 8 can be drawn in 'C,,

ways.'. Probability of drnwing 2 white balls

= .c,

= -2!to c, 9l

Similarl.y 2 rcd balls out of 6 can be clrawn in''C,, u'a1's'. Probability of'drawing 2 red balls

ttc, l-5

= ,ac" = s,

.'. Probability of drawing 2 balls of the samecolortr (eitlrcr both white ol both red)

28 15 43=91*91=91

27. Probability of drar.r'ing an ace from a deck of 52

.4Cal^OS =

- '

52Similtrrly the probtrbility ot'drawing a card of

_13SPaOL'S =

-'52

probability of'rlrawing an ace of spades = 1'52Since two events (i.e. a card bcing an ace and acerrd being o{'spades) are ntlt mutually cxclusive,thereftrreProbability ofdrarving alt ace or a spade

1)-ll-+=-+=--\2 52 .s2 l.l

5.25

22. P(Ar) . P(r\rrA,). P(A,,/A, n.\,,)

- p /a ., . P(A, tn,) l(A$'^UP(A ) l(n,,:l,'r.)

-P(ArnA,nA,)=P(A.re-rrz\l=1-P(A, r,-,11. t-,A,)

23.l'}(AnC) P(tloC)

P (c) P (c)

= P(An0) >P(BnC)r,/r ^FI plR,^'-\\"' '', '\'/^/

allo -

- > -----:=i-r,((') P((')

= P(A. C) >P(Bn C)Adding

P(AnC)+P(Ar', g) >P(BnC)+P(Bn C)

P (A) > P(B) sincc (An Ct irnciAn Cl are disjoint.

P(C) =0

= C=0P(AnIlnC) =P(AnBnQ)

=P(t|)=0P(A)xPtBl,rP(C) =0 ...sinceP(C)=0.. P(AnBr-rC; =P(A)-P(B)-P(C)l{encc A, B. C are inclependent.

llP(lI) = 1,P(Tr = 1

Lct A bc tlre evcnt of getting 5 r:r 6.

I.'. P (A)= , and B bt' the evcnt of gctting

1 q a A.Lt a, 'J, a,

z.'. P(B)= -tIr.r ordcr to get a 'Hctrcl' belirrc "5 or 6", thefollowng cvents havc ttl ()ccul':

H, TI]H, TBTBFI, TBTBTBII. ....

26. Nurnber of ways differcnt numbcrs can occuron six dice = 6l

Total number of'ways of occurrencc = 65

6: 5!.'. Required probability, P = ^s - ---'---'--rr-5rr-64

24.

25.

I I 2 | rt)' I tl\' Ip = r*J. r "r*[iJ :.[l,l z

I

=-2 =3rl4

_l

5.26

p(A) = +

and P(B/A) = n!1?l)

= I- P(A) 2

II--' PtAnB) = t.PtA/B)= +

e(nne) I

P(B) 4

I

= P(B)=2

(o) IfAis a sutrevent ofB, Ac B, then A n B = A

.'. P (A n B) should be equai to P (A)' This is false'aJ

tbr P(A/B) = ; is false.'4

(c) P(An B) -P(A)-P(An")=;1n(nne) s r

.'. P(A/ll)=-+=t=;p(u) t- 2 a

.. P(A/B)+P(A/e,= i * i =:Hence given statement is false'

Probability of drawing a white ball from box B

will depend on whether transferred ball is black

or white.Ifa black ball is transferred, then its probability

A

t" 6'There are now 5 white and 8 black balls in the

box B. Then probability of drawing white ball

5from boxB = ,,Hencc probability of drawing a white ball fromurn B, if the transferred ball is black

= ! * a = 1q6 13 39

Similarly, probability of drawing a white bail

from urn B, if transferred ball is white262

=;"13=13l0 2 16

.'. Requiredprobability = , * 13

= 39'P(A) = 0

-- A=0. ArB = 0.'. P(Ar-rB)=P(Q)=0


The sum 6 can be obtained as follorvs :

(1,5), (2,4), (3,3), (4,2),(5,1), i e' in 5 ways'5

Probability ofA's throwing 6 with 2 dicc= jOjl

.'. Probability ofA's not throwing 6 = i66

Similarly, probability of B's throwiug 7 = -- ' i'e'

ProbabilitY of B's not throwing ? + :lo

NowA can win, if he throws 6 in thc first, thircl'

frfth, seventh etc' throws..'. Chance ofA's winning

1-(31/ 36) x(5 / 6)

5 36x6 30-36 61 61

(Av B)=A

(qt P (At-, Bt> P (Al =

ft) AaBcB.'. P(An B)

Also P(AnB)

32.27.

1188

33.

izD

1

-1>1 (B)

31-

=Ift

28.

(c) An B c B

.'. P(An ll)<P(B)

AIso An B

.'. P(An B )

3

8

= (o., o)

=1-P(AuIlt>1-{P(A)+P(B)

ft ,51 I= 1- j -F:f = :

[4 6J E

34. Let the eveut of drawilrg a faulty item from

any of the machines bc A, and the cvent thatan item drau'n at random was produced by /t/,

be B . To P (B,lAl Proceed as {bllorvs :

M, M, M, Remarks

P (B.) 0.25 030 045 'sum=1

P (A/8,) 0.05 004 0.03

P (8,)P (A/8, 0.0125 0.012 0.0135 ;rrnt = 0.ljt

P (B,iA) 0.0125 0.0125 0.0r iJ5 Ry Baye'slheorem0.038 0.038 0.0:lu


The highest output being from M, required

probability = t= o^t"T

= o. 355.0.038

35. Sample space, S = {HHHH, HHHT, HHTH,

HHTT' ... , TTTTI

Clearly, lsl :l=16.

36. (a) B =(AnB)u(A nB)

P(B) =P(AnB)+P(R nB)as

A n B and A n B are mutually exclusive.

= P(n nB) =P(B)-P(AnB).(b) p(AuB) =plAu(A nB)l

(c)

=p(A)+p(A nB)=P(A)+P(B)-P(AnB)

P(A,ts) =P(A)P(B/A)< P (A) as P (B/A) < 1.

(o) P (Au B) = P (A) + P (B) -P (AnB).'. P((An C) u(B nC)) = P(AnC) + P(B nC)

-P(enCnBnC)and P((AuB)nC) =P(AnC)+P(BnC)

-P(AnBnC)P((a u B) n c) p(A n c)

- P(C) P(C)

.p(BnC) P(AnBnC)P(C) P(C)

= P (A u B/C) = P (A/C) + P (B/C) - P (An B/C)

fal P (A., nlc)+ P (A n B/C)

_ P(AnBnC) * P(AnBnC)P(c) P(c)

_ P(AnBnC)+P(AnBnC)P(C)

_ P(AnBnC)u(AnBnC)P(C)

P(AnC)P(C)

= P (NC)

icr P (A.- ll )

=P(A)-P(AnB)=P(A)-P(A)P(B)= P (A) (1 -P (B)) = P(A) P ( B )

Hence A and B are independent.

5.27

38. (o) P((AuB) n C) = P((AnC) u(B nC))

= P (An C) + P (B n C)-P (An B n C)

= P (A) P (C) + P (B) P (C) - P (A) P (B) P (C)

= P (C) (P (A) + P (B) - P (A n B))

=P(C)P(AvB)

(b)

if

if

P(A/E)=P(BlE)

P(AnE)P(E)

P(AnE)=(n\

e len[Jn, | =\ i=r )

P(BnE)P(E)

P(BnE)(,)

P lB.rt JE ItYt\

,=j /

(, \PIU(BnE)l

\i=r .)

37.

if P U (AnE,)=

t P(AnE.)t'n

'''

= I p(BnE,)

ThisistruesinceP(AnE,)= P (B n E) foreach i.

39. Let E, be the event of a person to get into anaccident. Then

P(E)=p y I

P (at least one man meet with an accident)

= P (Er u E, r'-' ... v E,,)

=1-p(E, t-,E,-...rq,)=1-p(E, nE,^....,q,)= 1-Pt n, i P (E.t ...et n, t

= 1 _ (1 _ p) (t _p) ... (, _ o)= 1_ (1_p),,

Hence P (at least one man meets rvith an

accidcnt/a person is chosen) = -(I-p)").

Expectation of X = E(X) = avel-age = mean

=1i*nan\-= /Pr xt

L-I

where p, denote probability of occurrence ofr .

Letq=1-p.Then probabilities of succcss in lst. 2nd, 3rdtrials are

P, QP, q2P, -....

1

- (1IT

N.

5.28

.' E(n - (o) -p+(1) . qp+\2). q,p+...

= qp (L + 2q + 3q'+ ...+ nq " *t + ...)

= qp _qp_q

(l-q)z p2 p

Ratio ofAand B = 60 x 80 + 40 x 70 : 60 x 20+40x30

= 76:24=19:6

P (A/Au B) =P(An(AuR))

P(A u B)

= P(A) = P(A)P(AuB) P(A) + P(B)

P(A)=P(B)P(A)=i-P(B)P(A)+P(B)=1P(AuB)-P(AnB)=1=P(S)AUB =SonlyifAnB=gA=BonlyifAnB=0

P(A)=0pf4l =o\s/n(A)'=0n(S)tz(A)=0A=0


51. Given regression limit is8r-10y+66=0

866y - _x + _ ="10104

-54

D

9xl..J

52.S4ren each ibem is rlivicled tr-y a, i.c. ib becomcs Ia

When each item is increased bv 10. mean alsoincreased by 10.

Newmean= -t - ,, = "t-loo0-cI

53. Sirrce, x * ! = 6 for all tlre pairs of observations,b"u=br.r=-1

- p(X,Y) =-154. Mode =SMndian-2Mean

= Nlean = 1t"o,or,- 1onoo"z21'

= -ti3 Median - Mode)

55. Standard Deviation ofnew data

r; E': (,,,:=i)". V , -l;j =

Mode =3Median-2MeanSMcdian =Mode+2Mean

1

Median = 5(Mode + 2 N{can)

1

= M=:J

57. If we take y = 3x - 5 as regression equation ofYon X and Y = 2x - 4 as that ofX on Y, then

brr=3b"^,=

= brrbr, = ],which is not possible-

Hence, equation ! = 3x - S, i.e.r = |, . Ithe regression equation ofX on Yand equation, Y = 2x - 4 is that Y on X.

Hence 6u.

cov(X,Y)

ofcov(X, Y)

.l J,i-I+ --55

4t.

43.

M.

Hence

i.e.

4.5. When each item is increased by i", the mean isalso increased by )..

46.When each item is increased by ).; the S.D. ismultiplied bV I f I and variance is multiplied by12

47.18=3Median-2x24

*-2fi0, - r) = f, ! i - Y1 Ef, = Lf, t, - Lf,Y, = o

49. b,,b*;=+ttrtr t2

- ] covff'Y) | = to x, Yl,[' o'o,

I

50.Cov (X, Y), b.r b., and p(X, Y) all are eitherpositive, or 0 or negative simultaneously.

56.


Then

=58. Solving equations

3r-v -5=0 and2,l -Y-4=O,we get r-1andy=-2.

59. Here 6r-. = L a.td b', = I '

Since 0 ! bnrb*n < 1, therefore we must have

M. P(X,Y =

65. p(X,Y) - cov(X'Y) = -J- ?' o"ou 3x4 3

a + (a + d) + @ + 2d) +" "+ (o + 2nd)66.Mean,V=- Zn+t--

2tt+l= ila+@+(a+2ndt)l=a+nd'

Sum of numerical deviations from the mean

ztl+ |

= )tt"+(t-l)dl -la+ ndlli=1

= la-a-ndl+lo' + d-a -n'dl+"'+ lu + nd, -a -nrJl + lo + (rz +'1) d -a - rull

+...+lc +2nd-o-ndl= ld,lln + (n -1) + (rz - 2) +"'+ 1 + 0

+ 1' + 2 +...+ rzl

= ldlx2ll + 2 + 3 +...+ nl

= n(n + l)ldln(n+1)ldl

M.D. (about mean) = --%, *l67. Iftwo variables X and Y are in perfect correl ation

(direct or indirect), ttren there is a linearrelationship between the two variables'

68.Given : lower quartile, Qr = 35

and upper quartile Qs = 75

Quartile d".ri.tior, = 9* =Ef =zo

69. Since p > 0, therefore bn* and br" are also positive

b..- +b r:------Hence "Jr _ "r.r, > Jby., b..J

2

b.,- + b"^, G= > r/p=2

bur+b-

- tJ >o2

- m>P

70.If frrst time r + 2y = 7 is taken as regressron

equation of Y on X, then we write it as

I7\]=--X+-'22

h =-l-yx 2

1b),, -- 2 and br, = 5

3b,rbr, = z

o.J{o(X,Y)}z = 5

Eo(XY) = 1-

1

4

A*s

f1\o<r[7J <t

0<).<4b"n + 0,

brt+ o

- ),+0.Hence, we must have 0 < l' < 4'

60. Cov (X, Y) = 0

br, =brn-=O

So, tw<t lines of regression ate y - y = 0 and

x - i = 0, which are al right angles'

61. The two lines of regression are at right angles

only when

. (r )u_ [,*,J=_t

:1 brt =-bt!

But both bn, and b* are of same sign, thereforewe must have

br, = br, __ o

= P(X,Y) = 0

62.Given : b^,, = - 0.9 is negative and b', = 0'4 ispositive *ftlcft is not possible as bn' and b* are

"l*uy. of same sign. Hence given data isinconsistent.

ffl.We know

.-

But b... andf^

same srgn

brr bt, = P2

p = !\Er' 4,

b..,, p and cov (X, Y) are all of the

by bop = (sgn brr)

s.30

Also tbc second equation (which will be regressionequation X on Y) can be u'ritten as

t7)c = - - v + -2"2

b]-i

Hence regression equation of Y on X isx+2Y =f

71. Since X and Y connected by linear relationship,therefore they are in perfect correlatiott.

Al-qo ob < 0

a;<0.b

/ .r\ c' shows that when r increases,"' Y= l, OJt-ay also incrcascs; so X and Y are in perfect directcorrelation+ p(X,10 = 1.

7Z.Here,

:= cov(u,r) = 12(u- u)2(u- u In

t_=::{9(.r -r)2-40-y)2n

+ek_i )Cy_ t))/>rr-ti2) (z -o\=el- r-4r w-Yr

I

\n/\n)^( ztx- rXy - y) )+ Jl- I\n)

- s o'..' - 4<r,,2 + 9 cov. (r, y)

=9x4-4x9+9(0)-0

lsince X, Y trre independent,therefore cor'(X, Y) = 0l

cov(tl, rr)

= p(u., L') = -o ,., = 0

73. Required mean


fl2"-1

2"

= + (..' C, + 2oo + 3c" +... nCn- n2,-r1

7:!.Occurrence of evcnt of ch<losirtg a ball occursonly after a box is choscn. Hence to gct theprobability of choosing a rt'd hall in box B. rve

must consider box B as sirttrple spacc.

P(brown) = P(brown ir hox A afterchoosing A) + I' (trrown in box Bafter choosing'B)

75.P (green or blue)

P(A,uAr) - P(Ar t + P(A,t - P(A,nAr)

)p(Ar) = r

t ,(,r,)ylr(A,) -1P(ArnAr)= ,- uttd I 3

P(A,nAr) = P(Ar) r P(A,,)

Hence A, and A, arc independent.

79.There are six possible ways in rvlrich the diecan fall and of these therc is onll' onc way ofthrowing 4.

Required.h^.,"" = I .

o

80.There are six possible ways in which the die canfall. Of these there are only 3 r,r,avs of getting 2.

4or6.

,1o.y =- 2

h =(-t)il) =1=1+ i2t(21 4

(t o) lt r) |- |

- V - | +l

- y

- | _- --(2":/'[z s/ r)

-t = '+P(A.t ]623

(t 2) (r l)- l-x- l+t- x- |-\.2 3) \2 5)

1 I 13=_*-=-31030

i =3r+4y andu =3i-tLt-i =3@-Jc)+4$t-Y1u-6 =3(x-r)+a0-y)

(t 2) (t 2\76. prredorblue)= lJ*:].l;"l]-/

= 1*1= 8

3 5 15

77. Ptn)=I

=P(A1 ')= i

78.

rlRcquiredchance= I = -''62

5.31PROBABILITY & STATISTICS

81, Number of ways of choosing 3 students out 9=eC", i.e.84

A student can be removed from 1st year studentsin 2 ways, from 2nd year in B ways and from Brdyear in 4 ways, so that the total number of waysof removing three students, one from eachgroupis2x3x4.

85.Probability of drawing a king =If card is not51 cards only

A.aqueen ts _;,)l

I

13'replaced, then pack will haveso that chance of drawing a

4

663(- 1)"

n!

Required chance2x3x4

nc,

82.Number of ways of removing two from 1st yearstudents and one from others =tCrrrCtNumber of ways of removing two from ind yearstudents and one from others =

rC, x oCr.

Number of ways of removing 2 from Brd yearstudents and one from others =

rC, "

nC'Hence total number ofways in which two studentsof the same class and third from the othersmay be removed

= zCrxTC r+

sCrx 6C, + aC rx

sC,

=7+18+30=55

Required chance = !1.84

83.Three students can be removed from 2nd yeargroup in tC' i.e., 1 way and from3rdyeargroup in aC., i.e. L way from 81fl ys31group in aC' i.e.4ways.Hence total number of ways in which threestudents Outo"*=rnr rl?.j;r" .to.,

.'. Required chance = 5

84

S4.Probability of'dra 'l\mngaKlllg= - =

If card is replaced, the pack will again have b2

cards, so that plobability of drawing a queen is a .

l3The two events beingindependent, the probabilityof drawing b<-rth cards in succession

.'. Probability of drawing both cardst413 5l

I+; -............ +

a:

tl242847

86. Probabil ity, p =

1

87. -t, since there is only one way in which all thenl'

letters to the correct envelopes.

88.We solve the problem using a tree diagram ; Dand N stand fo defective and non-defective cases.The various probabilities are marked on thebranches ofthe tree.Following the branches

P(D) = Probability of a defective bulb

13 12153103638I 202_ _x_3 240

302720

89. Probability that defective bulb was frombox I

= P (A,/D)

(where A is the event of choosing box I)

_ p(Anp)P(D)

13-x-310- 302

720('.' An D = defective from box I

+ P(AnD)= 1,1 =fL3 10 302101. Probability of technocrat manaser

20i= 1oo=s=P

Probability of non-technocrat mana ger

804= 1oo

=E="

I

l3

lt=- x

-13 13

I169 '

Probability of a random committee of 5 withexactly 2 technocrats

= 5c2ll2q3

:ox4(r)2f+)3=-t

-l1x2\5/ \5/=0.2048

102. Probability that frrst friend is born in any month= I00Vo = 1

Probability that second friend is born in the samemonth as that of frrst friend

,11= t'r2=E

107. Probability of drawing a red ball = frIf ball is not replaced, then box will have a ball,so probability of drawing the red ball in nextchance

4

9Hence probability of drawing 2 balls

542= 10 "t = 9

109. Probability ofboth cards beingkings

431= - - =

-52 51 227

111. After frrst heads in first tossProbability of tails in 2nd and 3rd toss

=1.1= 1

22 4.'. Probability of exactly two heads

=1-1='44Il2. Let lower limit be r.

Then, upperlimit = r+4

t9l9 = rsz

= ;r=13113. Range = Difference between largest values

= 95-62=33


7+9+11+13+15 55_ -_

11f

114. m=

f, d': =17 - 111'z + le - 11|'z + lrt - ttl' + lr: - t tl' + ltS - ttl'=40

Eto=\, =

= 2.8

@Y)

115. Requiredprobability

= P [(A dies and B is alive)or (A is alive and B dies)l

= p(7-q)+(1-p)q

= p+q-2pq116. We have np= 4vyllnpq=J

aJ+ 4 = + andp=(1-q)=

Mode is an integer such thatnp+p>x>np-q

13+ 4+->x>4-144

13 77

= Tt,tT+ 3.25<x<4.25+ x=4

ll7. tn+6'= 24 and nrc2=128

Solving we getm =76or8.

Ifm=16then02 =8

Ifm=8then) P^

O- = OO.

Case I '. ftp = 16 and np7 = 8

11+ P=;,1=;andn=32.zzCase II ;11p = g v1f, npq = gg

+ 4 = 7, which is not possible.Hence distribution is

(q+p)" =

1

At

(r r\"\r-, )

gk math 2.pdf

Documents