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Probability Theory and Random Processes
Title : Probability Theory and Random Processes Course Code : 07B41MA106 (3-1-0), 4 CreditsPre-requisite : NilObjectives : To study● Probability: its applications in studying the outcomes of random experiments● Random variables: types, characteristics, modeling random data● Stochastic systems: their reliability ● Random Processes: types, properties and characteristics with special
reference to signal processing and trunking theory.
Learning Outcomes :students will be able to (i) model real life random processes using appropriate statistical distributions;(ii) compute the reliability of different stochastic systems;(iii) apply the knowledge of random processes in signal processing and trunking
theory.
Evaluation Scheme
Evaluation Components
Weightage ( in percent)
Teacher Assessment
25(based on assignments, quizzes,attendence etc.)
T1 (1 hour)
20T2 (1hour 15 min)
25T3 (1 hour 30 min)
30 ---------------------------------------------------------------------- Total 100
Reference Material :1. T. Veerarajan. Probability, Statistics and Random processes. Tata McGraw-Hill.2. J. I. Aunon & V. Chandrasekhar. Introduction to Probability and Random Processes. McGraw-Hill International Ed. 3. A. Papoulis & S. U. Pillai. Probability, Random Variables and Stochastic Processes. Tata WcGraw-Hill.4. Stark, H. and Woods, J.M. Probability and Random Processes with Applications to Signal Processing..
The study of probabilities originallycame from gambling!
Origins of Probability
Why are Probabilities Important?
• They help you to make good decisions, e.g.,– Decision theory
• They help you to minimize risk, e.g.,– Insurance
• They are used in average-case time complexity analyses of - Computer algorithms.• They are used to model processes in - Engineering.
Random Experiments• An experiment whose outcome or result can
be predicted with certainty is called a deterministic experiment.
•Although all possible outcomes of an experiment may be known in advance, the outcome of a particular performance of the experiment cannot be predicted owing to a number of unknown causes. Such an experiment is called a random experiment.
•A random experiment is an experiment that can be repeated over and over, giving different results. •e.g A fair 6-faced cubic die, the no. of telephone calls received in a board in a 5-min. interval.
Probability theory is a study of random orunpredictable experiments and is helpful in investigating the important features of these random experiments.
Probability Definitions
• For discrete math, we focus on the discrete version of probabilities.
• For each random experiment, there is assumed to be a finite set of discrete possible results, called outcomes. Each time the experiment is run, one outcome occurs. The set of all possible outcomes is called the sample space.
Example.
If the experiment consists of flipping two coins, then the sample space is:
S = {(H, H), (H, T), (T, H), (T, T)}
Example.
If the experiment consists of tossing two dice, then the sample space is:
S = {(i, j) | i, j = 1, 2, 3, 4, 5, 6}
More Probability Definitions
• A subset (say E) of the sample space is called an event. In other words, events are sets of outcomes.
( If the outcome of the experiment is contained in E, then we say E has occurred.)
For each event, we assign a number between 0 and 1, which is the probability that the event occurs.
Example.
If the experiment consists of flipping two
coins, and E is the event that a head
appears on the first coin, then E is:
E = {(H, H), (H, T)}
Example.
If the experiment consists of tossing two
dice, and E is the event that the sum of the
two dice equals 7, then E is:
E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
Union: E F
Intersection:E F also denoted as EF
(If E F = , then E and F are said to be mutual exclusive.)
NOTE
It may or may not be true that all outcomes are equally likely. If they are, then we assume their probabilities are the same.
Definition of Probability in Text
The probability of an event E is the sum of the probabilities of the outcomes in E
p(E) = n(E)/n(S)
Sin cases ofnumber eexhasustiv
E tofavourable cases ofnumber
Example
• Rolling a die is a random experiment.
• The outcomes are: 1, 2, 3, 4, 5, and 6, presumably each having an equal probability of occurrence (1/6).
• One event is “odd numbers”, which consists of outcomes 1, 3, and 5. The probability of this event is 1/6 + 1/6 + 1/6 = 3/6 = 0.5.
Example• An urn contains 4 green balls and 6 red balls.
What is the probability that a ball chosen from the urn will be green?
• There are 10 possible outcomes, and all are assumed to be equally likely. Of these, 4 of them yield a green ball. So probability is 4/10 = 0.4.
Example• What is the probability that a person wins the
lottery by picking the correct 6 lucky numbers out of 40? It is assumed that every number has the same probability of being picked (equally likely).
Using combinatorics, recall that the total number of ways we can choose 6 numbers out of 40 is:
C(40,6) = 40! / (34! 6!) = 3,838,380.Therefore, the probability is 1/3,838,380.
Examples• Consider an experiment in which a coin is tossed twice.• Sample space: { HH, HT, TH, TT }• Let E be the event that at least one head shows up on
the two tosses. Then E = { HH, HT, TH }• Let F be the event that heads occurs on the first toss.
Then F = { HH, HT }• A natural assumption is that all four possible events in
the sample space are equally likely, i.e., each has probability ¼.Then the P(E) = ¾ and P(F) = ½.
Probability as a FrequencyLet a random experiment be repeated n times andlet an event A occur times out of the n trials.An
Aevent the
offrequency relative thecalled is n
n ratio The A
.n
nlimP(A) i.e., A,event theofy probabilit thecalled is
P(A),by denoted value,This alue.constant v aapproach
toand stabilise to tendency a shows n
n increases,n As
An
A
Frequency Definition of Probability
• Consider probability as a measure of the frequency of occurrence.– For example, the probability of “heads” in a
coin flip is essentially equal to the number of heads observed in T trials, divided by T, as T approaches infinity.
T
heads ofnumber lim)headsPr( T
Probability as a Frequency• Consider a random experiment with possible outcomes
w1, w2, …,wn. For example, we roll a die and the possible outcomes are 1,2,3,4,5,6 corresponding to the side that turns up. Or we toss a coin with possible outcomes H (heads) or T (tails).
• We assign a probability p(wj) to each possible outcome wj in such a way that:
p(w1) + p(w2) +… + p(wn) = 1• For the dice, each outcome has probability 1/6. For the
coin, each outcome has probability ½.
ExampleTo find the probability that a spare part produced by a machine is defective.
If , out of 10,000 items produced , 500 are defective, it is assumed that the probability of a defective itemis 0.05
Axioms of Probability
A B
A B
A B
A B
A
A
Axioms (where A and B are events):• 0 <= P(A) <= 1• P(S) = 1; P({}) = 0• P(A B) = P(A) + P(B) – P(A B)• If A and B are disjoint then P(A B) = P(A) + P(B) (mutually exclusive events)
•
Example• Rolling a die is a random experiment.• The outcomes are: 1, 2, 3, 4, 5, and 6. Suppose the die
is “loaded” so that 3 appears twice as often as every other number. All other numbers are equally likely. Then to figure out the probabilities, we need to solve:
p(1) + p(2) + p(3) + p(4) + p(5) + p(6) = 1 and p(3) = 2*p(1) and p(1) = p(2) = p(4) = p(5) = p(6). Solving, we get p(1) = p(2) = p(4) = p(5) = p(6) = 1/7 and p(3) = 2/7.
• One event is “odd numbers”, which consists of outcomes 1, 3, and 5. The probability of this event is: p(odd) = p(1) + p(3) + p(5) = 1/7 + 2/7 + 1/7 = 4/7.
0.)P( point,
sample no containing (event)subset the is if , i.e.
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1 Theorem
)P(P(S))P(S Hence
exclusive. mutually are
event impossible the and Sevent certain The
oofPr
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events, 2 any are B and AIf
Theorem
AB.and BA events exclusive
mutually the of union the is B and ABand BA
events exclusive mutually the of union the is A
:oofPr
)AB(P)BA(P)B(P&
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BA AB BA
S
A B
)AB(P)BA(P)AB(P)BA(P
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)BA(P)BA(P
Hence proved.
BA AB BA
S
A B
If event F occurs, what is the probability
that event E also occurs?
This probability is called conditional
probability and denoted as p(E|F).
Conditional Probability
Definition of Conditional Probability
If p(F) > 0, then
Fp
FEpFEp
|
Example.
An urn contains 8 red balls and 4 white
balls. We draw 2 balls from the urn
without replacement. What is the
probability that both balls are red?
Solution: Let E be the event that both balls drawn arered. Then p(E) = C(8, 2)/C(12, 2)
or, we can solve the problem using conditionalprobability approach,
Let E1 and E2 denote, respectively, the eventsthat the first and second balls drawn are red.Then
p(E1E2) = p(E1) p(E2 | E1 ) = (8/12) (7/11)
Multiplication Rule
nEEEEp ...321
121213121 ...|...|| nn EEEEpEEEpEEpEp
Example.
A deck of 52 playing cards is randomly
divided into 4 piles of 13 cards each.
Compute the probability that each pile
has exactly one ace.
Solution: Define events
E1 = the first pile has exactly one ace
E2 = the second pile has exactly one ace
E3 = the third pile has exactly one ace
E4 = the fourth pile has exactly one ace
p(E1E2E3E4) = p(E1)p(E2| E1)p(E3|E1E2)p(E4|E1E2E3)
p(E1) = C(4,1)C(48,12)/C(52,13)
p(E2|E1) = C(3,1)C(36,12)/C(39,13)
p(E3|E1E2) = C(2,1)C(24,12)/C(26,13)
p(E4|E1E2 E3) = C(1,1)C(12,12)/C(13,13)
p(E1E2E3E4) 0.1055
Let E and F be events. We can express E as
E = EF EFc
Where Fc is the complementary event of F.
Therefore, we have
p(E) = p(EF) +p(EFc) = p(E|F)p(F) +p(E|Fc)p(Fc)
Independent Events
Two events E and F are independent if
p(EF)=p(E)p(F)
Two events are not independent are said to bedependent.
p(EF) = p(E)p(F) if and only if p(E|F) = p(E).
If E and F are independent, then so are E and
Fc.
ts.independen also are B&A
events the,t independen are B andA events theIf
Theorem
heorem)addition tby )(B(P)BA(PB)P(Asuch that
exclusivemutually are BA&BA events The
oofPr
)BA(P)B(P)BA(P eorem)product th by)(B(P)A(P)B(P
)B(P)A(P)A(P1)B(P
Three events E, F, and G are independent if
p(EFG) = p(E)p(F)p(G)p(EF) = p(E)p(F)p(EG) = p(E)p(G)p(FG) = p(F)p(G)
Example. Two fair dice are thrown. Let E denotethe event that the sum of the dice is 7.Let F denote the event that the firstdie is 4 and let G be the event that thesecond die is 3. Is E independent ofF? Is E independent of G? Is Eindependent of FG?
p(E) = 6/36 = 1/6 p(F) = 1/6 p(G) = 1/6 p(EF) = 1/36
E and F are independent.
p(EG) = 1/36 E and G are independent. p(FG) = 1/36 F and G are independent. p(EFG) = 1/36 but, E and FG are NOT independent.
k1,2,.....,i )B(P)B/A(P
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thenk, to1 ifor
iesprobabilit lconditiona thebe )A/B(P&)P(A/BLet
S. space sample theofpartition a be B,.......B,BLet
Theorem sBaye'
k
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ii
k21
.jiBB (i) i.e.,
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4B
A
S
.i0)B(P)iii(SB)ii( iik
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)B(A.......)B(A)B(A
Then S. with associatedevent thebeA Let
k321
k21
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A
ji),BB(A)B(A.......)B(A)B(A
For, exclusive.mutually pairwise are
)B(A),.......,B(A),B(A events theall Also
jik21
k21
)BA(P
...)BA(P)BP(AP(A) Then
k
21
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expressed bemay )BP(A each termHowever
jj
j
k
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kk
2211
)B(P)B/A(P
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...)B(P)B/A(P)B(P)B/A(P)A(P
)A/B(P)A(P)B/A(P)B(P)AB(P iiii
)A(P
)B/A(P)B(P)A/B(P ii
i
k
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)B/A(P)B(P
)B/A(P)B(P
Example. In answering a question on a multiple-choicetest, a student either knows the answer orguesses. Let p be the probability that thestudent knows the answer and 1p theprobability that the student guesses.Assume that a student who guesses at theanswer will be correct with probability 1/m,where m is the number of multiple-choicealternatives. What is the (conditional)probability that a student knew the answer to aquestion, given that his answer is correct?
Solution:
C = the student answers the question correctly,
K = the student actually knows the answer
KpKCpKpKCp
KpKCpCKp
~|~|
||
mpp
p
/11
pm
mp
11
Example.
When coin A is flipped it comes up heads
with probability ¼, whereas when coin B
is flipped it comes up heads with probability ¾.
Suppose that one coin is randomly chosen and
is flipped twice. If both flips land heads, what is
the probability that coin B was the one chosen?
Solution:
C = coin B is chosen
H = both flips show head
CpCHpCpCHp
CpCHpHCp
~|~|
||
21
41
41
21
43
43
21
43
43
10
9 9.0
Example. A laboratory test is 95 percent correct in detecting a certain
disease when the disease is actually present. However, the test also yields a “false” result for 1 percent of the healthy people tested. If 0.5 percent of the population has the disease, what is the probability a person has the disease given that his test
result is positive?
Example.
A suspect is believed 60 percent guilty. Suppose now a new piece of evidence shows the criminal is left-handed. If 20
percent of the population is left-handed,and it turns out that the suspect is also left-handed, then does this change the guilty
probability of the suspect? By how much?
Solution:
D = the person has the disease
E = the test result is positive
995.001.0005.095.0
005.095.0
294
95
323.0
DpDEpDpDEp
DpDEpEDp
~|~|
||
Example.
A suspect is believed 60 percent guilty. Suppose now a new piece of evidence shows the criminal is left-handed. If 20
percent of the population is left-handed,and it turns out that the suspect is also left-handed, then does this change the
guilty probability of the suspect? By how much?
Solution:
G = the suspect is guilty
LH = the suspect is left-handed
GpGLHpGpGLHp
GpGLHpLHGp
~|~|
||
4.02.06.00.1
6.00.1
68
60
88.0
Random Variable
Definition:A random variable (RV) is a function that assigns a real number X(s) to every element
,Ss where S is the sample space corresponding to a random experiment E.
Discrete Random VariableIf X is a random variable (RV) which can take a finitenumber or countably infinite number of values, X is called a discrete RV.
Eg. 1. The number shown when a die is thrown 2. The number of alpha particles emitted by a radioactive source are discrete RVs.
RS:X
ExampleSuppose that we toss two coins and consider the sampleSpace associated with this experiment. Then S={HH,HT,TH,TT}. Define the random variable X as follows: X is the number of heads obtained in the two tosses. Hence X(HH) = 2, X(HT) = 1 = X(TH) & X(TT) = 0.
Note that to every s in S there corresponds exactlyone value X(s). Different values of x may lead to the same value of S.
Eg. X(HT) = X(TH)
Probability FunctionIf X is a discrete RV which can take the values
,....x,x,x 321 such that
,...)3,2,1i(p provided function,y probabilit
point or function massy probabilitor function
y probabilit thecalled is p then ,p)xX(P
i
iii
satisfy the following conditions:
ii
i
1p)ii(
,&i,0p)i(
Example of a Discrete PDF
• Suppose that 10% of all households have no children, 30% have one child, 40% have two children, and 20% have three children.
• Select a household at random and let X = number of children.
• What is the pmf of X?
Example of a Discrete PDF
• We may list each value.– P(X = 0) = 0.10– P(X = 1) = 0.30– P(X = 2) = 0.40– P(X = 3) = 0.20
Example of a Discrete PDF
• Or we may present it as a chart.
x P(X = x)
0 0.10
1 0.30
2 0.40
3 0.20
Example of a Discrete PDF
• Or we may present it as a stick graph.
x
P(X = x)
0 1 2 3
0.10
0.20
0.30
0.40
Example of a Discrete PDF
• Or we may present it as a histogram.
x
P(X = x)
0 1 2 3
0.10
0.20
0.30
0.40
2)0)(ii)P(XP(X (i) Find value.
positive some is where,.....,2,1,0ii!
cp(i)
by given is X variablerandom a of pmf. The
Example
i
1!i
c have we,1p(i) Since
.Solution
0i 0i
i
.1ce have we,!
e As0
i
i
i
e!0
e0)(XP Hence
0-
)2X(P1)2X(P
2
eee1
2
!
e)(X
-
xxP
x
If X represents the total number of heads obtained,when a fair coin is tossed 5 times, find the probability distribution of X.
32
1
32
5
32
10
32
10
32
5
32
1 :P
5 4 3 2 1 0 :X
Continuous Random VariableIf X is an RV which can take all values (i.e., infinite number of values ) in an interval, then X is called a continuous RV.
1x,1
,1x0,x
,0x,0
)x(X
Probability Density FunctionIf X is a continuous RV,then f is said to be the probability density function (pdf) of X , if it satisfies the following conditions:
1dx)x(f)ii( and,Rx,0)x(f)i( x
b
adx)x(fb)XP(a
,ba- with ba,any For (iii)
When X is a continuous RV
a
a0dx)x(f)aXa(P)aX(P
This means that it is almost impossible that a continuousRV assumes a specific value.
b)XP(ab)XP(ab)XP(a Hence
Probability Density Function
0.00.10.20.30.40.50.6
-3 0 3 6 9 12 15x
f X (x )
xe
xxf xX 0if
10if0
)( /
not.or function density y probabilit a is
1x0,4x f(x)function he whether tCheck
Example3
.1dxx4dx)x(f Also
[0,1].x0f(x)Clearly .Solution1
0
1
0
3
)53X2(P)iii)(1X1)(ii)P(P(X (i) Obtain
elsewhere 0
2x2- 4/1f(x)
functiondensity thehas X variablerandom A
(¾,1/2,1/4)
Find the formula for the probability distributionof the number of heads when a fair coin istossed 4 times.
Cumulative Distribution Function (cdf)If X is an RV, discrete or continuous , then P(X<=x)is called the cumulative distribution function of X or distribution function of X and denoted as F(x).
xX
j j
j
p F(x) , discrete is X If
X
-
f(x)dxx)XP(-F(x) ,continuous is X If
Probability Density Function
0.00.10.20.30.40.50.6
-3 0 3 6 9 12 15x
f X (x )
xe
xxf xX 0if
10if0
)( /
Cumulative Distribution Function
0.00.20.40.60.81.01.2
-3 0 3 6 9 12 15x
F X (x )
xe
xxF xX 0if1
0if0)( /
Probbility Density Function
0.0000
0.0004
0.0008
0.0012
0.0016
-100 0 100 200 300 400 500 600 700 800
x
f X (x )
0 if 0
1( ) if 0
0 if
X
x
f x x uu
u x
Cumulative Distribution Function
0.0
0.2
0.4
0.6
0.8
1.0
-100 0 100 200 300 400 500 600 700 800
x
F X (x )
0 if 0
( ) if 0
1 if
X
x
xF x x u
uu x
variable.random the
offunction on distributi cumulative the(ii) K, (i) Find
otherwise 0
1x0),x-K(1 f(x)by given
is variablerandom a ofdensity y probabilit theIf2
function.on distributi theandK eminDeter
otherwise 0
xifx1
1K.
f(x)
fuctiondensity thehas X variablerandom A
2
otherwise 0
1x0 ]3/xx[2/3 3
otherwise,0
x),x(tan/1)x(F
/1K1
Properties of the cdf F(x)
. )F(x)F(xthen
,x xif , i.e. x,ofunction f decreasing-non a is F(x).1
21
21
.1)(F&0)(F.2
).x(F)x(F)xP(X then .....,xx....xxx
where,....,x, x values takingRV discrete a is X If .3
1iiii1i321
21
able.differenti is F(x) wherepoints
all at),x(f)x(Fdx
dthen RV, continuous a is X If.4
X. offunction massy probabilit the
)xX(Pp where,p)g(xE{g(X)}
as defined is g(X) of mean value the
or valueexpected then theRV, discrete a is X If
sDefinition
iiiii
xR
g(x)f(x)dxE{g(X)}
thenf(x), pdf with RV continuous a is X If
2xx iancevar&mean its are X RV a singcharacterifor used
commonlymost are which valuesexpected Two
xR
iiix
continuous is X if xf(x)dx,
discrete is X if,px)X(E
xR
2x
ii
2xi
2x
2x
continuous is X if,dx)x(f)-(x
discrete is X if ,p)x(
)X(E)x(Var
The square root of variance is called the standard
deviation.
22 )}X(E{)X(E)X(Var
))X(Ece(sin)X(E
constant) a is ce(sin)X(E2)X(E
}X2X{E})X{(E)X(Var
x2
x2
x2
xx2
2xx
22x
22 )}X(E{)X(E)X(Var
ExampleFind the expected value of the number on a die when thrown.(7/2)
X. ofmean theevaluate (d) and X
of cdf thefind (c) 2),XP(-2 and 2)P(X Evaluate (b) K, Find (a)
3K 0.32K 0.2K 0.1 :p(x)
3 2 1 0 1- 2- :x
ondistributiy probabilit following thehas X variablerandom A
Example
3).by divisible is P(X&5)P(X even), is P(X also Find
on.distributi theof varianceandmean thefind
1and isy probabilit total that theVerify
,..).3,2,1j(1/2j)P(Xby given ison distributi
discrete infinitean offunction y probabilit Thej
P(x).function on distributi its find (b)
X.) RV continuous a pdf(of a is p(x) that show)a(
0 x0
0xxep(x) If
2/-x2
variance.andmean find and k ttancons the eminDeter
otherwise0
0,0xkxef(x)by definedfunction density
y probabilit thehas variablerandom continuous A
Example
x -
)/6,/2,k( 22
0x,e1)x(F 2/x2
MomentsIf X is the discrete or continuous RV, is called rthorder raw moment of S about the origin and denotedby .
)X(E r
r'
-
r
x
r
rr
continuous is X if dx)x(fx
discrete is X if )x(fx
)X(E'
.)X(EVar(X)
origin about themoment first mean
),X(E&)X(Eby given
areorigin about the moments second andfirst theSince
211
12
212
212
11
)X(Var))X(EX(E
0)X(E)X(E))X(EX(E2
2
1
.by denoted and X of
moment centralorder nth thecalled is )X{(E
n
nx
X. of moments absolute called are }X{E&}X{En
x
n
X. of moments dgeneralise called are }aX{E&})aX{(Enn
Two-Dimensional Random Variables
variable.random ldimensiona- twoa called
is Y)(X,Then S.s outcomeseach number to real a assigning
each functions twobe Y(s) Y and X(s) XLet E. experiment
random a with associated space sample thebe SLet :sDefinition
Two-dimensional continuous RV.Two-dimensional discrete RV.
,.....n,...,3,2,1j,......;m,....,3,2,1i),y,x()Y,X( ii
Probability Function of (X,Y)
Y)(X, ofon distributiy probabilitjoint thecalled is
...3,2,1j,...,2,1i},p,y,x triplets{ofset The
1p)ii(
j&i,0p(i)
provided Y)(X, offunction massy probabilit thecalled is p then
p)yy,xP(xsuch that RV discrete ldimensiona- twoa is Y)(X, If
ijii
j iij
ij
ij
ijii
• If (X,Y) is a two-dimensional continuous RV. The joint probability density function (pdf) f is a function satisfying the following conditions
1dxdy)y,x(f
yx,0)y,x(f
2
1
2
1
xx
yy2121 dydx)y,x(f]yYy,xXxPr[
x y
j iij
dvdu)v,u(f
p)y,x(F
Cumulative Distribution Function
Y)(X, of cdf thecalled is y}Y&xP{Xy)F(x,then
),continuousor eRV(discret ldimensiona- twoa is Y),(X If
)(),()(),(
increase. both,or y,or either x asfunction ingnondecreas a is ),(
1),(
0),(),(),(
,1),(0
xFxFyFyF
yxF
F
FxFyF
yxyxF
XY
•Properties of joint PDF
yx
yxFyxf
yYxXyxF
),(),(
),Pr(),(2
Examplestossing two coins
headfor 1 for tail, 0 )}1,1(),0,1(),1,0(),0,0{( space sample
coin second with theassociated variablerandom ...
coinfirst with theassociated variablerandom ...
Y
X
1)1,1(F4
3)0,1(F
4
2)1,0(F
4
1)0,0(F
1/21/4
1 x
y
F(x,y)
ExampleThree balls are drawn at random without replacementfrom a box containing 2 white,3red and 4 black balls.If X denotes the number of white balls drawn and Ydenotes the no of red balls drawn, find the joint probability distribution of (X,Y).
SolutionsAs there are only 2 white balls in the box, X can take the values 0,1,2and Y can take the values 0,1,2,3.
P(X=0,Y=0)=P(drawing 3 balls none of which is white or red)=P(all the 3 balls drawn are black)
21/1C/C 39
34
P(X=0,Y=1)=3/14,P(X=0,Y=2)=1/7………
X Y
0 1 2 3
0
1
2
X Y
0 1 2 3
0 1/21 3/14 1/7 1/84
1 1/7 2/7 1/14 0
2 1/21 1/28 0 0
4).YP(X&1)3/XP(Y
3)1/YP(X3),Y1,P(X1),P(X find below,given
Y)(X, ofon distributiy probabilit bivariate For the
1 2 3 4 5 6
0 0 0 1/32 2/32 2/32 3/32
1 1/16 1/16 1/8 1/8 1/8 1/8
2 1/32 1/32 1/64 1/64 0 2/64
X
Y
(ans.7/8,9/32,18/32,9/28,13/32)
a21-
y2x-
e1,3/1),e1(e ans.
a)(iii)P(XY),(ii)P(X1),Y1,P(X (i) Compute
otherwise0
y0,x0e2ey)f(x,
bygiven is Y)(X, offunction density joint The
Example
Marginal probability density function
For every fixed j
p(xj, y1) + p(xj, y2) + p(xj, y3) + … = p{X= xj} = f(xj)
and for every fixed k
p(x1, yk) + p(x2, yk) + p(x3, yk) + … = p{Y= yk} = f(yk)
The probability functions f(xj) and g(yk) are also
called marginal probability density functions.
dx)y,x(f)y(f,dy)y,x(f)x(f YX
As an illustrative example, consider a joint pdf of the form
-Integrating this wrt y alone and wrt x alone gives the two marginal pdf
-
elsewhere 0
10,10for )1(5
6),( 2
yxyxyxf
1y0)3
y1(
5
6)y(f
1x0)2
x1(
5
6)x(f
Y
2
X
Independent Random Variables
Two random variables X and Y with joint probability density function f(x,y) and marginal probability functions and
If
F(x,y) = F(x) G(y)
Or p(x,y) =
for all x, y, then X and Y are independent.
)x(fX
)x(fX
)y(fY
)y(fY
Example A machine is used for a particular job in the forenoonand for a different job in the afternoon. The joint probability of (X,Y), where X and Y represent the number of times the machine breaks down in the forenoon and in the afternoon respectively, is givenin the following table. Examine if X and Y are independent RVs.
X Y 0 1 2
0 0.1 0.04 0.06
1 0.2 0.08 0.12
2 0.2 0.08 0.12
4.0P;4.0P;2.006.004.01.0)0(fP
j,iPPP if ,t independen are Y&X
*2*1*0
ijj*i*
3.0P;2.0P;5.0P 2*1*0*
011**0
00*00*
P04.02.02.0PP
P1.05.02.0PP
Hence the RVs X and Y are independent
X Y 0 1 2
0 0.1 0.04 0.06
1 0.2 0.08 0.12
2 0.2 0.08 0.12
t.independen are Y and X that ovePr
otherwise0
0y,x,eee-1y)F(x,
bygiven is Y)(X, variablerandom continuous
theoffunction on distributi cumulative The
y)(x-y-x-
otherwise0
0y,xe
yx
)y,x(F)y,x(f
)yx(2
otherwise0
0ye)y(f
otherwise0
0xe)x(f
y
2
x
1
0ye1
0y0)y(F
0xe1
0x0)x(F
x2x1
)y,x(F)e1)(e1()y(F)x(F yx21
tindependen are Y and X then
0y,0x,ey)f(x,by given is pdfjoint r that theiSuppose
devices. electronic twoof lifetimes thebe Y and X Let
Example
y)(x-
Y. and X of ceindependen Check the
1.yx8xy,0y)f(x, that Suppose
Example
Expectation of Product of random variables
If X and Y are mutually independent random variables, then the expectation of their product exists and is
E(XY) = E(X) E(Y)
y0,x0,ey)f(x,by given be pdf Let the
variablesrandom continuous as modeled being are lightbulb
a of Y brightness theand X lifetime that theAssuming
)yx(21
21
Find the joint distribution function
Example
A line of length a units is divided into two parts. If the first part is of length X, find E(X), Var(X) and E{X(a-X)}.
Expectation of Sum of random variables
If X1, X2, …, Xn are random variables, then the expectation of their sum exists and is
E(X1+ X2+…+ Xn) = E(X1) + E(X2) +… + E(Xn)
kj
kjkkjkj
j yxpyyxpxYEXE,,
,,
kjkj
kj yxpyx ,,
YXE
Example What is the mathematical expectation of the sum of points on n dice?
A box contains tickets among which tickets bear the number r (r = 0,1,2,…,n). A group of m tickets is drawn . Let S denote the sum of their numbers. Find E(S) and Var S.
n2 rn C
Ans. (7/2)n
Ans. (n/2)m
kj
kjkj yxpyxXYE,
,
kjkkj
j ygxfyx,
kkk
jjj ygyxfx
YEXE
E(XY) Find
1xyx),0-24y(1y)f(x,by given is Y)(X, of pdfjoint theIf
Example
y=x
0 x
y
1
0
1
ydxdy)y,x(xyf)XY(E
Ans. 4/15
Binomial Distribution (re-visit)
Suppose that n Bernoulli trials, each of which results in a success with probability p and results in a failure with 1–p, are performed. If Sn represents the number
of successes that occur in the n Bernoulli trials, then Sn is said to be a binomial random variable with
parameter n and p.
Let Xk be the number successes scored at the kth
trial. Since Xk assumes only the values 0 and 1 with
corresponding probabilities q and p, we have
E(Xk) = 0 q + 1 p = p
Since
Sn = X1 + X2+…+ Xn
We have
E(Sn) = E(X1+ X2+…+ Xn)
= E(X1) + E(X2) +… + E(Xn) = np
Conditional Probability
Now, consider the case where the event M depends on some other random variable Y.
0)Pr()Pr(
,Pr
Pr
MM
Mxx
MxXMxF
Conditional Probability Density Function
or
- the continuous version of Bayes’ theorem
- another expression of the marginal pdf
)y(f
)y,x(f)y|x(f
Y
)x(f
)y,x(f)x|y(f
X
)(
)()|()|(
xf
yfyxfxyf
X
Y
dxxfxyfdxyxfyf
dyyfyxfdyyxfxf
XY
YX
)()|(),()(
)()|(),()(
4).YP(X&1)3/XP(Y
3)1/YP(X3),Y1,P(X1),P(X find below,given
Y)(X, ofon distributiy probabilit bivariate For the
1 2 3 4 5 6
0 0 0 1/32 2/32 2/32 3/32
1 1/16 1/16 1/8 1/8 1/8 1/8
2 1/32 1/32 1/64 1/64 0 2/64
X
Y
(ans.7/8,9/32,18/32,9/28,13/32)
Suppose that p(x,y) the joint probability mass function of X and Y , is given by p(0,0) =.4p(0,1)=.2,p(1,0)=.1,p(1,1)=.3 Calculate the conditional probability mass function of X given that Y = 1
Ans. 2/5,3/5
ExampleSuppose that 15 percent of the families in a certain community have no children, 20% have 1, 35% have2, & 30% have 3 children; suppose further that eachchild is equally likely (and independently) to be a boy or a girl. If a family is chosen at random from this community, then B, the number of boys, and G , the number of girls, in this family will have the joint probability mass function .
j
i
0 1 2 3
0 .15 .10 .0875 .0375
1 .10 .175 .1125 0
2 .0875 .1125 0 0
3 .0375 0 0 0
1125.8
9.
32
1
2
1
2
130.)GBB(P)BGB(P)BBG(P
)GBBBBGorBGBor(P)1G,2B(P
175.22
1
2
135.
)GB(P)BG(P
)BGorGB(P)1G,1B(P
If the family chosen has one girl, compute the conditional probability mass function of the number of boys in the family
Ans.8/31,14/31,9/31,0
1y0 where,y Y
given that , X ofdensity lconditiona thecompute
otherwise0
1y0,1x0,yx2x5
12 y)f(x,
bygiven is Y and X ofdensity joint The
Example
)y(f
)y,x(f)y|x(f
Y
y34
yx2x6ans
)1YX(P&)YX(P
)1X/2
1Y(P),
2
11/YP(X),
2
1P(Y1),P(X Compute
.1y0,2x0,8
xxyy)f(x,
bygiven is RV ldimensiona- twoa of pdfjoint The
Example
22
(ans. 19/24,1/4,5/6,5/19,53/480,13/480)
Variance of a Sum of random variables
If X and Y are random variables, then the variance of their sum is
Var(X + Y) = E({(X+Y) – (X + Y)}2)
YX2
Y2
X YXE2YEXE
))((2)()( YX YXEYVarXVar
YXXYEYVarXVar 2
The covariance of X and Y is defined by
YXYX XYEYXEYXCov ,
Var(X + Y) = E({(X+Y) – ( + )} )2
x y
• If X and Y are mutually independent, then Cov(X,Y) = 0.
Q: Is the reverse of the above true?
• If X and Y are mutually independent, then
Var(X + Y) = Var(X) + Var(Y)
• If X1, …, Xn are mutually independent, and
Sn = X1 + …+ Xn, then
Var(Sn) = Var(X1) + … + Var(Xn)
Q: Let Sn be a binomial random variable with parameter n and p. Show that
Var(Sn) = np(1-p)
ExampleCompute Var(X) when X represents the outcomewhen we roll a fair die.
SolutionSince P(X=i)=1/6, i = 1,2,3,4,5,6, we obtain
6
1i
22 ]iX[Pi)X(E
)6
1(6)
6
1(5)
6
1(4)
6
1(3)
6
1(2)
6
1(1 222222
=91/6
22 )X(E)X(E)X(Var
12/352
7
6
912
Compute the variance of the sum obtained when 10independent rolls of a fair die are made.
Ans 175/6
Compute the variance of the number of heads resultingfrom 10 independent tosses of a fair coin.
yx
xyxy
xy
C
as defined is ,by denoted
Y and Xbetween n correlatio oft coefficien The
The correlation co-efficient is a measure of dependencebetween RV’s X and Y.
If = 0 , we say that X and Y are uncorrelated If E(XY) = 0 , X and Y are said to be orthogonal RV’s.
xy
yxxyxy C or 1
Example Calculate the correlation coefficient for the followingheights (in inches) of fathers (X) & their sons (Y):
X Y
65 67
66 68
67 65
67 68
68 72
69 72
70 69
72 71
(Ans .603)
yx
xyxy
C
2222
ny
ny
nx
nx
ny
nx
nxy
Example If X,Y and Z are uncorrelated RVs with zero means and standard deviations 5, 12 and 9 respectivelyand if U=X+Y and V=Y+Z find the correlation coefficient between U and V.
(ans 48/65)
1/11)- (ans.
Y. and Xbetween t coefficienn correlatio theFind
elsewhere 0
1y1,0x0y xy)f(x, pdfjoint
thehave Y and X variablesrandom Let the
Example
)Y,X(g
p Y)(X,
Values Expected lConditiona
ij
i
jijij )yY/xX(P)y,x(g}yY/)Y,X(g{E
ij*
ijji
j
jiji
i p
p)y,x(g
}yY{P
}yYxX{P)y,x(g
dy)x/y(f)y,x(g}X/)Y,X(g{E
dx)y/x(f)y,x(g}Y/)Y,X(g{E
dy)x/y(yf)X/Y(E
means lConditiona
y/x
dy)x/y(f)y()Y(E
are variancelConditiona
2x/y
2x/y
2y/x
Example The joint pdf of (X,Y) is given by f(x,y)=24xy, x>0,y>0,x+y<=1,and 0, elsewhere, find the conditionalmean and variance of Y, given X.
E(Y/X) and E(X/Y) find , 0y&x-1 y by bounded
semicircle over the ddistributeuniformly is Y)(X, If
2 E(Y) E{E(Y/X)} and E(X)E{E(X/Y)}at verify thAlso
2
2
)x1(18/1var
),x1(3/2)X/Y(E,)x1/(y2)x/y(f
Properties(1)If X and Y are independent RV’s, then E(Y/X)=E(Y)
and E(X/Y)=E(X).(2)E[E{g(X,Y)/X)=E{g(X,Y)} in particular E{E(X/Y)}=E(X)(3)E(XY)=E[X.E(Y/X)]
)]X/Y(EX(E)YX(E 2222
ExampleThree coins are tossed. Let X denote the number of
heads on the first two coins,Y denote the no of tailson the last two, and z denote the number of headson the last two. Find
(a)The joint distribution of (i) X and Y (ii) X and Z(b) Conditional distribution of Y given X = 1(c) Find covariance of x,y and x,z(d) Find E(Z/X=1)(e)Give a joint distribution , that is not the joint
distribution of X and Z in (a), but has the same marginals as of (b)
)x(f...)x(f)x(f)x,...,x,x(f
tindependen)X,...X,(X RVs
n21n21
n21
)x,f(x
)x,x,f(x)x,x/f(x
)xf(
)x,x,f(x)x/x,f(x
density lconditiona
32
321321
3
321321
DefinitionLet X denote a random variable with probability density function f(x) if continuous (probability mass function p(x) if discrete)
Then
M(t) = the moment generating function of X
tXE e
if is continuous
if is discrete
tx
tx
x
e f x dx X
e p x X
XY. offunction generating-moment theFind
otherwise0
0y,0xxe)y,x(f If
Example)1y(x
0 0
)t1(xyx
0 0
)y(xxytx
0 0
)1y(xxytXY
dx}dye{xe
dx}dyee{xe
dydxxeeM:Solution
=1/1-t
Properties
MX(0) = 1
)x(fe)t(M txX
)x(f.....)
!2
txtx1(
2
10tX )X(E)x(fx)t('M
)x(f...)!2
tx2x()t('M
2
X
22
0tX2
32
X2
)x(fx)t(M
)x(f...)!3
tx6x()t(M
0 derivative of at 0.k thX Xm k m t t
2 33211 .
2! 3! !kk
Xm t t t t tk
continuous
discrete
k
kk k
x f x dx XE X
x p x X
Let X be a random variable with moment generating function MX(t). Let Y = bX + a
Then MY(t) = MbX + a(t) = E(e [bX + a]t) = eatMX (bt)
Let X andY be two independent random
variables with moment generating functionMX(t)and MY(t) .
Then MX+Y(t) = MX (t) MY (t)
6. Let X and Y be two random variables with moment generating function MX(t) and MY(t) and two distribution functions FX(x) and FY(y) respectively.
Let MX (t) = MY (t) then FX(x) = FY(x).
This ensures that the distribution of a random variable can be identified by its moment generating function
Example If X represents the outcome, when a fair die is tossed, find the MGF of X and hence find E(X) and Var(X).
If a RV X has the MGF M(t) = 3/(3-t), obtain the standard deviation of X. (ans, 1/3)
(Ans. 7/2,35/12)
t4t3t2t e10
4e
10
3e
10
2e
10
1)t(M
Find p.d.f.
t4t3t2t e10
4e
10
3e
10
2e
10
1)t(M
x
tx )x(fe)t(M
...e)b(fe)a(fe10
4e
10
3e
10
2e
10
1 btatt4t3t2t
otherwise,0
4,3,2,1x,10
x)x(f
X. ofmean thefind
otherwise,0
x1,x
1f(x) p.d.f thehas X If 2
)e(E)w(
by defined is X variablerandom a offunction sticcharacteri Theiwx
X
continuous is X if,dx)x(fe
discrete is X if),x(fe
iwx
x
iwx
11
wxsinwxcoswxsiniwxcose2/1222/1iwx
dx)x(fe)e(E)w( iwxiwx
x
1dx)x(fdx)x(feiwx
Hence the characteristic function always exist even when moment-generating function may not exist.
Properties of Characteristic Function
.i of powers asending of seriesin )( ofexpansion
in the !n
i ofefficient -co the)X(E.1
nnn'
n
x
iwxiwxX )x(fe)e(E)w(
)x(f...
!3
iwx
!2
iwxiwx1
x
32
x
22
x x.....)x(fx
!2
iw)x(xfiw)x(f
.de)(2
1f(x) then ),( is f(x)function density
withX RV continuous a offunction sticcharacteri theIf 5.
).()()(
then RVs,t independen are Y and X If .4
)a(e)( then b,aXY if and )( is
X RV a offunction sticcharacteri theIf.3
ix
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n
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i
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bygiven
is X variablerandom a offunction sticcharacteri The
x
dwe)w(2
1f(x)
is X of pdf The
iwxx
dwe)w1(
2
1 iwx1
1
dwe)w1(dwe)w1(
2
1 iwx1
0
iwx0
1
xcos1x
1)ee2(
x2
12
ixix2
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2
12
x,x1
11f(x)function
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2
-
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1)x(f xi
).(parameter on with distributiPoisson
a follows also )X(X that prove ,&
parameterson with distributiPoisson follow
that RVst independen twoare X&X If
21
2121
21
Reproductive property of Poisson distribution
)1e()t(x
)1e()t(x
i2
2
i1
1
e
e
RVs,t independen are X&X since 21
)1e()t(xx
i21
21e
Joint Characteristic Function
).,(by denoted and Y)(X, offunction
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then RV, ldimensiona- twoa is Y)(X, If
21xy
YiXi 21
0,0
21xyn2
m1
nm
nmnm
xy
21
),(i
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1)0,0()i(
i jji
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21
21
.conversely and
)()(),(
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),0()(&)0,()()iii(
2y1x21xy
xyyxyx
.else,0
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0yx,3/1
P
is PMFjoint theif Y and X sr.v.' discrete
theoffunction sticcharacteri theCompute
XY
1
1k
1
1lXY
)lwkw(i21XY Pe)w,w( 21
21121121 iwiwiwiwiwiw0iw0iw e6
1e
6
1e
6
1e
6
1e
3
1
212111
212111
wwsiniwwcos6
1wsiniwcos
6
1
wwsiniwwcos6
1wsiniwcos
6
1
3
1
211 wwcos3
1wcos
3
1
3
1
eduncorrelat arethey
that also and RVsmean zeroboth are Y and X
thatShow .e),(function
sticcharacterijoint thehave Y and X RVs Two
Example
22
21 82
21xy
0,0
82
121
22
21e
i
1E(X)
CFjoint ofproperty By the
0i4e 0,0182
21
22
21
0,0
21xyn2
m1
nm
nmnm
21
),(i
1}YX{E)ii(
0i16eE(Y) 0,0282
21
22
21
0
}e64{
16e
ei
1E(XY)
0,082
21
0,0
282
1
0,0
82
21
2
2
21
22
21
21
22
21
21
22
21
0)Y(E)X(E)XY(ECxy
Compute the joint characteristic function of X and Y if
)yx(
2
1exp
2
1f 22
xy
dxdyee
2
1)w,w(
.Ans
yixi)yx(2
1
21xy21
22
Random Variable
Binomial DistributionThe Bernoulli probability mass function is the densityfunction of a discrete variable X having 0 and 1 as the only possible values
The pmf of X is given by P(0) = P{X = 0} = 1-p P(1) = P{P = 1} = p where p, 0<=p<=1 is the probability that the trial is a success.
(0,1).p somefor equation above satisfies pmf if
variablerandom Bernoulli be tosaid is X variablerandomA
An experiment consists of performing a sequence of subexperiments. If each subexperiment is identical, then the subexperiments are called trials.
Bernoulli Trials
• Each trial of an experiment that has only two possible outcomes (success or failure) is called a “Bernoulli trial.”
• If p is the probability of success, then (1-p) is the probability of failure.
• The probability of exactly k successes in n independent Bernoulli trials, with probability of success p and probability of failure q = 1-p, is given by a formula called the Binomial Distribution:
C(n, k) pk q(n-k)
Example of Bernoulli Trials• Suppose I roll a die and I consider a 3 to be
success and any other number to be a failure.• What is the probability of getting exactly 5
successes if I roll the die 20 times?• Solution: C(20, 5) (1/6)5 (5/6)15
• What is the probability of getting 5 or more successes?
event. theof failure ofy probabilit thep,-1 q
where,pqnC toequal is t trialsindependenn ofout r timesexactly occurs
event y that theprobabilit then thep, is experiment sBernoulli' a of trialsingle
ain success) ofty (probabilievent an of occurrance ofy probabilit theIf
Theorem
rrnr
rnrqp
failures)r -n and successesr P(getting
usly.simultaneo failures r)-(n and successes
r getting means successesr exactly Getting
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choosing of waysnC are There specified.not are
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r
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P(A&B get the same no. of heads)=P(they get no head each or 1 head each or 2 headseach or 3 heads each)
= P(A gets 0 head) P(B gets o head)+------
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Example.
A game consists of 5 matches and two players, A and B. Any player who firstly wins 3 matches will be the winner of the game.
If player A wins a match with probability 2/3. Suppose matches are independent. What will be the probability for player A to win the game?
Solution:
Player A wins 3 matches or more out of the 5 matches.
This event has probability equal to:
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Example.
A sequence of independent trials is to be performed. Each trial results in a success with probability p and a failure with probability 1–p. What is the probability that
(a) at least 1 success occurs in the first n trials;
(b) exactly k success occurs in the first n trials;
Solution:
(a)The probability of no success in the
first n trials is (1-p)n. Thus, the answer
is 1–(1–p)n.
(b) knk ppk
n
1
Assuming that p remains the same for all repetitions,if we consider n independent repetitions (or trials) of E and if the random variable X denotes the number of times the event A has occurred, then X is called a binomial random variable with parameters n and p
The pmf of a binomial random variable having parameters (n,p) is given by
1 qp where,n,......,1,0i,qpnC)i(P inii
ExampleIt is known that car produced by an automobile companywill be defective with probability 0.01 independently ofeach other. The company sells the cars in packages of 10 and offers a money-back guarantee that atmost 1 of the10 cars is defective. What proportion of packages sold must the company replace?
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Example
The average number of radioactive particles passingthrough a counter during 1 millisecond is in a laboratory experiment is 4. What is the probability that6 particles enter the counter in a given millisecond?
If the probability of a defective fuse from a manufacturingunit is 2%, in a box of 200 fuses, find the probability that exactly 4 fuses are defective.
!4
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Example
At a busy traffic intersection the probability p of an Individual car having an accident is very small say p=0.0001. However, during a certain peak hours of the day say between 4 p.m. and 6 p.m., a large number of cars (say 1000) pass through the intersection. Under these conditions what is the probability of two or more accidents occurring duringthat period.In a component manufacturing industry, there is a smallprobability of 1/500 for any component to be defective. The components are supplied in packets of 10. Use Poisson distribution to calculate the app no of packets containing (i) no defective (ii)one defective componentsin a consignment of 10,000 packets. 100000196,.100009802.
Binomial Distribution
n,...,2,1,0r;qpC)rX(P rnrr
n
If we assume that n trials constitute a set and if weconsider N sets, the frequency function of the binomial distribution is given by f(r)=N p(r)
rnrr
n qpCN
ExampleFit a binomial distribution for the following data and hence find the theoretical frequencies:x : 0 1 2 3 4f: 5 29 36 25 5
Ans. 7,26,37,34,66.89,19.14,23.94,17.74,8.63,2.88,0.67,0.1,0.01,0,0
The following data are the number of seeds germinating out of 10 on damp filter paper for 80 set of seeds. Fit a binomial distribution to these data: x 0 1 2 3 4 5 6 7 8 9 10
y 6 20 28 12 8 6 0 0 0 0 0
Fit a Poisson distribution for the following distribution:x: 0 1 2 3 4 5f: 142 156 69 27 5 1
Ans. 147 147 74 25 6 1107,141,93,41,4,0,0,0,0,0
Fit a Poisson distribution to the following data which gives the number of yeast cells per square for 400 squaresNo. of cells per square (x)
0 1 2 3 4 5 6 7 8 9 10
No. of squares (f) 103 143 98 42 8 4 2 0 0 0 0
GEOMETRIC DISTRIBUTION
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In a chemical engineering process industry it is known that , on the average, 1 in every 100 items is defective. What is the probability that the fifthitem inspected is the first defective item found.
0096.)99)(.01(.ans 4
At busy time, a telephone exchange will be working busy with full capacity . So people cannot get a line to use immediately. It may be of interest to know the number of attempts necessary in order to get a connection. Suppose that p = 0.05, then find the probability that 5 attempts are necessary for a successful call connection.
Ans .041
Mean
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Negative Binomial Distribution
Trials repeated until a fixed number of success occur.
Instead of finding the probability of r success in n trialswhen n is fixed
Probability that rth success occurs on the xth trial.
Negative binomial experiment
The number of x trials to produce r success in a negativebinomial experiment is called a negative binomialrandom variable and its probability distribution is called the negative binomial distribution
The negative binomial distribution is used when the number of successes is fixed and we're interested in the number of failures before reaching the fixed number of successes. An experiment which follows a negative binomial distribution will satisfy the following requirements: 1.The experiment consists of a sequence of independent trials. 2.Each trial has two possible outcomes, S or F. 3.The probability of success,is constant from one trial to another. 4.The experiment continues until a total of r successesare observed, where r is fixed in advance
Suppose we repeatedly throw a die, and consider a "1" to be a "success". The probability of success on each trial is 1/6. The number of trials needed to get three successes belongs to the infinite set { 3, 4, 5, 6, ... }. That number of trials is a (displaced) negative-binomially distributed random variable. The number of failures before the third success belongs to the infinite set { 0, 1, 2, 3, ... }. That number of failures is also a negative-binomially distributed random variable.
A Bernoulli process is a discrete time process, and so the number of trials, failures, and successes are integers. For the special case where r is an integer, the negative binomial distribution is known as the Pascal distribution.
A further specialization occurs when r = 1: in this case we get the probability distribution of failures before the first success (i.e. the probability of success on the (k+1)th trial), which is a geometric distribution.
Let the random variable y denotes the no of failures before the occurrence of the rth success. Then y+rdenotes the number of trials necessary to produce exactly r success and y failures with the rth successoccurring at the (y+r)th trial.
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Pat is required to sell candy bars to raise money for the 6th grade field trip. There are thirty houses in the neighborhood, and Pat is not supposed to return home until five candy bars have been sold. So the child goes door to door, selling candy bars. At each house, there is a 0.4 probability of selling one candy bar and a 0.6 probability of selling nothing.
Example
What’s the probability that Pat finishes on the tenth house?
A fair die is cast on successive independent trials untilthe second six is observed. The probability of observing exactly ten non-sixes before the second six is cast is ……..
32
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Find the probability that a person tossing three coins will get either all heads or all tails for the second timeon the fifth toss.
The probability that an experiment will succeed is 0.8.If the experiment is repeated until four successful outcomes have occurred, what is the expected number of repetitions required?
Ans. 1
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PROBABILITY DISTRIBUTIONS
Binomial Distribution
Discrete Distributions
n,...,2,1,0r;qpC)rX(P rnrr
n
If we assume that n trials constitute a set and if weconsider N sets, the frequency function of the binomial distribution is given by f(r)=N p(r)
rnrr
n qpCN
ExampleFit a binomial distribution for the following data and hence find the theoretical frequencies:x : 0 1 2 3 4f: 5 29 36 25 5
Ans. 7,26,37,34,66.89,19.14,23.94,17.74,8.63,2.88,0.67,0.1,0.01,0,0
The following data are the number of seeds germinating out of 10 on damp filter paper for 80 set of seeds. Fit a binomial distribution to these data: x 0 1 2 3 4 5 6 7 8 9 10
y 6 20 28 12 8 6 0 0 0 0 0
Fit a Poisson distribution for the following distribution:x: 0 1 2 3 4 5f: 142 156 69 27 5 1
Ans. 147 147 74 25 6 1107,141,93,41,4,0,0,0,0,0
Fit a Poisson distribution to the following data which gives the number of yeast cells per square for 400 squaresNo. of cells per square (x)
0 1 2 3 4 5 6 7 8 9 10
No. of squares (f) 103 143 98 42 8 4 2 0 0 0 0
Probbility Density Function
0.0000
0.0004
0.0008
0.0012
0.0016
-100 0 100 200 300 400 500 600 700 800
x
f X (x )
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0.2
0.4
0.6
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If X is uniformly distributed over the interval [0,10]compute the probability (a) 2<X<9(b) 1<X<4(c)X<5 (d) X>6
Ans. 7/10,3/10,5/10,4/10
b
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Buses arrive at a specific stops at 15min. Intervalsstarting at 7 A.M., that is , they arrive at 7,7:15,7:30 and so on. If a passenger arrives at the stop at a random time that is uniformly distributed between7 and 7:30A.M., find the probability that he waits(a) less than 5 min (b) at least 12 min. for a bus.
Ans. 1/3,1/5
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a = 2
Example:
The total time it takes First Bank to process a loan application is uniformly distributed between 3 and 7 days. What is the probability that the application will be processed in less than 4 days?
What is the probability that it will take more than 6.5 days?
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Exponential Distribution
If the occurances of events over nonoverlapping intervals are independent, such as arrival times of telephone calls or bus arrival times at a bus stop, then the waiting time distribution of these events can be shown to be exponential
• Time between arrivals to a queue (e.g. time between people arriving at a line to check out in a department store. (People, machines, or telephone calls may wait in a queue)
• Lifetime of components in a machine
0 parameter
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Example Let X have an exponential distribution with mean of 100 . Find the probability that X<90
Ans. 0.593
Customers arrive in a certain shop according to anapproximate Poisson process at a mean rate of 20per hour. What is the probability that the shopkeeper will have to wait for more than 5 minutes for his firstcustomer to arrive?
-15e .ans
Memoryless Property of the Exponential Distribution
0t s,any for t),P(Xs)t/XsP(X
thend,distributelly exponentia is X If
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If x represents the lifetime of an equipment then above property states that if the equipment has been working for time s, then the probability that itwill survive an additional time t depends only on t(not on s) and is identical to the probability of survivalfor time t of a new piece of equipment.
Equipment does not remember that it has been in use for time s.
A crew of workers has 3 interchangeable machines, of which 2 must be working for the crew to its job. When in use, each machine will function for an exponentially distributed time having parameters before breaking down. The workers decide to initiallyuse machines A and B and keep machine C inreserve to replace whichever of A or B breaks downfirst. They will then be able to continue working untilone of the remaining machines breaks down. Whenthe crew is forced to stop working because onlyone of the machines has not yet broken down, whatis the probability that the still operable machine is machine C?
Suppose the life length of an appliance has an exponential distribution with years. 10
A used appliance is bought by someone. What is the probability that it will not fail in the next 5 years?
Suppose that the amount of waiting time a customerspends at a restaurant has an exponential distributionwith a mean value of 5 minutes Then find theprobability that a customer will spend more than 10 minutes in the restaurant
0.368
0.1353
ExampleSuppose that the length of a phone call in minutes is an exponenetial random variable with parameter
.10
1
If A arrives immediately ahead of B at a publictelephone booth, find the probability that B will haveto wait (i) more than 10 minutes, and (ii) between 10 and 20 minutes.
(ans. 0.368,0.233)
Erlang distribution or General Gamma distribution
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Suppose that an average of 30 customers per hourarrive at a shop in accordance with a Poisson ProcessThat is, if a minute is our unit , then . What is the probability that the shopkeeper wait more than5 minutes before both of the first two customer arrive?
2/1
Solution.If X denotes the waiting time in minutes until the second customer arrives, then X has Erlang(Gamma)distribution with k = 2, 2/1
dxe
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Mean and Variance of Erlang Distribution
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The sum of a finite number of Erlang variables is also an Erlang variable.
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Weibull Distribution
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Properties of aNormal Distribution
• Continuous Random Variable
• Symmetrical in shape (Bell shaped)
• The probability of any given range of numbers is represented by the area under the curve for that range.
• Probabilities for all normal distributions are determined using the Standard Normal Distribution.
Probability for aContinuous Random Variable
Probability Density Function for Normal Distribution
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Figure 6.5
Determining the Probability for a Standard Normal Random Variable• P(- Z 1.62) = .5 + .4474 = .9474
• P(Z > 1.62) = 1 - P(- Z 1.62) =1 - .9474 = .0526
z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0190 0.0239 0.0279 0.0319 0.0359
0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753
0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141
0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517
0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879
0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224
0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549
0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852
0.8 0.2881 0.2910 0.2939 0.2969 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133
0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389
1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3513 0.3554 0.3577 0.3529 0.3621
1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830
1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015
1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177
1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319
1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441
1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545
1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633
1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706
1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767
2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817
2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857
2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890
2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916
2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936
2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952
2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964
2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974
2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981
2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986
3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990
3.1 0.4990 0.4991 0.4991 0.4991 0.4992 0.4992 0.4992 0.4992 0.4993 0.4993
3.2 0.4993 0.4993 0.4994 0.4994 0.4994 0.4994 0.4994 0.4995 0.4995 0.4995
3.3 0.4995 0.4995 0.4995 0.4996 0.4996 0.4996 0.4996 0.4996 0.4996 0.4997
3.4 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4998
Determining the probability of any Normal Random Variable
Interpreting Z
• In figure Z = -0.8 means that the value 360 is .8 standard deviations below the mean.
• A positive value of Z designates how many standard deviations () X is to the right of the mean ().
• A negative value of Z designates how may standard deviations () X is to the left of the mean ().
Example: A group of achievement scores are normally distributed with a mean of 76 and a standard deviation of 4. If one score is randomlyselected what is the probability that it is at least 80.
76 80
4
1587.3413.5.
)1z0(P5.)1z(P)80x(P
14
7680xZ
0 1
.341
3
.1587
76 80
4
1587.3413.5.
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14
7680
zPzPxP
uxZ
Continuing, what is the probability that it is less than 70.
70 76
.06684332.5.
)0.05.1(5.)5.1()70(
5.14
7670
zPzPxP
uxZ
-1.5 0
.433
2
.0668
What proportion of the scores occur within 70 and 85.
9210.4878.4332.
)25.20.0()0.05.1(
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25.24
7685
5.14
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332
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8
-1.5 0 2.25
Time required to finish an exam is known to be normally distributed with a mean of 60 Min. and a Std Dev. of 12 minutes. How much time should be allowed in order for 90% of the students to finish?
12
60 x
.9
36.75
60)12(28.1
x
x
xz
xz
xz
An automated machine that files sugar sacks has an adjusting device to change the mean fill per sack. It is now being operated at a setting that results in a mean fill of 81.5 oz. If only 1% of the Sacks filled at this setting contain less than 80.0 oz, what is the value of the variance for this population of fill weights. (Assume Normality).
-2.33 0
.01
4144.
6437.33.2
)5.810.80(
5.810.8033.2
01.805.81
2
ux
Z
PROBx
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Let X be the number of times that a fair coin, flipped 40times, land heads. Find P(X=20). Use normalapproximation and compare it to the exact solution.
P(X=20)=P(19.5<X<20.5)
10
205.20
10
20X
10
205.19P
1272.)16.()16(.
0 1 2 3 4 5 6 7 8 9 10 11 12
0 1 2 3 4 5 6 7 8 9 10 11 12
If 20% of the momory chips made in a certain plantare defective, what are the probabilities that in a lot of 100 randomly chosen for inspection (a) at most 15 will be defective? (b) exactly 15 will be defective ?
4,20)20(.100
1292.0)4
205.15(F
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205.14F
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205.15F
Fit a normal distribution to the following distributionand hence find the theoretical frequencies:
Class Freq 60-65 365-70 2170-75 15075-80 33580-85 336 85-90 13590-95 2695-100 4 ------------- 1000
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8 1.34 1.65 2.18 2.73 3.49 13.36 15.51 17.53 20.09 21.96
9 1.73 2.09 2.70 3.33 4.17 14.68 16.92 19.02 21.67 23.59
10 2.16 2.56 3.25 3.94 4.87 15.99 18.31 20.48 23.21 25.19
11 2.60 3.05 3.82 4.57 5.58 17.28 19.68 21.92 24.73 26.76
12 3.07 3.57 4.40 5.23 6.30 18.55 21.03 23.34 26.22 28.30
13 3.57 4.11 5.01 5.89 7.04 19.81 22.36 24.74 27.69 29.82
14 4.07 4.66 5.63 6.57 7.79 21.06 23.68 26.12 29.14 31.32
15 4.6 5.23 6.26 7.26 8.55 22.31 25 27.49 30.58 32.80
16 5.14 5.81 6.91 7.96 9.31 23.54 26.30 28.85 32.00 34.27
18 6.26 7.01 8.23 9.39 10.86 25.99 28.87 31.53 34.81 37.16
20 7.43 8.26 9.59 10.85
12.44 28.41 31.41 34.17 37.57 40.00
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15.66 33.20 36.42 39.36 42.98 45.56
30 13.79 14.95 16.79 18.49
20.60 40.26 43.77 46.98 50.89 53.67
40 20.71 22.16 24.43 26.51
29.05 51.81 55.76 59.34 63.69 66.77
60 35.53 37.48 40.48 43.19
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0.0190
0.0239
0.0279
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0.0438
0.0478
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0.0596
0.0636
0.0675
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0.1406
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0.1480
0.1517
0.40.1554
0.1591
0.1628
0.1664
0.1700
0.1736
0.1772
0.1808
0.1844
0.1879
0.50.1915
0.1950
0.1985
0.2019
0.2054
0.2088
0.2123
0.2157
0.2190
0.2224
0.60.2257
0.2291
0.2324
0.2357
0.2389
0.2422
0.2454
0.2486
0.2517
0.2549
0.70.2580
0.2611
0.2642
0.2673
0.2704
0.2734
0.2764
0.2794
0.2823
0.2852
0.80.2881
0.2910
0.2939
0.2969
0.2995
0.3023
0.3051
0.3078
0.3106
0.3133
0.90.3159
0.3186
0.3212
0.3238
0.3264
0.3289
0.3315
0.3340
0.3365
0.3389
1.00.3413
0.3438
0.3461
0.3485
0.3508
0.3513
0.3554
0.3577
0.3529
0.3621
1.10.3643
0.3665
0.3686
0.3708
0.3729
0.3749
0.3770
0.3790
0.3810
0.3830
1.20.3849
0.3869
0.3888
0.3907
0.3925
0.3944
0.3962
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0.3997
0.4015
1.30.4032
0.4049
0.4066
0.4082
0.4099
0.4115
0.4131
0.4147
0.4162
0.4177
1.40.4192
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0.4222
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0.4292
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0.4345
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0.4370
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0.4418
0.4429
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1.60.4452
0.4463
0.4474
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0.4505
0.4515
0.4525
0.4535
0.4545
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0.4573
0.4582
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0.4599
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0.4616
0.4625
0.4633
1.80.4641
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0.4664
0.4671
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0.4699
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1.90.4713
0.4719
0.4726
0.4732
0.4738
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0.4750
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0.4761
0.4767
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0.4778
0.4783
0.4788
0.4793
0.4798
0.4803
0.4808
0.4812
0.4817
2.10.4821
0.4826
0.4830
0.4834
0.4838
0.4842
0.4846
0.4850
0.4854
0.4857
2.20.4861
0.4864
0.4868
0.4871
0.4875
0.4878
0.4881
0.4884
0.4887
0.4890
2.30.4893
0.4896
0.4898
0.4901
0.4904
0.4906
0.4909
0.4911
0.4913
0.4916
2.40.4918
0.4920
0.4922
0.4925
0.4927
0.4929
0.4931
0.4932
0.4934
0.4936
2.50.4938
0.4940
0.4941
0.4943
0.4945
0.4946
0.4948
0.4949
0.4951
0.4952
2.60.4953
0.4955
0.4956
0.4957
0.4959
0.4960
0.4961
0.4962
0.4963
0.4964
2.70.4965
0.4966
0.4967
0.4968
0.4969
0.4970
0.4971
0.4972
0.4973
0.4974
2.80.4974
0.4975
0.4976
0.4977
0.4977
0.4978
0.4979
0.4979
0.4980
0.4981
2.90.4981
0.4982
0.4982
0.4983
0.4984
0.4984
0.4985
0.4985
0.4986
0.4986
3.00.4987
0.4987
0.4987
0.4988
0.4988
0.4989
0.4989
0.4989
0.4990
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0.4991
0.4991
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0.4992
0.4992
0.4992
0.4992
0.4993
0.4993
3.20.4993
0.4993
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0.4994
0.4994
0.4994
0.4994
0.4995
0.4995
0.4995
3.30.4995
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0.4995
0.4996
0.4996
0.4996
0.4996
0.4996
0.4996
0.4997
3.40.4997
0.4997
0.4997
0.4997
0.4997
0.4997
0.4997
0.4997
0.4997
0.4998
