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ONE DIMENSIONAL
RANDOM VARIABLES
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SYLLABUS:
Random variables – Probability function –
Moments – Moment generating functions and
their properties – Binomial, Poisson, Geometric,
Uniform, Exponential, Gamma and Normal
Distributions – Functions of a Random Variable.
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PROBABILITY BASIC DEFINITIONS
PROBABILITY:
It is the science of decision-making with calculated
risks in the face of uncertainty.
EXPERIMENT:
It is a process which results in some well defined
outcomes.
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RANDOM EXPERIMENT:
Eg:
Tossing a fair coin, Throwing a die ,Taking a card
from a pack of cards are Random Experiment.
SAMPLE SPACE:
The set of all possible outcomes of a random
experiment is called its Sample Space. It is denoted by S.
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Eg:
Random Experiment
Tossing a coin once
Tossing a coin Twice
Throwing a die once
Sample Space S
{H,T}
{HH,TH,HT,TT}
{1,2,3,4,5,6}
OUTCOME:
The result of a random experiment is called an outcome.
Eg: Head or tail
TRIAL:
Any particular performance of a random experiment is
called a trial.
Eg: Tossing a coin
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EVENT:
The outcomes of a trial are called events.
Eg: Getting head or tail
MATHEMATICAL PROBABILITY:
The Probability of the occurrence of a event A is
denoted by P[A] and defined as
P[A] = n[A] / n[S]
0 ≤ P[A] ≤ 1
n[A] = number of favourable cases to A.
n[S] = number of possible cases in S.
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INDEPENDENT EVENTS:
If the Occurrence of any one of them does not
depend on the occurrence of the other.
Eg: Tossing a coin, the event of getting a head in the first
toss is independent of getting a head in the second and
subsequent throws.
CONDITIONAL PROBABILITY:
The Conditional Probability of an event B ,assuming that
the event A has already happened , is denoted by P[B/A]
and defined as
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EXAMPLE FOR CONDITIONAL PROBABILITY:
When a fair die is tossed ,what is the probability of
getting 1 given that an odd number has been obtained.
SOLUTION:
Let S={1,2,3,4,5,6} ; n[S]=6
Let A be the probability of getting a odd number
A={1,3,5} ; n[A]=3
P[A]=n[A] / n[S] =3/6
Let B be the probability of getting 1. B={1}
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PROBABILITY DISTRIBUTION FUNCTION OF X :
If X is a random variable ,then the function F[x] defined as
F[ x]  P[ X  x]
is called the distribution function of X.
PROBABILITY DISTRIBUTION FUNCTION OF Y :
If X is a random variable ,then the function F[y] defined as
F[ y]  P[Y  y]
is called the distribution function of Y.
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RANDOM VARIABLES:
If E is an experiment having sample space S, and X is a
function that assigns a real number X(s) to every outcome s є S,
then X(s) is called a random variable (r.v.)
Random variables are two types:
i) Discrete Random variable
ii) Continuous Random variable
Eg:
Someone continuously shoot on the same target, until really shot
and then stop. s is the number of shots，so
s
One shooting Two shooting.... n shooting......
X(s)
1
2
....
n
......
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Random Variable
 Let S be the sample space.
 A random variable X is a function
X: SReal
Suppose we toss a coin twice. Let X be the random variable
number of heads
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Random Variable
(Number of Heads in two coin tosses)
S
X
TT
0
TH
1
HT
1
HH
2
We also associate a probability with X attaining that value.
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Random Variable
(Number of Heads in two coin tosses)
S
Prob
X
X
P(X=x)
TT
1/4
0
0
1/4
TH
1/4
1
1
1/2
HT
1/4
1
2
1/4
HH
1/4
2
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Discrete Random variable:
If X is a random variable which can take a finite number
or countably infinite number of values, X is called a Discrete
Random variable.
Eg:
1. The number shown when a die is thrown.
2. Number of transmitted bits received in error.
Continuous Random variable:
If X is a random variable which can take all values in an
interval, then is called a Continuous Random variable.
Eg:
The length of time during which a vacuum tube installed in a
circuit functions is a continuous RV.
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Probability Mass Function (or) Probability Function:
If X is a discrete RV with distinct values x1 x2 x3 ,…
xn… then P ( X = xi) = pi, then the function p(x) is called
the Probability Mass Function.
Provided p(i = 1, 2, 3, …) satisfy the following
conditions:
1.
2.
pi  0,
for all i, and
 p i 1
i
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PROBABILITY DENSITY FUNCTION FOR
CONTINUOUS CASE:
If X is a continuous r.v., then f(x) is defined the
probability density function of X.
Provided f(x) satisfies the following conditions,
1.
f ( x) ³ 0
¥
2.
ò
-¥
3.
f ( x ) dx = 1
b
P(a £ x £ b) =ò f ( x)dx
a
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CUMULATIVE DISTRIBUTION FUNCTION (OR)
DISTRIBUTION FUNCTION OF D.R.V.:
The cumulative distribution function F(x) of a discrete r.v.
X with probability distribution p(x) is given by
F ( x) = P( X £ x) =
å
Xi £ x
p( xi ) for - ¥ < x < ¥
CUMULATIVE DISTRIBUTION FUNCTION OF C.R.V.:
The cumulative distribution function of a continuous r.v.
X is
x
F ( x) = P ( X £ x) = ò f ( x)dx, for - ¥ < x < ¥
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EXPECTED VALUE OF X FOR D.R.V:
The mean or expected value of the discrete random
variable of X, denoted as μ or E(X) is
m = E ( X ) = å xp ( x)
x
The variance of X denoted as σ2 or V(X) is
s 2 = V ( X ) = E ( X 2 ) - [ E ( X )]2
or V ( X ) = E ( X - m) 2 = å ( x - m) 2 p( x)
x
= å x 2 p( x) - m2
x
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PROBLEMS UNDER DISCRETE RANDOM
VARIABLES:
TYPES
GIVEN TO FIND
TYPE I
P[x]
F[x]
TYPE II
F[x]
P[x]
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PROBLEM 1: [TYPE I]
A r. v. X has the following probability functions
X
0
1
2
3
4
5
6
7
p(x)
0
k
2k
2k
3k
k2
2k2
7k2 + k
Find (i) value of k
(ii) P(1.5 < X < 4.5 / X > 2)
(iii) if P(X ≤ a) > ½, find the minimum value of a.
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Solution:
7
(i ) w.k .t.å p ( x ) = 1
x =0
Þ 10k 2 + 9k = 1
Þ
(10k - 1)( k +1) = 0
1
Þ k =
,- 1
10
p ( x ) cannot be negative,
k = - 1 is neglected.
1
\ k =
10
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(ii ) P (1.5 < X < 4.5 / x > 2)
P( A Ç B)
P( A / B) =
P( B)
P[(1.5 < X < 4.5) Ç ( X > 2)]
P (1.5 < X < 4.5 / X > 2) =
P ( X > 2)
P ( X = 3) + P ( X = 4)
=
7
å P( X = r )
r =3
5
5
10
=
=
7
7
10
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X
0
p(x)
0
1
2
3
4
5
1/10 2/10 2/10 3/10 1/100
6
7
2/100
17/100
(iii ) By trials,
P ( X £ 0) = 0
P( X
P( X
P( X
P( X
1
£ 1) =
10
3
£ 2) =
10
5
£ 3) =
10
8
£ 4) =
10
\ the minimum value of 'a' satisfying the condition P( X £ a ) >
1
is 4
2
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PROBLEM 2
A random Variable X has the following probability
distribution.
x
-2
p(x) 0.1
-1
k
0
0.2
1
2k
2
0.3
3
3k
Find
(i) the value of k
(ii) Evaluate P[X<2] and P[-2<X<2]
(iii)Find the cumulative Distribution of X.
(iv)Evaluate the Mean and Variance of X.
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Solution:
( i)
P [ X < 2] = P [ X = - 2] + P [ X = - 1]
+P [ X = 0] + P [ X = 1]
1
2
= 0.1 + + 0.2 +
15
15
3 1
= 0.3 + =
15 2
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( ii)
P [ - 2 < X < 2]
= P [ X = - 1] + P [ X = 0] + P [ X = 1]
1
2
= + 0.2 +
15
15
3
= 0.2 +
15
2
=
5
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(iii) The Cumulative Distribution of X:
X
-2
-1
0
F(x) = P(X ≤ x)
F(-2)=P(X≤-2)=P(X=-2)=0.1=1/10
F(-1)=P(X≤-1)=F(-2)+P(-1)=1/10+k=1/10+1/15=0.17
F(0)=P(X≤0)=F(-1)+P(0)=1/6+2/10=1/6+1/5=0.37
1
2
3
F(1)=P(X≤1)=F(0)+P(1)=11/30+2/15=15/30=1/2
F(2)=P(X≤2)=F(1)+P(2)=1/2+3/10=(5+3)/10=8/10=4/5
F(3)=P(X≤3)=F(2)+P(3)=4/5+3/15=(12+3)/15=1
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(iv) Mean:
3
Mean =E ( x) = å xp ( x)
x =2
= - 2 p (- 2) + (- 1) p(- 1) + 0 p(0)
+1 p (1) + 2 p(2) + 3 p(3)
æ1 ö
3
3
= - 2ç
+ 3.
ç ÷
÷ + ( - 1) .k + 0 + 2k + 2.
10
15
è 10 ø
16
=
15
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Variance:
2
var( X ) = E ( X ) 2
3
E( X ) =
å
[ E ( X )]
2
x 2 p( x)
x =- 2
= ( - 2)
2
p ( - 2) + ( - 1)
+ (1)
2
p (1) + ( 2)
2
2
p ( - 1) + ( 0)
p ( 2) + ( 3)
2
2
p ( 0)
p ( 3)
æ1 ö
æ3 ö
÷
÷
= 4ç
ç
÷ +1( k ) + 0 +1.(2 k ) + 4 ç
ç
÷ + 9(3k )
è 10 ø
è 10 ø
16 30 18
=
+
=
10 15
5
2
æ
ö
18
16
2
÷
\ var X = E ( X 2 ) - [ E ( X )] =
- ç
ç
÷ = 2.46
5 è 15 ø
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PROBLEM 3 : [TYPE II]
Obtain the probability function or probability
distribution function from the following distribution
x
F[x]
0
0.1
1
0.4
2
0.9
3
1.0
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Problem 3:
The diameter, say X of an electric cable, is assumed to be
a continuous r.v. with p.d.f. :
f(x) = 6x(1-x), 0 ≤ x ≤ 1
(i) Check that the above is a p.d.f.
(ii) Compute P(X ≤ ½ | ⅓ ≤ X ≤ ⅔)
(iii) Determine the number k such that
P(X< k) = P(X > k)
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Solution:
(i )
1
ò
0
1
f ( x ) dx =ò 6 x (1 - x )dx
0
éx
x ù
ú
=6ê
ê
3 ú
ë 2
û0
=1
2
3
1
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(ii )
1
1
P( £ X £
)
1 1
2
3
2
P( X £
|
£ X £
)=
1
2
2 3
3
P( £ X £
)
3
3
1
2
ò 6 x(1 =
x ) dx
1
3
2
3
ò 6 x(1 -
x ) dx
1
3
11
11
= 54 =
13
26
27
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(iii )
We have P ( X < k ) = P ( X > k )
k
Þ
ò 6 x(1 -
1
x )dx =ò 6 x (1 - x )dx
0
k
Þ 3k 2 - 2k 3 = 3(1 - k 2 ) - 2(1 - k 3 )
Þ 4k 3 - 6k 2 +1 = 0
Þ k=
1 1± 3
,
2
2
The only possible value of k in the given range is
1
.
2
1
\ k=
2
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rth MOMENT ABOUT ORIGIN
The rth moment about origin of a r.v. X is defined
as the Expected value of the rth power of the X
E ( X r ) = mr¢ = å x r p ( x), if X is discrete
x
¥
E ( X r ) = mr¢ = ò x r f ( x ), if X is continuous
-¥
If r = 1 then
Mean = E ( X ) = m1¢= å x p( x)
x
If r = 2 then
E ( X 2 ) = m2¢ = å x 2 p( x)
x
Variance = m2¢ - [ m1¢]2
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MOMENT GENERATING FUNCTION
The moment generating function of a r.v. X
(about origin) whose probability function f(x) is given
by
ì ¥ tx
ï ò e f ( x), for continuous probability distribution
ï -¥
tx
M X (t ) = E (e ) = í
ï ¥ tx
ï å e P( x), for discrete probability distribution
î -¥
t is a real parameter and the integration or summation
being extended to the entire range of x.
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Moment Generating Function
Problem: Let the random variable X has the p.d.f
ì 1 -x
ï e 2, x >0
f ( x) = í 2
ï
ïî 0, otherwise
Find the Moment Generating Function,
(i) mean,
(ii) variance of X and also
(iii) find P(x>1/2)
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Solution:
1 - 2x
Given f ( x ) =
e ,
x >0
2
The m.g.f is given by
¥
MX
( t)
ù
=Eé
ëe û = ò e
tx
-¥
¥
tx
f ( x ) dx =ò e
0
tx
1 - 2x
e dx
2
é
ù¥
æ1
ö
ê - ççè 2 - t ÷
ú
÷x
æ1
ö
¥
ø
ç
÷
t
x
ç
÷
1
1ê e
ú
= ò e è 2 ø dx =
ê æ
öú
1
2 0
2ê
ú
ç
t
÷
ç
÷
ê
ú
ø û0
ë è2
é
ù
1ê
1 ú
1
ê
ú= 1. 2
=0
=
ú
1
2ê
2 1 - 2t
1 - 2t
ê
- tú
ê
ú
2
ë
û
M X (t ) =
1
1 - 2t
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(i ) Mean :
E ( X )  Mean  M X ' (0)
 d  1 
 

 dt  1  2t  t 0
 1


(2)  2
2
 (1  2t )
 t 0
Mean  2
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(ii) VARIANCE:
d

E ( X )  M X '' (0)  
M X ' (t ) 
 dt
 t 0
2
d 
  4


2

 
( 2)  

2 
3



 dt  1  2t   
 t 0
  1  2t 
 t 0
 4


( 2) 
8
3
 1  2t 
 t 0
Variance  E ( X )  E ( x )  8  ( 2)  4
2
2
2
Variance  4
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(iii )
x
¥
æ
1ö
1 - 2x
1¥
2
÷
Pç
x
³
=
e
dx
=
e
dx
ò
ò
ç
÷
2ø
2 1
è
1 2
2
2
é -x
2
1 ê
e
ê
=
-1
2ê
ê
ê
ë 2
é
= ê- e
ê
ë
\
x
2
ù¥
ú
ú
ú
ú
ú
û1
2
1
ù¥
ú =e 4
ú
û1
2
æ
1ö
pç
çx > ÷
÷= 0
2ø
è
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
AND STATISTICS /UNIT–I/PPT/VER1.2
41
STANDARD DISTRIBUTIONS
DISCRETE
DISTRIBUTION
CONTINUOUS
DISTRIBUTION
1.Binomial Distribution
1. Uniform Distribution
2. Poisson Distribution
2. Exponential Distribution
3. Geometric Distribution
3. Gamma Distribution
4. Negative Binomial
Distribution
4. Normal Distribution
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
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42
DISCRETE DISTRIBUTION
BINOMIAL DISTRIBUTION
A random variable X is said to follow
Binomial Distribution if it assumes only nonnegative values and its probability mass
function is given by
P(X=x) =P(x) =
ncxpxqn-x, x=0, 1, 2, 3…..n, q=1-p
0
, otherwise
Where n and p are called parameter of the
Binomial distribution.
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
AND STATISTICS /UNIT–I/PPT/VER1.2
43
Mean of Binomial Distribution.
Mean = E(X) = np
Variance of Binomial Distribution.
Variance = Var(X) = npq.
Moment Generating Function (M.G.F) of a
Binomial Distribution.
M G F = Mx ( t ) = E ( etx ) = ( p et + q )n
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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44
PROBLEM
Out of 800 families with 4 children each, how many
families would be expected to have
(i) two boys and 2 girls,
(ii) atleast 1 boy
(iii) atmost 2 girls and
(iv) children of both sexes.
Assume equal probabilities for boys and girls.
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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45
Considering each child as a trial, n = 4.
Asumming that birth of a boy is a success,
1
1
p = and q = .
2
2
Let X denote the number of success ( boys) .
(i ) P (2 boys and 2 girls ) = P ( X = 2)
æ1 ö
= 4C2 ç
ç ÷
÷
è2ø
2
æ1 ö
ç
ç ÷
÷
è2ø
4- 2
æ1 ö
3
= 6´ ç
ç ÷
÷ =
8
è2ø
4
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
STATISTICS /UNIT–I/PPT/VER1.2
46
\ No. of families having 2 boys and 2 girls
= N * P( X = 2)
(where N is the total no. of families considered)
3
= 800´
8
= 300
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(ii ) P ( at least 1 boy )  P ( X  1)
 P ( X  1)  P ( X  2)  P ( X  3)  P ( X  4)
0
1 1
 1  P ( X  0)  1  4C0    
2 2
1 15
 1

16 16
No.of families having at least 1boy
4
15
 800 
16
 750
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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(iii ) P (atmost 2 girls ) = P ( X = 4) + P ( X = 3) + P ( x = 2)
= 1 - {P ( X = 0) + P ( X = 1)}
4
4ü
ìï
æ1 ö
æ1 ö ï
= 1 - í 4C0 çç ÷
÷ + 4C1 çç ÷
÷ý
ïî
è2ø
è 2 ø ïþ
11
=
16
\ No. of families having atmost 2 girls
11
= 800´
16
= 550
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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49
(iv) P(Childrenof both sexes)  1  P(Children of the same sexes)
 1  {P(all are boys)  P(all are girls)}
 1  { P( X  4)  P( X  0)}
4
  1 4
 1  
 1  4C4    4C0   
 2  
  2 
1 7
 1 
8 8
 No. of families having children of both sexes
7
 800 
8

700
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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50
POISSON DISTRIBUTION
A random variable X is said to follow Poison
distribution if it assumes only non -negative values
and its probability mass function is given by
ì e- l l
ïï
P( X = x) = P( x) = í x !
ï
ïî 0
x
; x = 0,1, 2,...l > 0
; otherwise
Where λ is parameter of the Poisson distribution.
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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Mean of Poisson Distribution.
Mean=E(X) = λ
Variance of Poisson Distribution.
Variance=Var(X) = λ
Moment Generating Function (M.G.F) of a Poisson
Distribution.
M.G.F = Mx ( t ) = E ( etx ) =
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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52
Problem:
The number of monthly breakdowns of a
computer is random variable having a Poisson
distribution with mean equal to 1.8. Find the probability
that this computer will function for a month
(i) Without a breakdown
(ii) With only one breakdown and
(iii) With atleast one breakdown.
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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53
Solution:
Let X denotes the number of break downs of the computer in a
month.
X follows a Poisson distribution with mean λ = 1.8
e- l l
P( X = x) =
x!
x
e - 1.8 (1.8) x
=
x!
(i )
P ( without a breakdown) = P( X = 0)
e - 1.8 (1.8) 0
=
0!
= e - 1.8 = 0.1653
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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(ii)
P( with only breakdown) = P( X = 1)
e- 18 (18)1
=
= 0.2975
1!
(iii)
P( with atleast one breakdown) = P( X ³ 1)
= 1 - P( X < 1)
= 1 - P( X = 0)
= 1 - 0.1653
= 0.8347
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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55
GEOMETRIC DISTRIBUTION
Suppose that independent trails, each
having a probability p, 0 < p < 1, of being
a success, are performed until a success
occurs. If we let X equal the number of
trails required, then
P(X = x) = (1-p)x-1p = qx-1 p,
x=1,2,3…
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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Mean of Geometric Distribution.
Mean =E(X) = 1/p
Variance of Geometric Distribution.
Variance = Var(X) = q/p2
Moment Generating Function (M.G.F) of a Geometric
Distribution.
M.G.F = Mx ( t ) = p/e-t-q
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
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AND STATISTICS /UNIT–I/PPT/VER1.2
58
PROBLEM :
If the probability that a target is destroyed on any one shot is
0.5. What is the probability that it would be destroyed on 6th
attempt?
SOLUTION:
Let `X’ be the R.V denoting the number of attempts required
for the first success.
Given p = 0.5 ∴ q= 1 – 0.5 = 0.5.
We know that P(X = x) = qx-1p
P(X = 6) = q6-1p = q5p = (0.5)5(0.5)
[∵p=0.5, q=0.5, n=6]
= (0.5)6 =0.0156.
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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59
Problem:
A die is tossed until 6 appears. What is the probability that
it must be tossed more than 4 times.
Solution:
Let X be a Random variable denoting the number
of times 6 appears.
Here p = 1/6, q = 1- p = 1-1/6 = 5/6
w.k.t.
P(X = x) = qx-1p x=1,2,3…..
 P(X = x) = (5/6)x-1(1/6)
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
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60
Here P(x> 4) = 1-P(x ≤ 4)
= 1{P(x = 1) + P( x= 2) + P(x = 3) + P(x = 4)}
= 1{(1/6) + (5/6)(1/6) + (5/6)2(1/6)
+ (5/6)3(1/6)}
= 1 - 1/6 (3.1065)
= 0.4823.
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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61
NEGATIVE BINOMIAL DISTRIBUTION
A random variable X is said to follow Negative
Binomial Distribution if its probability mass function is
given by
 x  1 r
 p 1  p x r , x  r , r  1,...
P(X = x) = P(x) = 
 r  1
(or)
 x  r  1 r
x
 p 1  P  , x  0,1,2..
P(X = x) = P(x) = 
 r 1 
Where p is the probability of success in a trail.
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
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Mean of Negative Binomial Distribution .
Mean = E(X) = rP
Variance of Negative Binomial Distribution .
Variance = Var(X) = rPQ
Moment Generating Function (M.G.F) of a Negative
Binomial Distribution.
M.G.F = Mx ( t ) = ( Q-P et ) -r
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
AND STATISTICS /UNIT–I/PPT/VER1.2
63
CONTINUOUS DISTRIBUTION
Uniform Distribution
A random variable X is said to have a
continuous Uniform Distribution over a interval
(a,b) if its probability density function is given by
ì 1
,a < x <b
(i.e.) f ( x) = ï
í b- a
ï
î 0 , otherwise
Where `a’ and `b’ are the two parameters of the
uniform distribution.
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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Mean of Uniform Distribution
a +b
Mean = m1 ' =
2
Variance of Uniform Distribution
2
(b - a)
Variance = Var ( X ) =
12
Moment Generating Function (M.G.F) of a
Uniform Distribution
bt
at
e -e
M .G.F = M X (t ) =
(b - a)t
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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65
PROBLEM :
Buses arrive at a specified stop at 15 min. interval
starting at 7 AM, That is, they arrive at 7, 7.15, 7.30, 7.45
and so on. If a passenger arrives at the stop at random time
that is uniformly distributed between 7 and 7.30 A.M. find
the probability that he waits.
(a) less than 5 min for a bus and
(b) at least 12 min for a bus.
Solution:
Let X denotes the time that a passenger arrives
between 7 and 7.30 a.m. Then X∼U(0,30)
Then
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
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(a) Passenger waits less than 5 minutes,
(i.e.) he arrives between 7.10-7.15 or7.25-7.30
P (waiting time less than 5minutes)
= P(10 £ X £ 15) + P(25 £ X £ 30)
15
30
1
1
=ò
dx +ò
dx
10 30
25 30
1
15
30
=
{[ x]10 +[ x] 25 }
30
1
=
3
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
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67
(b) Passenger waits at least 12 minutes
(i.e.) he arrives between 7-7.03 or7.15-7.18.
P (waiting time at least 12 minutes)
= P (0 £ X £ 3) + P (15 £ X £ 18)
3
18
1
1
=ò
dx +ò
dx
0 30
15 30
1
3
18
=
x] 0 +[ x]15
[
30
1
=
5
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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THE EXPONENTIAL DISTRIBUTION
A continuous random variable X is said to
follow an Exponential Distribution with
parameter λ > 0 if its probability density
function is given by
 e
f ( x)  
0,
 x
, x0
otherwise
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
AND STATISTICS /UNIT–I/PPT/VER1.2
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Mean of Exponential Distribution
1
Mean = m1 ' =
l
Variance of Exponential Distribution
Variance = Var(X) =
1

Moment Generating Function (M.G.F) of
Exponential Distribution
æt
M .G.F = M X ( t ) = å ç
ç
r =0 è l
¥
2
ö
÷
÷
ø
r
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
STATISTICS /UNIT–I/PPT/VER1.2
70
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
AND STATISTICS /UNIT–I/PPT/VER1.2
71
PROBLEM
The time (in Hours) required to repair a
machine is exponentially distributed with
parameter λ = ½. What is the probability
that the repair time exceeds 2 h? What is the
conditional probability that a repair takes at
least 10 h given that its duration exceeds
9h?
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
AND STATISTICS /UNIT–I/PPT/VER1.2
72
Solution:
The conditional probability that a repair takes at least 10
h given that its duration exceeds 9h is given by,
P ( X ³ 10 | X > 9) = P ( X ³ 9 +1| X > 9)
= P ( X > 1)
¥
=ò
1
1 -21 x
1 ¥ -21 x
e dx = ò e dx
2
21
é - 1 x ù¥
- 1 ù¥
ê
2 ú
é
x
1 êe ú
1
2
êe ú
= ê
=
(
2)
êë
ú
2 ê -1 ú
2
û2
ú
êë 2 ú
û1
-1
2
= ( - 1)[0 - e ] = e
-1
2
= 0.6065
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
AND STATISTICS /UNIT–I/PPT/VER1.2
73
THE GAMMA DISTRIBUTION
 A continuous random variable X is said to follow an
Gamma Distribution with parameter λ > 0 if its
probability function is given by
 e  x x  1
;   0,0  x  

f ( x )    
0
; otherwise

IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND
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Mean of Gamma Distribution
Mean = m1 ' = l
Variance of Gamma Distribution
Variance = Var ( X ) = l
Moment Generating Function (M.G.F) of
Gamma Distribution
-l
M .G.F = M X (t ) = (1- t ) , t <1
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
AND STATISTICS /UNIT–I/PPT/VER1.2
75
Problem:
Find the p.d.f of Gamma distribution Find the M.G.F,
Mean and variance.
Solution: A continuous random variable X taking
non-negative values is said to follow gamma
distribution, if its probability density function is given
by
 e  x x  1
;   0,0  x  

f ( x )    
0
; otherwise

Where α is the parameter of the distribution.
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
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Moment generating function

M x t   E e

xt
  e
tx
f
 x dx


0
1

( x)

e e
tx
x
x
0
 1
0
1
dx 
( x)
put
(1  t ) x  u
(1  t ) dx  du , if
1

( x)



 u 
e
0  1  t 
u
1

1  t  ( x ) 0
1
1  t   
e
tx
 1
e  x x 1
dx
( x)

(1 t ) x
 1
e
x
dx

0
if x  0, u  0.
x  , u  
du
1

1 t
( x)
u  1e u du
.   
 1

 u 
e
0  1  t 
u
du
1 t
 Gamma function



x
n 1
dx
  ( n)   e x
0

1
 .
1  t 
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
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Mean and variance of Gamma Distribution
1
w.k .t
M (t ) =
= (1 - t )
(1 - t )
M ¢ (t ) = - a (1 - t )
( - 1) = a (1 - t )
Mean = m = M ¢ (0) = a (1 - 0)
=a
d
d
M ¢¢(t ) =
M ¢ (t ) =
a (1 - t )
= a ( - a - 1) (1 - t )
dt
dt
a
X
a
- a-1
- a-1
X
- a-1
'
1
X
- a-1
X
X
= a ( a +1) (1 - t )
.( - 1)
-a- 2
¢
m2' = M ¢
X (t ) = a ( a +1)
\
-a- 2
t =0
'
Variance = m2 = m 2 -
( )
'
1
m
2
= a ( a +1) - a 2 = a
var X = a
Hence Mean and var iance of Gamma distribution = a .
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
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Normal Distribution
A random variable X is said to have a normal
distribution with parameters μ and σ2, if its p.d.f is
given by the probability law:

1
f (X ) 
(e)
 2
(X  ) 2
2 2
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY
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Mean of Normal Distribution
¥
Mean = m = ò
-¥
1
x
e
s 2p
2
1 æx - mö
÷
- ç
ç
÷
2è s ø
dx
Variance of Normal Distribution
¥
Variance = Var ( X ) = ò
-¥
2
1
x
e
s 2p
2
1 æx - mö
÷
- çç
÷
2è s ø
dx - m2
Moment Generating Function (M.G.F) of
Gamma Distribution
M .G.F = M X ( t ) = e
mt +t 2s
2
/2
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NORMAL DISTRIBUTION CURVE
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Problem:
X is a normal variate with mean 30 and S.D. 5.
Find
(i) P(26 ≤ X ≤ 40)
(ii) P(X ≥ 45)
(iii) P(|X -30| > 5)
Solution:
Here m = 30 and s = 5
X - m 20 - 30
(i) When X = 26, Z =
=
= - 0.8
s
5
40 - 30
and When X = 40, Z =
=2
5
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\ P ( 26 £ X £ 40) = P ( - 0.8 £ Z £ 2)
= P ( - 0.8 £ Z £ 0) + P ( 0 £ Z £ 2)
= P ( 0 £ Z £ 0.8) + P ( 0 £ Z £ 2)
= 0.2881 + 0.4772 = 0.7653
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45 - 30
(ii ) When X = 45, Z =
=3
5
\ P ( X ³ 45) = P ( Z ³ 3)
= 0.5 - P ( 0 £ Z £ 3)
= 0.5 - 0.49865 = 0.00135
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(iii ) P (| X - 30 |£ 5) = P ( 25 £ X £ 35)
= P ( -1 £ Z £ 1)
= 2P ( 0 £ Z £ 1)
= 2(0.3413)
= 0.6826
\ P (| X - 30 |> 5) = 1 - P (| X - 30 |£ 5)
= 1 - 0.6826
= 0.3174
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