Notesale: Turn your study into money

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Lecture Notes

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Prepared by

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15MA301-Probability & Statistics

S ATHITHAN

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Assistant Professor

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Department of of Mathematics
Faculty of Engineering and Technology
SRM UNIVERSITY

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Kattankulathur-603203, Kancheepuram District
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Probability & Statistics

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1 Defintion of Probability and Basic Theorems
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2 The Approaches to Define the Probability
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3 Total Probability and Bayes’ Theorem
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1 Properties of Expectation
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2 Variance and its properties
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1 Moments
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2 Moment Generating Function (MGF)
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google
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ATHITHAN

Probability and Random Variables
T OPICS :
? Probability Concepts and Problems
? Random Variables (Discrete and Continuous)
? General Probability Distributions and Problems

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? Moments and Moment Generating Functions

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Probability & Statistics

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1
1
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1

Defintion of Probability and Basic Theorems
Random Experiment

An Experiment whose outcome or result can be predicted with certainty is called a deterministic
experiment
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1
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Definition 1
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Set of all possible outcomes which are assumed to be equally
likely is called the sample space and is usually denoted by S
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Definition 1
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Let S be a sample space and
A be an event associated with a random experiment
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Then the probability of event A occurring, denoted by P(A) and defined as

S

n(A)
Number of cases favorable to A
=
n(S)
Exhaustive number of cases in S(Set of all possible cases)

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P (A) =

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Definition 1
...
Let a random experiment
n1
be repeated n times and let an event A occur n1 times out of n trials
...
As n increases,
shows a tendency to stabilize and to
n
approach a constant value and is denoted by P(A) is called the probability of the event A
...
e
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4 (Axiomatic Approach/Definition of Probability)
...
Then the probability of the event A is denoted
by P(A) is defined as a real number satisfying the following conditions/axioms:
(i) 0 ≤ P (A) ≤ 1
(ii) P(S)=1
(iii) If A and B are mutually exclusive events, then P (A ∪ B) = P (A) + P (B)
Axiom (iii) can be extended to arbitrary number of events
...
e
...
, An ,
...
) = P (A1 ) + P (A2 ) + P (A2 ) + P (A3 ) + · · · +
P (An ) +
...
The probability of the impossible event is zero
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e
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Probability & Statistics

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If A be the complimentary event of A, then P (A) = 1 − P (A) ≤ 1
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If A and B be any two events, then P (A ∪ B) =
P (A) + P (B) − P (A ∩ B) ≤ P (A) + P (B)
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2

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Theorem 4
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P (A ∩ B)
, provided P (A) 6= 0
P (A)

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P (B/A) =

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The conditional probability of an event B, assuming that the event A has happened, is denoted
by P (B/A) and is defined as

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Theorem 5 (Product Theorem of Probability)
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Please go through it in the textbook
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The converse of this statement is also
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e
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This result can be extended to any number of events
...
e
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, An be n no
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If the events A and B are independent, then the events A and B (and similarly A
and B, A and B) are also independent
...
3

Total Probability and Bayes’ Theorem

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Probability & Statistics

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If B1 , B2 ,
...
, n
P (Bi /A) = P
n
P (Bi )P (A/Bi )

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Theorem 8 (Bayes’ Theorem of Probability of Causes)
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, Bn be the set of exhaustive
and mutually exclusive events associated with a random experiment and A is another event
associated with(or caused by)Bi , then

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i=1

Discrete and Continuous Distributions

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2

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For the problems on Discrete and Continuous Random Variables(RV’s) we have to note down
the following points from the table for understanding
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0
P (X = x)

f (x)

F (X = x) = P (X ≤ x)

F (X = x) = P (X ≤ x)

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Takes Values of the form (Eg
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1 (Expectation)
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google
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ATHITHAN

by E(X)
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2 (Expectation for Discrete Random Variable)
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with probabilities P (X = x1 ) = p(x1 ), P (X = x2 ) =
∞
X
p(x2 ), P (X = x3 ) = p(x3 ),
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Definition 3
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Let X be a continuous random variable with pdf f (x) defined in (−∞, ∞), then the expected value of X is defined as
Z∞
E(X) =
xf (x)dx
...
1

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Note: E(X) is called the mean of the distribution or mean of X and is denoted by X or µ
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If c is any constant, then E(c) = c
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If a, b are constants, then E(aX + b) = aE(X) + b
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If (X, Y ) is two dimensional random variable, then E(X + Y ) = E(X) + E(Y )
...
2

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Definition 3
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Let X be a random variable with mean E(X), then the variance of X is defined
2
as E(X 2 ) − [E(X)]2 and id denoted by V ar(X) or σX

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V ar(aX) = a2 V ar(X)

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2
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4
4
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1 (Moments about origin)
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Moments about origin are known as raw
moments
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google
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S
...
µ01 = E(X)=Mean
2
...
µ03 = E(X 3 )
4
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Definition 4
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The rth moment of a random variable
X about the mean µ is defined as E[(X − µ)r ] and is denoted by µr
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µ1 = E(X − µ) = E(X) − E(µ) = µ − µ = 0

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2
...
µ3 = E[(X − µ)3 ]
4
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3 (Moments about any point a)
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The first four moments about a point ‘a’ are given by
1
...
m02 = E[(X − a)2 ]

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4
...
m03 = E[(X − a)3 ]

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Relation between moments about the mean and moments about any arbitrary point a

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Let µr be the rth moment about mean and m0r be the rth moment about any point a
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Probability & Statistics

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2

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µ1 = m01 − m00 m01 = 0
µ2 = m02 − 2 C1 m01 · m01 + (m01 )2
= m02 − (m01 )2
µ3 = m03 − 3 C1 m02 · m01 + 3 C2 m01 · (m01 )2 − (m01 )3 ·
= m03 − 3m02 · m01 + 2(m01 )3
µ4 = m04 − 4 C1 m03 · m01 + 4 C2 m02 · (m01 )2 − 4 C3 m01 · (m01 )3 + (m01 )4 ·
= m04 − 4m03 · m01 + 6m02 · (m01 )2 − 3(m01 )4

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Definition 4
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The moment generating function of a
random variable X is defined as E(etX ) for allt ∈ (−∞, ∞)
...


MX (t) = E(etX ) =

 n
X



etx P (X = x) if X is discrete




 x=0

Z∞




etx f (x)dx if X is continuous



x=0

If X is a random variable
...

1!
2!
3!
r!
t
(t)2
(t)3
(t)r
= 1 + E(X) +
E(X 2 ) +
E(X 3 ) + · · · +
E(X r ) +
...

1!
2!
3!
r! r
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Probability & Statistics

S
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e
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of t in the expansion of MX (t)
t2
µ02 = Coefft
...


...
of in the expansion of MX (t)
r!

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∴ MX (t) generates all the moments about the origin
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Another phenomenon which we often use to find the moments is given below:
(r)

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µ0r = MX (0)

(t)2 0
(t)3 0
(t)r 0
t 0
µ1 +
µ2 +
µ3 + · · · +
µ +
...

2!
(r − 1)! r
(t)r−2 0
(t)2 0
µ4 + · · · +
µ +
...

(r)
MX (t) = µ0r + terms of higher powers of t

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MX0 (t) = µ01 + tµ02 +

Putting t = 0, we get
MX0 (0) = µ01
MX00 (0) = µ02

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(r)
MX (0) = µ0r

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The Maclaurin’s series expansion given below will give all the moments
...

1!
2!
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Probability & Statistics


The MGF of X about its mean µ is MX−µ (t) = E et(X−µ)
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ATHITHAN



Similarly, the MGF of X about any point a is MX−a (t) = E et(X−a)
...
McX (t) = MX (ct)
2
...
MaX+b (t) = ebt MX (at)

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Example Worked out Problems

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Example: 1
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What is the probability that the product is negative?

F

Hints/Solution: If the product is negative(-ve), any one number is -ve (or) any 3 numbers are
-ve
...
of ways choosing 1 negative no
...

No
...
= 6 C1 ·8 C3 = 6 · 56 = 336
...

1001
1001

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∴ P (the product is -ve) =

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Total no
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out of 14 nos
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Example: 2
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Two are drawn out from the box at a time
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What is the probability that the other one is also
good
...

∴ The required probability is P (B/A) =

4
P (A ∩ B)
C2 /10 C2
6/45
1
=
=
=
4
P (A)
4/10
3
10

4
Aliter: Probability of the first drawn apple is good=
...

3
1
∴ Probability of the second drawn apple is also good= =
...
google
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ATHITHAN
Example: 3
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The probabilities that the bonus scheme will be introduced in the company if A, B
and C become general manager is 0
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7 and 0
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Find the probability that the
bonus scheme is introduced
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P (Ai )P (B/Ai ) =

IT

i=1

50
12 14 24
+
+
=
90 90 90
90

A

n
X

H

P (B) =

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Hints/Solution: Let B be an event of the bonus scheme is introduced
...

4
2
3
Given, P (A1 ) = , P (A2 ) = , P (A3 ) = and
9
9
9
7
8
3
P (B/A1 ) = 0
...
7 = , P (B/A3 ) = 0
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A bag contains 5 balls and if it not known that how many of them are white
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What is the probability
that the bag contains at least 2 white balls and all the balls in the bag are white?

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Hints/Solution: Let B be an event that two balls are drawn, they turns to be white
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Let A1 , A2 , A3 , A4 be the event that the bag contains 2,3,4,5
white balls respectively
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If we choose the bag that may be any one of these bags, so all the events A1 , A2 , A3 , A4
are equally likely
...

C2
10
C2
∴ By Total Probability Theorem and Bayes’ Theorem, the required probabilities are


n
X
1 1
3
6
10
1
P (B) =
P (Ai )P (B/Ai ) =
+
+
+
=
4 10 10 10 10
2
i=1
and

P (A2 )P (B/A2 )
1/4
1
P (A2 /B) = P
=
=
n
1/2
2
P (Ai )P (B/Ai )
i=1

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Probability & Statistics

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1% of women at age forty who participate in routine screening have breast cancer
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9
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A woman in this age group had a positive
mammography in a routine screening
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80 of every 100 women with breast cancer will get a positive mammography
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If 10,000 women in this age
group undergo a routine screening, about what fraction of women with positive mammographies
will actually have breast cancer?

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Hints/Solution:

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Now we pass both groups through the sieve; note that both sieves are the same; they just behave
differently depending on which group is passing through
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Let B be an event of showing a positive mammography
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8%
=
(1% ∗ 80%) + (99% ∗ 9
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Probability & Statistics

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8

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P(B/A) = 0
...
6%

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1% ∗ 80%

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Example: 6
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5 < X <
1
4
...

2

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Hints/Solution:

(i) We know that

∞
X

P [X = x] = 1
...
e
...
e
...
e
...

10
Page 13 of 36

=
=
=
=

1
1
0
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Probability & Statistics
∴ The probability distribution is given by
X
P [X = x]

0
0

1
2
3
k 2k 2k
1
2
2
P [X = x] 0
10 10 10
(ii) The Cumulative Distribution Function (CDF)

4
3k
3
10

S
...


F

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H


...

F (7) = P (X ≤ 7) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 2)
+P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7)
100
=
=1
100

1

P [X = x]

0

k

P [X = x]

0

3

4

5

6

7

2k

2k

3k

k2

2k 2

7k 2 + k

2
10
3
10

2
10
5
10

3
10
8
10

1
100
81
100

2
100
83
100

17
100
100
=1
100

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0

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F [X = x]

1
10
1
10

2

S

0

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X

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(iii)

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P (1
...
5/X > 2) =
P (1
...
5) =
P (X > 2) =
=
=
P (1
...
5/X > 2) =

P [(1
...
5) ∩ (X > 2)]
P (X > 2)
5
P (X = 3) + P (X = 4) =
10
1 − P (X ≤ 2)
1 − {P (X = 0) + P (X = 1) + P (X = 2)}
3
7
1−
=
10
10
5
10
7
10

=

5
7

(iv) The smallest value of α for which P (X ≤ α) >

Page 14 of 36

1
= 0
...

2

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Probability & Statistics
S
...
5
...
8 > 0
...
∴ α = 4 satisfies the given
is more than 0
...


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Example: 7
...

2
3
7k
7
81

4
9k
9
81

6
13k
13
81

7
15k
15
81

8
17k
17
81

F

1
81

5
11k
11
81

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2
5k
5
81

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(i) The value of k =

1
3k
3
81

H

P [X = x]

0
k
1
81

AT

X
P [X = x]

H

Hints/Solution:

(ii) the Distribution Function (CDF)
3
7k
7
81
16
81

S

2
5k
5
81
9
81

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1
3k
3
81
4
81

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0
k
1
P [X = x]
81
1
F [X = x]
81
(iii) P (0 < X < 3/X > 2)

4
9k
9
81
25
81

5
11k
11
81
36
81

6
13k
13
81
49
81

7
15k
15
81
64
81

8
17k
17
81
81
=1
81

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X
P [X = x]

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P (0 < X < 3/X > 2) =

P [(0 < X < 3) ∩ (X > 2)]
P (X > 2)

P (0 < X < 3) = P (X = 1) + P (X = 2) =

P (X > 2) = 1 − P (X ≤ 2)
= 1 − {P (X = 0) + P (X = 1) + P (X = 2)}
9
72
= 1−
=
81
81

P (1
...
5/X > 2) =

8
81
72
81

=

8
72

(iv) The smallest value of α for which P (X ≤ α) >

Page 15 of 36

8
81

1
= 0
...

2

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Probability & Statistics
S
...
5
...
5
...

more than 0
...
Find the value of k for the pdf f (x) =

3k − kx,



0,
(a) the CDF of X (b) P (1
...
2/0
...
8)

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when 0 ≤ x ≤ 1
when 1 ≤ x ≤ 2

...
e
...
e
...
e
...
e
...
google
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ATHITHAN

−∞

When x < 0, F (x) = 0, since f (x) = 0 for x < 0
Zx
When 0 ≤ x < 1, F (x) =
f (x)dx
−∞

=

Zx
f (x)dx =
0

N

0

x2
x
dx =
2
4

S

Zx

When 1 ≤ x < 2, F (x) =

A

Zx

H

f (x)dx

F
x
dx +
2

O

Z1

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S

f (x)dx
0

N
R

E

0

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O

Zx

=

Zx

1
dx +
2

0

Zx

1
1
(3 − x) dx = (−x2 + 6x − 5)
2
4

0

When x ≥ 3, F (x) = 1
whenx < 0

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C



0






x2



,


2



1
F (x) =
(2x − 1),

4





1


(−x2 + 6x − 5),


4




1,

1
1
dx = (2x − 1)
2
4

0

0

When 2 ≤ x < 3, F (x) =

Zx

AT

=

x
dx +
2

H

Z1

IT

0

∴

when 0 ≤ x ≤ 1
when 1 ≤ x ≤ 2
when 2 ≤ x ≤ 3
when x ≥ 3

(b) P (1
...
2/0
...
8) =

Page 17 of 36

P [(1
...
2) ∩ (0
...
8)]
P (0
...
8)

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Probability & Statistics

S
...
5 < X < 3
...
5 < X < 1
...
5 < X < 1
...
8
1
3
dx =
=
2
20
1
...
5 < X < 1
...
5

1

N

S

P [(1
...
2) ∩ (0
...
8)]
P (0
...
8)
16
20

AT

H

IT

H

=

1
3
dx =
2
16

A

P (1
...
2/0
...
8) =

Z1
...
Experience has shown that while walking in a certain park, the time X (in minutes)
duration between seeing two people smoking has a density function
(
kxe−x , when x > 0
f (x) =
0,
otherwise

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Find the (a) value of k (b) distribution function of X (c) What is the probability that a person,
who has just seen a person smoking will see another person smoking in 2 to 5 minutes, in at least
7 minutes?

C

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Hints/Solution: Given the pdf of X is

LE

∴ f (x) ≥ 0 ∀ x

=⇒

(
kxe−x , when x > 0
f (x) =
0,
otherwise

k > 0
...
e
...
e
...
google
...
ATHITHAN

(b) The distribution function (CDF) is given by
Zx
F (x) = P (X ≤ x) =

f (x) dx

Here we have x > 0, F (x) =

xe−x dx = 1 − (1 + x)e−x , x > 0

N

Zx

S

−∞

A

0

H

(c)
f (x) dx

=

O

2

xe−x dx

F

Z5

AT

2

H

P (2 < X < 5) =

IT

Z5

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S

= −6e−5 + 3e−2
= −0
...
406 = 0
...
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Probability & Statistics
S
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The sales of convenience store on a randomly selected day are X thousand
dollars, where X is a random variable with a distribution function


0,
when x < 0



2

x
,
when 0 ≤ x < 1
F (x) = 2


k(4x − x2 ) − 1, when 1 ≤ x < 2



1,
when x ≥ 2

A

N

S

Suppose that this convenience store’s total sales on any given day is less than 2000 units(Dollars
or Pounds or Rupees)
...
Find P (A) and P (B)
...
e
...
e
...
e
...
google
...
ATHITHAN
Since the total sales X is in thousands of units, the sales between 500 and 1500 units is the event
1
3
A which stands for = 0
...
5 and the sales over 1000 units is the event B which
2
2
stands for X > 1
...
5
Now
Z1
...
5 < X < 1
...
5

1

H

A

0
...
5
3
x dx + (2 − x) dx =
4

N

Z1

f (x)dx

H

P (B) = P (X > 1) =

IT

Z2

AT

1

Z2

1
2

F

(2 − x) dx =

=

O

1

1

N

O

TE

S

Z1
...
5) = f (x)dx
Z1
...


Example: 11
...

p(x) 0
...
1 0
...
2
Hints/Solution: Here X is a discrete RV
...
3 + 0 × 0
...
4 + 2 × 0
...
3 + 0 + 0
...
4 = 0
...
google
...
ATHITHAN

x2i p(xi )

i=−∞
2

= (−1) × 0
...
1 + 12 × 0
...
2
= 0
...
4 + 0
...
5

S

V ar(X) = E(X 2 ) − [E(X)]2
= (1
...
5)2

IT

AT

= 1+1= 2

H

E(2X + 1) = 2E(X) + 1
= 2 × (0
...
5 − 0
...
25

F

V ar(2X + 1) = 22 V ar(X)

O

TE

S

O

= 4 × (1
...
If a random variable X has the following probability distribution, find
x
0
1
2
3
4
2
E(X), E(X ), V ar(X), E(3X − 4), V ar(3X − 4)
...
∴
E(X) =

∞
X

xi p(xi )

i=−∞

=

E(X 2 ) =

14
5

∞
X

x2i p(xi )

i=−∞

=

Page 22 of 36

46
5

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Probability & Statistics
V ar(X) = E(X 2 ) − [E(X)]2
 2
14
46
−
=
5
5

S
...
A test engineer found that the distribution function of the lifetime X (in years) of
an equipment follows is given by
(
0,
when x < 0
F (x) =
− x5
1 − e , when x ≥ 0

N

Find the pdf, mean and variance of X
...
google
...
ATHITHAN

−∞

Z0

1 x
x2 e− 5 dx
5

=
−∞

V ar(X) = 50 − [5]2 = 25

H

A

N

∴

x ∞
1 2
(5x + 50x + 250)e− 5 0 = 50
5

S

= −

AT

H

IT

Example: 14
...
e
...
e
...
e
...
google
...
ATHITHAN

Z∞
The Mean E(X) =

xf (x)dx
−∞

Z0
=

3
x x(2 − x)dx
4

−∞

N
H

x2 f (x)dx

3
x2 x(2 − x)dx
4

O

−∞

F

=

AT

−∞

Z0

IT

Now E(X ) =

H

Z∞

A

V ar(X) = E(X 2 ) − [(X)]2

2

S


2
3 x 3 x4
= 1
= − 2 −
4
3
4 0

TE

S


2
3 x4 x5
6
= − 2 −
=
4
4
5 0
5

O

N

r
=⇒

S
...
=

1
5

TU

R

E

∴

1
6
V ar(X) = − [1]2 =
5
5

LE

C

Example: 15
...


Page 25 of 36

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S
...


H

A

i = 1, 2, 3,
...
∴ The probability distribution takes its values as
follows:

O

F

P (X = 1) = F (1) − F (0) =

TE

S

P (X = 4) = F (4) − F (1) =

1
1 1
− =
2 3
6
5 1
1
− =
6 2
3

O

P (X = 6) = F (6) − F (4) =

1
1
−0=
3
3

5
1
=
6
6

E

N

P (X = 10) = F (10) − F (6) = 1 −

TU

R

x
p(x)

4
1
6

6
1
3

10
1
6

C

∴ The probability distribution is given by

1
1
3

LE

14
Mean=E(X) = Σxp(x) =
...

3
285 196
89
2
V ar(X) = E(X ) − [E(X)]2 =
−
=
...
google
...
ATHITHAN

S

Probability & Statistics
Example: 16
...


IT

when x < 0

AT

H

when 0 ≤ x < 2
when 2 ≤ x < 4

O

F

when 4 ≤ x < 6

TE

S

when x ≥ 6

We know that P (X = xi ) = F (xi ) − F (xi−1 ),
xi−1 ≤ x ≤ xi
...
, where F is constant in

P (X = 0) =

1
6

LE

C

TU

R

E

The CDF changes its values at x = 1, 4, 6, 10
...
google
...

12
101
2
2
E(X ) = Σx p(x) =

...
ATHITHAN

 2
101
37
V ar(X) = E(X ) − [E(X)] =
−
= 16
...
51 = 7
...

6
12
2

2

S

Example: 17
...
Find the first four central
moment
...
Given m01 =
1, m02 = 5, m03 = 12, m04 = 48 be the first four moments about X = 5 m01 = E(X − 5) =
1 =⇒ E(X) − 5 = 1 =⇒ E(X) = µ1 = 6

TE

S

O

F

AT

H

IT

µ2 = m02 − (m01 )2 = 5 − 1 = 4
µ3 = m03 − 3m02 · m01 + 2(m01 )3
= 12 − 3 × 5 × 1 + 2(1) = −1
µ4 = m04 − 4m03 · m01 + 6m02 · (m01 )2 − 3(m01 )4
= 48 − 4(12)(1) + 6(5)(1) − 3(1) = 27

O

Example: 18
...
Find the first four moments
about the mean
...
Given m01 = 1, m02 = 4, m03 =
10, m04 = 45 be the first four moments about X = 4 m01 = E(X − 4) = 1 =⇒ E(X) − 4 =
1 =⇒ E(X) = µ1 = 5
µ2 = m02 − (m01 )2 = 4 − 1 = 3
µ3 = m03 − 3m02 · m01 + 2(m01 )3
= 10 − 3 × 4 × 1 + 2(1) = 0
µ4 = m04 − 4m03 · m01 + 6m02 · (m01 )2 − 3(m01 )4
= 45 − 4(10)(1) + 6(4)(1) − 3(1) = 26

Example: 19
...
Find the rth moment
and hence find the first four moments
...
google
...
ATHITHAN

Hints/Solution: We know that
Z∞
f (x) dx = 1
−∞

Z∞
i
...
,

kx2 e−x dx = 1

0

Z∞
i
...
, k

e−x x3−1 dx = 1

i
...
, kΓ(3) = 1

∵ Γ(n) =


e−x xn−1 dx

H

A

0

IT

[∵ Γ(n) = (n − 1)!]

H

1
i
...
, k · 2! = 1 =⇒ k =
...
A random variable X has the pdf f (x) = e− 2 , x ≥ 0
...

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Probability & Statistics
Hints/Solution: The MGF of X is given by
Z∞

MX (t) = E(etX ) =

S
...

1 − 2t
2

H

∴ MX (t) =

S

1

1 e− 2 (1−2t)x
=
2 − 21 (1 − 2t)

N

"0

1

e− 2 (1−2t)x dx

A

1
2

IT

=

Z∞

AT

= (1 − 2t)−1 = 1 + 2t + 4t2 + 8t3 +
...

8 + 48t +
...


R

E

N

O

TE

S

O

MX0 (t)
MX00 (t)
Now, First Moment=Mean= E(X) = µ01
Second Moment= E(X 2 ) = µ02
∴ V ar(X)

F

Now, Differentiating w
...
to t, we get

LE

C

TU

1
Example: 21
...
Find the
2
MGF(Moment Generating Function) and hence find its mean and variance
...
google
...

2
2
2

 t −1
e
et
1−
=
2
2

N
H

A

et
if et 6= 2
...
ATHITHAN

AT

Now, Differentiating w
...
to t, we get

2et
(2 − et )2

O

F

MX0 (t) =

TE

S

MX00 (t) =

2[(2 − et )et + 2e2t ]
(2 − et )3

TU

R

E

N

O

Now, First Moment=Mean= E(X) = µ01 = MX0 (0) = 2
Second Moment= E(X 2 ) = µ02 = MX00 (0) = 6
∴ V ar(X) = E(X 2 ) − [E(X)]2 = 6 − 4 = 2
...
A random variable X has the rth moment of the form µ0r = (r + 1)!2r
...


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Probability & Statistics
Hints/Solution: The MGF of X is given by
(r + 1)!2r
2!2
3!22
4!23

t
t3
t2
2 + 3!22 + 4!23 +
...

1!
2!
3!
r!

S

∴ MX (t) = E(etX ) = 1 +

N

=
=
=
=

...


A

Given µ0r
∴ µ01
µ02
µ03

S
...

∴ MX (t) = (1 − 2t)−2
...
r
...


LE

C

TU

R

E

N

O

TE

S

MX0 (t)
MX00 (t)
Now, First Moment=Mean= E(X) = µ01
Second Moment= E(X 2 ) = µ02
∴ V ar(X)

Page 32 of 36

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Probability & Statistics

6

S
...
Find the probability of drawing two red balls in succession from a bag containing 3 red
and 6 black balls when (i) the ball that is drawn first is replaced, (ii) it is not replaced
...
A bag contains 3 red and 4 white balls
...


S

3
...
Box 2 contains 2000 bulbs of
which 5% are defective
...
(i) Find the probability that both balls are defective and (ii) assuming that
both are defective, find the probability that they came from box 1
...
The chance of a doctor D will diagnose a disease Z correctly is 60%
...
Find the probability of the patient survival
...
What is the chance that his disease was
diagnosed wrongly?

LE

C

TU

R

E

N

O

TE

S

O

F

AT

5
...
1 k 0
...
3 3k
Find (i) the value of k,(ii) the Distribution Function (CDF) (iii) P (0 < X < 3/X < 2)
1
and (iv) the smallest value of α for which P (X ≤ α) >
...
A random variable X has the following distribution
X
0 1
2
3
4
P[X=x] k 2k 5k 7k 9k
Find (i) the value of k,(ii) the Distribution Function (CDF) (iii) P (0 < X < 3/X < 2)
1
and (iv) the smallest value of α for which P (X ≤ α) >
...
A random variable X has a pdf
(
3x2 , when 0 ≤ x ≤ 1
f (x) =
0,
otherwise
Find the (i) Find a, b such that P (X ≤ a) = P (X > a) and P (X > b)r= 0
...
5/X < 0
...
Ans: a =
,b=
2
20
8
...
google
...
ATHITHAN
Find the (i) the value of A,(ii) the Distribution Function (CDF) (iii) P (X > 5/X <
5), P (X > 5/2
...
5)
...
Ans: A =
25
9
...
5 < X < 7
...


AT

H

IT

11
...
A coin is tossed until a head appears
...

Also find its variance
...
5/X <
0
...
The first three moments about the origin are 5,26,78
...
Ans: 2,5,-48

O

13
...
Find the mean and variance
...
The pdf of a random variable X is given by
(
k(1 − x), when 0 ≤ x ≤ 1
f (x) =
0,
otherwise

C

Find the (i) the the value of k,(ii) the rth moment about origin (iii) mean and variance
...
An unbiased coin is tossed three times
...

16
...

17
...

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Probability & Statistics
S
...
The pdf of a random variable X is given by f (x) = ke , −∞ < x < ∞ Find the (i) the
the value of k,(ii) the rth moment about origin (iii) the MGF and hence mean and variance
...
Find the MGF of the RV whose moments are given by µ0r = (2r)!
...


H

AT

Figure 1: Values of e−λ
...
s@ktr
...
ac
...
google
...
google
...
ATHITHAN

LE

C

TU

R

E

N

O

TE

S

O

F

AT

H

IT

H

A

N

S

Probability & Statistics

Figure 2: Area under Standard Normal Curve
...
google

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Title: Probability and random variables
Description: The pdf is a chapter from the subject probability and statistics .Many engineering colleges have this subject in their cirriculum.

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