Probability & Statistics
Dr. Santosh Kumar Yadav
Department of Mathematics
Lovely Professional University, Phagwara,
Punjab.
Dr. S. K. Yadav (LPU)
1 / 53
Motivation
Statistics is fundamentally based on the theory of probability.
There are two types of statistics: Descriptive Statistics and Inferential Statistics.
Descriptive Statistics belongs to the data analysis where
the data set size is manageable and can be analysed analytically or graphically.
Dr. S. K. Yadav (LPU)
2 / 53
Motivation
Statistics is fundamentally based on the theory of probability.
There are two types of statistics: Descriptive Statistics and Inferential Statistics.
Descriptive Statistics belongs to the data analysis where
the data set size is manageable and can be analysed analytically or graphically.
Some of these methods are graphical in nature; the construction of histograms, boxplots, and scatter plots are primary examples.
Other descriptive methods involve calculation of numerical
summary measures, such as means, standard deviations,
and correlation coefficients etc.
Dr. S. K. Yadav (LPU)
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Inferential Statistics is applied where the entire data set
(population) can not be analysed as a whole. So we draw
a sample (a small or manageable portion) from the population. Then we analyse the sample for the characteristic of
interest and try to infer the same about the population.
Dr. S. K. Yadav (LPU)
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Inferential Statistics is applied where the entire data set
(population) can not be analysed as a whole. So we draw
a sample (a small or manageable portion) from the population. Then we analyse the sample for the characteristic of
interest and try to infer the same about the population.
For example: when you cook rice, you take out few grains
and crush them to see whether the rice is properly cooked.
Similarly, survey polls prior to voting in elections, TRP ratings
of TV channel, shows etc are samples based and therefore
belong to the inferential statistics.
Dr. S. K. Yadav (LPU)
3 / 53
Why first the theory of Probability?
The discipline of probability forms a bridge between the descriptive and inferential techniques.
The knowledge of probability leads to a better understanding
of how inferential procedures are developed and used, how
statistical conclusions can be translated into everyday
language and interpreted.
Dr. S. K. Yadav (LPU)
4 / 53
Why first the theory of Probability?
The discipline of probability forms a bridge between the descriptive and inferential techniques.
The knowledge of probability leads to a better understanding
of how inferential procedures are developed and used, how
statistical conclusions can be translated into everyday
language and interpreted.
As probability deals with the uncertainty, thus before understanding what a particular sample can tell us about the
population, we should first understand the uncertainty associated with taking a sample from population. This is why we
study theory of probability before statistics.
Dr. S. K. Yadav (LPU)
4 / 53
Definition of Probability
If a random experiment results in n mutually exclusive, and equally
likely outcomes, where m are favourable to an event A. Then
P(A) =
m
n
=
number of cases favorable to A
total number of possible cases
since m ≥ 0 and n ≥ 0 thus, we always have 0 ≤ P(A) ≤ 1.
Dr. S. K. Yadav (LPU)
5 / 53
Definition of Probability
If a random experiment results in n mutually exclusive, and equally
likely outcomes, where m are favourable to an event A. Then
P(A) =
m
n
=
number of cases favorable to A
total number of possible cases
since m ≥ 0 and n ≥ 0 thus, we always have 0 ≤ P(A) ≤ 1.
The probability of non-happening of event A is denoted by Ā and
is given by
Dr. S. K. Yadav (LPU)
5 / 53
Definition of Probability
If a random experiment results in n mutually exclusive, and equally
likely outcomes, where m are favourable to an event A. Then
P(A) =
m
n
=
number of cases favorable to A
total number of possible cases
since m ≥ 0 and n ≥ 0 thus, we always have 0 ≤ P(A) ≤ 1.
The probability of non-happening of event A is denoted by Ā and
is given by
n−m
m
=1−
n
n
Thus, for any event A, we have
P(Ā) =
= 1 − P(A)
P(A) + P(Ā) = 1
Dr. S. K. Yadav (LPU)
5 / 53
Some Basic Terminology
Random experiment: An experiment whose outcome or result is random, that is, is not known before the experiment,
is called random experiment.
Dr. S. K. Yadav (LPU)
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Some Basic Terminology
Random experiment: An experiment whose outcome or result is random, that is, is not known before the experiment,
is called random experiment.
Examples:
Tossing a fair coin
Rolling an unbaised die
Drawing a card from a well-shuffled pack of cards.
Dr. S. K. Yadav (LPU)
6 / 53
Some Basic Terminology
Random experiment: An experiment whose outcome or result is random, that is, is not known before the experiment,
is called random experiment.
Examples:
Tossing a fair coin
Rolling an unbaised die
Drawing a card from a well-shuffled pack of cards.
Sample space:
Dr. S. K. Yadav (LPU)
6 / 53
Some Basic Terminology
Random experiment: An experiment whose outcome or result is random, that is, is not known before the experiment,
is called random experiment.
Examples:
Tossing a fair coin
Rolling an unbaised die
Drawing a card from a well-shuffled pack of cards.
Sample space: Set of all possible outcomes of a random
experiment is called sample space, usually denoted by S.
For example,
Tossing of a fair coin, S= {H, T}
Throwing an unbaised die, S = {1,2,3,4,5,6}
Dr. S. K. Yadav (LPU)
6 / 53
Some Basic Terminology
Random experiment: An experiment whose outcome or result is random, that is, is not known before the experiment,
is called random experiment.
Examples:
Tossing a fair coin
Rolling an unbaised die
Drawing a card from a well-shuffled pack of cards.
Sample space: Set of all possible outcomes of a random
experiment is called sample space, usually denoted by S.
For example,
Tossing of a fair coin, S= {H, T}
Throwing an unbaised die, S = {1,2,3,4,5,6}
Event: Any subset of the sample space is called an event.
The event {H, T } is sure events in tossing of a fair coin.
Dr. S. K. Yadav (LPU)
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Conti...
Equally Likely Events:
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Conti...
Equally Likely Events: Any two elementary events of a sample space are called equally likely if both of them have equal
chance of occurrence, i.e., there is no reason to preffered
one in comparison to other event.
Example: for S = {H, T }, the events {H} and {T } are
equally likely.
Dr. S. K. Yadav (LPU)
7 / 53
Conti...
Equally Likely Events: Any two elementary events of a sample space are called equally likely if both of them have equal
chance of occurrence, i.e., there is no reason to preffered
one in comparison to other event.
Example: for S = {H, T }, the events {H} and {T } are
equally likely.
Mutually Exclusive and Exhastive Events:
Dr. S. K. Yadav (LPU)
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Conti...
Equally Likely Events: Any two elementary events of a sample space are called equally likely if both of them have equal
chance of occurrence, i.e., there is no reason to preffered
one in comparison to other event.
Example: for S = {H, T }, the events {H} and {T } are
equally likely.
Mutually Exclusive and Exhastive Events:
Two events are said to be mutually exclusive if happening of
one event precludes the happening of the other. For example, the events {H} and {T } in the sample space of the toss
of a fair coin are mutually exclusive.
The events {H} and {T} in the sample space of the toss of
a fair coin are mutually exclusive and exhaustive.
Dr. S. K. Yadav (LPU)
7 / 53
Some Problems
Q1: Find the probability of getting exactly two heads in tossing of two fair coins?
Sol: P(A) = 1/4.
Q2: If three fair coin are tossed simultaneously, find the probability that at least two tails occur.
Sol: P(A) = 4/8.
Dr. S. K. Yadav (LPU)
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Some Problems
Q3: If two unbaised dice are rolled, find the probability that
(a) sum is greater than 8
(b) sum is neither 7 nor 11
Dr. S. K. Yadav (LPU)
9 / 53
Some Problems
Q3: If two unbaised dice are rolled, find the probability that
(a) sum is greater than 8
(b) sum is neither 7 nor 11
Sol: Let S be the sample space and event M denote the
sum of numbers on upper face of dice. Then
(a) In this case, if
M = 9, sample points: {(6, 3), (5, 4), (4, 5), (3, 6)}
M = 10, sample points: {(6, 4), (5, 5), (4, 6)}
M = 11, sample points: {(6, 5), (5, 6)}
M = 12, sample points: {(6, 6)}
Dr. S. K. Yadav (LPU)
9 / 53
Cont...
Then probability that sum will be greater than 8 is
3
2
1
5
4
+ 36
+ 36
+ 36
= 18
P(M > 8) = 36
Dr. S. K. Yadav (LPU)
10 / 53
Cont...
Then probability that sum will be greater than 8 is
3
2
1
5
4
+ 36
+ 36
+ 36
= 18
P(M > 8) = 36
(b) Let N denote that sum is neither 7 nor 11, then as we can
6
2
see that P(M = 7) = 36
= 16 and also P(M = 11) = 36
.
Hence the probability that sum is neither 7 nor 11 is given
by,
P(N) = 1 − [P(M = 7) + P(M = 11)]
= 1 − [1/6 + 2/36]
7
=
9
Dr. S. K. Yadav (LPU)
10 / 53
Counting sample points
Evaluating of number of chance associated with the occurrence
of certain events in an experiment is one of the problems of the
statistician. These problems belong in the field of probability. In
many cases, we shall be able to solve a probability problem by
counting the number of points in the sample space without actually listing each element.
Multiplication rule: If work A can be done in m ways and B
can be done in n ways, then both work can be done
simultaneously in m × n ways. This is called fundamental
principle of counting.
Dr. S. K. Yadav (LPU)
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Q: How many sample points are there in the sample space
when a pair of dice is thrown once?
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Q: How many sample points are there in the sample space
when a pair of dice is thrown once?
Sol: The first die can land face-up in any one of n1 = 6
ways. For each of these 6 ways, the second die can also
land face-up in n2 = 6 ways. Therefore, the pair of dice can
land in n1 n2 = (6)(6) = 36 possible ways. So the sample
space carries 36 points given by
S = {(1, 1), (1, 2), ....., (1, 6), (2, 1), (2, 2), .....(2, 6), ..........., (6, 6
Remark: The above principle can be generalized to any
number of events.
Dr. S. K. Yadav (LPU)
12 / 53
Example: Suppose a customer wishes to buy a new cell
phone and can choose from 5 brands, 5 sets of capability,
and 4 colors. Then in how many ways customer can order
one of these phones?
Dr. S. K. Yadav (LPU)
13 / 53
Example: Suppose a customer wishes to buy a new cell
phone and can choose from 5 brands, 5 sets of capability,
and 4 colors. Then in how many ways customer can order
one of these phones?
These three classifications result in (5)(5)(4) = 100 different
ways for a customer to order one of these phones.
Dr. S. K. Yadav (LPU)
13 / 53
Example: Suppose a customer wishes to buy a new cell
phone and can choose from 5 brands, 5 sets of capability,
and 4 colors. Then in how many ways customer can order
one of these phones?
These three classifications result in (5)(5)(4) = 100 different
ways for a customer to order one of these phones.
Q1: A person is going to assemble a computer by himself.
He has the choice of chips from two brands, a hard drive
from four, memory from three, and an accessory bundle from
five local stores. How many different ways can the person
order the parts?
Dr. S. K. Yadav (LPU)
13 / 53
Example: Suppose a customer wishes to buy a new cell
phone and can choose from 5 brands, 5 sets of capability,
and 4 colors. Then in how many ways customer can order
one of these phones?
These three classifications result in (5)(5)(4) = 100 different
ways for a customer to order one of these phones.
Q1: A person is going to assemble a computer by himself.
He has the choice of chips from two brands, a hard drive
from four, memory from three, and an accessory bundle from
five local stores. How many different ways can the person
order the parts?
Ans:120.
Dr. S. K. Yadav (LPU)
13 / 53
Permutation (arrangment in definite order)
Many times, we are interested in a sample space that contains
as elements all possible orders or arrangements of a group of
objects. For example, we may want to know “how many different
arrangements are possible for sitting 6 people around a table”
or “how many different orders are possible for drawing 2 lottery
tickets from a total of 20”. The different arrangements are called
permutations.
The number of total permutations of n distinct objects taken
r at a time is denoted by n Pr and defined as
n
Dr. S. K. Yadav (LPU)
Pr =
n!
(n − r )!
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Permutation of identical objects:
Suppose in total n objects, some are of p1 type, some are
of p2 type, some are of p3 type, and remaining are of pk
type such that n = p1 + p2 + p3 + ... + pk . Then, number of
arrangments is
n!
=
(1)
p1 !.p2 !.p3 !...pk !
Dr. S. K. Yadav (LPU)
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Q: In how many ways can 7 graduate students be assigned
to 1 triple and 2 double hotel rooms during a conference?
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16 / 53
Q: In how many ways can 7 graduate students be assigned
to 1 triple and 2 double hotel rooms during a conference?
7!
Ans: 3!2!2!
Q1: Find the number of arrangment of the letters of the
word “COMMERCE” such that all vowels comes together
and hence the probability.
Dr. S. K. Yadav (LPU)
16 / 53
Q: In how many ways can 7 graduate students be assigned
to 1 triple and 2 double hotel rooms during a conference?
7!
Ans: 3!2!2!
Q1: Find the number of arrangment of the letters of the
word “COMMERCE” such that all vowels comes together
and hence the probability.
8!
Sol: Total ways of arranging the letters = 2!2!2!
.
6! 3!
. 2! .
Out of these favourable (vowels comes together) are 2!2!
favourable cases
So the probability that all vowels come together = total cases .
Dr. S. K. Yadav (LPU)
16 / 53
Q: In how many ways can 7 graduate students be assigned
to 1 triple and 2 double hotel rooms during a conference?
7!
Ans: 3!2!2!
Q1: Find the number of arrangment of the letters of the
word “COMMERCE” such that all vowels comes together
and hence the probability.
8!
Sol: Total ways of arranging the letters = 2!2!2!
.
6! 3!
. 2! .
Out of these favourable (vowels comes together) are 2!2!
favourable cases
So the probability that all vowels come together = total cases .
Q2:How many different letter arrangements can be made
from the letters in the word “STATISTICS” ?
Dr. S. K. Yadav (LPU)
16 / 53
Q: In how many ways can 7 graduate students be assigned
to 1 triple and 2 double hotel rooms during a conference?
7!
Ans: 3!2!2!
Q1: Find the number of arrangment of the letters of the
word “COMMERCE” such that all vowels comes together
and hence the probability.
8!
Sol: Total ways of arranging the letters = 2!2!2!
.
6! 3!
. 2! .
Out of these favourable (vowels comes together) are 2!2!
favourable cases
So the probability that all vowels come together = total cases .
Q2:How many different letter arrangements can be made
from the letters in the word “STATISTICS” ?
10!
Ans: 3!3!2!
= 50400
Dr. S. K. Yadav (LPU)
16 / 53
Combination (arrangment without order)
In many problems, we are interested in the number of ways of
selecting r objects from n objects without regard to order or the
r objects are of same kind in the sense of permutations. These
selections are called combinations.
Thus, the number of combinations of n objects taking r at a
time is denoted by n Cr and defined by
n
Dr. S. K. Yadav (LPU)
Cr =
n!
r !(n − r )!
17 / 53
Q1: A flushlight operates on two batteries. Total 8 batteries
are available in which three are dead. In a random selection
of batteries, what is the probability that exactly one dead
battery will be selected.
Dr. S. K. Yadav (LPU)
18 / 53
Q1: A flushlight operates on two batteries. Total 8 batteries
are available in which three are dead. In a random selection
of batteries, what is the probability that exactly one dead
battery will be selected.
3
5
Sol: P(A) = C81C. 2C1 = 15/28.
Dr. S. K. Yadav (LPU)
18 / 53
Q1: A flushlight operates on two batteries. Total 8 batteries
are available in which three are dead. In a random selection
of batteries, what is the probability that exactly one dead
battery will be selected.
3
5
Sol: P(A) = C81C. 2C1 = 15/28.
Q2: A firm offers a choice of 10 software packages to
buyers. There are 25 package from which to choose. There
are 5 computer games in the packages. Find the probability
that exactly three games are selected.
Dr. S. K. Yadav (LPU)
18 / 53
Q1: A flushlight operates on two batteries. Total 8 batteries
are available in which three are dead. In a random selection
of batteries, what is the probability that exactly one dead
battery will be selected.
3
5
Sol: P(A) = C81C. 2C1 = 15/28.
Q2: A firm offers a choice of 10 software packages to
buyers. There are 25 package from which to choose. There
are 5 computer games in the packages. Find the probability
that exactly three games are selected.
5
20
Sol: P(A) = C253 C. 10C7 =?
Dr. S. K. Yadav (LPU)
18 / 53
BLANK
Dr. S. K. Yadav (LPU)
19 / 53
Axioms1 of Probability
Let S be sample space and A is any event in S.Then, we have
(a) P(S) = 1, i.e., sum of total probability will be one.
(b) P(A) ≥ 0 i.e., probability can never be negative.
(c) P[A1 ∪ A2 ∪ A3 ... ∪ An ] = P(A1 ) + P(A2 ) + P(A3 )... + P(An ),
where Ai ’s are mutually exclusive events.
i.e., Probability of atleast one event can be found by adding the
individual probabilities.
1
Axioms are mathematical statements or assumptions without proof,
which form the basis of the logical development of a theory.
Dr. S. K. Yadav (LPU)
20 / 53
Algebra on Events
For any two events, A and B (not necessary mutually exclusive).
Addition Rule:
(i)
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
(ii) If exactly A or B occur, then
P(A∆B) = P(A ∪ B) − P(A ∩ B)
= P(A) + P(B) − 2P(A ∩ B)
(iii) If none of A and B occur, then
P(A ∪ B) = 1 − P(A ∪ B)
Dr. S. K. Yadav (LPU)
21 / 53
(iv) Event A but not B occur, then
P(A ∩ B̄) = P(A) − (A ∩ B)
(v) Event B but not A occur, then
P(Ā ∩ B) = P(B) − (A ∩ B)
(vi) For any three events A ,B and C,
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C)
− P(C ∩ A) + P(A ∩ B ∩ C)
Dr. S. K. Yadav (LPU)
22 / 53
Problems
Example 1: What is the probability of getting a total of 7 or
11 when a pair of fair dice is tossed?
Dr. S. K. Yadav (LPU)
23 / 53
Problems
Example 1: What is the probability of getting a total of 7 or
11 when a pair of fair dice is tossed?
Ans: P(A ∪ B) = P(A) + P(B) = 2/9
Example 3: A card is drawn from a well shuffled pack of
cards. Find the probability that the card is either a king or an
ace.
Dr. S. K. Yadav (LPU)
23 / 53
Problems
Example 1: What is the probability of getting a total of 7 or
11 when a pair of fair dice is tossed?
Ans: P(A ∪ B) = P(A) + P(B) = 2/9
Example 3: A card is drawn from a well shuffled pack of
cards. Find the probability that the card is either a king or an
ace.
Sol: 4/52+4/52 = 2/13 (since both the events are mutually
exclusive)
Dr. S. K. Yadav (LPU)
23 / 53
Problems
Example 2: A die is loaded in such a way that an even number is twice as likely to occur as an odd number. If E is the
event that a number less than 4 occurs on a single toss of
the die, find P(E).
Sol: P(E) = 4/9
Dr. S. K. Yadav (LPU)
24 / 53
Example 5: Two dice are rolled once. Find the probability of
getting an even number on the first die or a total of 8.
Ans: 5/9
Q4: If the probabilities are, respectively, 0.09, 0.15, 0.21,
and 0.23 that a person purchasing a new automobile will
choose the color green, white, red, or blue, what is the probability that a given buyer will purchase a new automobile that
comes in one of those colors?
Dr. S. K. Yadav (LPU)
25 / 53
Example 5: Two dice are rolled once. Find the probability of
getting an even number on the first die or a total of 8.
Ans: 5/9
Q4: If the probabilities are, respectively, 0.09, 0.15, 0.21,
and 0.23 that a person purchasing a new automobile will
choose the color green, white, red, or blue, what is the probability that a given buyer will purchase a new automobile that
comes in one of those colors?
Sol: (try yourself, Ans: 0.68)
Dr. S. K. Yadav (LPU)
25 / 53
Q5: The probability that at least one events A or B occurs is
0.8 and probability that both occurs simultaneously is 0.25.
Find P(Ā) + P(B̄)
Sol: (try yourself, Ans: 0.95)
Dr. S. K. Yadav (LPU)
26 / 53
Q5: The probability that at least one events A or B occurs is
0.8 and probability that both occurs simultaneously is 0.25.
Find P(Ā) + P(B̄)
Sol: (try yourself, Ans: 0.95)
Q6: Sohan is going to graduate from an industrial engineering department by the end of the semester. After being interviewed at two companies he likes, he assesses that his
probability of getting an offer from company A is 0.8, and his
probability of getting an offer from company B is 0.6. If he
believes that the probability that he will get offers from both
companies is 0.5, what is the probability that he will get at
least one offer from these two companies?
(try yourself, Ans: 0.90)
Dr. S. K. Yadav (LPU)
26 / 53
Q7: When a computer goes down, there is a 75% chances
that it is due to an overload and a 15% that it is due to a
software prblem. There is an 85% chance that it is due to an
overload or a software problem. What is the probability that
both of these problem are at fault. What is the probability
that there is a software problem but no overload?
Dr. S. K. Yadav (LPU)
27 / 53
Dr. S. K. Yadav (LPU)
28 / 53
Conditional Probability
Here we shall see the law to find the probability of an event
under the condition that another event has already occured.
The probability of an event B occurring when it is known that
some event A has occurred is called a conditional probability,
denoted by P(B|A) and defined as
P(B|A) =
Dr. S. K. Yadav (LPU)
P(A ∩ B)
,
P(A)
provided
P(A) ̸= 0
29 / 53
Cont...
The notion of conditional probability provides the capability
of reevaluating the idea of probability of an event in light of
additional information. The probability P(B|A) is an updation
of P(B) based on the knowledge that event A has occurred.
If the advance information that some event A occured has
no effect on B, then we can write
P(B|A) = P(B)
Dr. S. K. Yadav (LPU)
30 / 53
Example 1 The probability that a regularly scheduled flight departs on time is P(D) = 0.83; the probability that it arrives on
time is P(A) = 0.82; and the probability that it departs and arrives on time is P(D ∩ A) = 0.78. Find the probability that a plane
(a) arrives on time, given that it departed on time
(b) departed on time, given that it has arrived on time.
Dr. S. K. Yadav (LPU)
31 / 53
Example 1 The probability that a regularly scheduled flight departs on time is P(D) = 0.83; the probability that it arrives on
time is P(A) = 0.82; and the probability that it departs and arrives on time is P(D ∩ A) = 0.78. Find the probability that a plane
(a) arrives on time, given that it departed on time
(b) departed on time, given that it has arrived on time.
Sol: (a) P(A|D) = P(D∩A)
=
P(D)
Dr. S. K. Yadav (LPU)
31 / 53
Example 1 The probability that a regularly scheduled flight departs on time is P(D) = 0.83; the probability that it arrives on
time is P(A) = 0.82; and the probability that it departs and arrives on time is P(D ∩ A) = 0.78. Find the probability that a plane
(a) arrives on time, given that it departed on time
(b) departed on time, given that it has arrived on time.
Sol: (a) P(A|D) = P(D∩A)
= 0.78/0.83 = 0.94
P(D)
Dr. S. K. Yadav (LPU)
31 / 53
Example 1 The probability that a regularly scheduled flight departs on time is P(D) = 0.83; the probability that it arrives on
time is P(A) = 0.82; and the probability that it departs and arrives on time is P(D ∩ A) = 0.78. Find the probability that a plane
(a) arrives on time, given that it departed on time
(b) departed on time, given that it has arrived on time.
Sol: (a) P(A|D) = P(D∩A)
= 0.78/0.83 = 0.94
P(D)
(b) P(D|A) =
P(D∩A)
P(A)
Dr. S. K. Yadav (LPU)
= 0.78/0.82 = 0.95
31 / 53
Independent events
If this condtion holds we say that events are independent.
Hence by the definition of conditional probability, we have
P(A ∩ B) = P(A)P(B)
This is the mathematical condition to check the independence of the events A and B. If this condition holds then
events will be independent and vice-versa. This is necessary and sufficient condition for events to be independent.
Note that three events A, B, and C are independent provided
that these are pairwise independent and
P(A ∩ B ∩ C) = P(A)P(B)P(C)
.
Dr. S. K. Yadav (LPU)
32 / 53
Multiplication Rule of Probability
The simultaneous occurance of two events can be found by two
ways,
P(A ∩ B) = P(A)P(B), if events are independent.
P(A ∩ B) = P(A) + P(B) − P(A ∪ B), if events need not be
independent.
We have drawn another way to find P(A ∩ B)
P(A ∩ B) = P(B|A)P(A)
Dr. S. K. Yadav (LPU)
33 / 53
Example 2 Consider a experiment that consists of rolling a single
fair die once and then tossing a fair coin once, Let first member
of each order pair denote nmber on die and second, the face
showing on coin (H, T ). Let A is the event that the die shows
one or two and B is the event that coin shows heads. Check the
independence of events A and B.
Ans: Events are independent
Dr. S. K. Yadav (LPU)
34 / 53
Q. Suppose that we have a fuse box containing 20 fuses, of
which 5 are defective. If 2 fuses are selected at random and
removed from the box in succession without replacing the first,
what is the probability that both fuses are defective?
Dr. S. K. Yadav (LPU)
35 / 53
Q. Suppose that we have a fuse box containing 20 fuses, of
which 5 are defective. If 2 fuses are selected at random and
removed from the box in succession without replacing the first,
what is the probability that both fuses are defective?
Sol.: Let A be the event that the first fuse is defective and B the
event that the second fuse is defective; then we interpret A∩B as
the event that A occurs and then B occurs after A has occurred.
The probability of first removing a defective fuse is P(A) = 1/4.
Then the probability of removing a second defective fuse from
the remaining 4 is P(B|A) = 4/19.
Therefore
Dr. S. K. Yadav (LPU)
35 / 53
Q. Suppose that we have a fuse box containing 20 fuses, of
which 5 are defective. If 2 fuses are selected at random and
removed from the box in succession without replacing the first,
what is the probability that both fuses are defective?
Sol.: Let A be the event that the first fuse is defective and B the
event that the second fuse is defective; then we interpret A∩B as
the event that A occurs and then B occurs after A has occurred.
The probability of first removing a defective fuse is P(A) = 1/4.
Then the probability of removing a second defective fuse from
the remaining 4 is P(B|A) = 4/19.
Therefore P(A ∩ B) = P(A)P(B|A) = (1/4)(4/19) = 1/19.
Dr. S. K. Yadav (LPU)
35 / 53
Q. Consider the drawing a card from well shuffled pack of 52
cards. Let A is the event that a Spade card is drawn and B is
the event that a honor card (10, J, K, Q, A) is drawn. Check the
independence of events A and B. (Try Yourself)
Dr. S. K. Yadav (LPU)
36 / 53
Important Theorems
Theorem
Let A and B are two independent events, then
Events A and B̄ are independent.
Events Ā and B are independent.
Events Ā and B̄ are independent.
(Try to proof these)
Dr. S. K. Yadav (LPU)
37 / 53
Q2 A problem is given to three students in a class. The probabilities of the solution from the three students are 0.5, 0.7 and
0.8 respectively. What is the probability that the problem will be
solved.
Dr. S. K. Yadav (LPU)
38 / 53
Q2 A problem is given to three students in a class. The probabilities of the solution from the three students are 0.5, 0.7 and
0.8 respectively. What is the probability that the problem will be
solved.
Sol.: Let A, B and C are the events that problem is solved by
student X , Y , and Z respectively. Then we have to find P(A ∪ B ∪
C).
Dr. S. K. Yadav (LPU)
38 / 53
Q2 A problem is given to three students in a class. The probabilities of the solution from the three students are 0.5, 0.7 and
0.8 respectively. What is the probability that the problem will be
solved.
Sol.: Let A, B and C are the events that problem is solved by
student X , Y , and Z respectively. Then we have to find P(A ∪ B ∪
C).
(a) 0.28
(b) 0.72
(c) 0.97
(d) 0.82.
Dr. S. K. Yadav (LPU)
38 / 53
Q2 A problem is given to three students in a class. The probabilities of the solution from the three students are 0.5, 0.7 and
0.8 respectively. What is the probability that the problem will be
solved.
Sol.: Let A, B and C are the events that problem is solved by
student X , Y , and Z respectively. Then we have to find P(A ∪ B ∪
C).
(a) 0.28
(b) 0.72
(c) 0.97
(d) 0.82.
Ans: 0.97
Dr. S. K. Yadav (LPU)
38 / 53
Example3 A box contains 4 bad and 6 good tubes. Two are
drawn out from the box at a time. One of them is tested and
found to be good. What is the probability that the other one is
also good?
Dr. S. K. Yadav (LPU)
39 / 53
Example3 A box contains 4 bad and 6 good tubes. Two are
drawn out from the box at a time. One of them is tested and
found to be good. What is the probability that the other one is
also good?
1/3
Soln. P(B|A) = 6/10
= 59
Dr. S. K. Yadav (LPU)
39 / 53
Q. 4 In studying the causes of power failures, it have been found
that 5% are due to transformer damage, 80% are due to line damage, 1% involve both problems. Based on these percentages,
approximate the probability that a given power failure involves:
(a) line damage given that there is transformer damage
(b) transformer damage but not line damage
(c) transformer damage given that there is no line damage
(d) transformer damage or line damage
Dr. S. K. Yadav (LPU)
40 / 53
Q. 4 In studying the causes of power failures, it have been found
that 5% are due to transformer damage, 80% are due to line damage, 1% involve both problems. Based on these percentages,
approximate the probability that a given power failure involves:
(a) line damage given that there is transformer damage
(b) transformer damage but not line damage
(c) transformer damage given that there is no line damage
(d) transformer damage or line damage
Soln. Let events A and B denote that failure is due to transformer
damage and line damage, respectively. Then we have, P(A) =
0.05, P(B) = 0.80, and P(A ∩ B) = 0.01. We have to find
Dr. S. K. Yadav (LPU)
40 / 53
Q. 4 In studying the causes of power failures, it have been found
that 5% are due to transformer damage, 80% are due to line damage, 1% involve both problems. Based on these percentages,
approximate the probability that a given power failure involves:
(a) line damage given that there is transformer damage
(b) transformer damage but not line damage
(c) transformer damage given that there is no line damage
(d) transformer damage or line damage
Soln. Let events A and B denote that failure is due to transformer
damage and line damage, respectively. Then we have, P(A) =
0.05, P(B) = 0.80, and P(A ∩ B) = 0.01. We have to find
(a) P(B|A) =
(b) P(A ∩ B̄)
(c) P(A|B̄) =
(d) P(A ∪ B) = (Try yourself)
Dr. S. K. Yadav (LPU)
40 / 53
BLANK
Dr. S. K. Yadav (LPU)
41 / 53
Theorem of Total Probability
Theorem
Let Ei , i = 1, 2, 3, ....n are n mutually exclusive and exhaustive
events in the sample space of a random experiment with the
probabilities P(Ei ) such that P(Ei ) ̸= 0. Let A be an event in the
sample space such that P(A) ̸= 0. Then
P(A) =
n
X
P(Ei )P(A|Ei )
i=1
Dr. S. K. Yadav (LPU)
42 / 53
Proof: Since events Ei ’s are exhaustive, hence S = E1 ∪ E2 ∪
E3 ... ∪ En . We have
⇒
A=A∩S
= A ∩ (E1 ∪ E2 ∪ E3 ... ∪ En )
= (A ∩ E1 ) ∪ (A ∩ E2 ) ∪ (A ∩ E3 )... ∪ (A ∩ En )
P(A) = P(A ∩ E1 ) + P(A ∩ E2 ) + P(A ∩ E3 )... + P(A ∩ En )
= P(E1 )P(A|E1 ) + P(E2 )P(A|E2 ) + ... + P(En )P(A|En )
n
X
=
P(Ei )P(A|Ei )
i=1
Dr. S. K. Yadav (LPU)
43 / 53
Example1 In a certain assembly plant, three machines, M1 , M2 ,
and M3 , make 30%, 45%, and 25%, respectively, of the products. It is known from past experience that 2%, 3%, and 2% of
the products made by each machine, respectively, are defective.
Now, suppose that a finished product is randomly selected. What
is the probability that it is defective?
Dr. S. K. Yadav (LPU)
44 / 53
Example1 In a certain assembly plant, three machines, M1 , M2 ,
and M3 , make 30%, 45%, and 25%, respectively, of the products. It is known from past experience that 2%, 3%, and 2% of
the products made by each machine, respectively, are defective.
Now, suppose that a finished product is randomly selected. What
is the probability that it is defective?
Sol.: Let A be the event that product is defective. Let B1 , B2 , and
B3 are the events that product is made by machine M1 , M2 and
M3 , respectively. Given that
Dr. S. K. Yadav (LPU)
44 / 53
Example1 In a certain assembly plant, three machines, M1 , M2 ,
and M3 , make 30%, 45%, and 25%, respectively, of the products. It is known from past experience that 2%, 3%, and 2% of
the products made by each machine, respectively, are defective.
Now, suppose that a finished product is randomly selected. What
is the probability that it is defective?
Sol.: Let A be the event that product is defective. Let B1 , B2 , and
B3 are the events that product is made by machine M1 , M2 and
M3 , respectively. Given that
P(B1 ) = 0.3, P(B2 ) = 0.45, P(B3 ) = 0.25 P(A|B1 ) = 0.02, P(A|B2 ) =
0.03, P(A|B3 ) = 0.02
Dr. S. K. Yadav (LPU)
44 / 53
Example1 In a certain assembly plant, three machines, M1 , M2 ,
and M3 , make 30%, 45%, and 25%, respectively, of the products. It is known from past experience that 2%, 3%, and 2% of
the products made by each machine, respectively, are defective.
Now, suppose that a finished product is randomly selected. What
is the probability that it is defective?
Sol.: Let A be the event that product is defective. Let B1 , B2 , and
B3 are the events that product is made by machine M1 , M2 and
M3 , respectively. Given that
P(B1 ) = 0.3, P(B2 ) = 0.45, P(B3 ) = 0.25 P(A|B1 ) = 0.02, P(A|B2 ) =
0.03, P(A|B3 ) = 0.02
From the total probability theorem,
P(A) = P(B1 )P(A|B1 ) + P(B2 )P(A|B2 ) + P(B3 )P(A|B3 )
= (0.3)(0.02) + (0.45)(0.03) + (0.25)(0.02)
Ans:0.0245.
Dr. S. K. Yadav (LPU)
44 / 53
Q.1 A person has undertaken a construction job. The probabilities are 0.65 that there will be strike, 0.80 that the construction
job will be completed on time if there is no strike, and 0.32 that
the construction job will be completed on time if there is a strike.
Determine the probability that the construction job will be completed on time.
Dr. S. K. Yadav (LPU)
45 / 53
Q.1 A person has undertaken a construction job. The probabilities are 0.65 that there will be strike, 0.80 that the construction
job will be completed on time if there is no strike, and 0.32 that
the construction job will be completed on time if there is a strike.
Determine the probability that the construction job will be completed on time.
Sol: Let A be the event that the construction job will be completed on time, and B be the event that there will be a strike.
Dr. S. K. Yadav (LPU)
45 / 53
Q.1 A person has undertaken a construction job. The probabilities are 0.65 that there will be strike, 0.80 that the construction
job will be completed on time if there is no strike, and 0.32 that
the construction job will be completed on time if there is a strike.
Determine the probability that the construction job will be completed on time.
Sol: Let A be the event that the construction job will be completed on time, and B be the event that there will be a strike.
By Total probability theorem
P(A) = P(B)P(A|B) + P(B̄)P(A|B̄)
Dr. S. K. Yadav (LPU)
45 / 53
Q.1 A person has undertaken a construction job. The probabilities are 0.65 that there will be strike, 0.80 that the construction
job will be completed on time if there is no strike, and 0.32 that
the construction job will be completed on time if there is a strike.
Determine the probability that the construction job will be completed on time.
Sol: Let A be the event that the construction job will be completed on time, and B be the event that there will be a strike.
By Total probability theorem
P(A) = P(B)P(A|B) + P(B̄)P(A|B̄)
= (0.65)(0.32) + (1 − 0.65)(0.8)
Ans: 0.488.
Dr. S. K. Yadav (LPU)
45 / 53
Q.2 A factory has four independent units which produce 40%,
30%,20%, and 10% of identical items, respectively. The percentage of defective items produced by theseu units are 2%, 1%, 0.5%,
and 0.25%, respevtively. If an item is selected at random, find the
probability that the item is defective.
Sol: Let M be the event that item is defective.
Ans: P(M) = 0.01225.
Dr. S. K. Yadav (LPU)
46 / 53
Baye’s Theorem: (Thomas Bayes, 1761)
Theorem
Let Ei are n mutually exclusive and exhaustive events in the
sample space of a random experiment with the probabilities
P(Ei ) such that P(Ei ) ̸= 0. Let A be an event in the sample
space such that P(A) ̸= 0, then
P(Ei |A) =
P(Ei ).P(A|Ei )
n
P
P(Ei )P(A|Ei )
i=1
Remark: This gives a rule to update prior probability of some
event on the basis of new information by the event.
Dr. S. K. Yadav (LPU)
47 / 53
Example 1. The probabilities of X , Y and Z becoming managers
are 4/9, 2/9 and 1/3 respectively. The probabilities that the bonus
scheme will be introduced if X , Y and Z become managers are
3/10, 1/2 and 4/5 respectively.
(a) What is the probability that bonus scheme will be introduced.
(b) If bonus scheme has been introduced, what is the probability
that the manager appointed was X .
Soln. Let A1 , A2 and A3 denote the events that X , Y and Z
become manager respectively. Let B denote that bonus scheme
is introduced. Then, we have to find P(B), where given that
Dr. S. K. Yadav (LPU)
48 / 53
Example 1. The probabilities of X , Y and Z becoming managers
are 4/9, 2/9 and 1/3 respectively. The probabilities that the bonus
scheme will be introduced if X , Y and Z become managers are
3/10, 1/2 and 4/5 respectively.
(a) What is the probability that bonus scheme will be introduced.
(b) If bonus scheme has been introduced, what is the probability
that the manager appointed was X .
Soln. Let A1 , A2 and A3 denote the events that X , Y and Z
become manager respectively. Let B denote that bonus scheme
is introduced. Then, we have to find P(B), where given that
P(A1 ) = 4/9, P(A2 ) = 2/9, P(A3 ) = 1/3,
and P(B|A1 ) = 3/10, P(B|A2 ) = 1/2, P(B|A3 ) = 4/5.
Dr. S. K. Yadav (LPU)
48 / 53
Example 1. The probabilities of X , Y and Z becoming managers
are 4/9, 2/9 and 1/3 respectively. The probabilities that the bonus
scheme will be introduced if X , Y and Z become managers are
3/10, 1/2 and 4/5 respectively.
(a) What is the probability that bonus scheme will be introduced.
(b) If bonus scheme has been introduced, what is the probability
that the manager appointed was X .
Soln. Let A1 , A2 and A3 denote the events that X , Y and Z
become manager respectively. Let B denote that bonus scheme
is introduced. Then, we have to find P(B), where given that
P(A1 ) = 4/9, P(A2 ) = 2/9, P(A3 ) = 1/3,
and P(B|A1 ) = 3/10, P(B|A2 ) = 1/2, P(B|A3 ) = 4/5.
(a) B = (A1 ∩ B) ∪ (A2 ∩ B) ∪ (A3 ∩ B).
⇒ P(B) = P(A1 ∩ B) + P(A2 ∩ B) + P(A3 ∩ B).
= 4/9.3/10 + 2/9.1/2 + 1/3.4/5
Dr. S. K. Yadav (LPU)
48 / 53
Example 1. The probabilities of X , Y and Z becoming managers
are 4/9, 2/9 and 1/3 respectively. The probabilities that the bonus
scheme will be introduced if X , Y and Z become managers are
3/10, 1/2 and 4/5 respectively.
(a) What is the probability that bonus scheme will be introduced.
(b) If bonus scheme has been introduced, what is the probability
that the manager appointed was X .
Soln. Let A1 , A2 and A3 denote the events that X , Y and Z
become manager respectively. Let B denote that bonus scheme
is introduced. Then, we have to find P(B), where given that
P(A1 ) = 4/9, P(A2 ) = 2/9, P(A3 ) = 1/3,
and P(B|A1 ) = 3/10, P(B|A2 ) = 1/2, P(B|A3 ) = 4/5.
(a) B = (A1 ∩ B) ∪ (A2 ∩ B) ∪ (A3 ∩ B).
⇒ P(B) = P(A1 ∩ B) + P(A2 ∩ B) + P(A3 ∩ B).
= 4/9.3/10 + 2/9.1/2 + 1/3.4/5
Ans:
23/45
Dr. S. K. Yadav (LPU)
48 / 53
(b) If bonus scheme has been introduced, what is the probability
that the manager appointed was X .
We have to find P(A1 |B).
Dr. S. K. Yadav (LPU)
49 / 53
(b) If bonus scheme has been introduced, what is the probability
that the manager appointed was X .
We have to find P(A1 |B).
P(A1 ).P(B|A1 )
P(B)
12/90
=
23/45
= 6/23
P(A1 |B) =
Dr. S. K. Yadav (LPU)
49 / 53
Example1 In a certain assembly plant, three machines, M1 , M2 ,
and M3 , make 30%, 45%, and 25%, respectively, of the products. It is known from past experience that 2%, 3%, and 2% of
the products made by each machine, respectively, are defective.
If a product was chosen randomly and found to be defective, what
is the probability that it was made by machine M3 ??
Dr. S. K. Yadav (LPU)
50 / 53
Example1 In a certain assembly plant, three machines, M1 , M2 ,
and M3 , make 30%, 45%, and 25%, respectively, of the products. It is known from past experience that 2%, 3%, and 2% of
the products made by each machine, respectively, are defective.
If a product was chosen randomly and found to be defective, what
is the probability that it was made by machine M3 ??
Sol.: We have already find P(A) = 0.0245. Here, we need to
10
3 ).P(A|E3 )
= 0.25×0.02
= 49
find P(E3 |A) = PP(E
P(Ei ).P(A|Ei )
0.0245
Dr. S. K. Yadav (LPU)
50 / 53
Example 2. The chances that doctor will diagnose a disease
X correctly is 60%. The chances that a patient will die by his
treatment after correct diagnose is 40% and the chances of death
by wrong diagnose is 70%. A patient of doctor, who had disease
X died. what is the probability that his disease was diagnose
correctly.
Soln. Let A1 denotes that disease is diagnose correctly by doctor. Then A2 denote that disease is not diagnose correctly.
Let B denotes that a patient who had disease X died. P(A1 |B) =
P(A1 ).P(B|A1 )
= 6/13
P(B)
Dr. S. K. Yadav (LPU)
51 / 53
BLANK
Dr. S. K. Yadav (LPU)
52 / 53
THANK YOU
Dr. S. K. Yadav (LPU)
53 / 53
Probability & Statistics
Dr. Santosh Kumar Yadav
Assistant Professor
Department of Mathematics
Lovely Professional University, Phagwara,
Punjab.
Dr. Santosh Yadav, LPU Punjab
1 / 71
Probability Distribution
Functions
In this Lecture we will discuss about the random variables, their
types and illustrate the properties of typical probability distribution functions, joint probability distributions. The mathematical
expectation, variance, standard deviation, and Covariance will
also be discussed for these random variables.
Dr. Santosh Yadav, LPU Punjab
2 / 71
What is Random Variable?
A random variable (RV) is function which assign real value
to each outcome of a random experiment.
Dr. Santosh Yadav, LPU Punjab
3 / 71
What is Random Variable?
A random variable (RV) is function which assign real value
to each outcome of a random experiment.
It is mostly denoted by capital functions, X , Y , Z etc. and
their argument by small letters, x, y , z etc. Therefore, if X
be a random variable then
X :S→R
Example 1: Consider the experiment of tossing a fair coin.
Let X denotes the number of heads. Then X is a random
variable defined as
X : S = {HH, HT , TH, TT } → {0, 1, 2} ⊂ R
Dr. Santosh Yadav, LPU Punjab
3 / 71
Discrete and Continuous Random
Variables
Discrete random variables assume a finite or a countable
number of values.
Examples:
Number of calls per hours in an office.
Number of accident per month in a metro city.
Number of successes in n total trials
number of trials in tossing a coin untill a head appears.
Dr. Santosh Yadav, LPU Punjab
4 / 71
Cont...
Continuous random variables have an infinite continuum of
possible values.
Examples:
blood pressure of a patient
height of persons below age 25 in India
the speed of a car on expressway
the real numbers from 1 to 3.
Dr. Santosh Yadav, LPU Punjab
5 / 71
Probability Desity Function
A function f is said to be probability density function of a
discrete random variable X if it satisfies the following two
conditions:
(i) f (x) ≥ 0 for each value x of X .
P
(ii)
f (x) = 1, that is, sum of probabilities of all values x of X is
x
equal to one. We will use to find the probability at a certain
point, say x = xi .
Dr. Santosh Yadav, LPU Punjab
6 / 71
Probability Desity Function
A function f is said to be probability density function of a
discrete random variable X if it satisfies the following two
conditions:
(i) f (x) ≥ 0 for each value x of X .
P
(ii)
f (x) = 1, that is, sum of probabilities of all values x of X is
x
equal to one. We will use to find the probability at a certain
point, say x = xi .
Density Function (Notation): Let X be a discrete random
variable, then the function f defined by
f (x) = p(x)
= P[X = x]
for real x is called the probability density function for X .
Dr. Santosh Yadav, LPU Punjab
6 / 71
Cummulative Distribution Function
Let X be a discrete random variable, then a function F defined by F (x) = P[X ≤ x], i.e.,
X
F (x) =
f (x)
X ≤x
is called cummulative distribution function.
Dr. Santosh Yadav, LPU Punjab
7 / 71
Cummulative Distribution Function
Let X be a discrete random variable, then a function F defined by F (x) = P[X ≤ x], i.e.,
X
F (x) =
f (x)
X ≤x
is called cummulative distribution function.
Therefore, F (x) is the sum of probabilities of all the values
of X starting from its lowest value to the value x.
In particular,
X
F (x0 ) =
f (x)
X ≤x0
is the sum of probabilities up to x0 .
Dr. Santosh Yadav, LPU Punjab
7 / 71
Example 1
Consider the experiment of tossing two fair coins. Let X
denote the number of heads, then find the pdf and cdf of
descrete random variable X .
Sol: The Sample space is,
S = {HH, HT , HT , TT }
We can see that
P(X = 0) = 1/4, P(X = 1) = 2/4 = 1/2, P(X = 2) = 1/4. It
is easy to see that the function given by
X =x
0
1
2
f (x) = P(X = x) 1/4 1/2 1/4
is the pdf of the descrete random variable X . This is the
probability distribution of X .
Dr. Santosh Yadav, LPU Punjab
8 / 71
Cont...
The cummulative function of the X can be given as
X =x
0
1
2
F (x) = P(X ≤ x) 1/4 3/4 1
Ex2. A random variable X has the following probalility function
X =x 0 1 2 3 4
f (x) 0 k 3k k k
Find (i) value of k (ii) Evaluate P(X < 3) and P(X ≥ 3)
(Try yourself)
Dr. Santosh Yadav, LPU Punjab
9 / 71
Dr. Santosh Yadav, LPU Punjab
10 / 71
Problems
Q1. Let X denotes the number of heads in the sample
space in the tossing of three fair coins simultaneously.
Find the pdf and cdf of random variable
Dr. Santosh Yadav, LPU Punjab
11 / 71
Continuous Probability Distributions
A random variable which assumes values in some interval
is called continuous.
If X be a continuous RV then probability at any specific
value of X is zero.
i.e.,
P(X = x0 ) = 0
Dr. Santosh Yadav, LPU Punjab
12 / 71
Continuous Probability Distributions
A random variable which assumes values in some interval
is called continuous.
If X be a continuous RV then probability at any specific
value of X is zero.
i.e.,
P(X = x0 ) = 0
PDF: The function f (x) is a probability density function for the
continuous random variable X, over the set of reals, if
(i) f (x) ≥ 0, for each real x.
R∞
(ii) −∞ f (x)dx = 1
Rb
(iii) P(a ≤ X ≤ b) = a f (x)dx
Dr. Santosh Yadav, LPU Punjab
12 / 71
Cummulative Distribution Function
Let X be a continuous random variable with pdf f (x), then a
function F defined by F (x) = P[X ≤ x], x is real.
i.e.,
Z x
F (x) =
f (t)dt
−∞
is called cummulative distribution function.
P(a < X < b) = F (b) − F (a) and f (x) =
tive exists.
Dr. Santosh Yadav, LPU Punjab
dF (x)
,
dx
if the deriva-
13 / 71
Example 1
Let X be is a continuous random variable with distribution
1
; 0≤x ≤8
8
f (x) =
0
; elsewhere
Find
(i) P(2 ≤ X ≤ 5)
(ii) P(X ≥ 6)
Dr. Santosh Yadav, LPU Punjab
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Dr. Santosh Yadav, LPU Punjab
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Example 2
Suppose that the error in the reaction temperature, in 0C ,
for a controlled laboratory experiment is a continuous
random variable X having the probability density function
x2
; −1 < x < 2
3
f (x) =
0
; elsewhere
Then
(i) Verify that f (x) is a density function.
(ii) Find P(0 < X ≤ 1)
(iii) find F (x), and use it to evaluate P(0 < X ≤ 1).
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Dr. Santosh Yadav, LPU Punjab
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Example 3
The diameter of an electric cable X is a continuous random
variable with PDF
kx(1 − x) ; 0 < x < 1
f (x) =
0
; elsewhere
Find
(i) the constant k .
(ii) the cumulative distribution function (CDF) of X .
Dr. Santosh Yadav, LPU Punjab
18 / 71
Problem 4
Let S be the sample space when a pair of fair dice is rolled.
Let X denote the sum of the numbers on upturned face.
Then
(i) find P(X < 4)
(ii) find P(X = 7)
(iii) find P(X ≥ 4)
Dr. Santosh Yadav, LPU Punjab
19 / 71
Problem5
Q. A shipment of 20 similar laptop computers to a retail
outlet contains 3 that are defective. If a school makes a
random purchase of 2 of these computers, find the
probability distribution for the number of defectives.
Dr. Santosh Yadav, LPU Punjab
20 / 71
Problem5
Q. A shipment of 20 similar laptop computers to a retail
outlet contains 3 that are defective. If a school makes a
random purchase of 2 of these computers, find the
probability distribution for the number of defectives.
Sol: Let X be a random variable denotes the possible
numbers of defective computers purchased by the school.
Then x can only take the numbers 0, 1, and 2.
Dr. Santosh Yadav, LPU Punjab
20 / 71
Problem5
Q. A shipment of 20 similar laptop computers to a retail
outlet contains 3 that are defective. If a school makes a
random purchase of 2 of these computers, find the
probability distribution for the number of defectives.
Sol: Let X be a random variable denotes the possible
numbers of defective computers purchased by the school.
Then x can only take the numbers 0, 1, and 2.
X =x
0
1
2
f (x) = P(X = x) 68/95 91/190 3/190
Dr. Santosh Yadav, LPU Punjab
20 / 71
Dr. Santosh Yadav, LPU Punjab
21 / 71
Problem 6
Let the probability density function of a random variable X is
given as
2f (x1 ) = 3f (x2 ) = f (x3 ) = 5f (x4 )
Then find f (x3 ) and F (x3 )
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BLANK
Dr. Santosh Yadav, LPU Punjab
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Joint Probability Distributions
In the discrete case, joint pdf
f (x, y ) = P(X = x, Y = y ) orP(X = x ∩ Y = y )
The values f (x, y) give the probability that outcomes x and y
occur at the same time.
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Dr. Santosh Yadav, LPU Punjab
25 / 71
Discrete Probability Functions
The function f (x, y ) is a joint probability distribution of the
discrete random variables X and Y if
(i) f (x, y ) ≥ 0, for all (x, y)
PP
(ii)
f (x, y ) = 1
x
y
(iii) P(X = x, Y = y ) = f (x, y )
P
For any region A in the xy plane, P[(X , Y ) ∈ A] = f (x, y ).
A
Dr. Santosh Yadav, LPU Punjab
26 / 71
Continuous Probability Functions
The function f (x, y) is a joint density function of the continuous random variables X and Y if
(i) f (x, y ) ≥ 0, for all (x, y ).
R∞ R∞
(ii) −∞ −∞ f (x, y )dx dy = 1
RR
(iii) P[(X , Y ) ∈ A] =
f (x, y )dx dy , for any region A in the xy
A
plane.
Dr. Santosh Yadav, LPU Punjab
27 / 71
Marginal Density for X and Y
Marginal density of X and Y are
X
fX (x) =
f (x, y )
y
fY (x) =
X
f (x, y )
x
in case of discrete random variable.
And,
Z
fX (x) =
f (x, y ) dy
y
Z
fY (x) =
f (x, y ) dx
x
in case of continuous random variable.
Dr. Santosh Yadav, LPU Punjab
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Example 1
In tossing of two fair coins Let X denote the number of
heads and Y denote the number of tails. Find joint
probability distribution of (X, Y)
Values of random variable X and Y:
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Cont..
Joint probability distribution of X and Y:
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Example 2
In rolling a pair of dice. Let X denotes the maximum of the
numbers and Y denote the sum of numbers. Find joint
probability distribution of (X, Y) as well as marginal of X and
Y.
Values of random variables X and Y:
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Cont..
Joint probability distribution of X and Y:
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Example 3 ( Ex 3.14)
Two ballpoint pens are selected at random from a box that
contains 3 blue pens, 2 red pens, and 3 green pens. If X is
the number of blue pens selected and Y is the number of
red pens selected, find the joint probability density function f
(x, y )
(Try it!)
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Example 4 (Ex 3.15)
A privately owned business operates both a drive-in facility
and a walk-in facility. On a randomly selected day, let X and
Y , respectively, be the proportions of the time that the drivein and the walk-in facilities are in use, and suppose that the
joint density function of these random variables is
f (x, y ) =
2
(2x
5
+ 3y ) ; 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
0
; elsewhere
(a) Verify that f(x,y) is pdf.
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Example 4 (Ex 3.15)
A privately owned business operates both a drive-in facility
and a walk-in facility. On a randomly selected day, let X and
Y , respectively, be the proportions of the time that the drivein and the walk-in facilities are in use, and suppose that the
joint density function of these random variables is
f (x, y ) =
2
(2x
5
+ 3y ) ; 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
0
; elsewhere
(a) Verify that f(x,y) is pdf.
(b) Find P[(X , Y ) ∈ A], A = {(x, y )|0 < x < 12 , 14 < y < 12 }.
(c) Find marginal of X and Y.
Dr. Santosh Yadav, LPU Punjab
34 / 71
13
(b) Ans: 160
(c) The marginal density of X and Y is given as
Dr. Santosh Yadav, LPU Punjab
35 / 71
Dr. Santosh Yadav, LPU Punjab
36 / 71
Expectation of X or Mean of X
Let X be a discrete random variable with pdf f (x). Then, the
expectation of X is denoted by E(X ), and defined as
E(X ) =
X
xf (x)
∀x
i.e.,
E(X ) =
X
(value of x)(Probability)
∀x
Dr. Santosh Yadav, LPU Punjab
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Expectation of X or Mean of X
Let X be a discrete random variable with pdf f (x). Then, the
expectation of X is denoted by E(X ), and defined as
E(X ) =
X
xf (x)
∀x
i.e.,
E(X ) =
X
(value of x)(Probability)
∀x
More generally, if H(X ) is function of the random variable X,
then we define
X
E[H(X )] =
H(x)f (x).
∀x
Dr. Santosh Yadav, LPU Punjab
37 / 71
Cont...
Let X be a continuous random variable with pdf f (x). Then,
the expectation or mean of X is defined as
Z
∞
E(X ) =
xf (x) dx
−∞
More generally,
Z
∞
E[H(X )] =
H(x)f (x) dx.
−∞
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Examples
Ex1 Let X denotes the number of heads in a toss of two fair coins.
Then find the expectation of variable X .
Ans: E(X ) = 1
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Examples
Ex1 Let X denotes the number of heads in a toss of two fair coins.
Then find the expectation of variable X .
Ans: E(X ) = 1
Ex2 Let X denotes the number of heads in the experiment of
tossing three fair coins simultaneously. Find expected value
of the variable X .
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Examples
Ex1 Let X denotes the number of heads in a toss of two fair coins.
Then find the expectation of variable X .
Ans: E(X ) = 1
Ex2 Let X denotes the number of heads in the experiment of
tossing three fair coins simultaneously. Find expected value
of the variable X .
Ex3 In rolling a die, let X denotes the number on the die, then find
expected value of X.
Ans: E(X ) = 3.5
Dr. Santosh Yadav, LPU Punjab
39 / 71
Dr. Santosh Yadav, LPU Punjab
40 / 71
Problem 3
Let a random variable has the following probability
distribution
X =x
4
5
6
7
8
f (x) = P(X = x) 1/12 1/12 1/4 1/4 1/6
9
1/6
Find the expectation of g(X ) = 2X − 1.
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41 / 71
Problem 3
Let a random variable has the following probability
distribution
X =x
4
5
6
7
8
f (x) = P(X = x) 1/12 1/12 1/4 1/4 1/6
9
1/6
Find the expectation of g(X ) = 2X − 1.
Ans: 12.67
Dr. Santosh Yadav, LPU Punjab
41 / 71
Dr. Santosh Yadav, LPU Punjab
42 / 71
Rules for Expectation
If X and Y are random variables and c, d are constants,
then it is easy to verify the following:
(i)
(ii)
(iii)
(iv)
E(c) = c
E(cX ) = cE(X )
E(cX + d) = cE(X ) + d
E(X + Y ) = E(X ) + E(Y )
The last rule can be generalized for n random variables.
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Problem 2
Let X and Y are two random variable such that E(X ) = 7
and E(Y ) = −5, then find E(4X − 2Y + 6).
Ans: 44
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Problem 3
The probability density function of a random variable X is
given by
( 2
x
, −1 < x < 2
f (x) = 3
0, otherwise
find expected value of 4X + 3
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45 / 71
Problem 3
The probability density function of a random variable X is
given by
( 2
x
, −1 < x < 2
f (x) = 3
0, otherwise
find expected value of 4X + 3 Ans: 8
Dr. Santosh Yadav, LPU Punjab
45 / 71
Problem 4
The probability density function of a random variable X is
given by
(
x
(4 − x 2 ), 0 ≤ x ≤ 2
f (x) = 4
0,
otherwise
find the mean µX of random variable X
Dr. Santosh Yadav, LPU Punjab
46 / 71
Problem 4
The probability density function of a random variable X is
given by
(
x
(4 − x 2 ), 0 ≤ x ≤ 2
f (x) = 4
0,
otherwise
find the mean µX of random variable X
(A)
(B)
(C)
37
30
16
15
24
15
Dr. Santosh Yadav, LPU Punjab
46 / 71
Dr. Santosh Yadav, LPU Punjab
47 / 71
Problem 5
The probability density function of a random variable X is
given by
(
3
(3x − x 2 ), 0 ≤ x ≤ 2
f (x) = 10
0,
otherwise
(i) then the mean of random variable is
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Problem 5
The probability density function of a random variable X is
given by
(
3
(3x − x 2 ), 0 ≤ x ≤ 2
f (x) = 10
0,
otherwise
(i) then the mean of random variable is
(A) 1
(B) 56
(C) 65
(D) None of these
Dr. Santosh Yadav, LPU Punjab
48 / 71
Variance of X
Let X and Y be two random variables assuming the values
X = 1, 9 and Y = 4, 6. We observe that both the variables
have the same mean value, µX = 5 = µY . You can see
variability of X and Y .
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Variance of X
Let X and Y be two random variables assuming the values
X = 1, 9 and Y = 4, 6. We observe that both the variables
have the same mean value, µX = 5 = µY . You can see
variability of X and Y .
Thus, the mean value of a random variable does not
account for its variability. In this regard, we define a new
parameter known as variance.
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This slide is left blank
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Variance of X : (σx2)
If X is a random variable with mean µ, then its variance,
denoted by σ 2 (or Var (X )) and defined as
σ 2 = E(X − µ)2
That is, variance measures variability by considering the
difference between the variable and its mean, i.e., X − µ.
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Variance of X : (σx2)
If X is a random variable with mean µ, then its variance,
denoted by σ 2 (or Var (X )) and defined as
σ 2 = E(X − µ)2
That is, variance measures variability by considering the
difference between the variable and its mean, i.e., X − µ.
The difference is squared so that the negative values will
not cancel the positive ones in the process of finding
expected value.
Dr. Santosh Yadav, LPU Punjab
51 / 71
Computation Formula for σ 2
The variance can be calculated by a more simple formula.
σ 2 = Var (X ) = E(X 2 ) − [E(X )]2
Proof:
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Examples
Ex1 Let X denotes the number of heads in a toss of two fair
coins. Find the variance of X .
Ans: Var(X)= 1/2
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53 / 71
Examples
Ex1 Let X denotes the number of heads in a toss of two fair
coins. Find the variance of X .
Ans: Var(X)= 1/2
Ex2 Let X denotes the number of heads in the experiment of
tossing three fair coins simultaneously. Find variance of X .
Dr. Santosh Yadav, LPU Punjab
53 / 71
Examples
Ex1 Let X denotes the number of heads in a toss of two fair
coins. Find the variance of X .
Ans: Var(X)= 1/2
Ex2 Let X denotes the number of heads in the experiment of
tossing three fair coins simultaneously. Find variance of X .
Ex3 In rolling a die, let X denotes the number on the die. Find
variance of X .
Dr. Santosh Yadav, LPU Punjab
53 / 71
Rules for Variance
If X and Y are two random variables and c is constants,
then it is easy to verify the following:
(i) Var (cX + d) = c 2 Var (X )
1
Random variable X and Y are independent if the knowledge of values
assumed by X do not give any clue to the values assumed by Y .
Dr. Santosh Yadav, LPU Punjab
54 / 71
Rules for Variance
If X and Y are two random variables and c is constants,
then it is easy to verify the following:
(i) Var (cX + d) = c 2 Var (X )
If X and Y are independent1 , then
Var (X + Y ) = Var (X ) + Var (Y )
1
Random variable X and Y are independent if the knowledge of values
assumed by X do not give any clue to the values assumed by Y .
Dr. Santosh Yadav, LPU Punjab
54 / 71
Problems
Q1 Let X be
( a random variable with pdf
x
, 0≤x ≤2
f (x) = 2
0, otherwise
Find variance of X.
Ans: 92
Dr. Santosh Yadav, LPU Punjab
55 / 71
Problems
Q1 Let X be
( a random variable with pdf
x
, 0≤x ≤2
f (x) = 2
0, otherwise
Find variance of X.
Ans: 92
Q2 Let X and Y are two independent random variables such
that σx2 = 9 and σY2 = 3, then find the variance of
(4X − 2Y + 6).
Dr. Santosh Yadav, LPU Punjab
55 / 71
Problems
Q1 Let X be
( a random variable with pdf
x
, 0≤x ≤2
f (x) = 2
0, otherwise
Find variance of X.
Ans: 92
Q2 Let X and Y are two independent random variables such
that σx2 = 9 and σY2 = 3, then find the variance of
(4X − 2Y + 6).
Choose the correct option from following:
(A)
(B)
(C)
(D)
150
146
156
None of these
Dr. Santosh Yadav, LPU Punjab
55 / 71
Standard Deviation: (σ)
As variance carries squared units of the original data, hence
it is a pure number often without any physical meaning. To
overcome this problem, a second measure of variability is
employed known as standard deviation.
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56 / 71
Standard Deviation: (σ)
As variance carries squared units of the original data, hence
it is a pure number often without any physical meaning. To
overcome this problem, a second measure of variability is
employed known as standard deviation.
Let X be a random variable with variance σ 2 . Then the
standard deviation of X denoted by σ is the the non-negative
square root of X , that is,
σ=
p
Var (X )
SD always reported in physical measurement unit that matches
the original data, where variance is unitless.
Dr. Santosh Yadav, LPU Punjab
56 / 71
Dr. Santosh Yadav, LPU Punjab
57 / 71
Covariance of X, Y
Cov (X , Y ) = E(XY ) − E(X )E(Y )
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Rules on Covariance
Cov (aX , bY ) =?
Cov (X + a, Y + b) =?
If X and Y are independent RVs then Cov (X , Y ) =?
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Chebychev’s Theorem
Statement?
Types of Problems
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Dr. Santosh Yadav, LPU Punjab
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Problem 1
Let a fair coins is tossed three times. Let X equal to 0 or 1
accordig as a head or a tail occurs on the first toss and let
Y is the total number of heads that occurs. Then
(i) the value of E(X + Y ) is
(A) 3
(B) 2
(C) None of these
Dr. Santosh Yadav, LPU Punjab
62 / 71
Problem 1
Let a fair coins is tossed three times. Let X equal to 0 or 1
accordig as a head or a tail occurs on the first toss and let
Y is the total number of heads that occurs. Then
(i) the value of E(X + Y ) is
(A) 3
(B) 2
(C) None of these
(ii) the values of Var (X ) and Var (Y ) are respectively,
(A) 41 and 34 ,
(B) 34 and 13
(C) 34 and 14
(D) None of above
Dr. Santosh Yadav, LPU Punjab
62 / 71
Problem 2
Let X be is a continuous random variable with distribution
kx ; 0 ≤ x ≤ 5
f (x) =
0
; elsewhere
Find
(i) the value of k
(ii) P(1 ≤ X ≤ 3)
(ii) P(X ≤ 3)
Dr. Santosh Yadav, LPU Punjab
63 / 71
(i) The value of constant is 2/25
(ii) value of P(1 ≤ X ≤ 3) is
(A)
(B)
4
25
8
25
Dr. Santosh Yadav, LPU Punjab
64 / 71
(i) The value of constant is 2/25
(ii) value of P(1 ≤ X ≤ 3) is
(A)
(B)
4
25
8
25
(iii) value of P(X ≤ 3) is 9/25
Dr. Santosh Yadav, LPU Punjab
64 / 71
Problem 1
The probability density function of a random variable X is
given by
(
x
(4 − x 2 ), 0 ≤ x ≤ 2
f (x) = 4
0,
otherwise
Find the variance of random variable.
Ans: 0.195
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65 / 71
Problem 2
Let X continuous random variable X with pdf
1
; −2 < x < 2
4
f (x) =
0
; elsewhere
(i) the value of P(X < 1) is
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66 / 71
Problem 2
Let X continuous random variable X with pdf
1
; −2 < x < 2
4
f (x) =
0
; elsewhere
(i) the value of P(X < 1) is
(A)
(B)
1
4
3
4
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Problem 2
Let X continuous random variable X with pdf
1
; −2 < x < 2
4
f (x) =
0
; elsewhere
(i) the value of P(X < 1) is
(A)
(B)
1
4
3
4
(ii) the value of P(2x + 3 > 5) is
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66 / 71
Problem 2
Let X continuous random variable X with pdf
1
; −2 < x < 2
4
f (x) =
0
; elsewhere
(i) the value of P(X < 1) is
(A)
(B)
1
4
3
4
(ii) the value of P(2x + 3 > 5) is
(A) 34
(B) 14
(C) None of these
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66 / 71
Problem 3
The probability density function of a continuous random
variable X is given as

ax,
0≤x <1



a,
1≤x <2
f (x) =

−ax + 3a, 2 ≤ x < 3



0,
x ≥3
(i) Find the value of constant a
(ii) Find cummulative
 distribution F (x)

0,
x ≤0



x2


0<x ≤1
4,
Ans: F (x) = 2x−1
,
1<x ≤2
4


−x 2 −6x−5

, 2<x ≤3

4


1,
x ≥3
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This slide is intensionally left blank
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Dr. Santosh Yadav, LPU Punjab
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Problem 5
The probability density function of a random variable X is
given by
f (x) = 6x (1 − x), 0 ≤ X ≤ 1
Then, find P(X ≤
Dr. Santosh Yadav, LPU Punjab
1
2
| 13 ≤ X ≤ 23 ).
70 / 71
Problem 5
The probability density function of a random variable X is
given by
f (x) = 6x (1 − x), 0 ≤ X ≤ 1
Then, find P(X ≤
11
Ans: = 26
Dr. Santosh Yadav, LPU Punjab
1
2
| 13 ≤ X ≤ 23 ).
70 / 71
NOTE: This PPT is Just for quick revision for topic and practice
problems, hence not exhaustive. Refer the text book for more
problems.
THANK YOU
Dr. Santosh Yadav, LPU Punjab
71 / 71
Probability & Statistics
Dr. Santosh Kumar Yadav
Assistant Professor
Department of Mathematics
Lovely Professional University, Punjab.
Dr. S. K. Yadav, LPU Punjab
1 / 27
Some Probability
Distributions
In this Lecture series we will discuss about some continuous
probability distribution functions.
Dr. S. K. Yadav, LPU Punjab
2 / 27
Normal Distribution
The most important continuous probability distribution in the
entire field of statistics is the normal distribution.
Graph of normal pdf is called the normal curve, and is the
bell-shaped curve.
Dr. S. K. Yadav, LPU Punjab
3 / 27
Normal Distribution
The most important continuous probability distribution in the
entire field of statistics is the normal distribution.
Graph of normal pdf is called the normal curve, and is the
bell-shaped curve.
The normal distribution is often referred to as the Gaussian
distribution, in honor of Karl Friedrich Gauss(1777–1855),
who also derived its equation from a study of errors in repeated measurements of the same quantity.
Dr. S. K. Yadav, LPU Punjab
3 / 27
Normal Distribution
A continuous random variable X follows normal distribution if its
probability density function (pdf) is given by
(x−µ)2
1
f (x, µ, σ) = √ e− 2σ2 ,
σ 2π
x, µ ∈ R, σ > 0
where mean of X is mu and variance is σ 2 .
Notation: If X follows normal distribution , we generally
write X ∼ N(µ, σ 2 ).
Dr. S. K. Yadav, LPU Punjab
4 / 27
Properties of Normal Curve
The curve is symmetric about a vertical axis through the
mean µ.
The normal curve approaches the horizontal axis asymptotically as we proceed in either direction away from the mean.
The total area under the curve and above the horizontal axis
is equal to 1.
Dr. S. K. Yadav, LPU Punjab
5 / 27
Areas under the Normal Curve
The area under the curve bounded by the two ordinates x =
x1 and x = x2 equals the probability that the random variable
X assumes a value between x = x1 and x = x2 .
Z x2
P(x1 < X < x2) =
f (x, µ, σ 2 )dx
x1
We change this in terms of a new variable Z (called standard
normal variate). There area of normal curve for values of Z
is given in standard normal table (TABLE A.3. PAGE 736 in
textbook)
For example, P(Z < −1.4) = 0.0808 (check it!) gives the
entire area of normal curve which lies left side to Z = −1.4.
Dr. S. K. Yadav, LPU Punjab
6 / 27
Dr. S. K. Yadav, LPU Punjab
7 / 27
Standard Normal Variate
The difficulty encountered in solving integrals of normal density functions. Fortunately, we are able to transform all the
observations of any normal random variable X into a new set
of observations of a normal random variable Z defined as
Z =
X −µ
σ
with mean 0 and variance 1.
It is easy to verify that mean and variance of Z are 0 and 1.
(Do it), i.e Z ∼ N(0, 1) when X ∼ N(µ, σ 2 ).
Dr. S. K. Yadav, LPU Punjab
8 / 27
Area Property (1σ, 2σ, 3σ range)
P(µ − σ < X < µ + σ) = P(−1 < Z < 1) = 0.6826
P(µ − 2σ < X < µ + 2σ) = P(−2 < Z < 2) = 0.9544
P(µ − 3σ < X < µ + 3σ) = P(−3 < Z < 3) = 0.9973
Dr. S. K. Yadav, LPU Punjab
9 / 27
Example 1
Let X is normally distributed where mean of X is 12 and SD is 4.
Then find
(i) P(X ≥ 20)
(ii) P(0 ≤ X ≤ 12)
(i) Sol: Clearly Z = 2, Hence
Dr. S. K. Yadav, LPU Punjab
10 / 27
Example 1
Let X is normally distributed where mean of X is 12 and SD is 4.
Then find
(i) P(X ≥ 20)
(ii) P(0 ≤ X ≤ 12)
(i) Sol: Clearly Z = 2, Hence
P(X ≥ 20) = P(Z ≥ 2) =?
Dr. S. K. Yadav, LPU Punjab
10 / 27
Cont...
Let X is normally distributed and mean of X is 12 and SD is 4.
Then find
P(X ≥ 20)
P(0 ≤ X ≤ 12)
(ii) Sol: P(0 ≤ X ≤ 12) = P(−3 ≤ Z ≤ 0) =?
Dr. S. K. Yadav, LPU Punjab
11 / 27
Example 2
A certain type of storage battery lasts, on average, 3.0 years
with a standard deviation of 0.5 year. Assuming that battery
life is normally distributed, find the probability that a given
battery will last less than 2.3 years.
(Given that P(Z < −1.4) = 0.0808)
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12 / 27
Dr. S. K. Yadav, LPU Punjab
13 / 27
Problem 3
An electrical firm manufactures light bulbs that have a life,
before burn-out, that is normally distributed with mean equal
to 800 hours and a standard deviation of 40 hours. Find the
probability that a bulb burns between 778 and 834 hours.
Sol: P(778 < X < 834) = P(−0.55 < Z < 0.85) =?
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14 / 27
Example 4
Given a random variable X having a normal distribution with
mean 50 and S.D 10, find the probability that X assumes a
value between 45 and 62.
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15 / 27
Example 4
Given a random variable X having a normal distribution with
mean 50 and S.D 10, find the probability that X assumes a
value between 45 and 62.
Sol: P(45 < X < 62) = P(−0.5 < Z < 1.2) =?.
Dr. S. K. Yadav, LPU Punjab
15 / 27
Gamma Distribution
The continuous random variable X has a gamma distribution,
with parameters α and β, if its density function is given by
f (x, α, β) =
βα
x
1
e− β x α−1 , x > 0, α > 0, β > 0
Γ(α)
where X be the elapsed time before the αth event.
β is the average time between occurences, called rate parameter.
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16 / 27
Gamma Function and Properties
R∞
1
Gamma function is defined as Γ(α) =
2
Γ(α) = (α − 1)Γ(α − 1) for α > 1.
Γ(n) = (n − 1)! for a positive integer n.
Γ(1) = 1.
√
Γ( 12 ) = π
3
4
5
6
0
e−z z α−1 dz
Find Γ(5) and Γ( 23 )?
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17 / 27
Exponential Distribution
When α = 1, we have the exponential distribution which
models the waiting time between occrences in a Poisson
process.
The continuous random variable X has an exponential distribution with density function given by
f (x, β) =
Dr. S. K. Yadav, LPU Punjab
1 − βx
e , x > 0, β > 0
β
18 / 27
Mean and Variance
The mean and variance of the gamma distribution are
µ = αβ, σ 2 = αβ 2
The mean and variance of the exponential distribution are
µ = β, σ 2 = β 2
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19 / 27
Applications
The gamma distribution used to model the time that elapses
before α occurences of a randomly occuring event, like calls
to a pizza place. Such events are said to occur according to
a Poisson process.
Time between arrivals at service facilities and time to failure
of component parts and electrical systems are modeled by
the exponential distribution.
Consider a poisson process with parameter λ. Let X
denote the time of occurence of first event, then X has
an exponential distribution with parameter β = 1/λ.
Dr. S. K. Yadav, LPU Punjab
20 / 27
Applications
The gamma distribution used to model the time that elapses
before α occurences of a randomly occuring event, like calls
to a pizza place. Such events are said to occur according to
a Poisson process.
Time between arrivals at service facilities and time to failure
of component parts and electrical systems are modeled by
the exponential distribution.
Consider a poisson process with parameter λ. Let X
denote the time of occurence of first event, then X has
an exponential distribution with parameter β = 1/λ.
The exponential distribution is concerned with the amount
of time until some specific event occurs. For example, the
amount of time until an earthquake occurs.
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20 / 27
Example 1
Suppose that telephone calls arriving at a particular switchboard follow a Poisson process with an average of 5 calls
coming per minute. What is the probability that up to a
minute will elapse by the time 2 calls have come in to the
switchboard?
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21 / 27
Example 1
Suppose that telephone calls arriving at a particular switchboard follow a Poisson process with an average of 5 calls
coming per minute. What is the probability that up to a
minute will elapse by the time 2 calls have come in to the
switchboard?
Solution : The Poisson process applies, with time until 2
Poisson events following a gamma distribution with β = 1/5
and α = 2. The required probability is given by
Z 1
1 −x/β
P(X ≤ 1) =
xe
dx
2
0 β
Z 1
= 25
xe−5x = 1 − e−5 (1 + 5)
0
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21 / 27
Example 2
The time in hours to repair a machine is exponential distribution with parameter 1/3. What is the probability that repair
time exceeds 3 hours.
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22 / 27
Example 2
The time in hours to repair a machine is exponential distribution with parameter 1/3. What is the probability that repair
time exceeds 3 hours.
Ans: P(X > 3) = e−9
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22 / 27
Example 3
On a saturday morning, customers arrive at a bakery according to a posson process at an average rate of 15 customers
per hour.
(i) What is the probability that it takes less than 10 minutes
for first 3 customer to arrive.
Sol: Given that λ = 15/60 = 1/4 minutes, hence β = 1/λ =
4. Find P(X < 10).
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23 / 27
Example 3
On a saturday morning, customers arrive at a bakery according to a posson process at an average rate of 15 customers
per hour.
(i) What is the probability that it takes less than 10 minutes
for first 3 customer to arrive.
Sol: Given that λ = 15/60 = 1/4 minutes, hence β = 1/λ =
4. Find P(X < 10).
Ans: (i) 0.456
(ii) What is average amount of time in minutes that will elapse
before 3 customers arrive in a bakery.
Dr. S. K. Yadav, LPU Punjab
23 / 27
Example 3
On a saturday morning, customers arrive at a bakery according to a posson process at an average rate of 15 customers
per hour.
(i) What is the probability that it takes less than 10 minutes
for first 3 customer to arrive.
Sol: Given that λ = 15/60 = 1/4 minutes, hence β = 1/λ =
4. Find P(X < 10).
Ans: (i) 0.456
(ii) What is average amount of time in minutes that will elapse
before 3 customers arrive in a bakery.
Ans: (ii) 12min
Dr. S. K. Yadav, LPU Punjab
23 / 27
Review Problems
(4) The length of time a person speaks over phone follows exponential distribution with mean 6. Find the probability that
the person will talk for (i) more than 8 minutes, (ii) between
4 and 8 minutes.
Dr. S. K. Yadav, LPU Punjab
24 / 27
Review Problems
(4) The length of time a person speaks over phone follows exponential distribution with mean 6. Find the probability that
the person will talk for (i) more than 8 minutes, (ii) between
4 and 8 minutes.
Ans: (i) P(X > 8) = e−4/3 , (ii) P(4 < X < 8) = e−2/3 − e−4/3
Dr. S. K. Yadav, LPU Punjab
24 / 27
Cont..
(5) A certain type of device has an advertised fail- ure rate of
0.01 per hour. The failure rate is constant and the exponential distribution applies.
(a) What is the mean time to failure?
(b) What is the probability that 200 hours will pass before a
failure is observed?
Dr. S. K. Yadav, LPU Punjab
25 / 27
Cont...
(6) Given a normal distribution with µ = 40 and σ = 6, find the
value of x that has
(a) 45% of the area to the left and
(b) 14% of the area to the right.
Sol: (a) We require a z value that leaves an area of 0.45
to the left. From Table A.3 we find P(Z < −0.13) = 0.45, so
the desired z value is −0.13. Hence,
x = (6)(−0.13) + 40 = 39.22.
(b) we require a z value that leaves 0.14 of the area to the
right and hence an area of 0.86 to the left. Again, from
Table A.3, we find P(Z < 1.08) = 0.86, so the desired z
value is 1.08 and
x = (6)(1.08) + 40 = 46.48.
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26 / 27
THANK YOU
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27 / 27