Theory of probability

Dr. Brajesh Kumar Jha Page 1
Theory of Probability
By
Dr. Brajesh Kumar Jha
Department of Mathematics
School of Technology
Pandit Deendayal Petroleum University

L1-Theory of Probability
Introduction: If an experiment is repeated under essentially
homogeneous and similar conditions, we generally come
across two types of situations:
(a) The result or what is usually known as the ‘outcome’ is
unique or certain. The phenomena covered under this
trial are known as ‘deterministic’ or ‘predictable’
phenomena. By a deterministic phenomenon we mean
one in which the result can be predicted with certainty.

Example:
(i) For a perfect gas,
1
V
P
∝ , i.e., PV=constant, where V
is the volume and P is the pressure of the gas, provided
the temperature remains the same.
(ii) The velocity ‘v’ of a particle after time t is given by
v u at= + where u is the initial velocity and ‘a’ is the
acceleration. The equation uniquely determines v if the
right hand quantities are known.

(b) The result is not unique but may be one of the several
possible outcomes. The phenomena are ‘unpredictable’
or ‘probabilistic’.
(i) In a random toss of a uniform coin we are not sure of
getting the head or tail.
(ii) A manufacturer cannot ascertain the future demand of
his product with certainty.
(iii) A sales manager cannot predict with certainty about
the sales target next year.
(iv) If an electric tube has lasted for one year, nothing can
be predicted about its future life.

Probability is also used informally in day to day life. We
daily come across the sentences like:
(i) Possibly, it will rain today.
(ii) There is a high chance of my getting the job next
month.
(iii) This year’s demand for the product is likely to exceed
that of the last year’s.
(iv) The odds are 3:2 in favour of getting the contract
applied for.
“Probability is the science of decision making with
calculated risks in the face of uncertainty”.

Basic Terminology
1. Random Experiment: If in each trial of an experiment
conducted under identical conditions, the outcome is
not unique, but may be any one of the possible
outcomes, then such an experiment is called a random
experiment.
Example: tossing a coin, throwing a die, selecting a card
from a pack of playing cards, selecting a family out of a
given group of families, etc.
A pack of cards consists of four suits called spades,
Hearts, Diamonds and Clubs. Each suits consists of 13
cards, of which nine cards are numbered from 2 to 10, an

ace, king, a queen, and a jack. Spades and clubs are black
faced cards, while hearts and diamonds are red faced cards.
2. Outcome: The result of a random experiment will be
called an outcome.
3. Trial and Event: Any particular performance of a
random experiment is called a trial and outcome or
combination of outcomes are termed as events.
Example:
(i) If a coin is tossed repeatedly, the result is not unique.
We may get any of the two faces, head or tail. Thus
tossing of a coin is a random experiment or trial and
getting of a head or tail is event.

(ii) In an experiment which consists of the throw of a six
faced die and observing the number of points that
appear, the possible outcomes are 1,2,3,4,5,6.
4. Exhaustive Events or Cases: The total number of
possible outcomes of a random experiment is known as
the exhaustive events or cases.
Example:
(i) In tossing of a coin, there are two exhaustive cases,
viz., head and tail (the possibility of the coin standing
on an edge being ignored).
(ii) In throwing of a die, there are 6 exhaustive cases since
any one of the 6 faces 1,2,…6 may come uppermost.

(iii) In drawing two cards from a pack of cards, the
exhaustive number of cases is 52
C2, since 2 cards can
be drawn out of 52 cards in 52
C2 ways.
(iv) In throwing of two dice, the exhaustive number of
cases is 2
6 36= , since any of the numbers 1 to 6 on the
first die can be associated with any of the 6 numbers
on the other die. In general, in throwing n dice, the
exhaustive number of cases is 6n
.

5. Favourable Events of Cases: The number of cases
favourable to an event in a trial is the number of
outcomes which entail the happening of the event.
Example:
(i) In drawing a card from a pack of cards the number of
cases favourable to drawing of an ace is 4, for drawing
a spade is 13 and for drawing a red card is 26.
(ii) In throwing of two dice, the number of cases
favourable to getting the sum 5 is:
(1,4),(4,1),(2,3),(3,2), i.e., 4

6. Mutually Exclusive Events: Events are said to be
mutually exclusive or incompatible if the happening of
any one of them precludes the happening of all the
others, i.e., if no two or more of them can happen
simultaneously in the same trial. For example
(i) In throwing a die all the 6 faces numbered 1 to 6 are
mutually exclusive since if any one of these faces
comes, the possibility of others, in the same trial, is
ruled out.

(ii) Similarly in tossing a coin the events head and tail are
mutually exclusive.

7. Equally likely Events: Outcomes of trial are said to be
equally likely if taking into consideration all the
relevant evidences, there is no reason to expect one in
preference to the others. For example,
(i) In a random toss of an unbiased or uniform coin, head
and tail are equally likely events.
(ii) In throwing an unbiased die, all the six faces are
equally likely to come.

8. Independent Events: several events are said to be
independent if the happening (or non-happening) of an
event is not affected by the supplementary knowledge
concerning the occurrence of any number of the
remaining events. For Example
(i) In tossing an unbiased coin, the event of getting a head
in the first toss is independent of getting a head in the
second, third and subsequent throws.

(ii) When a die is thrown twice, the result of the first
throw is independent of getting a head in the second
throw.
(iii) If we draw a card from a pack of well-shuffled cards
and replace it before drawing the second card, the
result of the second draw is independent of the first
draw. But, however, if the first card drawn is not
replaced then the second draw is dependent on the first
draw.

Mathematical Probability
Definition: If a random experiment or a trial results in ‘n’
exhaustive, mutually exclusive and equally likely outcomes
(or cases), out of which m are favourable to the occurrence of
an event E, then the probability ‘p’ of occurrence (or
happening) of E, usually denoted by P(E), is given by:
n
m
casesexhausticeofnumberTotal
casesfavourableofNumber
EPp === )(
Remark:
(i) Since 0,0 ≥≥ nm and nm ≤ so

0)( ≥EP and 1)(01)( ≤≤⇒≤ EPEP
(ii) Sometimes m/n is expressed by saying that ‘the odds
in favour of E are m: (n-m) or the odds against E are
(n-m) : m’.
(iii) The non-happening of the event E is called the
complementary event of E and is denoted by c
EorE .
The number of cases favourable to c
EorE , i.e., non
happening of E is (n-m). then the probability q that E
will not happen is given by:

111)( =+⇒−=−=
−
== qpp
n
m
n
mn
EPq
(iv) Probability ‘p’ of the happening of an event is also
known as the probability of success and the probability
‘q’ of the non- happening of the event as the
probability of failure, i.e., (p+q=1)
(v) If 1)( =EP then E is called a certain event and if
0)( =EP , E is called an impossible event.

Q1. What is the chance that a leap year selected at random
will contain 53 Sundays?
Ans: 2/7
Q2. Two unbiased dice are thrown. Find the probability
that:
(i) Both the dice show the same number,
Ans: 1/6.
(ii) The first die shows 6,
Ans: 1/6.

(iii) The total of the numbers on the dice is 8,
Ans: 5/36
(iv) The total of the numbers on the dice is greater than 8,
Ans: 5/18.
(v) The total of the numbers on the dice is 13,
Ans: 0
(vi) The total of the numbers on the dice is any numer from
2 to 12, both inclusive.
Ans: 1.

Q3. Among the digits 1,2,3,4,5 at first one is chosen and
then a second selection is made among the remaining four
digits. Assuming that all twenty possible outcomes have
equal probabilities, find the probability that an odd digit will
be selected
(i) The first time
(ii) The second time
(iii) Both times
Ans: (i) 3/5 (ii) 3/5 (iii) 3/10

Q4. From 25 tickets, marked with first 25 numerals, one is
drawn at random. Find the chance that
(i) It is multiple of 5 or 7,
Ans: 8/25
(ii) It is multiple of 3 or 7,
Ans: 2/5

L2-Permutations and Combinations
Permutations: Given n different things (elements or
objects), we may arrange them in a row in any order. Each
such arrangement is called a permutations of the given
things. For example we have 6 permutations of the three
letters a, b, c, namely, abc, acb, bac, bca, cab, cba,
Theorem1. The number of permutations of n different
things taken all at a time is n!=1.2.3….n
Example 1. If there are 10 different screws in a box that
needed in a certain order for assembling a certain product,

and these screws are drawn at random from the box, the
probability P of picking them in the desired order is very
small, namely
%00003.0
3628800
1
10
1
===P
When not all given things are different, we obtain for their
number of permutations as follows

Theorem2. If n given things can be divided into c classes
such that things belonging to the same class are alike while
things belonging to different classes are different, then the
number of permutations of these things taken all at a time is,
!!........!
!
21 cnnn
n
nnnn c =+++ ........21
Where nj is the number of things in the jth
class.

Example: In how many ways can the letters of
REARRANGE be permuted?
Solution: There are nine letters with three R’s, two A’s and
two E’s. the number of permutations is.
15120
1.2.1.2.1.2.3
1.2.3.4.5.6.7.8.9
!2!2!3
!9
==
Example2: If a box contains 6 red and 4 blue balls, the
probability of drawing first the red and then the blue balls is.
9.10
4.6
=P

Theorem 3: The number of different permutations of n
different things taken k at a time without repetition is
)!(
!
)1).....(2)(1(
kn
n
knnnn
−
=+−−−
(3a)
And with repetitions is nk
. (3b)
Example: In a coded telegram the letters are arranged in
groups of five letters, called words, from (3b) we see that the
number of different such word is 1188137626 5
= . From (3a)
it follows that the number of different such words containing
each letter no more than once is, 7893600
)!526(
!26
=
− .

Combinations: In a permutation, the order of the selected
things is essential. In contrast, a combination of given things
means any selection of one or more things without regard to
order.
There are two kinds of combinations:
(i) The number of combinations of n different things,
taken k at a time without repetitions is the number
of sets that can be made up from the n given things,
each set containing k different things and no two sets
containing exactly the same k things.

(ii) The number of combinations of n different things,
taken k at a time, with repetitions is the number of
sets that can be made up of k things chosen from the
given n things, each being used as often as desired.
For Example: There are three combinations of the three
letters a, b, c, taken two letters at a time, without
repetitions, are ab, ac, bc, and six such combinations with
repetitions, are ab, ac, bc, aa, bb, cc.

Theorem 4: The number of different combinations of n
different things, k at a time, without repetition is
k
knnnn
knk
n
Ckn
......2.1
)1).....(2)(1(
)!(!
! +−−−
=
−
=
(4a)
The number of those combinations with repetitions is kkn C1−+
Ex 5: The number of samples of five light bulbs that can be
selected from a lot of 500 bulbs is
002552446876
5.4.3.2.1
496.497.498.499.500
!495!5
!500
5500 ===C

Theorem 5: knkn Ck
kn
n
P !
)!(
!
=
−
=
Or !k
P
C kn
kn =
Example 6: A student has seven books on his desk. In how
many different ways can he select a set of three?
Solution: Since the order is not important, this is a
combination problem:
35
1.2.3
5.6.7
!3
37
37 ===
P
C

Example 7. In how many ways can a committee of four be
selected from a group of ten people?
Solution: 210
!4
410
410 ==
P
C

Example 8. Four cards are drawn at random from a pack of
52 cards. Find the probability that
(i) They are a king, a queen, a jack and a ace.
(ii) Two are kings and two are queens.
(iii) Two are black and two are red.
(iv) There are two cards of hearts and two cards of
diamonds.
(8b) In shuffling a pack of cards, four are accidentally
dropped, find the chance that the missing cards should be
one from each suit.

Solution: Four cards can be drawn from a well-shuffled
pack of 52 cards in 52
C4 ways, which gives the exhaustive
number of cases.
(i) 1 king can be drawn out of the 4 king in 4
C1 ways.
Similarly, 1 queen, 1 jack and an ace can each be
drawn in 4
C1 = 4 ways. Since any one of the ways of
drawing a king can be associated with any one of the
ways of drawing a queen, a jack and an ace, the
favourable number of cases are 4
C1x4
C1x4
C1x4
C1.

Hence the required probability = 4
52
1
4
1
4
1
4
1
4
C
CCCC ×××
= 256/52
C4
(ii) Required probability = 4
52
2
4
2
4
C
CC ×
(iii) Since there are 26 black cards (of spades and clubs)
and 26 red cards (of diamonds and hearts) in a pack of
cards, the required probability = 4
52
2
26
2
26
C
CC ×
.
(iv) Required probability = 4
52
2
13
2
13
C
CC ×
(8b) Required probability = 4
52
1
13
1
13
1
13
1
13
C
CCCC ×××

Theorems on Probability
Theorem 1 If A is any event defined on finite sample space
U, then ( ) ( )APAP −= 1'
Where A’ is the complementary event of A.
Theorem 2 ( ) 10 ≤≤ AP
Theorem 3(a): Addition theorem of probability or
theorem of Total Probability.
If A and B are two events which are not disjoint defined on
a finite sample space U, then ( ) ( ) ( ) ( )BAPBPAPBAP IU −+=

Remark1. If A and B are mutually exclusive events (i.e.
disjoint events) then Φ=BAI
2. If A and B are mutually exclusive and exhaustive events,
then UBA =U and Φ=BAI .
Theorem 3(b): If A, B, C are three events defined on finite
sample space U, then
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )CBAPACPCBPBAPCPBPAPCBAP IIIIIUU −+−−−++=

Example1. Two people are selected at random from a group
of seven men and five women. Find the probability that both
men or both are women.
Ans: 31/66
Example2. A card is drawn at random from a pack of 52
cards. What is the probability that the card is a spade or a
king?
Ans: 4/13

Example 3. Two unbiased dice are tossed simultaneously.
Find the probability that sum of numbers on the upper face of
dice is 9 or 12.
Ans: 5/36

Conditional Probability
Let A and B be two events of finite sample space U. The
probability that B occurred given that A has occurred is
denoted by P(B/A) and is called conditional probability.
Theorem 1 If A and B are any two events of a finite sample
space U, then show that
( ) ( )
( )
( ) 0,/ >= AP
AP
BAP
ABP
I
( ) ( )
( )
( ) 0,/ >= BP
BP
BAP
BAP
I

Theorem 2 Multiplication Theorem of Probability or
Theorem of Compound Probability:
If the probability of an event A happening as a result of trial
is P(A) and after A has happened that probability of an
event B happening as a P(B/A), then the probability of both
the events A and B happening as a result of two trial is
P(AB) or
( ) ( ) ( )ABPAPBAP /.=I
OR
( ) ( ) ( )BAPBPBAP /.=I

Independent Events: If neither of two events A and B
affects the probability of happening of the other, we say that
they are independent.
Remarks: 1. The event A and B are independent if
( ) ( ) ( ) ( )././ BPABPorAPBAP ==
Otherwise A and B are dependent.
2. If A and B are independent, then
( ) ( ) ( )BPAPBAP .=I
( ) ( ) ( ) ( )nn APAPAPAAAP ................ 2121 =I

Example 1 Suppose a packet of 10 razor blades has 2
defective blades in it. Two blades are drawn from the packet
one after another without replacement. Find the probability
that both blades drawn are defective.
Solution: Let E1 be the event that first blade drawn is
defective and E2 the event that second blade drawn is
defective. Here E1 and E2 are independent events and
( ) ( ) .
9
1
/.
10
2
121 == EEPandEP
Hence ( ) ( ) ( ) .
45
1
9
1
.
10
2
/. 12121 === EEPEPEEP I

Example 2 Two cards are drawn from a bridge deck,
without replacement. What is the probability that the first is
an ace and the second is a king?
Solution: 663
4
51
4
52
4
=












Example 3: An urn contains 10 black and 10 white balls.
Find the probability of drawing two balls of the same
colour.
Ans: 9/19
Example 4: A bag contains four white and two black balls
and a second bag contains three of each colour. A bag is
selected at random, and a ball is then drawn at random from
the bag chosen. What is the probability that the ball drawn
is white?
Ans 7/12

Example 5: Three machines I,II and III manufacture
respectively 0.4,0.5 and 0.1, of the total production. The
percentage of defective items produced by I, II and III is 2,
4 and 1 percent respectively. For an item chosen at random,
what is the probability it is defective ?
Ans : 0.029

Example 3 Five salesmen A, B, C, D and E of a company
are considered for a three member trade delegation to
represent the company in an international trade conference.
Construct the sample space and find the probability that:
(i) A is selected.
(ii) A is not selected, and
(iii) Either A or B (not both) is selected.
(Assume the natural assignment of probability)

Solution: The sample space for selecting three salesmen out
of 5 salesmen A, B, C, D and E for the trade delegation is
given by:
S= {ABC, ABD, ABE,}

L4 - Bayes’s Theorem
If E, E, E,……….E are mutually disjoint events with
( ) ( )niEP i ,.......,3,2,1,0 =≠ then for any arbitrary event A which
is a subset of i
n
i
E
1−
U such that ( ) 0>AP , we have
( )
( ) ( )
( ) ( )∑=
= n
i
ii
ii
i
EAPEP
EAPEP
AEP
1
/
/
/

Example 1. Suppose that a product is produced in three
factories X, Y, and Z. It is known that factory X produces
thrice as many items as factory Y, and that factories Y and
Z produce the same number of items. Assume that it is
known 3 percent of the items produced by each of the
factories X and Z are defective while 5 percent of those
manufactured by factory Y are defective. All the items
produced in the three factories are stocked, and an item of
product is selected at random.
(i) What is the probability that this item is defective?

(ii) If an item selected at random is found to be defective,
what is the probability that it was produced by factory
X, Y and Z respectively ?

Example 2. In 2002 there will be three candidates for the
position of principal – Mr. Chatterji, Ayangar and Dr. Singh –
whose chances of getting the appointment are in the
proportion 4:2:3 respectively. The probability that Mr.
Chatterji if selected would introduce co-education in the
college is 0.3. The probabilities of Mr. Ayangar and Dr.
Singh doing the same are respectively 0.5 and 0.8.
(i) What is probability that there will be co-education in
the college in 2003?

(ii) If there is coeducation in the college in 2003, what is
the probability that Dr. Singh is the principal.

Example 3. The probability 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
becomes managers are 3/10, ½ and 4/5 respectively.

L5
Example 1. What is the probability of getting 9 cards of
the same suit in one hand at a game of bridge?
Ans: Required Probability = 13
52
4
39
9
13
4
C
CC ××
Example 2. A man is dealt 4 spade cards from an
ordinary pack of 52 cards. If he is given three more
cards, find the probability p that at least one of the
additional cards is also a spade.

48
3
39
1
C
C
−
.
Example 3. A committee of 4 people is to be appointed
from 3 officers of the production department, 4 officers
of the purchage department, 2 officers of the sales
department and 1 chartered accountant. Find the
probability of forming the committee in the following
manner:
(i) There must be one from each category.

(ii) It should have at least one from the purchase
department.
(iii) The chartered accountant must be in the committee.
Ans: (i) Required Probability = 210
241
4
10
1
2
1
3
1
4
=
×××
C
CCC
(ii) P(committee has no purchase officer) 14
1
210
15
==
P(committee has at least one purchase officer)
14
13
14
1
1 =−=
(iii) 841 3
9
=× C

Example 4: An urn contains 6 white, 4 red and 9 black
balls. If 3 balls are drawn at random, find the probability
that:
(i) Two of the balls drawn are white,
(ii) One is of each colour,
(iii) None is red,
(iv) At least one is white.

Ans:
(i) Required Probability = 3
19
1
13
2
6
C
CC ×
(ii) Required Probability = 3
19
1
9
1
4
1
6
C
CCC ××
(iii) Required Probability = 3
19
3
15
C
C
(iv) Required Probability = 3
19
3
13
1
C
C
−

Example 5. In a random arrangement of the letters of
the word ‘COMMERCE’, find the probability that all
the vowels come together.
3
!2!2!2
!8
!2!2!2
!3!6
=÷

Example 6. If the letters of the word
‘REGULATIONS’, be arranged at random, what is
the chance that there will be exactly 4 letters
between R and E ?

Ans: The word ‘REGULATIONS’ consists of 11 letters.
The two letters R and E can occupy 11010112
11
=×=P
positions.
The number of ways in which there will be exactly 4 letters
between R and E are enumerated below:
(i) R is in the 1st
place and E is in the 6th
place.
(ii) R is in the 2nd
place.
(iii) R is in the 3rd
place.
(iv) R is in the 4th
place.
(v) R is in the 5th
place.

(vi) R is in the 6th
place.
Since R and E can interchange their positions, the required
number of favourable cased is 12.
The required probability 55
6
110
12
=

Example 7. What is the probability that four S’s come
consecutively in the word ‘MISSISSIPPI’ ?
Ans: Total number of permutations of the 11 letters of the
word ‘MISSISSIPPI’ in which 4 are of one kind (S), 4 of
other kind (I), 2 of third kind (P) and 1 of fourth kind (M)
are !1!2!4!4
!11
Following are the 8 possible combinations of 4 S’s coming
consecutively:

(i) S S S S . . . . . . .
(ii) . S S S S . . . . . .
(iii) . . S S S S . . . . .
(iv) . . . S S S S . . . .
(v) . . . . S S S S
(vi) . . . . . S S S S
(vii) . . . . . . S S S S
(viii) . . . . . . . S S S S

Since each of the cases, the total number of arrangement of
the remaining 7 letters, MIIIPPI of which 4 are of one kind , 2
of other kind and one of third kind are 165
4
!1!2!4!4
!11
!1!2!4
!7
=÷
Example 8 Twenty-five books are placed at random in a shelf.
Find the probability that a particular pair of books shall be :
(i) Always together
(ii) Never together
Ans: (i) 25
2
!25
!2!24
=
×
(ii) 25
23
!25
23!24
=
×

Example 9 n persons are seated on n chairs at a round
table. Find the probability that two specified persons are
sitting next to each other.
Ans: 1
2
−n

L6 – Random Variable
Definition: A function whose domain is the set of possible
outcomes, and whose range is a subset of the set of reals.
Such a function is called a random variable.
A real number X connected with the outcome of a random
experiment E. For example, if E consisits of two tosses the
random variable which is the number of heads (0, 1 or 2).
Outcome: HH HT TH TT
Value of X 2 1 1 0

Let S be the sample space associate with a given random
experiment. A real- valued function defined on S and taking
values in R ( )∞∞− , is called a one – dimensional random
variable.
If the function values are ordered pairs of real numbers
(vectors in two- space), the function is said to be a two
dimensional random variable.
More generally, an n-dimensional random variable is
simply a function whose domain is S and whose range is a
collection of n-tuples of real numbers (vectors in n-space).

Mathematical and rigorous definition of the random
variable: Let us consider the probability space, the triplet (S,
B, P), where S is the sample space, B is the σ-field of subsets
in S, and P is a probability function on B.
A random variable is a function X(ω) with domain S and
range ( )∞∞− , such that for every real number a, the event
( )[ ] BaX ∈≤ωω : .
Example 1: If a coin is tossed, then
[ ]21,ωω=S , where ( )



=
=
===
Tif
Hif
XTH
ω
ω
ωωω
,0
,1
;, 21

X(w) is a Bernoulli random variable. Here X(w) takes
only two values.
Example 2: An experiment consists of rolling a die and
reading the number of points on the upturned face. The most
natural r.v. X to consider is ( ) 6....,2,1; == ωωωX
If we are interested in knowing whether the number of
points is even or odd, we consider a random variable Y
defined as follows: ( )



=
=
=
oddisif
evenisif
Y
ω
ω
ω
,1
,0

Example 3: If a pair of fair dice is tossed then
{ } { }6,5,4,3,2,16,5,4,3,2,1 ×=S and n(S) = 36. Let X be a random
variable with image set ( ) { }6,5,4,3,2,1=SX .
Remarks:
1. A function ( )ωX from S to R ( )∞∞− , is a r.v. if and only if for
real a, ( ){ } BaX ∈<= ωω .
2. If X1 and X2 are random variables and C is a constant then
CX1 , X1+X2, X1 .X2 are also random variables.
3. If X is a random variable then
(i) X
1
, where ( ) ∞=





ω
X
1
if ( ) 0=ωX ,

(ii) ( ) ( ){ }ωω XX ,0max=+
(iii) ( ) ( ){ }ωω XX ,0min−=−
(iv) X are random variables.
4. If X1 and X2 are random variables then [ ]21,max XX and
[ ]21,min XX are also random variables.
5.If X is a r.v. and f(.) is a continuous function, then f(X) is a r.v.
6.If X is a r.v. and f(.) is a increasing function, then f(X) is a r.v.
7. If f is a function of bounded variations on every finite interval
[a, b] and X is a r.v. then f(X) is a r.v.

Distribution Function : Let X be a r.v. The function F defined
for all real x by ( ) ( ) ( ){ } ∞<<∞−≤=≤= xxXPxXPxF ,: ωω
is called the distribution function (d.f) of the r.v. (X).
Remark: A distribution function is called the cumulative
distribution function. The domain of the distribution function
is ( )∞∞− , and its range is [0, 1].

Properties of Distribution Function:
1. If F is the d.f. of the r.v. X and if a < b, then
( ) ( ) ( ).aFbFbXaP −=≤<
(i) ( ) ( ) ( ) ( ).aFbFaXPbXaP −+==≤≤
(ii) ( ) ( ) ( ) ( )bXPaFbFbXaP =−−=<< .
(iii) ( ) ( ) ( ) ( ) ( )aXPbXPaFbFbXaP =+=−−=<≤ .
2. If F is d.f. of one-dimensional r.v. X, then
(i) ( ) 10 ≤≤ XF (ii) ( ) ( ) .yxifyFxF <<
3. If F is d.f. of one dimensional r.v. X , then
( ) ( ) .0lim ==∞−
∞−→
xFF
x and ( ) ( ) .1lim ==∞
∞→
xFF
x

Discrete Random Variable:
A variable which can assume only a countable number of real
values and for which the value which the variable takes
depends on chance, is called a discrete random variable.
A real valued function defined on a discrete sample space
is called a discrete random variable.
Example: marks obtained in a test,
number of accidents per month,
number of telephone calls per unit time,
number of successes in n trials, and so on.

Probability Mass Function: If X is a one – dimensional
discrete random variable taking at most a countably infinite
number of values x1, x2, …… then its probabilistic behavior
at each real point is described by a function called the
probability mass function (or discrete density function).
Definition: If X is a discrete random variable with distinct
values x1, x2, …… then the function p(x) defined as
( )
( )



=≠
===
=
..,2,1;,0 ixxif
xxifpxXP
xp
i
iii
X
is called the probability mass function of r.v. X.

Remarks: The number p(xi); I = 1, 2, … must satisfy the
following conditions
(i) ( ) ,0 ixp i ∀≥
(ii) ( )∑
∞
=
=
1
1
i
ixp
1. The set of values which X takes is called the spectrum
of the random variable.
2. For discrete r.v., a knowledge of the probability mass
function enables us to compute probabilities of arbitrary
events.

Example 1: A random variable X has the following
probability function:
Values of X,x: 0 1 2 3 4 5 6 7
p(x) 0 k 2k 2k 3k k2
2k2
7k2
+k
(i) Find k
(ii) Evaluate ( ) ( ) ( ),50,6,6 <<>< XPandXPXP
(iii) If ( )
2
1
,>≤ aXP , find the minimum value of a,
(iv) Determine the distribution function of X.

Solution:
(i) Since ( )∑=
=
7
0
1
x
xp 172322 222
=+++++++ kkkkkkkk
( )( ) 1
10
1
0111001910 2
−=⇒=+−⇒=−+⇒ orkkkkk
Since p(x) cannot be negative, k=-1 is rejected. Hence
k = 1/10.
Values of X,x: 0 1 2 3 4 5 6 7
p(x) 0
10
1
10
2
10
2
10
3
100
1
100
2
100
17

(ii) ( ) ( ) ( ) ( )
100
1
10
3
10
2
10
2
10
1
5.....1,0,6 ++++==++=+==< XPXPXPXP
100
81
=
( ) ( )
100
19
100
81
1616 =−=<−=> XPXP
( ) ( ) ( ) ( ) ( )
5
4
8432,1,50 kXPXPXPXPXP ==+=+=+==<<
(iii) ( ) 4,
2
1
=>≤ agetwetrialByaXP

(iv) The distribution function FX(x) of X is given by in
the adjoin Table.
Values of
X,x:
0 1 2 3 4 5 6 7
p(x) 0 K 3k 5k 8k 8k+k2
8k+3k2
10k2
+9k
FX(x) 0
10
1
10
3
10
5
5
4
100
81
100
83 1

Example 2: If,
( )




=
=
elsewhere
x
x
xp
,0
5,4,3,2,1;
15 find (i) ( )21 orXP = , and
(ii) 





><< 1
2
5
2
1
XXP
Solution:
(i) ( ) ( ) ( )2121 =+=== XPXPorXP
5
1
15
2
15
1
=+=
(ii) 





><< 1
2
5
2
1
XXP
( )
( )1
1
2
5
2
1
>






>





<<
=
XP
XXP I
( ) ( ){ }
( )
( )
( ) ( ) 7
1
15/11
15/2
11
2
1
121
=
−
=
=−
=
=
>
>=
=
XP
XP
XP
XorXP I

Example 3: Two dice are rolled. Let X denote the random
variable which counts the total number of points on the
upturned faces, Construct a table giving the non-zero values
of the probability mass function and draw the probability
chart. Also find the distribution function of X.

Solution: If both dice are unbiased and the two rolls are
independent, then each sample point of sample space S has
probability 1/36. Then
( ) ( ) ( ){ }
36
1
1,122 ==== PXPp
( ) ( ) ( ) ( ){ }
36
2
1,2,2,133 ==== PXPp
( ) ( ) ( ) ( ) ( ){ }
36
3
1,3,2,2,3,144 ==== PXPp
( ) ( ) ( ) ( ) ( ) ( ){ }
36
4
1,4,2,3,3,2,4,155 ==== PXPp

( ) ( ) ( ) ( ){ }
36
2
6,5,5,61111 ==== PXPp
( ) ( ) ( ){ }
36
1
6,61212 ==== PXPp
These values are summarized in the following probability
table:
X: p(x) X p(x)
2 1/36 8 5/36
3 2/36 9 4/36
4 3/36 10 3/36
5 4/36 11 2/36
6 5/36 12 1/36
7 6/36

L7-Continuous Random Variable
A random variable X is said to be continuous if it can take all
possible values (integral as well as fractional) between certain
limits.
A random variable is said to be continuous when its different
values cannot be put in 1-1 correspondence with a set of
positive integers.
A continuous random variable is a random variable that can
be measured to any desired degree of accuracy.
Example : age, height, weight, etc.

Probability Density Function: Consider the small interval (x,
x+dx) of length dx round the point x. let f(x) be any
continuous function of x so that f(x)dx represents the
probability that X falls in the infinitesimal interval (x, x+dx).
Symbolically, ( ) ( )dxxfdxxXxP X=+≤≤
p.d.f. fX(x) of the r.v. is
defined as
( ) ( )
x
dxxXxP
xf
x
X
δδ
+≤≤
=
→0
lim

The probability for a variate value to lie in the interval dx is
f(x)dx and hence the probability for a variate value to fall in
the finite interval [ ]βα, is :
( ) ( )∫=≤≤
β
α
βα dxxfXP
Which represents area between the curve ( )xfy = , x-axis and
the ordinates at α=x and β=x .
The total probability is unity: ( ) 1=∫
b
a
dxxf
, where [a, b] is the
range of the random variable X.

The range of the variable may be finite or infinite.
Important Remark: (Difference between Discrete and
Continuous Random Variable):

Various Measures of Central Tendency, Dispersion,
Skewness and Kurtosis for Continuous Probability
Distribution: Let fX(x) or f(x) be the p.d.f. of a r.v. X, where
X is defined from a to b. then
(i) ( )∫=
b
a
dxxfxMeanArithmetic
(ii) ( )∫=
b
a
dxxf
xH
MeanHarmonic
11
:
(iii) ( )∫=
b
a
dxxfxGMeanGeometric loglog

(iv) ( ) ( )∫=
b
a
r
r dxxfxoriginabout'
µ
(v)

Example 1: The diameter of an electric cable, say X, is
assumed to be a continuous random variable with p.d.f.:
( ) ( ) 10,16 ≤≤−= xxxxf
(i) Check that f(x) is p.d.f. and
(ii) Determine a number b such that ( ) ( )bXPbXP >=<
Example 2: A continuous random variable X has a p.d.f.
( ) 10,3 2
≤≤= xxxf
Find a and b such that
(i) ( ) ( )aXPaXP >=≤ (ii) ( ) 05.0=> bXP

Continuous Distribution Function:
If X is a continuous random variable with the p. d. f. f(x),
then the function
( ) ( ) ( ) ∞<<∞−=≤= ∫
∞
∞−
xdttfxXPxFX ,
is called the distribution function (d. f.) or sometimes the
cumulative distribution function (c. f. d.) of the random
variable X.

Properties of Distribution Function:
1. ( ) ∞<<∞−≤≤ xXF ,10
2. F(x) is non-decreasing function of x.
3. F(x) is a continuous function of x on the right.
4. The discontinuities of F(x) are at the most countable.
5. It is denoted as
( ) ( ) ( ) ( )
( ) ( ) ( )aFbFaXPbXP
dxxfdxxfdxxfbXaP
abb
a
−=≤−≤
−==≤≤ ∫∫∫ ∞−∞−
)(
6. Similarly ( ) ( ) ( ) ( ) ( )aFbFbXaPbXaPbXaP −=<≤=≤<=<<
7. ( ) ( ) ( ) ( ) ( )dxxfxdFxfxF
dx
d
xF ==='

Example: verify that following is a distribution function:
( )






>
≤≤−





+
−<
=
ax
axa
a
x
ax
XF
,1
,1
2
1
,0
Example: The diameter, say X, of an electric cable, is
assumed to be continuous random variable with p.d.f.
( ) ( ) 10,16 ≤≤−= xxxxf
(i) Check that the above is a p.d.f.
(ii) Obtain an expression for the c. d. f. of X.,

(iii) Compute 





≤≤≤
3
2
3
1
2
1
XXP
(iv) Determine the number k such that ( ) ( )kXPkXP >=< .

Mathematical Expectation
The expected value of a discrete random variable is a
weighted average of all possible values of the

Theory of probability

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