INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

JSPM’s
RAJARSHI SHAHU COLLEGE OF ENGINEERING
TATHAWADE, PUNE-33
An Autonomous Institute Affiliated to Savitribai Phule Pune University

F.Y. B. Tech Semester-I (Computer /IT Branch)

[ES1106]: Introduction to Statistics, Probability and Calculus

UNIT-II

Probability
Content: Experiments, sample space, event, Conditional Probability, Bayes Theorem.

1.1 Introduction

In our daily life, we often use phrases such as ‘It may rain today’, or India may win the match’ or
‘I may be selected for this post’. These phrases involve an element of uncertainty. How can we measure this
uncertainty? A measure of this uncertainty is provided by a branch of mathematics, called the theory of
probability. Probability theory began in the seventeenth century by the two great French mathematicians,
Blaise Pascal and Pierre de Fermat over two problems from games of chance.

They solved problems and influenced early researchers in establishing a mathematical theory of
probability. The probability of an event is determined by how likely that event is to happen. If an event is
impossible, it has a probability of 0. If an event is certain to happen, it has a probability of 1. Probabilities are
usually expressed as fractions, as decimals, or as percentages. Probability theory is designed to measure the
degree of uncertainty regarding the happening of a given event. Probability refers to the prediction of
occurring of an event and is easy to understand and calculate. It is measure of how likely it is that something
will happen or that a statement is true. The dictionary meaning of probability is ‘likely though not certain to
occur. Thus, when a coin is tossed, a head is likely to occur but may not occur. Similarly, when a die is
thrown, it may or may not show the number 4. Applications of probability can be seen in everyday life, risk
assessment, reliability etc. Due to its immense success and wide applications, theory of probability is viewed
as the most important area of mathematics. In this chapter, we shall study some basic concepts of probability,
addition and multiplication theorem of probability and applications of probability in our day to day life.
1.2 Permutations
The number of permutations of n different things taken r at a time is,
n (n – 1) (n – 2) , …….. , (n – r + 1)
It is denoted by nPr
 n
Pr = n (n – 1) (n – 2) ………. (n – r + 1)
n!
=
(n – r) !
1.3 Combinations
The number of combinations of n different objects. Taken r at a time is denoted by nCr.
n
Pr n!
 Cr =
n
=
r! r! (n – r) !
Note : (i) nCn = 1 (ii) n
C0 = 1 (iii) nCr = nCn – r
1.4 Basic Definitions
For probability we should know some definitions. Following are basic definitions.

Dice
It is small cube. On its face dots . .. …  .    are marked. On rolling of dice, the outcome is

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 1

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

the number of dots on its upper face.

Cards

A pack of cards consists of four suits, which are spades, Hearts, Diamonds and clubs. Each suit consists
of 13 cards, nine cards numbered 2, 3, 4, 5, 6, 7, 8, 9, 10 and Ace, a king, a Queen and a Jack (Knave).
Colour of spades and clubs is back and that of Hearts and Diamonds is red. Kings, Queens and Jacks are
known as face cards.
Experiment
A measurement process that produces quantifiable results.
Examples : (i) Rolling of two dice (ii) Dealing cards (iii) Tossing of a coin
Random Experiment
An experiment is called random experiment if it satisfies following conditions :
(i) It has more than one possible outcomes.
(ii) It has not possible to predict the outcome in advance.
Examples
(i) When a coin is tossed, either we get a head (H) or a tail (T)
(ii) Tossing of a die
We get either 1 or 2 or 3 or 4 or 5 or 6.
Outcomes
A possible result of a random experiment is called outcomes.
Examples
(i) Tossing of a coin (ii) Rolling of a die
The outcomes are H, T. The outcomes are 1, 2, 3, 4, 5, 6
Sample Space
The set of outcomes of a single performance of random experiment is called sample space of the
experiment. It is denoted by S.
Examples
(i) Tossing of a coin (ii) Rolling of a die
 The set of outcomes are H, T  The set of outcomes are 1, 2, 3, 4, 5, 6
 Sample space = S = {H, T}  Sample space = S = { 1, 2, 3, 4, 5, 6 }

Sample Points
Each element of the sample of space is called a sample point i.e. each outcome of the random
experiment is called sample point. Total number of sample points of the space S is denoted by n (S).
Examples
(i) Tossing of a coin (ii) Rolling of a die
 The set of outcomes are H, T  The set of outcomes are 1, 2, 3, 4, 5, 6
 Sample space = S = { H, T }  Sample space = S = { 1, 2, 3, 4, 5, 6 }
n(s) = 2 n(s) = 6
Trial and Event
Performing a random experiment is called trial and outcome is called an event.
Examples
(i) In a Experiment “Tossing of a coin”. Tossing of a coin is a trial turning up of head or tail is an event.
(ii) In a Experiment “Rolling of a die”. Rolling a die is a trial appearance of 1 or 2 or 3 or 4 or 5 or 6 is an
event. Event is subset of sample space.

1.5 Types of Events

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 2

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

Sample Event
An event, consisting of a single sample point is called a simple event.
Example: For an Experiment “Rolling a die”
Sample Space = S = { 1, 2, 3, 4, 5, 6 }
So each of { 1 } , { 2 }, { 3 }, { 4 }, { 5 } and { 6 } are simple events.
Compound Event
A subset of the sample space, which has more than one element is called a compound (mixed) event.
Example: In an Experiment “Rolling a die”, The event of appearing of odd numbers is a Compound Event,
because
Sample Space = S {1, 2, 3, 4, 5, 6}
Event = E = {1, 3, 5} which has ‘3’ elements.
E is compound Event.
Equally Likely Events
Events are said to be equally likely, if we have no reason to believe that one is more likely to occur than
the other.
Example: In an Experiment “A rolling of a die”, all the six faces { 1, 2, 3, 4, 5, 6 } are equally likely to come
up.
Exhaustive Events
When every possible outcome of an experiment is considered then the Event is called Exhaustive Event.
Example : In an Experiment “A rolling of a die”, cases 1, 2, 3, 4, 5, 6 form an exhaustive set of events.
Sure Event
Let ‘S’ be a sample space. If E is a subset of or equal to S then E is called a sure event.
Example : In an Experiment “A rolling of a die”,
Sample Space ={1, 2, 3, 4, 5, 6}
If E1 = Events of getting a number less than ‘7’
So ‘E1’ is a sure event. (Since all out comes are less than 7)
So, we can say, in a sure event
n(E) = n(S)
Mutually Exclusive or Disjoint Events
If two or more events cannot occur simultaneously, that is no two of them can occur together.
Example : In an Experiment “A coin is tossed”, We will get either head or tail not both together. The event
of occurrence of a head and the event of occurrence of a tail are mutually exclusive events.
Independent or Mutually Independent Events
Two or more events are said to be independent if occurrence or non-occurrence of any of them does not
affect the probability of occurrence or non-occurrence of the other event.

Example : In an Experiment “A coin is tossed twice”, The event of occurrence of head in the first throw. The
event of occurrence of head in the second throw are independent events.
Dependent Events
If the occurrence of one event in any trial affects the occurrence of other event in other trial.
E.g. If two cards are drawn successively from a well shuffled pack of cards, without replacement, then
the event of appearance of king at second draw will certainty depends upon the result of the first draw.
Let A - be the happening event ; m - number of favorable outcomes
n - Total number of outcomes (trials) (possible success and failure)

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 3

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

m
P (A) =
n
P (A) - be probability of event A
–
If A is non happening event
–
( ) m
Then P A = 1 – P(A) = 1 –
n
–
( )
 P (A) + P A = 1 (Note that 0  P(A)  1)
s = success ; f = failure
s f
P (success) = = p ; and P (failure) = =q
s+f s+f
 p +q = 1
n
The ‘r’ success in n trials, then number of ways = Cr
e. g. 3 cards are drawn = 52C3  ways.
Difference between Mutually Exclusive and Mutually Independent Events
Mutually exclusiveness is used when the events are taken from the same experiment, where as
independence is used when the events are taken from different experiments.
Consider a random experiment of tossing a coin three times.
(a) Find the sample space S1 if we wish to observe the exact sequences of heads and tails
obtained.
(b) Find the sample space S2 if we wish to observe the number of heads in the three tosses.
Soln. :
(a) The sampling space S1 is given by,
S1 = { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT }
Where, for example, HTH indicates a head on the first and third throws and a tail on the second throw.
There are eight sample points in S1.
n (S1) = 8
(b) The sample space S2 is given by,
S2 = { 0, 1, 2, 3 }
Where, for example, the outcome 2 indicates that two heads were obtained in the three losses.
The sample space S2 contains four sample points.
n (S2) = 4

1.6 Types of Probability

There are four types of probability as follows:

1. Classical Probability 2. Experimental Probability
3. Theoretical Probability 4. Subjective Probability
Classical Probability
Event and outcomes in sample space are determined by the rules of game.
Give n equally likely outcomes,
let ‘s’ represent the number of successful outcomes and
‘f’ represent the number of failure outcomes, then s + f = n.
s f
The probability of success is and The probability of failure is .
n n
Probability of success plus the probability of failure is equal to 100 % or 1.
s f
Therefore, + = 1.
n n
Experimental Probability
This is based on the number of possible outcomes by the total number of trials.

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 4

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

Example : For a coin tossing experiment, when a coin is tossed, the total possible outcomes are head and
tail. Total number of trials will be determined by the total number of times the coin is tossed. If a coin is
17
tossed 25 times and it lands on head 17 times, then the probability is .
25
Theoretical Probability
The theoretical probability P(E) of an event E is fraction of number of times we expect E to occur if we
repeat the same experiment over and over.
Example : Roll a fair die. The probability of rolling 1 is one sixth of the time.
1
P(1) =
6
1 1
Similarly, P(2) = , …….., P(6) =
6 6
Subjective Probability
Subjective probability is an individual person’s measure of belief that an event will occur. Subjective
probabilities contains no formal calculations. Only reflect the subjects opinions and past experience. They
differ from person to person. There will be a high degree of personal bias.
Axiomatic Probability
 The objective of probability is to assign to each event A a number P(A), called the probability of the
event A, which will give a precise measure of the chance that A will occur.
 Probability Axioms :
1. For any event A, P(A)  0.
2. P(S) = 1.
3. If A1, A2, A3,...... is an infinite collection of disjoint events, then

P(A1 ∪ A2 ∪ A3 ∪.....) =  P(Ai) 0  P(A)  1
i=1

1.7 Elementary Theorems (0  P(A)  1)

Proposition
P() = 0 where  is the null event. This is turn implies that the property contained in Axiom 3 is valid
for finite collection of events, i.e. if A1, A2, ......An is a finite collection of disjoint events, then P(A1 ∪ A2 ∪

..... ∪ A3) =  P(Ai)
i=1
 0  P(A)  1
Examples :
1. Consider the coin tossing experiment and we are only interested in tossing the coin one time
Then S = {H, T}.
Since P(S) = 1 (Axiom 1), and The event {H} and {T} are mutually disjoint.
By Axiom 3, we have.
P({H}) + P({T}) = P({H} ∪ {T}) = P(S) = 1
If the coin is fair, P(H) = 0.5, P(T) = 0.5. If the coin is more likely to give a Head,
Then 0.8 for P({H}) and 0.2 for P({T}) may be suitable.
In fact, if p is any fixed number between 0 and 1, then P({H}) = p, and P({T}) = 1 – p is an assignment
consistent with the axioms.

Addition Theorem on Probability

Statement: If A and B are any two events of the sample space S. Then the probability of happening
of at least one of the event is defined as, P (A  B) = P (A) + P (B) – P (A  B).

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 5

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

We can extend this theorem for three events, A,B, C of the sample space S
P(A  B  C) = P(A) + P(B) + P(C) – P(A  B) – P(A  C) – P(B  C) + P(A ∩ B ∩ C)
Note : If A and B are any two mutually exclusive events then P (A  B) = 0
Then, P (A  B) = P (A) + P (B)

De Morgan’s Laws
If A and B are two events of a sample space S
We know, (A  B) = A  B and (A  B) = A  B then
(i) P (A  B) = P (A  B) (ii) P (A  B) = P (A  B)
(iii) P (A  B) = 1 – P (A  B) (iv) P (A U B) = 1 – P (A  B)
Algebra of Sets
(i) Commutative Laws :
A  B = B  A and A  B = B  A
(ii) Associative Laws :
(A  B)  C = A  (B  C)
(A  B)  C = A  (B  C)
(iii) Distributive Laws :
(A  B)  C = (A  C)  (B  C)
(A  B)  C = (A  C)  (B  C)
1
Ex. 1. Professor X and Madam Y appear for an interview. The probability of professor X’s selection is and
7
1
that of Madam Y’s selection is . Find the probability that only one of them is selected. What is the
5
probability that at least one of them is selected.
Soln. :
Step I :
1 1
Given : P(X) = , P(Y) =
7 5
– 1 6 – 1 4
( )
P X = 1– = ; P Y = 1– =
7 7
( ) 5 5
–
( )
Step II : Here; P (X) means probability of professor’s selection. P X ; means probability of professor’s not
selected.
As only one out of them is selected.
A event  If X is selected, Y is not selected
B event  If Y is selected, X is not selected
Step III :
– 1 4 4
( )
P(A) = P(X)  P Y =  =
7 5 35
 independent events (mutually exclusive)
– 1 6 6
( )
P(B) = P(Y)  P X =  =
5 7 35
Required probability (only one selected)
Step IV :
(1) P (A or B) = P(A  B) = P(A) + P(B) – [P(A  B)]
= P(A) + P(B) (··· P (A  B) = 0)
4 6 10 2
= + = =
35 35 35 7
Step V :
(2) Probability that none of them is selected

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 6

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

– – 6 4 24
( ) ( )
= P X P Y =  = =q
7 5 35
24 11
P(at least one selected) = 1– = (∵ p+q=1 p=1–q)
35 35

Ex. 2: A spinner has 4 equal sectors coloured yellow, blue, green and red. What is the probability of landing
on each colour after spinning this spinner ?
Sol: Sample Space = S = {yellow, blue, green, red }
n(S) = 4
We know,
Probability of an Event A is,
Number of Favourable outcomes
P(A) =
Total number of Possible out comes
1 1
P (yellow) = P (blue) =
4 4
1 1
P (green) = P (red) =
4 4

Ex.3 : What is the probability of each outcome when a single 6-sided die is rolled ?
Sol: Sample Space = S = {1, 2, 3, 4, 5, 6 } , n(S) = 6
We know,
Probability of an Event A is,
Number of Favourable outcomes
P(A) =
Total number of Possible out comes
1 1 1
P (1) = , P (2) = , P (3) = ,
6 6 6
1 1 1
P (4) = , P (5) = , P (6) =
6 6 6
Ex.4 : A glass jar contains 1 red, 3 green, 2 blue and 4 yellow marbles. If a single marble is chosen at random
from the jar. What is the probability of each outcome ?
Sol: Sample Space = S = { Red, green, blue, yellow }
We know,
Probability of an Event A is,
Number of Favourable outcomes
P(A) =
Total number of Possible out comes
1 3
P (red) = P (green) =
10 10
1 4 2
P (blue) = P (yellow) = =
5 10 5

Ex. 5: Find the probability that if a card is drawn from an ordinary pack, it is a diamond.
Soln. :
Step I : A pack of card contains 52 cards.
One card drawn from a pack of 52 cards at random in 52 C1 ways.
52
Sample space = S = { C1 ways }
52
n(s) = C1 = 52 , n(s) = 52
Step II : In a pack 13 cards are diamond.
A = Event “A card is a diamond”
A = {13 cards of diamond}
n(A) = 13
Probability of a drawn card is diamond is

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 7

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

Number of favourable sample points to event A

P(A) =
Total number of sample points in a sample space of an experiment
n(A) 13
= =
n(S) 52
1 1
P(A) = Probability of a getting card is diamond =
4 4

Ex.6 :A bag contains 7 white balls, 5 black balls and 4 red balls. If two balls are drawn at random from the
bag. Find the probability that both the balls are white.
Soln. :
Step I : A bag contains 7 white balls, 5 black balls and 4 red balls.
 Total number of balls in a bag = 7 + 5 + 4 = 16
Out of 16 balls, 2 balls drawn at random in 16C2 ways.
16  15
 Exhaustive number of cases = 16C2 ways = = 120
12
 Sample space = S = {120 ways}
 n(S) = 120
Step II : A = Event “both balls are white”
 Out of 7 white balls, 2 can be drawn in 7C2 ways.
76
A = {7C2 ways } = = 21 ways
12
A = {21 ways}
 n(A) = 21
Step III : Probability that both the balls are white
Number of favourable sample points to event A
P(A) =
Total number of sample points in a sample space of an experiment
n(A) 21 7
= =
n(S) 120 40
7
 Probability that both the balls are white =
40
Ex.7 : A box contains 10 red, 5 white, 5 black balls. Two balls are drawn at random.
Find the probability that they are not of the same colour.
Soln. :
Step I : A box contains 10 red, 5 white, 5 black balls.
 Total number of balls in a box = 10 + 5 + 5 = 20
Out of 20 balls, 2 balls are drawn at random in 20C2 ways.
20  19
= = 190 ways
12
 Sample space = S = {190 ways}
 n(S) = 190
Step II : R = Event “both balls drawn are of the red colour”
Out of 10 red balls, 2 can be drawn in 10C2 ways.
10  9
R = {10C2 ways} = = 5  9 = 45 ways
12
 n(R) = 45
Step III : Probability that both balls are red colour
Number of favourable sample points to event R
P(R) =
Total number of sample points in a sample space of an experiment
n(R) 45 9
= =
n(S) 190 38
Step IV : W = Event “both balls drawn are of the white colour”

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 8

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

Out of 5 white balls, 2 can be drawn in 5C2 ways

54
W = {5C2 ways} =
12
n(W) = 10
Step V : Probability that both balls are white colour.
Number of favourable sample points to event W
P(W) =
Total number of sample points in a sample space of an experiment
n(W) 10 1
= =
n(S) 190 19
Step VI : B = Event “both balls drawn are of the black colour”
Out of 5 black balls, 2 can be drawn in 5C2 ways.
54
B = {5C2 ways} = = 10 ways
12
 n (B) = 10
Step VII : Probability that both balls are black colour
Number of favourable sample points to event B
P(B) =
Total number of sample points in a sample space of an experiment
n(B) 10
=
n(S) 190
1
P(B) =
19
Step VIII : A = Event “both balls drawn are of the same colour”
= RW B
P(A) = P (R  W  B)
= P(R) + P(W) + P(B)  ∵ R W B are mutually 
 exclusive events 
9 1 1 9 + 2(1) + 2(1) 9 + 2 + 2 13
= + + = = =
38 19 19 38 38 38
This is a probability of two balls drawn are of same colour
A = Event “ Both balls drawn are not of the same colour”.
 Probability of two balls drawn are not of the same colour
13 38 – 13 25
P(A) = 1 – P(A) = 1 – = =
38 38 38
25
 Probability of two balls drawn are not of the same colour =
38
Ex.8 : An urn contains 5 red and 10 black balls Eight of them are placed in another urn. What is the chance
that the latter than contains 2 red and 6 black balls ?
Soln. :
Step I : An urn contains 5 red and 10 black balls
Total balls = 5 + 10 = 15 balls.
Out of these eight are placed in another urn (drawn form an urn)
 Number of ways that 8 balls drawn from an urn
are 15C8 ways
Sample space = S = { 15C8 ways}
15  14  13  12  11  10  9  ╱
8
 n (S) = 15
C8 = , n (S) = 6435
1234567╱
8
Step II : Now A = Event is “out of these eight balls 2 red and 6 black balls”.
 Number of ways out of these eight balls
(i) 2 are red balls = 5C2 ways (Since there are 5 red balls)

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 9

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

(ii) 6 are black balls = 10C6 ways

 Total number of ways in which 2 red and 6 black balls are drawn
5  4 10  9  8  7  6  5
= 5C2  10C6 ways = 
12 123456
n(A) = 2100
Step III :
 Required probability
Number of favourable sample points to event A
P(A) =
Total number of sample points in a sample space of an experiment
n(A) 2100 140
= = =
n(S) 6435 429
140
 Probability of 2 red and 6 black balls =
429
Ex.9 : Two cards are drawn at random from a well shuffled pack of 52 cards. Find the probability that the two
cards drawn are king and a queen of the same suit.
Soln. :
Step I : In a pack of 52 cards,
52
Two cards are drawn at random in C2 ways
 Sample space = S = 52C2 ways
 Total number of sample points of an experiment
52  51
n(S) = 52C2 = = 26  51 = 1326
12
Step II : A = Event is “getting a king and a queen of same suit”
There are four suits and in each suit there is one pair of king and queen.
There are 4 pairs of king and queen of same suit.
 n (A) = 4
Step III : Probability of two cards drawn is a pair of king and queen of same suit is,
Number of favourable sample points to event A
P(A) =
Total number of sample points in a sample space of an experiment
n(A) 4 2
= = =
n(S) 26  51 13  51
2
P(A) =
663
2
 Probability of getting or pair of king and queen of same suit = P(A) =
663
Ex. 10 :Four cards are drawn from a pack of cards. Find the probability that :
(i) all are diamonds (ii) there is one card of each suit
(iii) there are two spades and two hearts
Soln. :
52
Step I : A pack of 52 cards, 4 cards are drawn in C4 ways
Exhaustive number of cases = 52C4 ways
52  51  50  49
= = 270725 ways
1234
 Sample space = S = {52C4 ways}
 n(S) = 270725 ways
(i) Step II : A pack contains 13 diamonds
A = Event “All four cards are diamonds”
 Out of 13 diamonds 4 cards are drawn in 13C4 ways.
13  12  11  10
A = {13C4 ways} = = 715 ways
1234

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 10

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

 n(A) = 715
Step III :  Probability that all four cards are diamonds
Number of favourable sample points to event A
P(A) =
Total number of sample points in a sample space of an experiment
n(A) 715 11
= =
n(S) 270725 4165
11
 Probability of all four cards are diamonds =
4165
(ii) Step IV : B = Event “There is one card of each suit”
There are 4 suits and each suit containing 13 cards
One card of each suit = B = {13C1  13C1  13C1  13C1 ways}
= {13  13  13  13 ways} = {28561 ways}
 n (B) = 28561
Step V : Probability that one card of each suit
Number of favourable sample points to event B
P(B) =
Total number of sample points in a sample space of an experiment
n(B) 28561 2197
= =
n(S) 270725 20825
2197
 Probability of one card of each suit =
20825
(iii) Step VI :
C = Event “There are two spades and two hearts”
13
2 spades out of 13 can be drawn in C2 ways
2 hearts out of 13 can be drawn in 13C2 ways.
 4 cards drawn contains two spades and two hearts in (13C2  13C2) ways
13  12 13  12 
C = {13C2  13C2 ways} =  ways 
 12 12 
= {13  6  13  6 ways} = {6084 ways}}
n(C) = 6084
Step VII :
 Probability of out of four, two spades and two hearts
Number of favourable sample points to event C
P(C) =
Total number of sample points in a sample space of an experiment
n(C) 6084 468
= =
n(S) 270725 20825
468
 Probability of two spades and two hearts =
20825
Ex. 11 : A bag contains 3 red 4 green, 2 black balls. Two balls are drawn at random find the probability that
getting one red and one green.
Sol:
Step I : A bag contains 3 red, 4 green, 2 black balls
 Total balls = 3 red + 4 green + 2 black = 9 balls
Out of 9 balls, 2 balls are drawn at random in 9C2 ways.
98
 Exhaustive number of cases = 9C2 ways = = 36 ways.
12
 Sample space = S = {36 ways }
n(s) = 36
Step II : A = Event “one ball is of red and one is of green colour”.
3
Out of 3 red colour ball, 1 ball can be drawn in C1 ways.
Out of 4 green balls, 1 ball can be drawn in 4C1 ways.

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 11

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

 A = { 3C1  4C1ways } = { 3  4 ways }

 n(A) = 12
Step III : Two balls are drawn of random, probability of getting 1 red and 1 green is,
Number of favourable sample points to event A
P(A) =
Total number of sample points in a sample space of an experiment
n(A) 12 1
= = =
n(s) 36 3
1
 Probability of getting one red and one green ball =
3
Ex. 12:A card is drawn from a pack of 52 cards find the probability that a card is a diamond or a face card.
Soln. :
Step I : A card is drawn from a pack of 52 cards in 52C1ways.
 Exhaustive number of cases = 52C1 ways = 52
 Sample space = S = { 52C1 ways }
n(S) = 52
Step II :  A = Event “A card is a diamond”. = { 13 cards of diamond }
 n(A) = 13
 Probability of a drawn card is diamond
Number of favourable sample points to event A
P(A) =
Total number of sample points in a sample space of an experiment
n(A) 13 1
= = =
n(S) 52 4
Step III : B = Event “A card is face card”.
In each suit 3 face cards Total suits are 4
Total Face Cards = 3  4 = 12

B = {12 face cards }

 n(B) = 12
 Probability of a drawn card is face card.
Number of favourable sample points to event B
P(B) =
Total number of sample points in a sample space of an experiment
n(B) 12 3
= = =
n(S) 52 13
Step IV : AB = Event “A face card is diamond”.
In a pack of 52 cards 3 diamond face cards.
= {3 diamond face cards}.
n(A  B) = 3
 Probability of a drawn card is face diamond card.
Number of favourable sample points to event (A  B)
P(A  B) =
Total number of sample points in a sample space of an experiment
n(A  B) 3
= =
n(S) 52
Step V : Probability that a card is a diamond or a face card = P( A  B).
 By addition theorem of probability,
1 3 3
P( A  B) = P(A) + P(B) – P (A  B) = + –
4 13 52
13 + 12 – 3 22 11
= = =
52 52 26
11
 Probability that a card is a diamond or a face card =
26

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 12

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

Ex:13 : X is registered contractor with the government .Recently ,X has submitted his tender for two
contractors, A and B. The Probability of getting the contract A is ,the contract B is and both contract A

and B is .Find the probability that X will get contract A or B

Solution ; As getting contract A and contract B are mutually non-exclusive event ,the required probability
will be

Ex:14 : Suppose we have a box with 3 red, 2 black and 5 white balls. Each time a ball is drawn it is returned
to the box. What is the probability of drawing

a) Either a red or a black ball ?

b) Either a white or a black ball?

Solution: The Probabilities of drawing the specific color ball are:

a)

b)

Ex:15 A salesman is known to sell a product in 3 out of 5 attempts while another salesman in 2 of 5 attempt
.Find the probability that

i) No sale will take place when they both try to sell the product
ii) Either of them will succeed in selling the product

Solution:
Let

It is given that
Therefore, =
The required probabilities are calculated as follows:

i)
= (By Multiplication Theorem)

(By Theorem of Total Probability)

(By Multiplication Theorem)

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 13

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

Ex:16 A candidate is selected for interviews for 3 posts. For the first post ,there are 3 candidates, for the
second 4 and for the third post there are 2 candidates .What is the Probability that the candidate is selected for
at least one post.

Solution : Let

In the absence of any other information , it is assumed that all candidates have equal chance of selection ,and
therefore the probability of a candidate being selected is

Accordingly, we can presume that it is given that

Therefore, the probability that a candidate is selected for at least one post is given by the Theorem of Total
Probability as

Assuming that selection for one post is independent of selection for other posts, we have, by Multiplication
Theorem Of Probability

Substituting the values of above Probabilities, we get

Conditional Probability :-
Let A and B be two events such that P(A) >0. Denote by P(B/A) the probability of B given that A has ocurred.
Since A is known to have occurred, It becomes the new sample space replacing the original S.From this we
are led to the definition

Or

In words, say that the probability that both A and B occur is equal to the probability that A occurs times the
probability that B occurs given that A has occurred. We call P(B/A) the conditional probability of B given A
i.e. the probability that B will occur given that A has occurred. It is easy to show that conditional probability
satisfies the axioms.

Ex. Find the probability that a single toss of a die will result in a number less than 4 if

(a) no other information is given and (b)it is given that the loss resulted in an odd number.

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 14

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

Ans.-(a) Let B denote the event {less than 4}.Since B is the union of the event 1,2,or3 turning up, we see by
theorem that

Assuming equal probabilities for the sample points.

(b)Letting A be the event {odd number}, we see that

Also

Then

Hence the added knowledge that toss results in an odd numbers raises the probability from 1/2 to 2/3.

Theorems on Conditional Probability:-

1) For any three events we have

In words, the probability that and and all occur is equal to the probability that occurs times the
probability that occurs given that has occurred times the probability that occurs given that both
and have occurred. The result is easily generalized to n events.

2) If an event A must result in one of the mutually exclusive events then

Independent Events :-

If i.e. the probability of B occurring is not affected by the occurrence or non-occurrence of

A,then we say that A and B are independent events. This is equivalent to

Conversely, if above equation holds, then A and B are independent.

We say three events are independent if they are pairwise independent:

And

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 15

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

Bayes’ theorem statement :-

Suppose are mutually exclusive events whose union is the sample space S,i.e. one of the
events must occur.Then if A is any event,we have the following important theorem:

This enables us to find the probabilities of the various events that can cause A to occur.
For this reason Bayes’ theorem is often referred to as a theorem on the probability of causes.

1.8 Applications of Bayes’ Theorem:

Ex:1 A restaurant is experiencing discontentment among its customer it analysis that there are three factors
responsible viz. food quality , service quality and interior décor .By conducting an analysis, it assesses the
probabilities of discontentment with the three factors as 0.40, 0.35,and 0.25,respectively .By conducting a
survey among customers ,it also evaluates the probabilities of a customer going away discontented on account
of these factors as : 0.6, 0.8, and 0.5,respectively with this information , the restaurant wants to know that if a
customer is discontented ,what are the probabilities that it is so due to food , service or interior decor

Solution ; This problem can be solved with the help of Bayes Theorem We can express the above events
/situations with the help of notations in Bayes Theorem as follows

We are given that

Also it is given that

We are required is to find out and

As per Bayes Theorem is given as:

Similarly, it can be worked out that

and 0.194

Ex:2 In a basin area Where oil is likely to be found underneath the surface , there are three locations with
three different types of earth composition , say , The probabilities for these three compositions

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 16

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

are 0.5 ,0.3 ,and 0.2 respectively .Further, it has been found from the past experience that after drilling of well
at these locations, the probabilities of finding oil is 0.2 ,0.4 and 0.3 respectively .Suppose a well drilled at a
location, and it yields oil , what is the probability that the earth composition was

Solution ; In a typical probability problem , one would be seeking the probability that under the given
conditions, what could be the probability that oil will be found .But ,here the situation is that oil has been
found and one is required to find the probability of its source i.e, the type of earth composition. This can be
found with the help of Bayes Theorem

Given

Also it is given that

What is required is to find out

As per Bayes Theorem is given as:

Thus the probability that the well that produce oil had type Composition beneath is
0.475.Incidentally,if it was required to find the probability that the oil was found in type Composition
I.e , then using Bayes Theorem ,we get

Similarly, the probability that the oil was found in type Composition, i.e , then using Bayes
Theorem we get

1.9 Exercise :

1. A fair coin is tossed five times. Find the probability obtaining (i) head in all the tosses (ii) head in at
least one of the tosses.
2. There are 20 persons. Five of them are graduates. Three persons are randomly selected from these 20
persons. Find the probability that at least one of the selected person is graduate.
3. In a college there are five Lecturers. Among them there are doctorates. If committee consisting three
lectures is formed, what is the probability that at least two of them are doctorates ?
4. A card is drawn at random From a pack of cards
(i) What is the probability that it is heart. (ii)If it is known that the card drawn is red,
5. From a bag containing 15 red and 10 blue balls, a ball is drawn ‘at random’. What is the probability of

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 17

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

drawing (i) a red ball ? (ii) a blue ball ?

6. If two dice are thrown, what is the probability that the sum of the numbers on the two faces is divisible
by 3 or by 4 ?
7. A pair of dice are thrown. Find the probability of getting
(i) a sum as a prime number (ii) a doublet, i.e. the same number on both dice
(iii) a multiple of 2 on one die and a multiple of 3 on the other.
8. A bag contains 3 red, 6 white and 7 blue balls. What is the probability that two balls drawn at
random are both white?
and P   =
3 1 A 2
9. If P(A) = , P(B) =
10 2  B 5
then find (i) P (A  B) (ii) P   (iii) P   (iv) P  
B A A
 A B  B 
7 9
10. If P (A  B) = and P (A  B) = Find P(A)
10 10
3
11. If A and B are independent with P (A) = 2 P(B) and P(A  B) =
20
Find P (A  B)
4 3
12. If A and B are independent with P (A  B) = and P(B) = then find P(A).
5 10
13. In a group of 10 outstanding students in a school, there are 6 boys and 4 girls. Three students are
to be selected out of these at random for a debate competition. Find the probability that,
14. (i) one of boy and two girls. (ii) all are boys (iii) all are girls

15. Two cards are drawn at random form 8 cards numbered 1 to 8. What is the probability that the
sum of the numbers is odd, if the cards are drawn together ?
(a) In a single throw of two dice, find the probability of total of 5 or 7.
(b) A and B are two mutually exclusive events such that P(A) = 0.3 and P(B) = 0.4
calculate P(A  B).
16. Two cards are drawn at random without replacement from a deck of 52 cards. What is the
17. probability that the first is a diamond and the second card is red ?
18. If A and B are event with P(A) = 0.4, P(B) = 0.2, P(A  B) = 0.1 find the probability of A given
B. Also P (B / A).
19. From a box containing 4 white balls, 3 yellow balls and 1 green ball, two balls are drawn one at a
20. tie without replacement. Find the probability that one white and one yellow ball is drawn.

21. In a simultaneous toss of four coins, what is the probability of getting

(a) exactly three heads ? (b) at least three heads ? (c) atmost three heads ?

22. For two events A and B it is given that P(A) = 0.4 ; P(B) = p and P (A  B) = 0.6
23. Find p so that A and B are independent events
(b) For what value of p, of A and B are mutually exclusive ?
– 1 – 2 1
24. Let A and B be the events such that P ( A ) = ,P(B)= ; P(A  B) =
2 3 4
Compute P (A | B) and P(B|A)
25. Two dice are thrown, Getting two numbers whose sum is divisible by 4 or 5 is considered a
success. Find the probability of success.
26. Suppose we have a box with 3 rd ball, 2 black and 5 white balls, Each time a balls drawn it is
returned to the box. What is the probability of drawing:
i. Either a red or a black ball?
ii. Either a white or a black ball?

27. X is a registered contractor with the government. Recently, X has submitted his tender for two
contracts, A and B. The probability of getting the contract A is ¼ ,the contract B is ½ and both
contracts A and B is 1/8 .Find the probability that X will get A or B.

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 18

INTRODUCTION TO STATISTICS , PROBABILITY AND CALCULUS F.Y. B. TECH.

28. In a group of 10 men, 6 are graduates. If 3 men are selected at random, what is the probability that
they consist of I) all graduates ii) at least one graduates iii) at most two graduates iv) no Graduate?

29. The letters of the word ‘SEMINAR’ are arranged at random. Find the probability that the vowels
occupy even places.

30. A salesman is known to sell a product in 3 out of 5 attempts while another salesman in 2 out of 5
attempts .Find the probability that

i) no sale will take place when they both try to sell the product.
ii) either of them will succeed in selling the product.

31. A restaurant is experiencing discontentment among its customers. It analyses that there are three
factors responsible viz. food quality, service quality and interior décor. By conducting an analysis, it
assesses the probabilities of discontentment with the three factors as 0.40, 0.35 and 0.25, respectively.
By conducting a survey among customers, it also evaluate the probabilities of a customer going away
discontented on account of these factors as :0.6, 0.8 and 0.5,respectively.With this information ,the
restaurant wants to know that if a customer is discontented ,what are the probabilities that it is so due
to food, service or interior décor.

32. In a random arrangement of the letters of the word “LAPTOP “, find the probability that all two
vowels come together.

33. A committee of four is to be formed from 3 engineers, 4 economists, 2 statisticians and 1 chartered
accountant.
34. What is the probability that each of the four categories of profession is included in the committee?
35. What is the probability that the committee consists of the charted accountant and at least one
engineer?

36. Four cards are drawn at random from a well shuffled pack of 52 cards. Find the probability that
Two cards of are red and two are black
All cards are of same suits
All cards are of same suits
One is king.
37. In an automobile, the shaft (if converts translator motion to rotator motion) can fail either due to
failure of bearing or failure of slider crankshaft mechanism. The probabilities of failure of bearing
and crankshaft are 0.2 and 0.3,respectively, and the probabilities of failure of shaft due to failure of
bearing an due to failure of crankshaft are 0.5 and 0.6,respectively.If the shaft has failed in an
automobile, what are the probabilities that it failed due to either failure of bearing or crankshaft?

JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 19