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Probability and Statistics 
 
  Basic Terms 
Random Experiment 
Consider an action which is repeated under essentially identical conditions. If it results in 
any one of the several possible outcomes, but it is not possible to predict which outcome 
will appear. Such an action is called as a Random Experiment. One performance of such 
an experiment is called as a Trial. 
 
Sample Space 
The set of all possible outcomes of a random experiment is called as the sample space. 
All the elements of the sample space together are called as ‘exhaustive cases’. The 
number of elements of the sample space i.e. the number of exhaustive cases is denoted 
by n(S) or N or n. 
 
Event 
Any subset of the sample space is called as an ‘Event’ and is denoted by some capital 
letter like A, B, C or A
1
, A
2
, A
3
,.. or B
1
, B
2
, ... etc. 
 
Favourable cases  
The cases which ensure the happening of an event A, are called as the cases favourable 
to the event A. The number of cases favourable to event A is denoted by n(A) or N
A
 or n
A
. 
 
Mutually Exclusive Events or Disjoint Events 
Two events A and B are said to be mutually exclusive or disjoint if A ? B = ?  
i.e. if there is no element common to A & B. 
 
Equally Likely Cases 
Cases are said to be equally likely if they all have the same chance of occurrence i.e. no 
case is preferred to any other case. 
 
Permutation 
A permutation is an arrangement of all or part of a set of objects. The number of 
permutations of n distinct objects taken r at a time is 
n
P
r
 = 
??
n!
nr! ?
 
 
 
Page 2


    
Probability and Statistics 
 
  Basic Terms 
Random Experiment 
Consider an action which is repeated under essentially identical conditions. If it results in 
any one of the several possible outcomes, but it is not possible to predict which outcome 
will appear. Such an action is called as a Random Experiment. One performance of such 
an experiment is called as a Trial. 
 
Sample Space 
The set of all possible outcomes of a random experiment is called as the sample space. 
All the elements of the sample space together are called as ‘exhaustive cases’. The 
number of elements of the sample space i.e. the number of exhaustive cases is denoted 
by n(S) or N or n. 
 
Event 
Any subset of the sample space is called as an ‘Event’ and is denoted by some capital 
letter like A, B, C or A
1
, A
2
, A
3
,.. or B
1
, B
2
, ... etc. 
 
Favourable cases  
The cases which ensure the happening of an event A, are called as the cases favourable 
to the event A. The number of cases favourable to event A is denoted by n(A) or N
A
 or n
A
. 
 
Mutually Exclusive Events or Disjoint Events 
Two events A and B are said to be mutually exclusive or disjoint if A ? B = ?  
i.e. if there is no element common to A & B. 
 
Equally Likely Cases 
Cases are said to be equally likely if they all have the same chance of occurrence i.e. no 
case is preferred to any other case. 
 
Permutation 
A permutation is an arrangement of all or part of a set of objects. The number of 
permutations of n distinct objects taken r at a time is 
n
P
r
 = 
??
n!
nr! ?
 
 
 
Notes on Probability and Statistics 
   
 
Note:  The number of permutations of n distinct objects is n!  i.e., 
n
P
n
 = n! 
 The number of permutations of n distinct objects arranged in a circle  
is (n ? 1)! 
 The number of distinct permutations of n things of which n
1
 are of one 
kind, n
2
 of a second kind … n
k
 of a k
th
 kind is 
12 k
n!
n !  n ! ......n !
 
 
Combination 
A combination is selection of all or part of a set of objects. The number of combinations of 
n distinct objects taken r at a time is 
    
??
n
r
n!
C=
r! n r ! ?
  
 
 
Note:  In a permutation, the order of arrangement of the objects is important. 
Thus abc is a different permutation from bca. 
In a combination, the order in which objects are selected does not matter.  
Thus abc and bca are the same combination. 
 
  Definition of Probability 
Consider a random experiment which results in a sample space containing n(S) cases 
which are exhaustive, mutually exclusive and equally likely. Suppose, out of n(S) cases, 
n(A) cases are favourable to an event A. Then the probability of event A is denoted by 
P(A) and is defined as follows. 
 
 
P(A) = 
n(A)
n(S)
 = 
number of cases favourable to event A
number of cases in the sample space S
 
 
 
 Complement of an event 
The complement of an event A is denoted by A and it contains all the elements of the 
sample space which do not belong to A.  
 
For example: Random experiment: an unbiased die is rolled. 
S = {1, 2, 3, 4, 5, 6} 
(i)  Let A: number on the die is a perfect square 
      ? A = {1, 4} ? A = {2, 3, 5, 6} 
Page 3


    
Probability and Statistics 
 
  Basic Terms 
Random Experiment 
Consider an action which is repeated under essentially identical conditions. If it results in 
any one of the several possible outcomes, but it is not possible to predict which outcome 
will appear. Such an action is called as a Random Experiment. One performance of such 
an experiment is called as a Trial. 
 
Sample Space 
The set of all possible outcomes of a random experiment is called as the sample space. 
All the elements of the sample space together are called as ‘exhaustive cases’. The 
number of elements of the sample space i.e. the number of exhaustive cases is denoted 
by n(S) or N or n. 
 
Event 
Any subset of the sample space is called as an ‘Event’ and is denoted by some capital 
letter like A, B, C or A
1
, A
2
, A
3
,.. or B
1
, B
2
, ... etc. 
 
Favourable cases  
The cases which ensure the happening of an event A, are called as the cases favourable 
to the event A. The number of cases favourable to event A is denoted by n(A) or N
A
 or n
A
. 
 
Mutually Exclusive Events or Disjoint Events 
Two events A and B are said to be mutually exclusive or disjoint if A ? B = ?  
i.e. if there is no element common to A & B. 
 
Equally Likely Cases 
Cases are said to be equally likely if they all have the same chance of occurrence i.e. no 
case is preferred to any other case. 
 
Permutation 
A permutation is an arrangement of all or part of a set of objects. The number of 
permutations of n distinct objects taken r at a time is 
n
P
r
 = 
??
n!
nr! ?
 
 
 
Notes on Probability and Statistics 
   
 
Note:  The number of permutations of n distinct objects is n!  i.e., 
n
P
n
 = n! 
 The number of permutations of n distinct objects arranged in a circle  
is (n ? 1)! 
 The number of distinct permutations of n things of which n
1
 are of one 
kind, n
2
 of a second kind … n
k
 of a k
th
 kind is 
12 k
n!
n !  n ! ......n !
 
 
Combination 
A combination is selection of all or part of a set of objects. The number of combinations of 
n distinct objects taken r at a time is 
    
??
n
r
n!
C=
r! n r ! ?
  
 
 
Note:  In a permutation, the order of arrangement of the objects is important. 
Thus abc is a different permutation from bca. 
In a combination, the order in which objects are selected does not matter.  
Thus abc and bca are the same combination. 
 
  Definition of Probability 
Consider a random experiment which results in a sample space containing n(S) cases 
which are exhaustive, mutually exclusive and equally likely. Suppose, out of n(S) cases, 
n(A) cases are favourable to an event A. Then the probability of event A is denoted by 
P(A) and is defined as follows. 
 
 
P(A) = 
n(A)
n(S)
 = 
number of cases favourable to event A
number of cases in the sample space S
 
 
 
 Complement of an event 
The complement of an event A is denoted by A and it contains all the elements of the 
sample space which do not belong to A.  
 
For example: Random experiment: an unbiased die is rolled. 
S = {1, 2, 3, 4, 5, 6} 
(i)  Let A: number on the die is a perfect square 
      ? A = {1, 4} ? A = {2, 3, 5, 6} 
    
(ii)  Let B: number on the die is a prime number 
      ? B = {2, 3, 5} ?B = {1, 4, 6} 
 
 
Note:  P(A) + P( A ) = 1 i.e. P(A) = 1 ? P( A ) 
  For any events A and B, ?? ? ? ? ?
?? PA=PAB+PAB 
 
 Independent Events 
Two events A & B are said to be independent if  
 
 
P(A ? B) = P(A).P(B) 
 
 
 
Note: If A & B are independent then 
  A & B are independent 
  
A & B are independent 
  
A &B are independent 
 
Theorems of Probability 
Addition Theorem 
If A and B are any two events then 
 
 
P(A ?B) = P(A) + P(B) – P(A ?B) 
 
 
 
 
Note: 1.    A ? B : either A or B or both i.e.  at least one of A & B 
   
AB ?  : neither A nor B  i.e.  none of A & B 
   A ?B & AB ? are complement to each other 
   ? P(AB ? ) = 1 – P(A ? B) 
   
  2. If A & B are mutually exclusive, P(A ? B) = 0 
   ?P(A ? B) = P(A) + P(B) 
 
  3. ?? ? ? ? ? ? ? ? ? ?? ?? ? ?
1 2 3 1 231 2
PA A A =P A +PA +PA P A A  
      ? ? ? ? ? ? ??? ? ? ? ?
23 3 1 1 23
PA A P A A +PA A A 
 
Page 4


    
Probability and Statistics 
 
  Basic Terms 
Random Experiment 
Consider an action which is repeated under essentially identical conditions. If it results in 
any one of the several possible outcomes, but it is not possible to predict which outcome 
will appear. Such an action is called as a Random Experiment. One performance of such 
an experiment is called as a Trial. 
 
Sample Space 
The set of all possible outcomes of a random experiment is called as the sample space. 
All the elements of the sample space together are called as ‘exhaustive cases’. The 
number of elements of the sample space i.e. the number of exhaustive cases is denoted 
by n(S) or N or n. 
 
Event 
Any subset of the sample space is called as an ‘Event’ and is denoted by some capital 
letter like A, B, C or A
1
, A
2
, A
3
,.. or B
1
, B
2
, ... etc. 
 
Favourable cases  
The cases which ensure the happening of an event A, are called as the cases favourable 
to the event A. The number of cases favourable to event A is denoted by n(A) or N
A
 or n
A
. 
 
Mutually Exclusive Events or Disjoint Events 
Two events A and B are said to be mutually exclusive or disjoint if A ? B = ?  
i.e. if there is no element common to A & B. 
 
Equally Likely Cases 
Cases are said to be equally likely if they all have the same chance of occurrence i.e. no 
case is preferred to any other case. 
 
Permutation 
A permutation is an arrangement of all or part of a set of objects. The number of 
permutations of n distinct objects taken r at a time is 
n
P
r
 = 
??
n!
nr! ?
 
 
 
Notes on Probability and Statistics 
   
 
Note:  The number of permutations of n distinct objects is n!  i.e., 
n
P
n
 = n! 
 The number of permutations of n distinct objects arranged in a circle  
is (n ? 1)! 
 The number of distinct permutations of n things of which n
1
 are of one 
kind, n
2
 of a second kind … n
k
 of a k
th
 kind is 
12 k
n!
n !  n ! ......n !
 
 
Combination 
A combination is selection of all or part of a set of objects. The number of combinations of 
n distinct objects taken r at a time is 
    
??
n
r
n!
C=
r! n r ! ?
  
 
 
Note:  In a permutation, the order of arrangement of the objects is important. 
Thus abc is a different permutation from bca. 
In a combination, the order in which objects are selected does not matter.  
Thus abc and bca are the same combination. 
 
  Definition of Probability 
Consider a random experiment which results in a sample space containing n(S) cases 
which are exhaustive, mutually exclusive and equally likely. Suppose, out of n(S) cases, 
n(A) cases are favourable to an event A. Then the probability of event A is denoted by 
P(A) and is defined as follows. 
 
 
P(A) = 
n(A)
n(S)
 = 
number of cases favourable to event A
number of cases in the sample space S
 
 
 
 Complement of an event 
The complement of an event A is denoted by A and it contains all the elements of the 
sample space which do not belong to A.  
 
For example: Random experiment: an unbiased die is rolled. 
S = {1, 2, 3, 4, 5, 6} 
(i)  Let A: number on the die is a perfect square 
      ? A = {1, 4} ? A = {2, 3, 5, 6} 
    
(ii)  Let B: number on the die is a prime number 
      ? B = {2, 3, 5} ?B = {1, 4, 6} 
 
 
Note:  P(A) + P( A ) = 1 i.e. P(A) = 1 ? P( A ) 
  For any events A and B, ?? ? ? ? ?
?? PA=PAB+PAB 
 
 Independent Events 
Two events A & B are said to be independent if  
 
 
P(A ? B) = P(A).P(B) 
 
 
 
Note: If A & B are independent then 
  A & B are independent 
  
A & B are independent 
  
A &B are independent 
 
Theorems of Probability 
Addition Theorem 
If A and B are any two events then 
 
 
P(A ?B) = P(A) + P(B) – P(A ?B) 
 
 
 
 
Note: 1.    A ? B : either A or B or both i.e.  at least one of A & B 
   
AB ?  : neither A nor B  i.e.  none of A & B 
   A ?B & AB ? are complement to each other 
   ? P(AB ? ) = 1 – P(A ? B) 
   
  2. If A & B are mutually exclusive, P(A ? B) = 0 
   ?P(A ? B) = P(A) + P(B) 
 
  3. ?? ? ? ? ? ? ? ? ? ?? ?? ? ?
1 2 3 1 231 2
PA A A =P A +PA +PA P A A  
      ? ? ? ? ? ? ??? ? ? ? ?
23 3 1 1 23
PA A P A A +PA A A 
 
Notes on Probability and Statistics 
   
Multiplication Theorem 
If A & B are any two events then 
 
 
P(A ? B) = P(A).P(B/A) = P(B).P(A/B) 
 
 
1. Conditional probability of occurrence of event B given that event A has already 
 occurred. 
  
 
P(B/A) = 
? ?
??
PA B
PA
?
 
 
2. Conditional probability of occurrence of event A given that event B has already 
 occurred  
 
 
P(A/B) = 
? ?
??
PA B
PB
?
 
 
Bayes’ Theorem 
Suppose that a sample space S is a union of mutually disjoint events B
1
, B
2
, B
3
, ..., B
n
, 
suppose A is an event in S, and suppose A and all the B
i
’s have nonzero probabilities. If 
k is an integer with 1 = k = n, then 
 
 
P(B
k
 / A) = 
? ? ? ?
? ??? ???? ????
kk
11 2 2 n n
PA/B P B
P A / B P B +P A / B P B +...+P A / B P B
 
 
 
 
Solved Example 1 : 
A single die is tossed. Find the probability 
of a 2 or 5 turning up. 
Solution : 
The sample space is S = {1, 2, 3, 4, 5, 6} 
 P(1) = P(2)  = … = P(6) = 
1
6
 
The event that either 2 or 5 turns up in 
indicated by 2 ? 5. Thus 
 ? ? ?? ??
11 1
P2 5 P 2 P 5
66 3
?? ? ? ? ? 
 
Solved Example 2 : 
A coin is tossed twice. What is the 
probability that at least one head occurs ? 
Solution : 
The sample space is 
 
 S = {HH, HT, TH, TT} 
 
Probability of each outcomes = 1/4 
 
Probability of atleast one head occurring is 
 P(A) = 
111 3
44 4 4
? ?? 
 
Page 5


    
Probability and Statistics 
 
  Basic Terms 
Random Experiment 
Consider an action which is repeated under essentially identical conditions. If it results in 
any one of the several possible outcomes, but it is not possible to predict which outcome 
will appear. Such an action is called as a Random Experiment. One performance of such 
an experiment is called as a Trial. 
 
Sample Space 
The set of all possible outcomes of a random experiment is called as the sample space. 
All the elements of the sample space together are called as ‘exhaustive cases’. The 
number of elements of the sample space i.e. the number of exhaustive cases is denoted 
by n(S) or N or n. 
 
Event 
Any subset of the sample space is called as an ‘Event’ and is denoted by some capital 
letter like A, B, C or A
1
, A
2
, A
3
,.. or B
1
, B
2
, ... etc. 
 
Favourable cases  
The cases which ensure the happening of an event A, are called as the cases favourable 
to the event A. The number of cases favourable to event A is denoted by n(A) or N
A
 or n
A
. 
 
Mutually Exclusive Events or Disjoint Events 
Two events A and B are said to be mutually exclusive or disjoint if A ? B = ?  
i.e. if there is no element common to A & B. 
 
Equally Likely Cases 
Cases are said to be equally likely if they all have the same chance of occurrence i.e. no 
case is preferred to any other case. 
 
Permutation 
A permutation is an arrangement of all or part of a set of objects. The number of 
permutations of n distinct objects taken r at a time is 
n
P
r
 = 
??
n!
nr! ?
 
 
 
Notes on Probability and Statistics 
   
 
Note:  The number of permutations of n distinct objects is n!  i.e., 
n
P
n
 = n! 
 The number of permutations of n distinct objects arranged in a circle  
is (n ? 1)! 
 The number of distinct permutations of n things of which n
1
 are of one 
kind, n
2
 of a second kind … n
k
 of a k
th
 kind is 
12 k
n!
n !  n ! ......n !
 
 
Combination 
A combination is selection of all or part of a set of objects. The number of combinations of 
n distinct objects taken r at a time is 
    
??
n
r
n!
C=
r! n r ! ?
  
 
 
Note:  In a permutation, the order of arrangement of the objects is important. 
Thus abc is a different permutation from bca. 
In a combination, the order in which objects are selected does not matter.  
Thus abc and bca are the same combination. 
 
  Definition of Probability 
Consider a random experiment which results in a sample space containing n(S) cases 
which are exhaustive, mutually exclusive and equally likely. Suppose, out of n(S) cases, 
n(A) cases are favourable to an event A. Then the probability of event A is denoted by 
P(A) and is defined as follows. 
 
 
P(A) = 
n(A)
n(S)
 = 
number of cases favourable to event A
number of cases in the sample space S
 
 
 
 Complement of an event 
The complement of an event A is denoted by A and it contains all the elements of the 
sample space which do not belong to A.  
 
For example: Random experiment: an unbiased die is rolled. 
S = {1, 2, 3, 4, 5, 6} 
(i)  Let A: number on the die is a perfect square 
      ? A = {1, 4} ? A = {2, 3, 5, 6} 
    
(ii)  Let B: number on the die is a prime number 
      ? B = {2, 3, 5} ?B = {1, 4, 6} 
 
 
Note:  P(A) + P( A ) = 1 i.e. P(A) = 1 ? P( A ) 
  For any events A and B, ?? ? ? ? ?
?? PA=PAB+PAB 
 
 Independent Events 
Two events A & B are said to be independent if  
 
 
P(A ? B) = P(A).P(B) 
 
 
 
Note: If A & B are independent then 
  A & B are independent 
  
A & B are independent 
  
A &B are independent 
 
Theorems of Probability 
Addition Theorem 
If A and B are any two events then 
 
 
P(A ?B) = P(A) + P(B) – P(A ?B) 
 
 
 
 
Note: 1.    A ? B : either A or B or both i.e.  at least one of A & B 
   
AB ?  : neither A nor B  i.e.  none of A & B 
   A ?B & AB ? are complement to each other 
   ? P(AB ? ) = 1 – P(A ? B) 
   
  2. If A & B are mutually exclusive, P(A ? B) = 0 
   ?P(A ? B) = P(A) + P(B) 
 
  3. ?? ? ? ? ? ? ? ? ? ?? ?? ? ?
1 2 3 1 231 2
PA A A =P A +PA +PA P A A  
      ? ? ? ? ? ? ??? ? ? ? ?
23 3 1 1 23
PA A P A A +PA A A 
 
Notes on Probability and Statistics 
   
Multiplication Theorem 
If A & B are any two events then 
 
 
P(A ? B) = P(A).P(B/A) = P(B).P(A/B) 
 
 
1. Conditional probability of occurrence of event B given that event A has already 
 occurred. 
  
 
P(B/A) = 
? ?
??
PA B
PA
?
 
 
2. Conditional probability of occurrence of event A given that event B has already 
 occurred  
 
 
P(A/B) = 
? ?
??
PA B
PB
?
 
 
Bayes’ Theorem 
Suppose that a sample space S is a union of mutually disjoint events B
1
, B
2
, B
3
, ..., B
n
, 
suppose A is an event in S, and suppose A and all the B
i
’s have nonzero probabilities. If 
k is an integer with 1 = k = n, then 
 
 
P(B
k
 / A) = 
? ? ? ?
? ??? ???? ????
kk
11 2 2 n n
PA/B P B
P A / B P B +P A / B P B +...+P A / B P B
 
 
 
 
Solved Example 1 : 
A single die is tossed. Find the probability 
of a 2 or 5 turning up. 
Solution : 
The sample space is S = {1, 2, 3, 4, 5, 6} 
 P(1) = P(2)  = … = P(6) = 
1
6
 
The event that either 2 or 5 turns up in 
indicated by 2 ? 5. Thus 
 ? ? ?? ??
11 1
P2 5 P 2 P 5
66 3
?? ? ? ? ? 
 
Solved Example 2 : 
A coin is tossed twice. What is the 
probability that at least one head occurs ? 
Solution : 
The sample space is 
 
 S = {HH, HT, TH, TT} 
 
Probability of each outcomes = 1/4 
 
Probability of atleast one head occurring is 
 P(A) = 
111 3
44 4 4
? ?? 
 
    
Solved Example 3 : 
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).  
Solution : 
 S = {1, 2, 3, 4, 5, 6 } 
We assign a probability of w to each odd 
number and a probability of 2w to each 
even number. Since the sum of the 
probabilities must be 1, we have 9w = 1 or 
w = 1/9.   
 
Hence probabilities of 1/9 and 2/9 are 
assigned to each odd and even number 
respectively. 
?  E = {1, 2, 3}  
and   P(E) = 
12 1 4
99 9 9
? ?? 
 
Solved Example 4 : 
In the above example let A be the event 
that an even number turns up and let B be 
the event that a number divisible by 3 
occurs.  Find P(A ? B) and P(A ? B). 
Solution : 
A = {2, 4, 6}  and  
B = {3, 6} 
We have, 
A ? B = {2, 3, 4, 6}   and   
A ? B = {6} 
By assigning a probability of 1/9 to each 
odd number and 2/9 to each even number 
??
21 2 2 7
PA B
99 9 9 9
?? ? ? ? ?  
 and   ??
2
PA B
9
? ? 
 
Solved Example 5 : 
A mixture of candies 6 mints, 4 toffees and 
3 chocolates. If a person makes a random 
selection of one of these candies, find the 
probability of getting (a) a mint, or 
(b) a toffee or a chocolate. 
Solution : 
(a) Since 6 of the 13 candies are mints, 
the probability of event M, selecting 
mint at random, is 
  P(M) = 
6
13
 
 
(b) Since 7 of the 13 candies are toffees 
or chocolates it follows that  
  ??
7
PT C
13
?? 
 
Solved Example 6 : 
In a poker hand consisting of 5 cards, find 
the probability of holding 2 aces and  
3 jacks. 
Solution : 
The number of ways of being dealt 2 aces 
from 4 is 
  
4
2
4!
C6
2!2!
??? 
The number of ways of being dealt 3 jacks 
from 4 is 
  
4
3
4!
C4
3!1!
? ? 
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