NCERT Textbook: Probability | NCERT Textbooks (Class 6 to Class 12) - CTET & State TET PDF Download

 Page 1


 406 MATHEMATICS
v
The theory of probabilities is simply the Science of logic
quantitatively treated. – C.S. PEIRCE v
13.1  Introduction
In earlier Classes, we have studied the probability as a
measure of uncertainty of events in a random experiment.
We discussed the axiomatic approach formulated by
Russian Mathematician, A.N. Kolmogorov (1903-1987)
and treated probability as a function of outcomes of the
experiment. W e have also established equivalence between
the axiomatic theory and the classical theory of probability
in case of equally likely outcomes. On the basis of this
relationship, we obtained probabilities of events associated
with discrete sample spaces. We have also studied the
addition rule of probability. In this chapter, we shall discuss
the important concept of conditional probability of an event
given that another event has occurred, which will be helpful
in understanding the Bayes' theorem, multiplication rule of
probability and independence of events. W e shall also learn
an  important concept of  random variable and its probability
distribution and also the mean and variance of a probability  distribution. In the last
section of the chapter, we shall study an important discrete probability distribution
called  Binomial distribution. Throughout this chapter, we shall take up the experiments
having equally likely outcomes, unless stated otherwise.
13.2  Conditional Probability
Uptill now in probability, we have discussed the methods of finding the probability of
events. If we have two events from the same sample space, does the information
about the occurrence of one of the events affect the probability of the other event? Let
us try to answer this question by taking up a random experiment in which the outcomes
are equally likely to occur.
Consider the experiment of tossing three fair coins. The sample space of the
experiment is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Chapter 13
PROBABILITY
Pierre de Fermat
(1601-1665)
Rationalised 2023-24
Page 2


 406 MATHEMATICS
v
The theory of probabilities is simply the Science of logic
quantitatively treated. – C.S. PEIRCE v
13.1  Introduction
In earlier Classes, we have studied the probability as a
measure of uncertainty of events in a random experiment.
We discussed the axiomatic approach formulated by
Russian Mathematician, A.N. Kolmogorov (1903-1987)
and treated probability as a function of outcomes of the
experiment. W e have also established equivalence between
the axiomatic theory and the classical theory of probability
in case of equally likely outcomes. On the basis of this
relationship, we obtained probabilities of events associated
with discrete sample spaces. We have also studied the
addition rule of probability. In this chapter, we shall discuss
the important concept of conditional probability of an event
given that another event has occurred, which will be helpful
in understanding the Bayes' theorem, multiplication rule of
probability and independence of events. W e shall also learn
an  important concept of  random variable and its probability
distribution and also the mean and variance of a probability  distribution. In the last
section of the chapter, we shall study an important discrete probability distribution
called  Binomial distribution. Throughout this chapter, we shall take up the experiments
having equally likely outcomes, unless stated otherwise.
13.2  Conditional Probability
Uptill now in probability, we have discussed the methods of finding the probability of
events. If we have two events from the same sample space, does the information
about the occurrence of one of the events affect the probability of the other event? Let
us try to answer this question by taking up a random experiment in which the outcomes
are equally likely to occur.
Consider the experiment of tossing three fair coins. The sample space of the
experiment is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Chapter 13
PROBABILITY
Pierre de Fermat
(1601-1665)
Rationalised 2023-24
PROBABILITY      407
Since the coins are fair, we can assign the probability  
1
8
  to each sample point. Let
E be the event ‘at least two heads appear’ and  F be the event ‘first coin shows tail’.
Then
E = {HHH, HHT, HTH, THH}
and F = {THH, THT, TTH, TTT}
Therefore P(E) = P ({HHH}) + P ({HHT}) + P ({HTH}) + P ({THH})
=
1 1 1 1 1
8 8 8 8 2
+ + + =
 (Why ?)
and P(F) = P ({THH}) + P ({THT}) + P ({TTH}) + P ({TTT})
=
1 1 1 1 1
8 8 8 8 2
+ + + =
Also E n F = {THH}
with P(E n F) = P({THH}) = 
1
8
Now, suppose we are given that the first coin shows tail, i.e. F occurs, then what is
the probability of occurrence of E? With the information of occurrence of F, we are
sure that the cases in which first coin does not result into a tail should not be considered
while finding the probability of E. This information reduces our sample space from the
set S to its subset F for the event E. In other words, the additional information really
amounts to telling us that the situation may be considered as being that of a new
random experiment for which the sample space consists of all those outcomes only
which are favourable to the occurrence of the event F.
Now, the sample point of F which is favourable to event E is THH.
Thus, Probability of E considering F as the sample space = 
1
4
,
or Probability of E given that the event F has occurred = 
1
4
This probability of the event E is called the conditional probability of E given
that F has already occurred, and is denoted by P (E|F).
Thus P(E|F) =
1
4
Note that the elements of F which favour the event E are the common elements of
E and F, i.e. the sample points of E n F.
Rationalised 2023-24
Page 3


 406 MATHEMATICS
v
The theory of probabilities is simply the Science of logic
quantitatively treated. – C.S. PEIRCE v
13.1  Introduction
In earlier Classes, we have studied the probability as a
measure of uncertainty of events in a random experiment.
We discussed the axiomatic approach formulated by
Russian Mathematician, A.N. Kolmogorov (1903-1987)
and treated probability as a function of outcomes of the
experiment. W e have also established equivalence between
the axiomatic theory and the classical theory of probability
in case of equally likely outcomes. On the basis of this
relationship, we obtained probabilities of events associated
with discrete sample spaces. We have also studied the
addition rule of probability. In this chapter, we shall discuss
the important concept of conditional probability of an event
given that another event has occurred, which will be helpful
in understanding the Bayes' theorem, multiplication rule of
probability and independence of events. W e shall also learn
an  important concept of  random variable and its probability
distribution and also the mean and variance of a probability  distribution. In the last
section of the chapter, we shall study an important discrete probability distribution
called  Binomial distribution. Throughout this chapter, we shall take up the experiments
having equally likely outcomes, unless stated otherwise.
13.2  Conditional Probability
Uptill now in probability, we have discussed the methods of finding the probability of
events. If we have two events from the same sample space, does the information
about the occurrence of one of the events affect the probability of the other event? Let
us try to answer this question by taking up a random experiment in which the outcomes
are equally likely to occur.
Consider the experiment of tossing three fair coins. The sample space of the
experiment is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Chapter 13
PROBABILITY
Pierre de Fermat
(1601-1665)
Rationalised 2023-24
PROBABILITY      407
Since the coins are fair, we can assign the probability  
1
8
  to each sample point. Let
E be the event ‘at least two heads appear’ and  F be the event ‘first coin shows tail’.
Then
E = {HHH, HHT, HTH, THH}
and F = {THH, THT, TTH, TTT}
Therefore P(E) = P ({HHH}) + P ({HHT}) + P ({HTH}) + P ({THH})
=
1 1 1 1 1
8 8 8 8 2
+ + + =
 (Why ?)
and P(F) = P ({THH}) + P ({THT}) + P ({TTH}) + P ({TTT})
=
1 1 1 1 1
8 8 8 8 2
+ + + =
Also E n F = {THH}
with P(E n F) = P({THH}) = 
1
8
Now, suppose we are given that the first coin shows tail, i.e. F occurs, then what is
the probability of occurrence of E? With the information of occurrence of F, we are
sure that the cases in which first coin does not result into a tail should not be considered
while finding the probability of E. This information reduces our sample space from the
set S to its subset F for the event E. In other words, the additional information really
amounts to telling us that the situation may be considered as being that of a new
random experiment for which the sample space consists of all those outcomes only
which are favourable to the occurrence of the event F.
Now, the sample point of F which is favourable to event E is THH.
Thus, Probability of E considering F as the sample space = 
1
4
,
or Probability of E given that the event F has occurred = 
1
4
This probability of the event E is called the conditional probability of E given
that F has already occurred, and is denoted by P (E|F).
Thus P(E|F) =
1
4
Note that the elements of F which favour the event E are the common elements of
E and F, i.e. the sample points of E n F.
Rationalised 2023-24
 408 MATHEMATICS
Thus, we can also write the conditional probability of E given that F has occurred as
P(E|F) =
Numberof elementaryeventsfavourableto E F
Numberof elementaryevents whicharefavourable to F
n
=
(E F)
(F)
n
n
n
Dividing the numerator and the denominator by total number of elementary events
of the sample space, we see that P(E|F) can also be written as
P(E|F) =
(E F)
P(E F) (S)
(F)
P(F)
(S)
n
n
n
n
n
n
=
... (1)
Note that (1) is valid only when P(F) ? 0 i.e., F ? f (Why?)
Thus, we can define the conditional probability as follows :
Definition 1 If  E and F are two events associated with the same sample space of a
random experiment, the conditional probability of the event E given that F has occurred,
i.e. P (E|F) is given by
P(E|F) =
P(E F)
P(F)
n
 provided P(F) ? 0
13.2.1  Properties of conditional probability
Let E and F be events of a sample space S of an experiment, then we have
Property 1 P(S|F) = P(F|F) = 1
We know that
P(S|F) =
P(S F) P(F)
1
P(F) P(F)
n
= =
Also P(F|F) =
P(F F) P(F)
1
P(F) P(F)
n
= =
Thus P(S|F) = P(F|F) = 1
Property 2 If A and B are any two events of a sample space S and F is an event
of S such that P(F) ? 0, then
P((A ? B)|F) = P(A|F) + P(B|F) – P((A n B)|F)
Rationalised 2023-24
Page 4


 406 MATHEMATICS
v
The theory of probabilities is simply the Science of logic
quantitatively treated. – C.S. PEIRCE v
13.1  Introduction
In earlier Classes, we have studied the probability as a
measure of uncertainty of events in a random experiment.
We discussed the axiomatic approach formulated by
Russian Mathematician, A.N. Kolmogorov (1903-1987)
and treated probability as a function of outcomes of the
experiment. W e have also established equivalence between
the axiomatic theory and the classical theory of probability
in case of equally likely outcomes. On the basis of this
relationship, we obtained probabilities of events associated
with discrete sample spaces. We have also studied the
addition rule of probability. In this chapter, we shall discuss
the important concept of conditional probability of an event
given that another event has occurred, which will be helpful
in understanding the Bayes' theorem, multiplication rule of
probability and independence of events. W e shall also learn
an  important concept of  random variable and its probability
distribution and also the mean and variance of a probability  distribution. In the last
section of the chapter, we shall study an important discrete probability distribution
called  Binomial distribution. Throughout this chapter, we shall take up the experiments
having equally likely outcomes, unless stated otherwise.
13.2  Conditional Probability
Uptill now in probability, we have discussed the methods of finding the probability of
events. If we have two events from the same sample space, does the information
about the occurrence of one of the events affect the probability of the other event? Let
us try to answer this question by taking up a random experiment in which the outcomes
are equally likely to occur.
Consider the experiment of tossing three fair coins. The sample space of the
experiment is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Chapter 13
PROBABILITY
Pierre de Fermat
(1601-1665)
Rationalised 2023-24
PROBABILITY      407
Since the coins are fair, we can assign the probability  
1
8
  to each sample point. Let
E be the event ‘at least two heads appear’ and  F be the event ‘first coin shows tail’.
Then
E = {HHH, HHT, HTH, THH}
and F = {THH, THT, TTH, TTT}
Therefore P(E) = P ({HHH}) + P ({HHT}) + P ({HTH}) + P ({THH})
=
1 1 1 1 1
8 8 8 8 2
+ + + =
 (Why ?)
and P(F) = P ({THH}) + P ({THT}) + P ({TTH}) + P ({TTT})
=
1 1 1 1 1
8 8 8 8 2
+ + + =
Also E n F = {THH}
with P(E n F) = P({THH}) = 
1
8
Now, suppose we are given that the first coin shows tail, i.e. F occurs, then what is
the probability of occurrence of E? With the information of occurrence of F, we are
sure that the cases in which first coin does not result into a tail should not be considered
while finding the probability of E. This information reduces our sample space from the
set S to its subset F for the event E. In other words, the additional information really
amounts to telling us that the situation may be considered as being that of a new
random experiment for which the sample space consists of all those outcomes only
which are favourable to the occurrence of the event F.
Now, the sample point of F which is favourable to event E is THH.
Thus, Probability of E considering F as the sample space = 
1
4
,
or Probability of E given that the event F has occurred = 
1
4
This probability of the event E is called the conditional probability of E given
that F has already occurred, and is denoted by P (E|F).
Thus P(E|F) =
1
4
Note that the elements of F which favour the event E are the common elements of
E and F, i.e. the sample points of E n F.
Rationalised 2023-24
 408 MATHEMATICS
Thus, we can also write the conditional probability of E given that F has occurred as
P(E|F) =
Numberof elementaryeventsfavourableto E F
Numberof elementaryevents whicharefavourable to F
n
=
(E F)
(F)
n
n
n
Dividing the numerator and the denominator by total number of elementary events
of the sample space, we see that P(E|F) can also be written as
P(E|F) =
(E F)
P(E F) (S)
(F)
P(F)
(S)
n
n
n
n
n
n
=
... (1)
Note that (1) is valid only when P(F) ? 0 i.e., F ? f (Why?)
Thus, we can define the conditional probability as follows :
Definition 1 If  E and F are two events associated with the same sample space of a
random experiment, the conditional probability of the event E given that F has occurred,
i.e. P (E|F) is given by
P(E|F) =
P(E F)
P(F)
n
 provided P(F) ? 0
13.2.1  Properties of conditional probability
Let E and F be events of a sample space S of an experiment, then we have
Property 1 P(S|F) = P(F|F) = 1
We know that
P(S|F) =
P(S F) P(F)
1
P(F) P(F)
n
= =
Also P(F|F) =
P(F F) P(F)
1
P(F) P(F)
n
= =
Thus P(S|F) = P(F|F) = 1
Property 2 If A and B are any two events of a sample space S and F is an event
of S such that P(F) ? 0, then
P((A ? B)|F) = P(A|F) + P(B|F) – P((A n B)|F)
Rationalised 2023-24
PROBABILITY      409
In particular, if A and B are disjoint events, then
P((A?B)|F) = P(A|F) + P(B|F)
We have
P((A?B)|F) =
P[(A B) F]
P(F)
? n
=
P[(A F) (B F)]
P(F)
n ? n
(by distributive law of union of sets over  intersection)
=
P(A F)+ P(B F) – P(A B F)
P(F)
n n n n
=
P(A F) P(B F) P[(A B) F]
P(F) P(F) P(F)
n n n n
+ -
= P(A|F) + P(B|F) – P((AnB)|F)
When A and B are disjoint events, then
P((A n B)|F) = 0
? P((A ? B)|F) = P(A|F) + P(B|F)
Property 3 P(E'|F) = 1 - P(E|F)
From Property 1, we know that P(S|F) = 1
? P(E ? E'|F) = 1   since  S = E ? E'
? P(E|F) + P (E'|F) = 1    since E and E' are disjoint events
Thus, P (E'|F) = 1 - P(E|F)
Let us now take up some examples.
Example 1 If P(A) = 
7
13
, P(B) = 
9
13
 and P(A n B) = 
4
13
, evaluate P(A|B).
Solution We have 
4
P(A B) 4
13
P(A|B)=
9
P(B) 9
13
n
= =
Example 2 A family has two children. What is the probability that both the children are
boys given that at least one of them is a boy ?
Rationalised 2023-24
Page 5


 406 MATHEMATICS
v
The theory of probabilities is simply the Science of logic
quantitatively treated. – C.S. PEIRCE v
13.1  Introduction
In earlier Classes, we have studied the probability as a
measure of uncertainty of events in a random experiment.
We discussed the axiomatic approach formulated by
Russian Mathematician, A.N. Kolmogorov (1903-1987)
and treated probability as a function of outcomes of the
experiment. W e have also established equivalence between
the axiomatic theory and the classical theory of probability
in case of equally likely outcomes. On the basis of this
relationship, we obtained probabilities of events associated
with discrete sample spaces. We have also studied the
addition rule of probability. In this chapter, we shall discuss
the important concept of conditional probability of an event
given that another event has occurred, which will be helpful
in understanding the Bayes' theorem, multiplication rule of
probability and independence of events. W e shall also learn
an  important concept of  random variable and its probability
distribution and also the mean and variance of a probability  distribution. In the last
section of the chapter, we shall study an important discrete probability distribution
called  Binomial distribution. Throughout this chapter, we shall take up the experiments
having equally likely outcomes, unless stated otherwise.
13.2  Conditional Probability
Uptill now in probability, we have discussed the methods of finding the probability of
events. If we have two events from the same sample space, does the information
about the occurrence of one of the events affect the probability of the other event? Let
us try to answer this question by taking up a random experiment in which the outcomes
are equally likely to occur.
Consider the experiment of tossing three fair coins. The sample space of the
experiment is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Chapter 13
PROBABILITY
Pierre de Fermat
(1601-1665)
Rationalised 2023-24
PROBABILITY      407
Since the coins are fair, we can assign the probability  
1
8
  to each sample point. Let
E be the event ‘at least two heads appear’ and  F be the event ‘first coin shows tail’.
Then
E = {HHH, HHT, HTH, THH}
and F = {THH, THT, TTH, TTT}
Therefore P(E) = P ({HHH}) + P ({HHT}) + P ({HTH}) + P ({THH})
=
1 1 1 1 1
8 8 8 8 2
+ + + =
 (Why ?)
and P(F) = P ({THH}) + P ({THT}) + P ({TTH}) + P ({TTT})
=
1 1 1 1 1
8 8 8 8 2
+ + + =
Also E n F = {THH}
with P(E n F) = P({THH}) = 
1
8
Now, suppose we are given that the first coin shows tail, i.e. F occurs, then what is
the probability of occurrence of E? With the information of occurrence of F, we are
sure that the cases in which first coin does not result into a tail should not be considered
while finding the probability of E. This information reduces our sample space from the
set S to its subset F for the event E. In other words, the additional information really
amounts to telling us that the situation may be considered as being that of a new
random experiment for which the sample space consists of all those outcomes only
which are favourable to the occurrence of the event F.
Now, the sample point of F which is favourable to event E is THH.
Thus, Probability of E considering F as the sample space = 
1
4
,
or Probability of E given that the event F has occurred = 
1
4
This probability of the event E is called the conditional probability of E given
that F has already occurred, and is denoted by P (E|F).
Thus P(E|F) =
1
4
Note that the elements of F which favour the event E are the common elements of
E and F, i.e. the sample points of E n F.
Rationalised 2023-24
 408 MATHEMATICS
Thus, we can also write the conditional probability of E given that F has occurred as
P(E|F) =
Numberof elementaryeventsfavourableto E F
Numberof elementaryevents whicharefavourable to F
n
=
(E F)
(F)
n
n
n
Dividing the numerator and the denominator by total number of elementary events
of the sample space, we see that P(E|F) can also be written as
P(E|F) =
(E F)
P(E F) (S)
(F)
P(F)
(S)
n
n
n
n
n
n
=
... (1)
Note that (1) is valid only when P(F) ? 0 i.e., F ? f (Why?)
Thus, we can define the conditional probability as follows :
Definition 1 If  E and F are two events associated with the same sample space of a
random experiment, the conditional probability of the event E given that F has occurred,
i.e. P (E|F) is given by
P(E|F) =
P(E F)
P(F)
n
 provided P(F) ? 0
13.2.1  Properties of conditional probability
Let E and F be events of a sample space S of an experiment, then we have
Property 1 P(S|F) = P(F|F) = 1
We know that
P(S|F) =
P(S F) P(F)
1
P(F) P(F)
n
= =
Also P(F|F) =
P(F F) P(F)
1
P(F) P(F)
n
= =
Thus P(S|F) = P(F|F) = 1
Property 2 If A and B are any two events of a sample space S and F is an event
of S such that P(F) ? 0, then
P((A ? B)|F) = P(A|F) + P(B|F) – P((A n B)|F)
Rationalised 2023-24
PROBABILITY      409
In particular, if A and B are disjoint events, then
P((A?B)|F) = P(A|F) + P(B|F)
We have
P((A?B)|F) =
P[(A B) F]
P(F)
? n
=
P[(A F) (B F)]
P(F)
n ? n
(by distributive law of union of sets over  intersection)
=
P(A F)+ P(B F) – P(A B F)
P(F)
n n n n
=
P(A F) P(B F) P[(A B) F]
P(F) P(F) P(F)
n n n n
+ -
= P(A|F) + P(B|F) – P((AnB)|F)
When A and B are disjoint events, then
P((A n B)|F) = 0
? P((A ? B)|F) = P(A|F) + P(B|F)
Property 3 P(E'|F) = 1 - P(E|F)
From Property 1, we know that P(S|F) = 1
? P(E ? E'|F) = 1   since  S = E ? E'
? P(E|F) + P (E'|F) = 1    since E and E' are disjoint events
Thus, P (E'|F) = 1 - P(E|F)
Let us now take up some examples.
Example 1 If P(A) = 
7
13
, P(B) = 
9
13
 and P(A n B) = 
4
13
, evaluate P(A|B).
Solution We have 
4
P(A B) 4
13
P(A|B)=
9
P(B) 9
13
n
= =
Example 2 A family has two children. What is the probability that both the children are
boys given that at least one of them is a boy ?
Rationalised 2023-24
 410 MATHEMATICS
Solution Let b stand for boy and g for girl. The sample space of the experiment is
S = {(b, b), (g, b), (b, g), (g, g)}
Let E and F denote the following events :
E : ‘both the children are boys’
F : ‘at least one of the child is a boy’
Then E = {(b,b)} and F = {(b,b), (g,b), (b,g)}
Now E n F = {(b,b)}
Thus P(F) =
3
4
 and P (E n F )= 
1
4
Therefore P(E|F) =
1
P(E F) 1
4
3
P( F) 3
4
n
= =
Example 3  Ten cards  numbered 1 to 10 are placed in a box, mixed up thoroughly and
then one card is drawn randomly. If it is known that the number on the drawn card is
more than 3, what is the probability that it is an even number?
Solution Let A be the event ‘the number on the card drawn is even’ and B be the
event ‘the number on the card drawn is  greater than 3’. We have to find P(A|B).
Now, the sample space of the experiment is S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Then A = {2, 4, 6, 8, 10},  B = {4, 5, 6, 7, 8, 9, 10}
and A n B = {4, 6, 8, 10}
Also P(A) =
5 7 4
, P(B) = and P(A B)
10 10 10
n =
Then P(A|B) =
4
P(A B) 4
10
7
P(B) 7
10
n
= =
Example 4 In a school, there are 1000 students, out of which 430 are girls. It is known
that out of 430, 10% of the girls study in class XII. What is the probability that a student
chosen randomly studies in Class XII given that the chosen student is a girl?
Solution Let E denote the event that a student chosen randomly studies in Class XII
and F be the event that the randomly chosen student is a girl. We have to find P (E|F).
Rationalised 2023-24