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1 | 6718-6721 | � Corner point method for solving a linear programming problem The method
comprises of the following steps:
(i) Find the feasible region of the linear programming problem and determine
its corner points (vertices) (ii) Evaluate the objective function Z = ax + by at each corner point Let M
and m respectively be the largest and smallest values at these points |
1 | 6719-6722 | The method
comprises of the following steps:
(i) Find the feasible region of the linear programming problem and determine
its corner points (vertices) (ii) Evaluate the objective function Z = ax + by at each corner point Let M
and m respectively be the largest and smallest values at these points (iii) If the feasible region is bounded, M and m respectively are the maximum
and minimum values of the objective function |
1 | 6720-6723 | (ii) Evaluate the objective function Z = ax + by at each corner point Let M
and m respectively be the largest and smallest values at these points (iii) If the feasible region is bounded, M and m respectively are the maximum
and minimum values of the objective function If the feasible region is unbounded, then
(i) M is the maximum value of the objective function, if the open half plane
determined by ax + by > M has no point in common with the feasible
region |
1 | 6721-6724 | Let M
and m respectively be the largest and smallest values at these points (iii) If the feasible region is bounded, M and m respectively are the maximum
and minimum values of the objective function If the feasible region is unbounded, then
(i) M is the maximum value of the objective function, if the open half plane
determined by ax + by > M has no point in common with the feasible
region Otherwise, the objective function has no maximum value |
1 | 6722-6725 | (iii) If the feasible region is bounded, M and m respectively are the maximum
and minimum values of the objective function If the feasible region is unbounded, then
(i) M is the maximum value of the objective function, if the open half plane
determined by ax + by > M has no point in common with the feasible
region Otherwise, the objective function has no maximum value (ii) m is the minimum value of the objective function, if the open half plane
determined by ax + by < m has no point in common with the feasible
region |
1 | 6723-6726 | If the feasible region is unbounded, then
(i) M is the maximum value of the objective function, if the open half plane
determined by ax + by > M has no point in common with the feasible
region Otherwise, the objective function has no maximum value (ii) m is the minimum value of the objective function, if the open half plane
determined by ax + by < m has no point in common with the feasible
region Otherwise, the objective function has no minimum value |
1 | 6724-6727 | Otherwise, the objective function has no maximum value (ii) m is the minimum value of the objective function, if the open half plane
determined by ax + by < m has no point in common with the feasible
region Otherwise, the objective function has no minimum value � If two corner points of the feasible region are both optimal solutions of the
same type, i |
1 | 6725-6728 | (ii) m is the minimum value of the objective function, if the open half plane
determined by ax + by < m has no point in common with the feasible
region Otherwise, the objective function has no minimum value � If two corner points of the feasible region are both optimal solutions of the
same type, i e |
1 | 6726-6729 | Otherwise, the objective function has no minimum value � If two corner points of the feasible region are both optimal solutions of the
same type, i e , both produce the same maximum or minimum, then any point
on the line segment joining these two points is also an optimal solution of the
same type |
1 | 6727-6730 | � If two corner points of the feasible region are both optimal solutions of the
same type, i e , both produce the same maximum or minimum, then any point
on the line segment joining these two points is also an optimal solution of the
same type © NCERT
not to be republished
530
MATHEMATICS
Historical Note
In the World War II, when the war operations had to be planned to economise
expenditure, maximise damage to the enemy, linear programming problems
came to the forefront |
1 | 6728-6731 | e , both produce the same maximum or minimum, then any point
on the line segment joining these two points is also an optimal solution of the
same type © NCERT
not to be republished
530
MATHEMATICS
Historical Note
In the World War II, when the war operations had to be planned to economise
expenditure, maximise damage to the enemy, linear programming problems
came to the forefront The first problem in linear programming was formulated in 1941 by the Russian
mathematician, L |
1 | 6729-6732 | , both produce the same maximum or minimum, then any point
on the line segment joining these two points is also an optimal solution of the
same type © NCERT
not to be republished
530
MATHEMATICS
Historical Note
In the World War II, when the war operations had to be planned to economise
expenditure, maximise damage to the enemy, linear programming problems
came to the forefront The first problem in linear programming was formulated in 1941 by the Russian
mathematician, L Kantorovich and the American economist, F |
1 | 6730-6733 | © NCERT
not to be republished
530
MATHEMATICS
Historical Note
In the World War II, when the war operations had to be planned to economise
expenditure, maximise damage to the enemy, linear programming problems
came to the forefront The first problem in linear programming was formulated in 1941 by the Russian
mathematician, L Kantorovich and the American economist, F L |
1 | 6731-6734 | The first problem in linear programming was formulated in 1941 by the Russian
mathematician, L Kantorovich and the American economist, F L Hitchcock,
both of whom worked at it independently of each other |
1 | 6732-6735 | Kantorovich and the American economist, F L Hitchcock,
both of whom worked at it independently of each other This was the well
known transportation problem |
1 | 6733-6736 | L Hitchcock,
both of whom worked at it independently of each other This was the well
known transportation problem In 1945, an English economist, G |
1 | 6734-6737 | Hitchcock,
both of whom worked at it independently of each other This was the well
known transportation problem In 1945, an English economist, G Stigler,
described yet another linear programming problem – that of determining an
optimal diet |
1 | 6735-6738 | This was the well
known transportation problem In 1945, an English economist, G Stigler,
described yet another linear programming problem – that of determining an
optimal diet In 1947, the American economist, G |
1 | 6736-6739 | In 1945, an English economist, G Stigler,
described yet another linear programming problem – that of determining an
optimal diet In 1947, the American economist, G B |
1 | 6737-6740 | Stigler,
described yet another linear programming problem – that of determining an
optimal diet In 1947, the American economist, G B Dantzig suggested an efficient method
known as the simplex method which is an iterative procedure to solve any
linear programming problem in a finite number of steps |
1 | 6738-6741 | In 1947, the American economist, G B Dantzig suggested an efficient method
known as the simplex method which is an iterative procedure to solve any
linear programming problem in a finite number of steps L |
1 | 6739-6742 | B Dantzig suggested an efficient method
known as the simplex method which is an iterative procedure to solve any
linear programming problem in a finite number of steps L Katorovich and American mathematical economist, T |
1 | 6740-6743 | Dantzig suggested an efficient method
known as the simplex method which is an iterative procedure to solve any
linear programming problem in a finite number of steps L Katorovich and American mathematical economist, T C |
1 | 6741-6744 | L Katorovich and American mathematical economist, T C Koopmans were
awarded the nobel prize in the year 1975 in economics for their pioneering
work in linear programming |
1 | 6742-6745 | Katorovich and American mathematical economist, T C Koopmans were
awarded the nobel prize in the year 1975 in economics for their pioneering
work in linear programming With the advent of computers and the necessary
softwares, it has become possible to apply linear programming model to
increasingly complex problems in many areas |
1 | 6743-6746 | C Koopmans were
awarded the nobel prize in the year 1975 in economics for their pioneering
work in linear programming With the advent of computers and the necessary
softwares, it has become possible to apply linear programming model to
increasingly complex problems in many areas —�
�
�
�
�—
© NCERT
not to be republished
PROBABILITY 531
�The theory of probabilities is simply the Science of logic
quantitatively treated |
1 | 6744-6747 | Koopmans were
awarded the nobel prize in the year 1975 in economics for their pioneering
work in linear programming With the advent of computers and the necessary
softwares, it has become possible to apply linear programming model to
increasingly complex problems in many areas —�
�
�
�
�—
© NCERT
not to be republished
PROBABILITY 531
�The theory of probabilities is simply the Science of logic
quantitatively treated – C |
1 | 6745-6748 | With the advent of computers and the necessary
softwares, it has become possible to apply linear programming model to
increasingly complex problems in many areas —�
�
�
�
�—
© NCERT
not to be republished
PROBABILITY 531
�The theory of probabilities is simply the Science of logic
quantitatively treated – C S |
1 | 6746-6749 | —�
�
�
�
�—
© NCERT
not to be republished
PROBABILITY 531
�The theory of probabilities is simply the Science of logic
quantitatively treated – C S PEIRCE �
13 |
1 | 6747-6750 | – C S PEIRCE �
13 1 Introduction
In earlier Classes, we have studied the probability as a
measure of uncertainty of events in a random experiment |
1 | 6748-6751 | S PEIRCE �
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 |
1 | 6749-6752 | PEIRCE �
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 |
1 | 6750-6753 | 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 |
1 | 6751-6754 | 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 We have also established equivalence between
the axiomatic theory and the classical theory of probability
in case of equally likely outcomes |
1 | 6752-6755 | N Kolmogorov (1903-1987)
and treated probability as a function of outcomes of the
experiment We 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 |
1 | 6753-6756 | Kolmogorov (1903-1987)
and treated probability as a function of outcomes of the
experiment We 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 |
1 | 6754-6757 | We 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 |
1 | 6755-6758 | 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 We shall also learn
an important concept of random variable and its probability
distribution and also the mean and variance of a probability distribution |
1 | 6756-6759 | 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 We 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 |
1 | 6757-6760 | 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 We 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 |
1 | 6758-6761 | We 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 |
1 | 6759-6762 | 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 |
1 | 6760-6763 | 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 |
1 | 6761-6764 | 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 |
1 | 6762-6765 | 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 |
1 | 6763-6766 | 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)
© NCERT
not to be republished
532
MATHEMATICS
Since the coins are fair, we can assign the probability 1
8 to each sample point |
1 | 6764-6767 | 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)
© NCERT
not to be republished
532
MATHEMATICS
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’ |
1 | 6765-6768 | 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)
© NCERT
not to be republished
532
MATHEMATICS
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 |
1 | 6766-6769 | The sample space of the
experiment is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Chapter 13
PROBABILITY
Pierre de Fermat
(1601-1665)
© NCERT
not to be republished
532
MATHEMATICS
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 ∩ F = {THH}
with
P(E ∩ F) = P({THH}) = 1
8
Now, suppose we are given that the first coin shows tail, i |
1 | 6767-6770 | 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 ∩ F = {THH}
with
P(E ∩ F) = P({THH}) = 1
8
Now, suppose we are given that the first coin shows tail, i e |
1 | 6768-6771 | 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 ∩ F = {THH}
with
P(E ∩ 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 |
1 | 6769-6772 | )
and
P(F) = P ({THH}) + P ({THT}) + P ({TTH}) + P ({TTT})
= 1
1
1
1
1
8
8
8
8
2
+
+
+
=
Also
E ∩ F = {THH}
with
P(E ∩ 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 |
1 | 6770-6773 | 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 |
1 | 6771-6774 | 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 |
1 | 6772-6775 | 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 |
1 | 6773-6776 | 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) |
1 | 6774-6777 | 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 |
1 | 6775-6778 | 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 |
1 | 6776-6779 | 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 ∩ F |
1 | 6777-6780 | 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 ∩ F © NCERT
not to be republished
PROBABILITY 533
Thus, we can also write the conditional probability of E given that F has occurred as
P(E|F) =
Numberof elementaryeventsfavourableto E
F
Numberof elementaryeventswhicharefavourableto F
∩
=
(E
(F)F)
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
nn
n
∩
∩
= |
1 | 6778-6781 | e the sample points of E ∩ F © NCERT
not to be republished
PROBABILITY 533
Thus, we can also write the conditional probability of E given that F has occurred as
P(E|F) =
Numberof elementaryeventsfavourableto E
F
Numberof elementaryeventswhicharefavourableto F
∩
=
(E
(F)F)
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
nn
n
∩
∩
= (1)
Note that (1) is valid only when P(F) ≠ 0 i |
1 | 6779-6782 | the sample points of E ∩ F © NCERT
not to be republished
PROBABILITY 533
Thus, we can also write the conditional probability of E given that F has occurred as
P(E|F) =
Numberof elementaryeventsfavourableto E
F
Numberof elementaryeventswhicharefavourableto F
∩
=
(E
(F)F)
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
nn
n
∩
∩
= (1)
Note that (1) is valid only when P(F) ≠ 0 i e |
1 | 6780-6783 | © NCERT
not to be republished
PROBABILITY 533
Thus, we can also write the conditional probability of E given that F has occurred as
P(E|F) =
Numberof elementaryeventsfavourableto E
F
Numberof elementaryeventswhicharefavourableto F
∩
=
(E
(F)F)
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
nn
n
∩
∩
= (1)
Note that (1) is valid only when P(F) ≠ 0 i e , F ≠ φ (Why |
1 | 6781-6784 | (1)
Note that (1) is valid only when P(F) ≠ 0 i e , F ≠ φ (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 |
1 | 6782-6785 | e , F ≠ φ (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 |
1 | 6783-6786 | , F ≠ φ (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)
∩
provided P(F) ≠ 0
13 |
1 | 6784-6787 | )
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)
∩
provided P(F) ≠ 0
13 2 |
1 | 6785-6788 | e P (E|F) is given by
P(E|F) = P(E
F)
P(F)
∩
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)
∩
=
=
Also
P(F|F) = P(F
F)
P(F) 1
P(F)
P(F)
∩
=
=
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 ∩ B)|F)
© NCERT
not to be republished
534
MATHEMATICS
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)
∪
∩
= P[(A
F)
(B
F)]
∩P(F)
∪
∩
(by distributive law of union of sets over intersection)
=
P(A
F)+P(B
F)–P(A
B
F)
P(F)
∩
∩
∩
∩
= P(A
F)
P(B
F)
P[(A
B)
F]
P(F)
P(F)
P(F)
∩
∩
∩
∩
+
−
= P(A|F) + P(B|F) – P((A ∩B)|F)
When A and B are disjoint events, then
P((A ∩ 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 |
1 | 6786-6789 | P (E|F) is given by
P(E|F) = P(E
F)
P(F)
∩
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)
∩
=
=
Also
P(F|F) = P(F
F)
P(F) 1
P(F)
P(F)
∩
=
=
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 ∩ B)|F)
© NCERT
not to be republished
534
MATHEMATICS
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)
∪
∩
= P[(A
F)
(B
F)]
∩P(F)
∪
∩
(by distributive law of union of sets over intersection)
=
P(A
F)+P(B
F)–P(A
B
F)
P(F)
∩
∩
∩
∩
= P(A
F)
P(B
F)
P[(A
B)
F]
P(F)
P(F)
P(F)
∩
∩
∩
∩
+
−
= P(A|F) + P(B|F) – P((A ∩B)|F)
When A and B are disjoint events, then
P((A ∩ 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 ∩ B) = 4
13 , evaluate P(A|B) |
1 | 6787-6790 | 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)
∩
=
=
Also
P(F|F) = P(F
F)
P(F) 1
P(F)
P(F)
∩
=
=
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 ∩ B)|F)
© NCERT
not to be republished
534
MATHEMATICS
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)
∪
∩
= P[(A
F)
(B
F)]
∩P(F)
∪
∩
(by distributive law of union of sets over intersection)
=
P(A
F)+P(B
F)–P(A
B
F)
P(F)
∩
∩
∩
∩
= P(A
F)
P(B
F)
P[(A
B)
F]
P(F)
P(F)
P(F)
∩
∩
∩
∩
+
−
= P(A|F) + P(B|F) – P((A ∩B)|F)
When A and B are disjoint events, then
P((A ∩ 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 ∩ B) = 4
13 , evaluate P(A|B) Solution We have
4
P(A
B)
4
13
P(A|B)=
9
P(B)
9
13
∩
=
=
Example 2 A family has two children |
1 | 6788-6791 | 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)
∩
=
=
Also
P(F|F) = P(F
F)
P(F) 1
P(F)
P(F)
∩
=
=
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 ∩ B)|F)
© NCERT
not to be republished
534
MATHEMATICS
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)
∪
∩
= P[(A
F)
(B
F)]
∩P(F)
∪
∩
(by distributive law of union of sets over intersection)
=
P(A
F)+P(B
F)–P(A
B
F)
P(F)
∩
∩
∩
∩
= P(A
F)
P(B
F)
P[(A
B)
F]
P(F)
P(F)
P(F)
∩
∩
∩
∩
+
−
= P(A|F) + P(B|F) – P((A ∩B)|F)
When A and B are disjoint events, then
P((A ∩ 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 ∩ B) = 4
13 , evaluate P(A|B) Solution We have
4
P(A
B)
4
13
P(A|B)=
9
P(B)
9
13
∩
=
=
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 |
1 | 6789-6792 | Example 1 If P(A) = 7
13 , P(B) = 9
13 and P(A ∩ B) = 4
13 , evaluate P(A|B) Solution We have
4
P(A
B)
4
13
P(A|B)=
9
P(B)
9
13
∩
=
=
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 © NCERT
not to be republished
PROBABILITY 535
Solution Let b stand for boy and g for girl |
1 | 6790-6793 | Solution We have
4
P(A
B)
4
13
P(A|B)=
9
P(B)
9
13
∩
=
=
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 © NCERT
not to be republished
PROBABILITY 535
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 ∩ F = {(b,b)}
Thus
P(F) = 3
4 and P (E ∩ F )= 1
4
Therefore
P(E|F) =
1
P(E
F)
1
34
P(F)
3
4
∩
=
=
Example 3 Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and
then one card is drawn randomly |
1 | 6791-6794 | What is the probability that both the children are
boys given that at least one of them is a boy © NCERT
not to be republished
PROBABILITY 535
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 ∩ F = {(b,b)}
Thus
P(F) = 3
4 and P (E ∩ F )= 1
4
Therefore
P(E|F) =
1
P(E
F)
1
34
P(F)
3
4
∩
=
=
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 |
1 | 6792-6795 | © NCERT
not to be republished
PROBABILITY 535
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 ∩ F = {(b,b)}
Thus
P(F) = 3
4 and P (E ∩ F )= 1
4
Therefore
P(E|F) =
1
P(E
F)
1
34
P(F)
3
4
∩
=
=
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’ |
1 | 6793-6796 | 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 ∩ F = {(b,b)}
Thus
P(F) = 3
4 and P (E ∩ F )= 1
4
Therefore
P(E|F) =
1
P(E
F)
1
34
P(F)
3
4
∩
=
=
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) |
1 | 6794-6797 | 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 ∩ B = {4, 6, 8, 10}
Also
P(A) = 5
7
4
, P(B) =
and P(A
B)
10
10
10
∩
=
Then
P(A|B) =
4
P(A
B)
4
710
P(B)
7
10
∩
=
=
Example 4 In a school, there are 1000 students, out of which 430 are girls |
1 | 6795-6798 | 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 ∩ B = {4, 6, 8, 10}
Also
P(A) = 5
7
4
, P(B) =
and P(A
B)
10
10
10
∩
=
Then
P(A|B) =
4
P(A
B)
4
710
P(B)
7
10
∩
=
=
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 |
1 | 6796-6799 | 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 ∩ B = {4, 6, 8, 10}
Also
P(A) = 5
7
4
, P(B) =
and P(A
B)
10
10
10
∩
=
Then
P(A|B) =
4
P(A
B)
4
710
P(B)
7
10
∩
=
=
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 |
1 | 6797-6800 | 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 ∩ B = {4, 6, 8, 10}
Also
P(A) = 5
7
4
, P(B) =
and P(A
B)
10
10
10
∩
=
Then
P(A|B) =
4
P(A
B)
4
710
P(B)
7
10
∩
=
=
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 |
1 | 6798-6801 | 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) |
1 | 6799-6802 | 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) © NCERT
not to be republished
536
MATHEMATICS
Now
P(F) = 430
1000 =0 |
1 | 6800-6803 | 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) © NCERT
not to be republished
536
MATHEMATICS
Now
P(F) = 430
1000 =0 43
and
43
P(E
F)=
0 |
1 | 6801-6804 | We have to find P (E|F) © NCERT
not to be republished
536
MATHEMATICS
Now
P(F) = 430
1000 =0 43
and
43
P(E
F)=
0 043
1000
(Why |
1 | 6802-6805 | © NCERT
not to be republished
536
MATHEMATICS
Now
P(F) = 430
1000 =0 43
and
43
P(E
F)=
0 043
1000
(Why )
Then
P(E|F) = P(E
F)
0 |
1 | 6803-6806 | 43
and
43
P(E
F)=
0 043
1000
(Why )
Then
P(E|F) = P(E
F)
0 043
0 |
1 | 6804-6807 | 043
1000
(Why )
Then
P(E|F) = P(E
F)
0 043
0 1
P(F)
0 |
1 | 6805-6808 | )
Then
P(E|F) = P(E
F)
0 043
0 1
P(F)
0 43
∩
=
=
Example 5 A die is thrown three times |
1 | 6806-6809 | 043
0 1
P(F)
0 43
∩
=
=
Example 5 A die is thrown three times Events A and B are defined as below:
A : 4 on the third throw
B : 6 on the first and 5 on the second throw
Find the probability of A given that B has already occurred |
1 | 6807-6810 | 1
P(F)
0 43
∩
=
=
Example 5 A die is thrown three times Events A and B are defined as below:
A : 4 on the third throw
B : 6 on the first and 5 on the second throw
Find the probability of A given that B has already occurred Solution The sample space has 216 outcomes |
1 | 6808-6811 | 43
∩
=
=
Example 5 A die is thrown three times Events A and B are defined as below:
A : 4 on the third throw
B : 6 on the first and 5 on the second throw
Find the probability of A given that B has already occurred Solution The sample space has 216 outcomes Now
A =
(1,1,4) (1,2,4) |
1 | 6809-6812 | Events A and B are defined as below:
A : 4 on the third throw
B : 6 on the first and 5 on the second throw
Find the probability of A given that B has already occurred Solution The sample space has 216 outcomes Now
A =
(1,1,4) (1,2,4) (1,6,4) (2,1,4) (2,2,4) |
1 | 6810-6813 | Solution The sample space has 216 outcomes Now
A =
(1,1,4) (1,2,4) (1,6,4) (2,1,4) (2,2,4) (2,6,4)
(3,1,4) (3,2,4) |
1 | 6811-6814 | Now
A =
(1,1,4) (1,2,4) (1,6,4) (2,1,4) (2,2,4) (2,6,4)
(3,1,4) (3,2,4) (3,6,4) (4,1,4) (4,2,4) |
1 | 6812-6815 | (1,6,4) (2,1,4) (2,2,4) (2,6,4)
(3,1,4) (3,2,4) (3,6,4) (4,1,4) (4,2,4) (4,6,4)
(5,1,4) (5,2,4) |
1 | 6813-6816 | (2,6,4)
(3,1,4) (3,2,4) (3,6,4) (4,1,4) (4,2,4) (4,6,4)
(5,1,4) (5,2,4) (5,6,4) (6,1,4) (6,2,4) |
1 | 6814-6817 | (3,6,4) (4,1,4) (4,2,4) (4,6,4)
(5,1,4) (5,2,4) (5,6,4) (6,1,4) (6,2,4) (6,6,4)
⎧
⎫
⎪
⎪
⎨
⎬
⎪
⎪
⎩
⎭
B = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)}
and
A ∩ B = {(6,5,4)} |
1 | 6815-6818 | (4,6,4)
(5,1,4) (5,2,4) (5,6,4) (6,1,4) (6,2,4) (6,6,4)
⎧
⎫
⎪
⎪
⎨
⎬
⎪
⎪
⎩
⎭
B = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)}
and
A ∩ B = {(6,5,4)} Now
P(B) =
6
216 and P (A ∩ B) = 1
216
Then
P(A|B) =
1
P(A
B)
1
216
6
P(B)
6
216
∩
=
=
Example 6 A die is thrown twice and the sum of the numbers appearing is observed
to be 6 |
1 | 6816-6819 | (5,6,4) (6,1,4) (6,2,4) (6,6,4)
⎧
⎫
⎪
⎪
⎨
⎬
⎪
⎪
⎩
⎭
B = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)}
and
A ∩ B = {(6,5,4)} Now
P(B) =
6
216 and P (A ∩ B) = 1
216
Then
P(A|B) =
1
P(A
B)
1
216
6
P(B)
6
216
∩
=
=
Example 6 A die is thrown twice and the sum of the numbers appearing is observed
to be 6 What is the conditional probability that the number 4 has appeared at least
once |
1 | 6817-6820 | (6,6,4)
⎧
⎫
⎪
⎪
⎨
⎬
⎪
⎪
⎩
⎭
B = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)}
and
A ∩ B = {(6,5,4)} Now
P(B) =
6
216 and P (A ∩ B) = 1
216
Then
P(A|B) =
1
P(A
B)
1
216
6
P(B)
6
216
∩
=
=
Example 6 A die is thrown twice and the sum of the numbers appearing is observed
to be 6 What is the conditional probability that the number 4 has appeared at least
once Solution Let E be the event that ‘number 4 appears at least once’ and F be the event
that ‘the sum of the numbers appearing is 6’ |
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