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1 | 7018-7021 | The probability of obtaining an even prime number on each die, when a pair of
dice is rolled is
(A) 0
(B) 1
3
(C)
121
(D)
1
36
18 Two events A and B will be independent, if
(A) A and B are mutually exclusive
(B) P(A′B′) = [1 – P(A)] [1 – P(B)]
(C) P(A) = P(B)
(D) P(A) + P(B) = 1
13 5 Bayes' Theorem
Consider that there are two bags I and II Bag I contains 2 white and 3 red balls and
Bag II contains 4 white and 5 red balls |
1 | 7019-7022 | Two events A and B will be independent, if
(A) A and B are mutually exclusive
(B) P(A′B′) = [1 – P(A)] [1 – P(B)]
(C) P(A) = P(B)
(D) P(A) + P(B) = 1
13 5 Bayes' Theorem
Consider that there are two bags I and II Bag I contains 2 white and 3 red balls and
Bag II contains 4 white and 5 red balls One ball is drawn at random from one of the
bags |
1 | 7020-7023 | 5 Bayes' Theorem
Consider that there are two bags I and II Bag I contains 2 white and 3 red balls and
Bag II contains 4 white and 5 red balls One ball is drawn at random from one of the
bags We can find the probability of selecting any of the bags (i |
1 | 7021-7024 | Bag I contains 2 white and 3 red balls and
Bag II contains 4 white and 5 red balls One ball is drawn at random from one of the
bags We can find the probability of selecting any of the bags (i e |
1 | 7022-7025 | One ball is drawn at random from one of the
bags We can find the probability of selecting any of the bags (i e 1
2 ) or probability of
drawing a ball of a particular colour (say white) from a particular bag (say Bag I) |
1 | 7023-7026 | We can find the probability of selecting any of the bags (i e 1
2 ) or probability of
drawing a ball of a particular colour (say white) from a particular bag (say Bag I) In
other words, we can find the probability that the ball drawn is of a particular colour, if
we are given the bag from which the ball is drawn |
1 | 7024-7027 | e 1
2 ) or probability of
drawing a ball of a particular colour (say white) from a particular bag (say Bag I) In
other words, we can find the probability that the ball drawn is of a particular colour, if
we are given the bag from which the ball is drawn But, can we find the probability that
the ball drawn is from a particular bag (say Bag II), if the colour of the ball drawn is
given |
1 | 7025-7028 | 1
2 ) or probability of
drawing a ball of a particular colour (say white) from a particular bag (say Bag I) In
other words, we can find the probability that the ball drawn is of a particular colour, if
we are given the bag from which the ball is drawn But, can we find the probability that
the ball drawn is from a particular bag (say Bag II), if the colour of the ball drawn is
given Here, we have to find the reverse probability of Bag II to be selected when an
event occurred after it is known |
1 | 7026-7029 | In
other words, we can find the probability that the ball drawn is of a particular colour, if
we are given the bag from which the ball is drawn But, can we find the probability that
the ball drawn is from a particular bag (say Bag II), if the colour of the ball drawn is
given Here, we have to find the reverse probability of Bag II to be selected when an
event occurred after it is known Famous mathematician, John Bayes' solved the problem
of finding reverse probability by using conditional probability |
1 | 7027-7030 | But, can we find the probability that
the ball drawn is from a particular bag (say Bag II), if the colour of the ball drawn is
given Here, we have to find the reverse probability of Bag II to be selected when an
event occurred after it is known Famous mathematician, John Bayes' solved the problem
of finding reverse probability by using conditional probability The formula developed
by him is known as ‘Bayes theorem’ which was published posthumously in 1763 |
1 | 7028-7031 | Here, we have to find the reverse probability of Bag II to be selected when an
event occurred after it is known Famous mathematician, John Bayes' solved the problem
of finding reverse probability by using conditional probability The formula developed
by him is known as ‘Bayes theorem’ which was published posthumously in 1763 Before stating and proving the Bayes' theorem, let us first take up a definition and
some preliminary results |
1 | 7029-7032 | Famous mathematician, John Bayes' solved the problem
of finding reverse probability by using conditional probability The formula developed
by him is known as ‘Bayes theorem’ which was published posthumously in 1763 Before stating and proving the Bayes' theorem, let us first take up a definition and
some preliminary results 13 |
1 | 7030-7033 | The formula developed
by him is known as ‘Bayes theorem’ which was published posthumously in 1763 Before stating and proving the Bayes' theorem, let us first take up a definition and
some preliminary results 13 5 |
1 | 7031-7034 | Before stating and proving the Bayes' theorem, let us first take up a definition and
some preliminary results 13 5 1 Partition of a sample space
A set of events E1, E2, |
1 | 7032-7035 | 13 5 1 Partition of a sample space
A set of events E1, E2, , En is said to represent a partition of the sample space S if
(a) Ei ∩ Ej = φ, i ≠ j, i, j = 1, 2, 3, |
1 | 7033-7036 | 5 1 Partition of a sample space
A set of events E1, E2, , En is said to represent a partition of the sample space S if
(a) Ei ∩ Ej = φ, i ≠ j, i, j = 1, 2, 3, , n
© NCERT
not to be republished
PROBABILITY 549
Fig 13 |
1 | 7034-7037 | 1 Partition of a sample space
A set of events E1, E2, , En is said to represent a partition of the sample space S if
(a) Ei ∩ Ej = φ, i ≠ j, i, j = 1, 2, 3, , n
© NCERT
not to be republished
PROBABILITY 549
Fig 13 4
(b) E1 ∪ Ε2 ∪ |
1 | 7035-7038 | , En is said to represent a partition of the sample space S if
(a) Ei ∩ Ej = φ, i ≠ j, i, j = 1, 2, 3, , n
© NCERT
not to be republished
PROBABILITY 549
Fig 13 4
(b) E1 ∪ Ε2 ∪ ∪ En= S and
(c) P(Ei) > 0 for all i = 1, 2, |
1 | 7036-7039 | , n
© NCERT
not to be republished
PROBABILITY 549
Fig 13 4
(b) E1 ∪ Ε2 ∪ ∪ En= S and
(c) P(Ei) > 0 for all i = 1, 2, , n |
1 | 7037-7040 | 4
(b) E1 ∪ Ε2 ∪ ∪ En= S and
(c) P(Ei) > 0 for all i = 1, 2, , n In other words, the events E1, E2, |
1 | 7038-7041 | ∪ En= S and
(c) P(Ei) > 0 for all i = 1, 2, , n In other words, the events E1, E2, , En represent a partition of the sample space
S if they are pairwise disjoint, exhaustive and have nonzero probabilities |
1 | 7039-7042 | , n In other words, the events E1, E2, , En represent a partition of the sample space
S if they are pairwise disjoint, exhaustive and have nonzero probabilities As an example, we see that any nonempty event E and its complement E′ form a
partition of the sample space S since they satisfy E ∩ E′ = φ and E ∪ E′ = S |
1 | 7040-7043 | In other words, the events E1, E2, , En represent a partition of the sample space
S if they are pairwise disjoint, exhaustive and have nonzero probabilities As an example, we see that any nonempty event E and its complement E′ form a
partition of the sample space S since they satisfy E ∩ E′ = φ and E ∪ E′ = S From the Venn diagram in Fig 13 |
1 | 7041-7044 | , En represent a partition of the sample space
S if they are pairwise disjoint, exhaustive and have nonzero probabilities As an example, we see that any nonempty event E and its complement E′ form a
partition of the sample space S since they satisfy E ∩ E′ = φ and E ∪ E′ = S From the Venn diagram in Fig 13 3, one can easily observe that if E and F are any
two events associated with a sample space S, then the set {E ∩ F′, E ∩ F, E′ ∩ F, E′ ∩ F′}
is a partition of the sample space S |
1 | 7042-7045 | As an example, we see that any nonempty event E and its complement E′ form a
partition of the sample space S since they satisfy E ∩ E′ = φ and E ∪ E′ = S From the Venn diagram in Fig 13 3, one can easily observe that if E and F are any
two events associated with a sample space S, then the set {E ∩ F′, E ∩ F, E′ ∩ F, E′ ∩ F′}
is a partition of the sample space S It may be mentioned that the partition of a sample
space is not unique |
1 | 7043-7046 | From the Venn diagram in Fig 13 3, one can easily observe that if E and F are any
two events associated with a sample space S, then the set {E ∩ F′, E ∩ F, E′ ∩ F, E′ ∩ F′}
is a partition of the sample space S It may be mentioned that the partition of a sample
space is not unique There can be several partitions of the same sample space |
1 | 7044-7047 | 3, one can easily observe that if E and F are any
two events associated with a sample space S, then the set {E ∩ F′, E ∩ F, E′ ∩ F, E′ ∩ F′}
is a partition of the sample space S It may be mentioned that the partition of a sample
space is not unique There can be several partitions of the same sample space We shall now prove a theorem known as Theorem of total probability |
1 | 7045-7048 | It may be mentioned that the partition of a sample
space is not unique There can be several partitions of the same sample space We shall now prove a theorem known as Theorem of total probability 13 |
1 | 7046-7049 | There can be several partitions of the same sample space We shall now prove a theorem known as Theorem of total probability 13 5 |
1 | 7047-7050 | We shall now prove a theorem known as Theorem of total probability 13 5 2 Theorem of total probability
Let {E1, E2, |
1 | 7048-7051 | 13 5 2 Theorem of total probability
Let {E1, E2, ,En} be a partition of the sample space S, and suppose that each of the
events E1, E2, |
1 | 7049-7052 | 5 2 Theorem of total probability
Let {E1, E2, ,En} be a partition of the sample space S, and suppose that each of the
events E1, E2, , En has nonzero probability of occurrence |
1 | 7050-7053 | 2 Theorem of total probability
Let {E1, E2, ,En} be a partition of the sample space S, and suppose that each of the
events E1, E2, , En has nonzero probability of occurrence Let A be any event associated
with S, then
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) + |
1 | 7051-7054 | ,En} be a partition of the sample space S, and suppose that each of the
events E1, E2, , En has nonzero probability of occurrence Let A be any event associated
with S, then
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) + + P(En) P(A|En)
=
1
P(E )P(A|E )
n
j
j
j=∑
Proof Given that E1, E2, |
1 | 7052-7055 | , En has nonzero probability of occurrence Let A be any event associated
with S, then
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) + + P(En) P(A|En)
=
1
P(E )P(A|E )
n
j
j
j=∑
Proof Given that E1, E2, , En is a partition of the sample space S (Fig 13 |
1 | 7053-7056 | Let A be any event associated
with S, then
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) + + P(En) P(A|En)
=
1
P(E )P(A|E )
n
j
j
j=∑
Proof Given that E1, E2, , En is a partition of the sample space S (Fig 13 4) |
1 | 7054-7057 | + P(En) P(A|En)
=
1
P(E )P(A|E )
n
j
j
j=∑
Proof Given that E1, E2, , En is a partition of the sample space S (Fig 13 4) Therefore,
S = E1 ∪ E2 ∪ |
1 | 7055-7058 | , En is a partition of the sample space S (Fig 13 4) Therefore,
S = E1 ∪ E2 ∪ ∪ En |
1 | 7056-7059 | 4) Therefore,
S = E1 ∪ E2 ∪ ∪ En (1)
and
Ei ∩ Ej = φ, i ≠ j, i, j = 1, 2, |
1 | 7057-7060 | Therefore,
S = E1 ∪ E2 ∪ ∪ En (1)
and
Ei ∩ Ej = φ, i ≠ j, i, j = 1, 2, , n
Now, we know that for any event A,
A = A ∩ S
= A ∩ (E1 ∪ E2 ∪ |
1 | 7058-7061 | ∪ En (1)
and
Ei ∩ Ej = φ, i ≠ j, i, j = 1, 2, , n
Now, we know that for any event A,
A = A ∩ S
= A ∩ (E1 ∪ E2 ∪ ∪ En)
= (A ∩ E1) ∪ (A ∩ E2) ∪ |
1 | 7059-7062 | (1)
and
Ei ∩ Ej = φ, i ≠ j, i, j = 1, 2, , n
Now, we know that for any event A,
A = A ∩ S
= A ∩ (E1 ∪ E2 ∪ ∪ En)
= (A ∩ E1) ∪ (A ∩ E2) ∪ ∪ (A ∩ En)
Also A ∩ Ei and A ∩ Ej are respectively the subsets of Ei and Ej |
1 | 7060-7063 | , n
Now, we know that for any event A,
A = A ∩ S
= A ∩ (E1 ∪ E2 ∪ ∪ En)
= (A ∩ E1) ∪ (A ∩ E2) ∪ ∪ (A ∩ En)
Also A ∩ Ei and A ∩ Ej are respectively the subsets of Ei and Ej We know that
Ei and Ej are disjoint, for i
≠j
, therefore, A ∩ Ei and A ∩ Ej are also disjoint for all
i ≠ j, i, j = 1, 2, |
1 | 7061-7064 | ∪ En)
= (A ∩ E1) ∪ (A ∩ E2) ∪ ∪ (A ∩ En)
Also A ∩ Ei and A ∩ Ej are respectively the subsets of Ei and Ej We know that
Ei and Ej are disjoint, for i
≠j
, therefore, A ∩ Ei and A ∩ Ej are also disjoint for all
i ≠ j, i, j = 1, 2, , n |
1 | 7062-7065 | ∪ (A ∩ En)
Also A ∩ Ei and A ∩ Ej are respectively the subsets of Ei and Ej We know that
Ei and Ej are disjoint, for i
≠j
, therefore, A ∩ Ei and A ∩ Ej are also disjoint for all
i ≠ j, i, j = 1, 2, , n Thus,
P(A) = P [(A ∩ E1) ∪ (A ∩ E2)∪ |
1 | 7063-7066 | We know that
Ei and Ej are disjoint, for i
≠j
, therefore, A ∩ Ei and A ∩ Ej are also disjoint for all
i ≠ j, i, j = 1, 2, , n Thus,
P(A) = P [(A ∩ E1) ∪ (A ∩ E2)∪ ∪ (A ∩ En)]
= P (A ∩ E1) + P (A ∩ E2) + |
1 | 7064-7067 | , n Thus,
P(A) = P [(A ∩ E1) ∪ (A ∩ E2)∪ ∪ (A ∩ En)]
= P (A ∩ E1) + P (A ∩ E2) + + P (A ∩ En)
Now, by multiplication rule of probability, we have
P(A ∩ Ei) = P(Ei) P(A|Ei) as P (Ei) ≠ 0∀i = 1,2, |
1 | 7065-7068 | Thus,
P(A) = P [(A ∩ E1) ∪ (A ∩ E2)∪ ∪ (A ∩ En)]
= P (A ∩ E1) + P (A ∩ E2) + + P (A ∩ En)
Now, by multiplication rule of probability, we have
P(A ∩ Ei) = P(Ei) P(A|Ei) as P (Ei) ≠ 0∀i = 1,2, , n
© NCERT
not to be republished
550
MATHEMATICS
Therefore,
P (A) = P (E1) P (A|E1) + P (E2) P (A|E2) + |
1 | 7066-7069 | ∪ (A ∩ En)]
= P (A ∩ E1) + P (A ∩ E2) + + P (A ∩ En)
Now, by multiplication rule of probability, we have
P(A ∩ Ei) = P(Ei) P(A|Ei) as P (Ei) ≠ 0∀i = 1,2, , n
© NCERT
not to be republished
550
MATHEMATICS
Therefore,
P (A) = P (E1) P (A|E1) + P (E2) P (A|E2) + + P (En)P(A|En)
or
P(A) =
1
P(E )P(A|E )
n
j
j
j=∑
Example 15 A person has undertaken a construction job |
1 | 7067-7070 | + P (A ∩ En)
Now, by multiplication rule of probability, we have
P(A ∩ Ei) = P(Ei) P(A|Ei) as P (Ei) ≠ 0∀i = 1,2, , n
© NCERT
not to be republished
550
MATHEMATICS
Therefore,
P (A) = P (E1) P (A|E1) + P (E2) P (A|E2) + + P (En)P(A|En)
or
P(A) =
1
P(E )P(A|E )
n
j
j
j=∑
Example 15 A person has undertaken a construction job The probabilities are 0 |
1 | 7068-7071 | , n
© NCERT
not to be republished
550
MATHEMATICS
Therefore,
P (A) = P (E1) P (A|E1) + P (E2) P (A|E2) + + P (En)P(A|En)
or
P(A) =
1
P(E )P(A|E )
n
j
j
j=∑
Example 15 A person has undertaken a construction job The probabilities are 0 65
that there will be strike, 0 |
1 | 7069-7072 | + P (En)P(A|En)
or
P(A) =
1
P(E )P(A|E )
n
j
j
j=∑
Example 15 A person has undertaken a construction job The probabilities are 0 65
that there will be strike, 0 80 that the construction job will be completed on time if there
is no strike, and 0 |
1 | 7070-7073 | The probabilities are 0 65
that there will be strike, 0 80 that the construction job will be completed on time if there
is no strike, and 0 32 that the construction job will be completed on time if there is a
strike |
1 | 7071-7074 | 65
that there will be strike, 0 80 that the construction job will be completed on time if there
is no strike, and 0 32 that the construction job will be completed on time if there is a
strike Determine the probability that the construction job will be completed on time |
1 | 7072-7075 | 80 that the construction job will be completed on time if there
is no strike, and 0 32 that the construction job will be completed on time if there is a
strike Determine the probability that the construction job will be completed on time Solution Let A be the event that the construction job will be completed on time, and B
be the event that there will be a strike |
1 | 7073-7076 | 32 that the construction job will be completed on time if there is a
strike Determine the probability that the construction job will be completed on time Solution Let A be the event that the construction job will be completed on time, and B
be the event that there will be a strike We have to find P(A) |
1 | 7074-7077 | Determine the probability that the construction job will be completed on time Solution Let A be the event that the construction job will be completed on time, and B
be the event that there will be a strike We have to find P(A) We have
P(B) = 0 |
1 | 7075-7078 | Solution Let A be the event that the construction job will be completed on time, and B
be the event that there will be a strike We have to find P(A) We have
P(B) = 0 65, P(no strike) = P(B′) = 1 − P(B) = 1 − 0 |
1 | 7076-7079 | We have to find P(A) We have
P(B) = 0 65, P(no strike) = P(B′) = 1 − P(B) = 1 − 0 65 = 0 |
1 | 7077-7080 | We have
P(B) = 0 65, P(no strike) = P(B′) = 1 − P(B) = 1 − 0 65 = 0 35
P(A|B) = 0 |
1 | 7078-7081 | 65, P(no strike) = P(B′) = 1 − P(B) = 1 − 0 65 = 0 35
P(A|B) = 0 32, P(A|B′) = 0 |
1 | 7079-7082 | 65 = 0 35
P(A|B) = 0 32, P(A|B′) = 0 80
Since events B and B′ form a partition of the sample space S, therefore, by theorem
on total probability, we have
P(A) = P(B) P(A|B) + P(B′) P(A|B′)
= 0 |
1 | 7080-7083 | 35
P(A|B) = 0 32, P(A|B′) = 0 80
Since events B and B′ form a partition of the sample space S, therefore, by theorem
on total probability, we have
P(A) = P(B) P(A|B) + P(B′) P(A|B′)
= 0 65 × 0 |
1 | 7081-7084 | 32, P(A|B′) = 0 80
Since events B and B′ form a partition of the sample space S, therefore, by theorem
on total probability, we have
P(A) = P(B) P(A|B) + P(B′) P(A|B′)
= 0 65 × 0 32 + 0 |
1 | 7082-7085 | 80
Since events B and B′ form a partition of the sample space S, therefore, by theorem
on total probability, we have
P(A) = P(B) P(A|B) + P(B′) P(A|B′)
= 0 65 × 0 32 + 0 35 × 0 |
1 | 7083-7086 | 65 × 0 32 + 0 35 × 0 8
= 0 |
1 | 7084-7087 | 32 + 0 35 × 0 8
= 0 208 + 0 |
1 | 7085-7088 | 35 × 0 8
= 0 208 + 0 28 = 0 |
1 | 7086-7089 | 8
= 0 208 + 0 28 = 0 488
Thus, the probability that the construction job will be completed in time is 0 |
1 | 7087-7090 | 208 + 0 28 = 0 488
Thus, the probability that the construction job will be completed in time is 0 488 |
1 | 7088-7091 | 28 = 0 488
Thus, the probability that the construction job will be completed in time is 0 488 We shall now state and prove the Bayes' theorem |
1 | 7089-7092 | 488
Thus, the probability that the construction job will be completed in time is 0 488 We shall now state and prove the Bayes' theorem Bayes’ Theorem If E1, E2 , |
1 | 7090-7093 | 488 We shall now state and prove the Bayes' theorem Bayes’ Theorem If E1, E2 , , En are n non empty events which constitute a partition
of sample space S, i |
1 | 7091-7094 | We shall now state and prove the Bayes' theorem Bayes’ Theorem If E1, E2 , , En are n non empty events which constitute a partition
of sample space S, i e |
1 | 7092-7095 | Bayes’ Theorem If E1, E2 , , En are n non empty events which constitute a partition
of sample space S, i e E1, E2 , |
1 | 7093-7096 | , En are n non empty events which constitute a partition
of sample space S, i e E1, E2 , , En are pairwise disjoint and E1∪ E2∪ |
1 | 7094-7097 | e E1, E2 , , En are pairwise disjoint and E1∪ E2∪ ∪ En = S and
A is any event of nonzero probability, then
P(Ei|A) =
1
P(E )P(A|E )
P(E )P(A|E )
i
i
n
j
j
j=∑
for any i = 1, 2, 3, |
1 | 7095-7098 | E1, E2 , , En are pairwise disjoint and E1∪ E2∪ ∪ En = S and
A is any event of nonzero probability, then
P(Ei|A) =
1
P(E )P(A|E )
P(E )P(A|E )
i
i
n
j
j
j=∑
for any i = 1, 2, 3, , n
Proof By formula of conditional probability, we know that
P(Ei|A) = P(A
E )
P(A)
i
∩
= P(E )P(A|E )
P(A)
i
i (by multiplication rule of probability)
=
1
P(E )P(A|E )
P(E )P(A|E )
i
i
n
j
j
j=∑
(by the result of theorem of total probability)
© NCERT
not to be republished
PROBABILITY 551
Remark The following terminology is generally used when Bayes' theorem is applied |
1 | 7096-7099 | , En are pairwise disjoint and E1∪ E2∪ ∪ En = S and
A is any event of nonzero probability, then
P(Ei|A) =
1
P(E )P(A|E )
P(E )P(A|E )
i
i
n
j
j
j=∑
for any i = 1, 2, 3, , n
Proof By formula of conditional probability, we know that
P(Ei|A) = P(A
E )
P(A)
i
∩
= P(E )P(A|E )
P(A)
i
i (by multiplication rule of probability)
=
1
P(E )P(A|E )
P(E )P(A|E )
i
i
n
j
j
j=∑
(by the result of theorem of total probability)
© NCERT
not to be republished
PROBABILITY 551
Remark The following terminology is generally used when Bayes' theorem is applied The events E1, E2, |
1 | 7097-7100 | ∪ En = S and
A is any event of nonzero probability, then
P(Ei|A) =
1
P(E )P(A|E )
P(E )P(A|E )
i
i
n
j
j
j=∑
for any i = 1, 2, 3, , n
Proof By formula of conditional probability, we know that
P(Ei|A) = P(A
E )
P(A)
i
∩
= P(E )P(A|E )
P(A)
i
i (by multiplication rule of probability)
=
1
P(E )P(A|E )
P(E )P(A|E )
i
i
n
j
j
j=∑
(by the result of theorem of total probability)
© NCERT
not to be republished
PROBABILITY 551
Remark The following terminology is generally used when Bayes' theorem is applied The events E1, E2, , En are called hypotheses |
1 | 7098-7101 | , n
Proof By formula of conditional probability, we know that
P(Ei|A) = P(A
E )
P(A)
i
∩
= P(E )P(A|E )
P(A)
i
i (by multiplication rule of probability)
=
1
P(E )P(A|E )
P(E )P(A|E )
i
i
n
j
j
j=∑
(by the result of theorem of total probability)
© NCERT
not to be republished
PROBABILITY 551
Remark The following terminology is generally used when Bayes' theorem is applied The events E1, E2, , En are called hypotheses The probability P(Ei) is called the priori probability of the hypothesis Ei
The conditional probability P(Ei |A) is called a posteriori probability of the
hypothesis Ei |
1 | 7099-7102 | The events E1, E2, , En are called hypotheses The probability P(Ei) is called the priori probability of the hypothesis Ei
The conditional probability P(Ei |A) is called a posteriori probability of the
hypothesis Ei Bayes' theorem is also called the formula for the probability of "causes" |
1 | 7100-7103 | , En are called hypotheses The probability P(Ei) is called the priori probability of the hypothesis Ei
The conditional probability P(Ei |A) is called a posteriori probability of the
hypothesis Ei Bayes' theorem is also called the formula for the probability of "causes" Since the
Ei's are a partition of the sample space S, one and only one of the events Ei occurs (i |
1 | 7101-7104 | The probability P(Ei) is called the priori probability of the hypothesis Ei
The conditional probability P(Ei |A) is called a posteriori probability of the
hypothesis Ei Bayes' theorem is also called the formula for the probability of "causes" Since the
Ei's are a partition of the sample space S, one and only one of the events Ei occurs (i e |
1 | 7102-7105 | Bayes' theorem is also called the formula for the probability of "causes" Since the
Ei's are a partition of the sample space S, one and only one of the events Ei occurs (i e one of the events Ei must occur and only one can occur) |
1 | 7103-7106 | Since the
Ei's are a partition of the sample space S, one and only one of the events Ei occurs (i e one of the events Ei must occur and only one can occur) Hence, the above formula
gives us the probability of a particular Ei (i |
1 | 7104-7107 | e one of the events Ei must occur and only one can occur) Hence, the above formula
gives us the probability of a particular Ei (i e |
1 | 7105-7108 | one of the events Ei must occur and only one can occur) Hence, the above formula
gives us the probability of a particular Ei (i e a "Cause"), given that the event A has
occurred |
1 | 7106-7109 | Hence, the above formula
gives us the probability of a particular Ei (i e a "Cause"), given that the event A has
occurred The Bayes' theorem has its applications in variety of situations, few of which are
illustrated in following examples |
1 | 7107-7110 | e a "Cause"), given that the event A has
occurred The Bayes' theorem has its applications in variety of situations, few of which are
illustrated in following examples Example 16 Bag I contains 3 red and 4 black balls while another Bag II contains 5 red
and 6 black balls |
1 | 7108-7111 | a "Cause"), given that the event A has
occurred The Bayes' theorem has its applications in variety of situations, few of which are
illustrated in following examples Example 16 Bag I contains 3 red and 4 black balls while another Bag II contains 5 red
and 6 black balls One ball is drawn at random from one of the bags and it is found to
be red |
1 | 7109-7112 | The Bayes' theorem has its applications in variety of situations, few of which are
illustrated in following examples Example 16 Bag I contains 3 red and 4 black balls while another Bag II contains 5 red
and 6 black balls One ball is drawn at random from one of the bags and it is found to
be red Find the probability that it was drawn from Bag II |
1 | 7110-7113 | Example 16 Bag I contains 3 red and 4 black balls while another Bag II contains 5 red
and 6 black balls One ball is drawn at random from one of the bags and it is found to
be red Find the probability that it was drawn from Bag II Solution Let E1 be the event of choosing the bag I, E2 the event of choosing the bag II
and A be the event of drawing a red ball |
1 | 7111-7114 | One ball is drawn at random from one of the bags and it is found to
be red Find the probability that it was drawn from Bag II Solution Let E1 be the event of choosing the bag I, E2 the event of choosing the bag II
and A be the event of drawing a red ball Then
P(E1) = P(E2) = 1
2
Also
P(A|E1) = P(drawing a red ball from Bag I) = 3
7
and
P(A|E2) = P(drawing a red ball from Bag II) = 5
11
Now, the probability of drawing a ball from Bag II, being given that it is red,
is P(E2|A)
By using Bayes' theorem, we have
P(E2|A) =
2
2
1
1
2
2
P(E )P(A|E )
P(E )P(A|E )+P(E )P(A|E ) =
1
5
35
2 11
1
3
1
5
68
2
7
2 11
×
=
×
+
×
Example 17 Given three identical boxes I, II and III, each containing two coins |
1 | 7112-7115 | Find the probability that it was drawn from Bag II Solution Let E1 be the event of choosing the bag I, E2 the event of choosing the bag II
and A be the event of drawing a red ball Then
P(E1) = P(E2) = 1
2
Also
P(A|E1) = P(drawing a red ball from Bag I) = 3
7
and
P(A|E2) = P(drawing a red ball from Bag II) = 5
11
Now, the probability of drawing a ball from Bag II, being given that it is red,
is P(E2|A)
By using Bayes' theorem, we have
P(E2|A) =
2
2
1
1
2
2
P(E )P(A|E )
P(E )P(A|E )+P(E )P(A|E ) =
1
5
35
2 11
1
3
1
5
68
2
7
2 11
×
=
×
+
×
Example 17 Given three identical boxes I, II and III, each containing two coins In
box I, both coins are gold coins, in box II, both are silver coins and in the box III, there
is one gold and one silver coin |
1 | 7113-7116 | Solution Let E1 be the event of choosing the bag I, E2 the event of choosing the bag II
and A be the event of drawing a red ball Then
P(E1) = P(E2) = 1
2
Also
P(A|E1) = P(drawing a red ball from Bag I) = 3
7
and
P(A|E2) = P(drawing a red ball from Bag II) = 5
11
Now, the probability of drawing a ball from Bag II, being given that it is red,
is P(E2|A)
By using Bayes' theorem, we have
P(E2|A) =
2
2
1
1
2
2
P(E )P(A|E )
P(E )P(A|E )+P(E )P(A|E ) =
1
5
35
2 11
1
3
1
5
68
2
7
2 11
×
=
×
+
×
Example 17 Given three identical boxes I, II and III, each containing two coins In
box I, both coins are gold coins, in box II, both are silver coins and in the box III, there
is one gold and one silver coin A person chooses a box at random and takes out a coin |
1 | 7114-7117 | Then
P(E1) = P(E2) = 1
2
Also
P(A|E1) = P(drawing a red ball from Bag I) = 3
7
and
P(A|E2) = P(drawing a red ball from Bag II) = 5
11
Now, the probability of drawing a ball from Bag II, being given that it is red,
is P(E2|A)
By using Bayes' theorem, we have
P(E2|A) =
2
2
1
1
2
2
P(E )P(A|E )
P(E )P(A|E )+P(E )P(A|E ) =
1
5
35
2 11
1
3
1
5
68
2
7
2 11
×
=
×
+
×
Example 17 Given three identical boxes I, II and III, each containing two coins In
box I, both coins are gold coins, in box II, both are silver coins and in the box III, there
is one gold and one silver coin A person chooses a box at random and takes out a coin If the coin is of gold, what is the probability that the other coin in the box is also of gold |
1 | 7115-7118 | In
box I, both coins are gold coins, in box II, both are silver coins and in the box III, there
is one gold and one silver coin A person chooses a box at random and takes out a coin If the coin is of gold, what is the probability that the other coin in the box is also of gold © NCERT
not to be republished
552
MATHEMATICS
Solution Let E1, E2 and E3 be the events that boxes I, II and III are chosen, respectively |
1 | 7116-7119 | A person chooses a box at random and takes out a coin If the coin is of gold, what is the probability that the other coin in the box is also of gold © NCERT
not to be republished
552
MATHEMATICS
Solution Let E1, E2 and E3 be the events that boxes I, II and III are chosen, respectively Then
P(E1) = P(E2) = P(E3) = 1
3
Also, let A be the event that ‘the coin drawn is of gold’
Then
P(A|E1) = P(a gold coin from bag I) = 2
2 = 1
P(A|E2) = P(a gold coin from bag II) = 0
P(A|E3) = P(a gold coin from bag III) = 1
2
Now, the probability that the other coin in the box is of gold
= the probability that gold coin is drawn from the box I |
1 | 7117-7120 | If the coin is of gold, what is the probability that the other coin in the box is also of gold © NCERT
not to be republished
552
MATHEMATICS
Solution Let E1, E2 and E3 be the events that boxes I, II and III are chosen, respectively Then
P(E1) = P(E2) = P(E3) = 1
3
Also, let A be the event that ‘the coin drawn is of gold’
Then
P(A|E1) = P(a gold coin from bag I) = 2
2 = 1
P(A|E2) = P(a gold coin from bag II) = 0
P(A|E3) = P(a gold coin from bag III) = 1
2
Now, the probability that the other coin in the box is of gold
= the probability that gold coin is drawn from the box I = P(E1|A)
By Bayes' theorem, we know that
P(E1|A) =
1
1
1
1
2
2
3
3
P(E )P(A|E )
P(E )P(A|E )+P(E )P(A|E )+P(E )P(A|E )
=
1 1
2
3
1
1
1
1
3
1
0
3
3
3
2
×
=
× + × + ×
Example 18 Suppose that the reliability of a HIV test is specified as follows:
Of people having HIV, 90% of the test detect the disease but 10% go undetected |
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