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1 | 6818-6821 | 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’ Then,
E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)}
and
F = {(1,5), (2,4), (3,3), (4,2), (5,1)}
We have
P(E) = 11
36 and P(F) = 5
36
Also
E∩F = {(2,4), (4,2)}
© NCERT
not to be republished
PROBABILITY 537
Therefore
P(E∩F) =
2
36
Hence, the required probability
P(E|F) =
2
P(E
F)
2
536
P(F)
5
36
∩
=
=
For the conditional probability discussed above, we have considered the elemen-
tary events of the experiment to be equally likely and the corresponding definition of
the probability of an event was used |
1 | 6819-6822 | 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’ Then,
E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)}
and
F = {(1,5), (2,4), (3,3), (4,2), (5,1)}
We have
P(E) = 11
36 and P(F) = 5
36
Also
E∩F = {(2,4), (4,2)}
© NCERT
not to be republished
PROBABILITY 537
Therefore
P(E∩F) =
2
36
Hence, the required probability
P(E|F) =
2
P(E
F)
2
536
P(F)
5
36
∩
=
=
For the conditional probability discussed above, we have considered the elemen-
tary events of the experiment to be equally likely and the corresponding definition of
the probability of an event was used However, the same definition can also be used in
the general case where the elementary events of the sample space are not equally
likely, the probabilities P(E∩F) and P(F) being calculated accordingly |
1 | 6820-6823 | 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’ Then,
E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)}
and
F = {(1,5), (2,4), (3,3), (4,2), (5,1)}
We have
P(E) = 11
36 and P(F) = 5
36
Also
E∩F = {(2,4), (4,2)}
© NCERT
not to be republished
PROBABILITY 537
Therefore
P(E∩F) =
2
36
Hence, the required probability
P(E|F) =
2
P(E
F)
2
536
P(F)
5
36
∩
=
=
For the conditional probability discussed above, we have considered the elemen-
tary events of the experiment to be equally likely and the corresponding definition of
the probability of an event was used However, the same definition can also be used in
the general case where the elementary events of the sample space are not equally
likely, the probabilities P(E∩F) and P(F) being calculated accordingly Let us take up
the following example |
1 | 6821-6824 | Then,
E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)}
and
F = {(1,5), (2,4), (3,3), (4,2), (5,1)}
We have
P(E) = 11
36 and P(F) = 5
36
Also
E∩F = {(2,4), (4,2)}
© NCERT
not to be republished
PROBABILITY 537
Therefore
P(E∩F) =
2
36
Hence, the required probability
P(E|F) =
2
P(E
F)
2
536
P(F)
5
36
∩
=
=
For the conditional probability discussed above, we have considered the elemen-
tary events of the experiment to be equally likely and the corresponding definition of
the probability of an event was used However, the same definition can also be used in
the general case where the elementary events of the sample space are not equally
likely, the probabilities P(E∩F) and P(F) being calculated accordingly Let us take up
the following example Example 7 Consider the experiment of tossing a coin |
1 | 6822-6825 | However, the same definition can also be used in
the general case where the elementary events of the sample space are not equally
likely, the probabilities P(E∩F) and P(F) being calculated accordingly Let us take up
the following example Example 7 Consider the experiment of tossing a coin If the coin shows head, toss it
again but if it shows tail, then throw a die |
1 | 6823-6826 | Let us take up
the following example Example 7 Consider the experiment of tossing a coin If the coin shows head, toss it
again but if it shows tail, then throw a die Find the
conditional probability of the event that ‘the die shows
a number greater than 4’ given that ‘there is at least
one tail’ |
1 | 6824-6827 | Example 7 Consider the experiment of tossing a coin If the coin shows head, toss it
again but if it shows tail, then throw a die Find the
conditional probability of the event that ‘the die shows
a number greater than 4’ given that ‘there is at least
one tail’ Solution The outcomes of the experiment can be
represented in following diagrammatic manner called
the ‘tree diagram’ |
1 | 6825-6828 | If the coin shows head, toss it
again but if it shows tail, then throw a die Find the
conditional probability of the event that ‘the die shows
a number greater than 4’ given that ‘there is at least
one tail’ Solution The outcomes of the experiment can be
represented in following diagrammatic manner called
the ‘tree diagram’ The sample space of the experiment may be
described as
S = {(H,H), (H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
where (H, H) denotes that both the tosses result into
head and (T, i) denote the first toss result into a tail and
the number i appeared on the die for i = 1,2,3,4,5,6 |
1 | 6826-6829 | Find the
conditional probability of the event that ‘the die shows
a number greater than 4’ given that ‘there is at least
one tail’ Solution The outcomes of the experiment can be
represented in following diagrammatic manner called
the ‘tree diagram’ The sample space of the experiment may be
described as
S = {(H,H), (H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
where (H, H) denotes that both the tosses result into
head and (T, i) denote the first toss result into a tail and
the number i appeared on the die for i = 1,2,3,4,5,6 Thus, the probabilities assigned to the 8 elementary
events
(H, H), (H, T), (T, 1), (T, 2), (T, 3) (T, 4), (T, 5), (T, 6)
are 1 1 1
1
1
1
1
1
,
,
,
,
,
,
,
4 4 12 12 12 12 12 12 respectively which is
clear from the Fig 13 |
1 | 6827-6830 | Solution The outcomes of the experiment can be
represented in following diagrammatic manner called
the ‘tree diagram’ The sample space of the experiment may be
described as
S = {(H,H), (H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
where (H, H) denotes that both the tosses result into
head and (T, i) denote the first toss result into a tail and
the number i appeared on the die for i = 1,2,3,4,5,6 Thus, the probabilities assigned to the 8 elementary
events
(H, H), (H, T), (T, 1), (T, 2), (T, 3) (T, 4), (T, 5), (T, 6)
are 1 1 1
1
1
1
1
1
,
,
,
,
,
,
,
4 4 12 12 12 12 12 12 respectively which is
clear from the Fig 13 2 |
1 | 6828-6831 | The sample space of the experiment may be
described as
S = {(H,H), (H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
where (H, H) denotes that both the tosses result into
head and (T, i) denote the first toss result into a tail and
the number i appeared on the die for i = 1,2,3,4,5,6 Thus, the probabilities assigned to the 8 elementary
events
(H, H), (H, T), (T, 1), (T, 2), (T, 3) (T, 4), (T, 5), (T, 6)
are 1 1 1
1
1
1
1
1
,
,
,
,
,
,
,
4 4 12 12 12 12 12 12 respectively which is
clear from the Fig 13 2 Fig 13 |
1 | 6829-6832 | Thus, the probabilities assigned to the 8 elementary
events
(H, H), (H, T), (T, 1), (T, 2), (T, 3) (T, 4), (T, 5), (T, 6)
are 1 1 1
1
1
1
1
1
,
,
,
,
,
,
,
4 4 12 12 12 12 12 12 respectively which is
clear from the Fig 13 2 Fig 13 1
Fig 13 |
1 | 6830-6833 | 2 Fig 13 1
Fig 13 2
© NCERT
not to be republished
538
MATHEMATICS
Let F be the event that ‘there is at least one tail’ and E be the event ‘the die shows
a number greater than 4’ |
1 | 6831-6834 | Fig 13 1
Fig 13 2
© NCERT
not to be republished
538
MATHEMATICS
Let F be the event that ‘there is at least one tail’ and E be the event ‘the die shows
a number greater than 4’ Then
F = {(H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
E = {(T,5), (T,6)} and E ∩ F = {(T,5), (T,6)}
Now
P(F) = P({(H,T)}) + P ({(T,1)}) + P ({(T,2)}) + P ({(T,3)})
+ P ({(T,4)}) + P({(T,5)}) + P({(T,6)})
= 1
1
1
1
1
1
1
3
4
12
12
12
12
12
12
4
and
P(E ∩ F) = P ({(T,5)}) + P ({(T,6)}) = 1
1
1
12
12
6
Hence
P(E|F) =
1
P(E
F)
2
36
P(F)
9
4
∩
=
=
EXERCISE 13 |
1 | 6832-6835 | 1
Fig 13 2
© NCERT
not to be republished
538
MATHEMATICS
Let F be the event that ‘there is at least one tail’ and E be the event ‘the die shows
a number greater than 4’ Then
F = {(H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
E = {(T,5), (T,6)} and E ∩ F = {(T,5), (T,6)}
Now
P(F) = P({(H,T)}) + P ({(T,1)}) + P ({(T,2)}) + P ({(T,3)})
+ P ({(T,4)}) + P({(T,5)}) + P({(T,6)})
= 1
1
1
1
1
1
1
3
4
12
12
12
12
12
12
4
and
P(E ∩ F) = P ({(T,5)}) + P ({(T,6)}) = 1
1
1
12
12
6
Hence
P(E|F) =
1
P(E
F)
2
36
P(F)
9
4
∩
=
=
EXERCISE 13 1
1 |
1 | 6833-6836 | 2
© NCERT
not to be republished
538
MATHEMATICS
Let F be the event that ‘there is at least one tail’ and E be the event ‘the die shows
a number greater than 4’ Then
F = {(H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
E = {(T,5), (T,6)} and E ∩ F = {(T,5), (T,6)}
Now
P(F) = P({(H,T)}) + P ({(T,1)}) + P ({(T,2)}) + P ({(T,3)})
+ P ({(T,4)}) + P({(T,5)}) + P({(T,6)})
= 1
1
1
1
1
1
1
3
4
12
12
12
12
12
12
4
and
P(E ∩ F) = P ({(T,5)}) + P ({(T,6)}) = 1
1
1
12
12
6
Hence
P(E|F) =
1
P(E
F)
2
36
P(F)
9
4
∩
=
=
EXERCISE 13 1
1 Given that E and F are events such that P(E) = 0 |
1 | 6834-6837 | Then
F = {(H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
E = {(T,5), (T,6)} and E ∩ F = {(T,5), (T,6)}
Now
P(F) = P({(H,T)}) + P ({(T,1)}) + P ({(T,2)}) + P ({(T,3)})
+ P ({(T,4)}) + P({(T,5)}) + P({(T,6)})
= 1
1
1
1
1
1
1
3
4
12
12
12
12
12
12
4
and
P(E ∩ F) = P ({(T,5)}) + P ({(T,6)}) = 1
1
1
12
12
6
Hence
P(E|F) =
1
P(E
F)
2
36
P(F)
9
4
∩
=
=
EXERCISE 13 1
1 Given that E and F are events such that P(E) = 0 6, P(F) = 0 |
1 | 6835-6838 | 1
1 Given that E and F are events such that P(E) = 0 6, P(F) = 0 3 and
P(E ∩ F) = 0 |
1 | 6836-6839 | Given that E and F are events such that P(E) = 0 6, P(F) = 0 3 and
P(E ∩ F) = 0 2, find P(E|F) and P(F|E)
2 |
1 | 6837-6840 | 6, P(F) = 0 3 and
P(E ∩ F) = 0 2, find P(E|F) and P(F|E)
2 Compute P(A|B), if P(B) = 0 |
1 | 6838-6841 | 3 and
P(E ∩ F) = 0 2, find P(E|F) and P(F|E)
2 Compute P(A|B), if P(B) = 0 5 and P (A ∩ B) = 0 |
1 | 6839-6842 | 2, find P(E|F) and P(F|E)
2 Compute P(A|B), if P(B) = 0 5 and P (A ∩ B) = 0 32
3 |
1 | 6840-6843 | Compute P(A|B), if P(B) = 0 5 and P (A ∩ B) = 0 32
3 If P(A) = 0 |
1 | 6841-6844 | 5 and P (A ∩ B) = 0 32
3 If P(A) = 0 8, P (B) = 0 |
1 | 6842-6845 | 32
3 If P(A) = 0 8, P (B) = 0 5 and P(B|A) = 0 |
1 | 6843-6846 | If P(A) = 0 8, P (B) = 0 5 and P(B|A) = 0 4, find
(i) P(A ∩ B)
(ii) P(A|B)
(iii) P(A ∪ B)
4 |
1 | 6844-6847 | 8, P (B) = 0 5 and P(B|A) = 0 4, find
(i) P(A ∩ B)
(ii) P(A|B)
(iii) P(A ∪ B)
4 Evaluate P(A ∪ B), if 2P(A) = P(B) = 5
13 and P(A|B) = 2
5
5 |
1 | 6845-6848 | 5 and P(B|A) = 0 4, find
(i) P(A ∩ B)
(ii) P(A|B)
(iii) P(A ∪ B)
4 Evaluate P(A ∪ B), if 2P(A) = P(B) = 5
13 and P(A|B) = 2
5
5 If P(A) = 6
11 , P(B) = 5
11 and P(A ∪ B)
117
, find
(i) P(A∩B)
(ii) P(A|B)
(iii) P(B|A)
Determine P(E|F) in Exercises 6 to 9 |
1 | 6846-6849 | 4, find
(i) P(A ∩ B)
(ii) P(A|B)
(iii) P(A ∪ B)
4 Evaluate P(A ∪ B), if 2P(A) = P(B) = 5
13 and P(A|B) = 2
5
5 If P(A) = 6
11 , P(B) = 5
11 and P(A ∪ B)
117
, find
(i) P(A∩B)
(ii) P(A|B)
(iii) P(B|A)
Determine P(E|F) in Exercises 6 to 9 6 |
1 | 6847-6850 | Evaluate P(A ∪ B), if 2P(A) = P(B) = 5
13 and P(A|B) = 2
5
5 If P(A) = 6
11 , P(B) = 5
11 and P(A ∪ B)
117
, find
(i) P(A∩B)
(ii) P(A|B)
(iii) P(B|A)
Determine P(E|F) in Exercises 6 to 9 6 A coin is tossed three times, where
(i) E : head on third toss , F : heads on first two tosses
(ii) E : at least two heads , F : at most two heads
(iii) E : at most two tails , F : at least one tail
© NCERT
not to be republished
PROBABILITY 539
7 |
1 | 6848-6851 | If P(A) = 6
11 , P(B) = 5
11 and P(A ∪ B)
117
, find
(i) P(A∩B)
(ii) P(A|B)
(iii) P(B|A)
Determine P(E|F) in Exercises 6 to 9 6 A coin is tossed three times, where
(i) E : head on third toss , F : heads on first two tosses
(ii) E : at least two heads , F : at most two heads
(iii) E : at most two tails , F : at least one tail
© NCERT
not to be republished
PROBABILITY 539
7 Two coins are tossed once, where
(i)
E : tail appears on one coin,
F : one coin shows head
(ii)
E : no tail appears,
F : no head appears
8 |
1 | 6849-6852 | 6 A coin is tossed three times, where
(i) E : head on third toss , F : heads on first two tosses
(ii) E : at least two heads , F : at most two heads
(iii) E : at most two tails , F : at least one tail
© NCERT
not to be republished
PROBABILITY 539
7 Two coins are tossed once, where
(i)
E : tail appears on one coin,
F : one coin shows head
(ii)
E : no tail appears,
F : no head appears
8 A die is thrown three times,
E : 4 appears on the third toss,
F : 6 and 5 appears respectively
on first two tosses
9 |
1 | 6850-6853 | A coin is tossed three times, where
(i) E : head on third toss , F : heads on first two tosses
(ii) E : at least two heads , F : at most two heads
(iii) E : at most two tails , F : at least one tail
© NCERT
not to be republished
PROBABILITY 539
7 Two coins are tossed once, where
(i)
E : tail appears on one coin,
F : one coin shows head
(ii)
E : no tail appears,
F : no head appears
8 A die is thrown three times,
E : 4 appears on the third toss,
F : 6 and 5 appears respectively
on first two tosses
9 Mother, father and son line up at random for a family picture
E : son on one end,
F : father in middle
10 |
1 | 6851-6854 | Two coins are tossed once, where
(i)
E : tail appears on one coin,
F : one coin shows head
(ii)
E : no tail appears,
F : no head appears
8 A die is thrown three times,
E : 4 appears on the third toss,
F : 6 and 5 appears respectively
on first two tosses
9 Mother, father and son line up at random for a family picture
E : son on one end,
F : father in middle
10 A black and a red dice are rolled |
1 | 6852-6855 | A die is thrown three times,
E : 4 appears on the third toss,
F : 6 and 5 appears respectively
on first two tosses
9 Mother, father and son line up at random for a family picture
E : son on one end,
F : father in middle
10 A black and a red dice are rolled (a) Find the conditional probability of obtaining a sum greater than 9, given
that the black die resulted in a 5 |
1 | 6853-6856 | Mother, father and son line up at random for a family picture
E : son on one end,
F : father in middle
10 A black and a red dice are rolled (a) Find the conditional probability of obtaining a sum greater than 9, given
that the black die resulted in a 5 (b) Find the conditional probability of obtaining the sum 8, given that the red die
resulted in a number less than 4 |
1 | 6854-6857 | A black and a red dice are rolled (a) Find the conditional probability of obtaining a sum greater than 9, given
that the black die resulted in a 5 (b) Find the conditional probability of obtaining the sum 8, given that the red die
resulted in a number less than 4 11 |
1 | 6855-6858 | (a) Find the conditional probability of obtaining a sum greater than 9, given
that the black die resulted in a 5 (b) Find the conditional probability of obtaining the sum 8, given that the red die
resulted in a number less than 4 11 A fair die is rolled |
1 | 6856-6859 | (b) Find the conditional probability of obtaining the sum 8, given that the red die
resulted in a number less than 4 11 A fair die is rolled Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5}
Find
(i) P(E|F) and P(F|E)
(ii) P(E|G) and P(G|E)
(iii) P((E ∪ F)|G) and P((E ∩ F)|G)
12 |
1 | 6857-6860 | 11 A fair die is rolled Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5}
Find
(i) P(E|F) and P(F|E)
(ii) P(E|G) and P(G|E)
(iii) P((E ∪ F)|G) and P((E ∩ F)|G)
12 Assume that each born child is equally likely to be a boy or a girl |
1 | 6858-6861 | A fair die is rolled Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5}
Find
(i) P(E|F) and P(F|E)
(ii) P(E|G) and P(G|E)
(iii) P((E ∪ F)|G) and P((E ∩ F)|G)
12 Assume that each born child is equally likely to be a boy or a girl If a family has
two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl, (ii) at least one is a girl |
1 | 6859-6862 | Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5}
Find
(i) P(E|F) and P(F|E)
(ii) P(E|G) and P(G|E)
(iii) P((E ∪ F)|G) and P((E ∩ F)|G)
12 Assume that each born child is equally likely to be a boy or a girl If a family has
two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl, (ii) at least one is a girl 13 |
1 | 6860-6863 | Assume that each born child is equally likely to be a boy or a girl If a family has
two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl, (ii) at least one is a girl 13 An instructor has a question bank consisting of 300 easy True / False questions,
200 difficult True / False questions, 500 easy multiple choice questions and 400
difficult multiple choice questions |
1 | 6861-6864 | If a family has
two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl, (ii) at least one is a girl 13 An instructor has a question bank consisting of 300 easy True / False questions,
200 difficult True / False questions, 500 easy multiple choice questions and 400
difficult multiple choice questions If a question is selected at random from the
question bank, what is the probability that it will be an easy question given that it
is a multiple choice question |
1 | 6862-6865 | 13 An instructor has a question bank consisting of 300 easy True / False questions,
200 difficult True / False questions, 500 easy multiple choice questions and 400
difficult multiple choice questions If a question is selected at random from the
question bank, what is the probability that it will be an easy question given that it
is a multiple choice question 14 |
1 | 6863-6866 | An instructor has a question bank consisting of 300 easy True / False questions,
200 difficult True / False questions, 500 easy multiple choice questions and 400
difficult multiple choice questions If a question is selected at random from the
question bank, what is the probability that it will be an easy question given that it
is a multiple choice question 14 Given that the two numbers appearing on throwing two dice are different |
1 | 6864-6867 | If a question is selected at random from the
question bank, what is the probability that it will be an easy question given that it
is a multiple choice question 14 Given that the two numbers appearing on throwing two dice are different Find
the probability of the event ‘the sum of numbers on the dice is 4’ |
1 | 6865-6868 | 14 Given that the two numbers appearing on throwing two dice are different Find
the probability of the event ‘the sum of numbers on the dice is 4’ 15 |
1 | 6866-6869 | Given that the two numbers appearing on throwing two dice are different Find
the probability of the event ‘the sum of numbers on the dice is 4’ 15 Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the
die again and if any other number comes, toss a coin |
1 | 6867-6870 | Find
the probability of the event ‘the sum of numbers on the dice is 4’ 15 Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the
die again and if any other number comes, toss a coin Find the conditional probability
of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’ |
1 | 6868-6871 | 15 Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the
die again and if any other number comes, toss a coin Find the conditional probability
of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’ In each of the Exercises 16 and 17 choose the correct answer:
16 |
1 | 6869-6872 | Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the
die again and if any other number comes, toss a coin Find the conditional probability
of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’ In each of the Exercises 16 and 17 choose the correct answer:
16 If P(A) = 1
2 , P(B) = 0, then P(A|B) is
(A) 0
(B) 1
2
(C) not defined
(D) 1
© NCERT
not to be republished
540
MATHEMATICS
17 |
1 | 6870-6873 | Find the conditional probability
of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’ In each of the Exercises 16 and 17 choose the correct answer:
16 If P(A) = 1
2 , P(B) = 0, then P(A|B) is
(A) 0
(B) 1
2
(C) not defined
(D) 1
© NCERT
not to be republished
540
MATHEMATICS
17 If A and B are events such that P(A|B) = P(B|A), then
(A) A ⊂ B but A ≠ B
(B) A = B
(C) A ∩ B = φ
(D) P(A) = P(B)
13 |
1 | 6871-6874 | In each of the Exercises 16 and 17 choose the correct answer:
16 If P(A) = 1
2 , P(B) = 0, then P(A|B) is
(A) 0
(B) 1
2
(C) not defined
(D) 1
© NCERT
not to be republished
540
MATHEMATICS
17 If A and B are events such that P(A|B) = P(B|A), then
(A) A ⊂ B but A ≠ B
(B) A = B
(C) A ∩ B = φ
(D) P(A) = P(B)
13 3 Multiplication Theorem on Probability
Let E and F be two events associated with a sample space S |
1 | 6872-6875 | If P(A) = 1
2 , P(B) = 0, then P(A|B) is
(A) 0
(B) 1
2
(C) not defined
(D) 1
© NCERT
not to be republished
540
MATHEMATICS
17 If A and B are events such that P(A|B) = P(B|A), then
(A) A ⊂ B but A ≠ B
(B) A = B
(C) A ∩ B = φ
(D) P(A) = P(B)
13 3 Multiplication Theorem on Probability
Let E and F be two events associated with a sample space S Clearly, the set E ∩ F
denotes the event that both E and F have occurred |
1 | 6873-6876 | If A and B are events such that P(A|B) = P(B|A), then
(A) A ⊂ B but A ≠ B
(B) A = B
(C) A ∩ B = φ
(D) P(A) = P(B)
13 3 Multiplication Theorem on Probability
Let E and F be two events associated with a sample space S Clearly, the set E ∩ F
denotes the event that both E and F have occurred In other words, E ∩ F denotes the
simultaneous occurrence of the events E and F |
1 | 6874-6877 | 3 Multiplication Theorem on Probability
Let E and F be two events associated with a sample space S Clearly, the set E ∩ F
denotes the event that both E and F have occurred In other words, E ∩ F denotes the
simultaneous occurrence of the events E and F The event E ∩ F is also written as EF |
1 | 6875-6878 | Clearly, the set E ∩ F
denotes the event that both E and F have occurred In other words, E ∩ F denotes the
simultaneous occurrence of the events E and F The event E ∩ F is also written as EF Very often we need to find the probability of the event EF |
1 | 6876-6879 | In other words, E ∩ F denotes the
simultaneous occurrence of the events E and F The event E ∩ F is also written as EF Very often we need to find the probability of the event EF For example, in the
experiment of drawing two cards one after the other, we may be interested in finding
the probability of the event ‘a king and a queen’ |
1 | 6877-6880 | The event E ∩ F is also written as EF Very often we need to find the probability of the event EF For example, in the
experiment of drawing two cards one after the other, we may be interested in finding
the probability of the event ‘a king and a queen’ The probability of event EF is obtained
by using the conditional probability as obtained below :
We know that the conditional probability of event E given that F has occurred is
denoted by P(E|F) and is given by
P(E|F) = P(E
P(F)F) ,P(F) 0
∩
≠
From this result, we can write
P(E ∩ F) = P(F) |
1 | 6878-6881 | Very often we need to find the probability of the event EF For example, in the
experiment of drawing two cards one after the other, we may be interested in finding
the probability of the event ‘a king and a queen’ The probability of event EF is obtained
by using the conditional probability as obtained below :
We know that the conditional probability of event E given that F has occurred is
denoted by P(E|F) and is given by
P(E|F) = P(E
P(F)F) ,P(F) 0
∩
≠
From this result, we can write
P(E ∩ F) = P(F) P(E|F) |
1 | 6879-6882 | For example, in the
experiment of drawing two cards one after the other, we may be interested in finding
the probability of the event ‘a king and a queen’ The probability of event EF is obtained
by using the conditional probability as obtained below :
We know that the conditional probability of event E given that F has occurred is
denoted by P(E|F) and is given by
P(E|F) = P(E
P(F)F) ,P(F) 0
∩
≠
From this result, we can write
P(E ∩ F) = P(F) P(E|F) (1)
Also, we know that
P(F|E) = P(F
P(E)E) ,P(E) 0
∩
≠
or
P(F|E) = P(E
P(E)F)
∩
(since E ∩ F = F ∩ E)
Thus,
P(E ∩ F) = P(E) |
1 | 6880-6883 | The probability of event EF is obtained
by using the conditional probability as obtained below :
We know that the conditional probability of event E given that F has occurred is
denoted by P(E|F) and is given by
P(E|F) = P(E
P(F)F) ,P(F) 0
∩
≠
From this result, we can write
P(E ∩ F) = P(F) P(E|F) (1)
Also, we know that
P(F|E) = P(F
P(E)E) ,P(E) 0
∩
≠
or
P(F|E) = P(E
P(E)F)
∩
(since E ∩ F = F ∩ E)
Thus,
P(E ∩ F) = P(E) P(F|E) |
1 | 6881-6884 | P(E|F) (1)
Also, we know that
P(F|E) = P(F
P(E)E) ,P(E) 0
∩
≠
or
P(F|E) = P(E
P(E)F)
∩
(since E ∩ F = F ∩ E)
Thus,
P(E ∩ F) = P(E) P(F|E) (2)
Combining (1) and (2), we find that
P(E ∩ F) = P(E) P(F|E)
= P(F) P(E|F) provided P(E) ≠ 0 and P(F) ≠ 0 |
1 | 6882-6885 | (1)
Also, we know that
P(F|E) = P(F
P(E)E) ,P(E) 0
∩
≠
or
P(F|E) = P(E
P(E)F)
∩
(since E ∩ F = F ∩ E)
Thus,
P(E ∩ F) = P(E) P(F|E) (2)
Combining (1) and (2), we find that
P(E ∩ F) = P(E) P(F|E)
= P(F) P(E|F) provided P(E) ≠ 0 and P(F) ≠ 0 The above result is known as the multiplication rule of probability |
1 | 6883-6886 | P(F|E) (2)
Combining (1) and (2), we find that
P(E ∩ F) = P(E) P(F|E)
= P(F) P(E|F) provided P(E) ≠ 0 and P(F) ≠ 0 The above result is known as the multiplication rule of probability Let us now take up an example |
1 | 6884-6887 | (2)
Combining (1) and (2), we find that
P(E ∩ F) = P(E) P(F|E)
= P(F) P(E|F) provided P(E) ≠ 0 and P(F) ≠ 0 The above result is known as the multiplication rule of probability Let us now take up an example Example 8 An urn contains 10 black and 5 white balls |
1 | 6885-6888 | The above result is known as the multiplication rule of probability Let us now take up an example Example 8 An urn contains 10 black and 5 white balls Two balls are drawn from the
urn one after the other without replacement |
1 | 6886-6889 | Let us now take up an example Example 8 An urn contains 10 black and 5 white balls Two balls are drawn from the
urn one after the other without replacement What is the probability that both drawn
balls are black |
1 | 6887-6890 | Example 8 An urn contains 10 black and 5 white balls Two balls are drawn from the
urn one after the other without replacement What is the probability that both drawn
balls are black Solution Let E and F denote respectively the events that first and second ball drawn
are black |
1 | 6888-6891 | Two balls are drawn from the
urn one after the other without replacement What is the probability that both drawn
balls are black Solution Let E and F denote respectively the events that first and second ball drawn
are black We have to find P(E ∩ F) or P (EF) |
1 | 6889-6892 | What is the probability that both drawn
balls are black Solution Let E and F denote respectively the events that first and second ball drawn
are black We have to find P(E ∩ F) or P (EF) © NCERT
not to be republished
PROBABILITY 541
Now
P(E) = P (black ball in first draw) = 10
15
Also given that the first ball drawn is black, i |
1 | 6890-6893 | Solution Let E and F denote respectively the events that first and second ball drawn
are black We have to find P(E ∩ F) or P (EF) © NCERT
not to be republished
PROBABILITY 541
Now
P(E) = P (black ball in first draw) = 10
15
Also given that the first ball drawn is black, i e |
1 | 6891-6894 | We have to find P(E ∩ F) or P (EF) © NCERT
not to be republished
PROBABILITY 541
Now
P(E) = P (black ball in first draw) = 10
15
Also given that the first ball drawn is black, i e , event E has occurred, now there
are 9 black balls and five white balls left in the urn |
1 | 6892-6895 | © NCERT
not to be republished
PROBABILITY 541
Now
P(E) = P (black ball in first draw) = 10
15
Also given that the first ball drawn is black, i e , event E has occurred, now there
are 9 black balls and five white balls left in the urn Therefore, the probability that the
second ball drawn is black, given that the ball in the first draw is black, is nothing but
the conditional probability of F given that E has occurred |
1 | 6893-6896 | e , event E has occurred, now there
are 9 black balls and five white balls left in the urn Therefore, the probability that the
second ball drawn is black, given that the ball in the first draw is black, is nothing but
the conditional probability of F given that E has occurred i |
1 | 6894-6897 | , event E has occurred, now there
are 9 black balls and five white balls left in the urn Therefore, the probability that the
second ball drawn is black, given that the ball in the first draw is black, is nothing but
the conditional probability of F given that E has occurred i e |
1 | 6895-6898 | Therefore, the probability that the
second ball drawn is black, given that the ball in the first draw is black, is nothing but
the conditional probability of F given that E has occurred i e P(F|E) = 9
14
By multiplication rule of probability, we have
P (E ∩ F) = P(E) P(F|E)
= 10
9
3
15
14
7
Multiplication rule of probability for more than two events If E, F and G are
three events of sample space, we have
P(E ∩ F ∩ G) = P(E) P(F|E) P(G|(E ∩ F)) = P(E) P(F|E) P(G|EF)
Similarly, the multiplication rule of probability can be extended for four or
more events |
1 | 6896-6899 | i e P(F|E) = 9
14
By multiplication rule of probability, we have
P (E ∩ F) = P(E) P(F|E)
= 10
9
3
15
14
7
Multiplication rule of probability for more than two events If E, F and G are
three events of sample space, we have
P(E ∩ F ∩ G) = P(E) P(F|E) P(G|(E ∩ F)) = P(E) P(F|E) P(G|EF)
Similarly, the multiplication rule of probability can be extended for four or
more events The following example illustrates the extension of multiplication rule of probability
for three events |
1 | 6897-6900 | e P(F|E) = 9
14
By multiplication rule of probability, we have
P (E ∩ F) = P(E) P(F|E)
= 10
9
3
15
14
7
Multiplication rule of probability for more than two events If E, F and G are
three events of sample space, we have
P(E ∩ F ∩ G) = P(E) P(F|E) P(G|(E ∩ F)) = P(E) P(F|E) P(G|EF)
Similarly, the multiplication rule of probability can be extended for four or
more events The following example illustrates the extension of multiplication rule of probability
for three events Example 9 Three cards are drawn successively, without replacement from a pack of
52 well shuffled cards |
1 | 6898-6901 | P(F|E) = 9
14
By multiplication rule of probability, we have
P (E ∩ F) = P(E) P(F|E)
= 10
9
3
15
14
7
Multiplication rule of probability for more than two events If E, F and G are
three events of sample space, we have
P(E ∩ F ∩ G) = P(E) P(F|E) P(G|(E ∩ F)) = P(E) P(F|E) P(G|EF)
Similarly, the multiplication rule of probability can be extended for four or
more events The following example illustrates the extension of multiplication rule of probability
for three events Example 9 Three cards are drawn successively, without replacement from a pack of
52 well shuffled cards What is the probability that first two cards are kings and the
third card drawn is an ace |
1 | 6899-6902 | The following example illustrates the extension of multiplication rule of probability
for three events Example 9 Three cards are drawn successively, without replacement from a pack of
52 well shuffled cards What is the probability that first two cards are kings and the
third card drawn is an ace Solution Let K denote the event that the card drawn is king and A be the event that
the card drawn is an ace |
1 | 6900-6903 | Example 9 Three cards are drawn successively, without replacement from a pack of
52 well shuffled cards What is the probability that first two cards are kings and the
third card drawn is an ace Solution Let K denote the event that the card drawn is king and A be the event that
the card drawn is an ace Clearly, we have to find P (KKA)
Now
P(K) = 4
52
Also, P (K|K) is the probability of second king with the condition that one king has
already been drawn |
1 | 6901-6904 | What is the probability that first two cards are kings and the
third card drawn is an ace Solution Let K denote the event that the card drawn is king and A be the event that
the card drawn is an ace Clearly, we have to find P (KKA)
Now
P(K) = 4
52
Also, P (K|K) is the probability of second king with the condition that one king has
already been drawn Now there are three kings in (52 − 1) = 51 cards |
1 | 6902-6905 | Solution Let K denote the event that the card drawn is king and A be the event that
the card drawn is an ace Clearly, we have to find P (KKA)
Now
P(K) = 4
52
Also, P (K|K) is the probability of second king with the condition that one king has
already been drawn Now there are three kings in (52 − 1) = 51 cards Therefore
P(K|K) = 3
51
Lastly, P(A|KK) is the probability of third drawn card to be an ace, with the condition
that two kings have already been drawn |
1 | 6903-6906 | Clearly, we have to find P (KKA)
Now
P(K) = 4
52
Also, P (K|K) is the probability of second king with the condition that one king has
already been drawn Now there are three kings in (52 − 1) = 51 cards Therefore
P(K|K) = 3
51
Lastly, P(A|KK) is the probability of third drawn card to be an ace, with the condition
that two kings have already been drawn Now there are four aces in left 50 cards |
1 | 6904-6907 | Now there are three kings in (52 − 1) = 51 cards Therefore
P(K|K) = 3
51
Lastly, P(A|KK) is the probability of third drawn card to be an ace, with the condition
that two kings have already been drawn Now there are four aces in left 50 cards © NCERT
not to be republished
542
MATHEMATICS
Therefore
P(A|KK) = 4
50
By multiplication law of probability, we have
P(KKA) = P(K) P(K|K) P(A|KK)
= 4
3
4
2
52
51
50
5525
13 |
1 | 6905-6908 | Therefore
P(K|K) = 3
51
Lastly, P(A|KK) is the probability of third drawn card to be an ace, with the condition
that two kings have already been drawn Now there are four aces in left 50 cards © NCERT
not to be republished
542
MATHEMATICS
Therefore
P(A|KK) = 4
50
By multiplication law of probability, we have
P(KKA) = P(K) P(K|K) P(A|KK)
= 4
3
4
2
52
51
50
5525
13 4 Independent Events
Consider the experiment of drawing a card from a deck of 52 playing cards, in which
the elementary events are assumed to be equally likely |
1 | 6906-6909 | Now there are four aces in left 50 cards © NCERT
not to be republished
542
MATHEMATICS
Therefore
P(A|KK) = 4
50
By multiplication law of probability, we have
P(KKA) = P(K) P(K|K) P(A|KK)
= 4
3
4
2
52
51
50
5525
13 4 Independent Events
Consider the experiment of drawing a card from a deck of 52 playing cards, in which
the elementary events are assumed to be equally likely If E and F denote the events
'the card drawn is a spade' and 'the card drawn is an ace' respectively, then
P(E) = 13
1
4
1
and P(F)
52
4
52
13
Also E and F is the event ' the card drawn is the ace of spades' so that
P(E ∩F) = 1
52
Hence
P(E|F) =
1
P(E
F)
1
152
P(F)
4
13
Since P(E) = 1
4 = P (E|F), we can say that the occurrence of event F has not
affected the probability of occurrence of the event E |
1 | 6907-6910 | © NCERT
not to be republished
542
MATHEMATICS
Therefore
P(A|KK) = 4
50
By multiplication law of probability, we have
P(KKA) = P(K) P(K|K) P(A|KK)
= 4
3
4
2
52
51
50
5525
13 4 Independent Events
Consider the experiment of drawing a card from a deck of 52 playing cards, in which
the elementary events are assumed to be equally likely If E and F denote the events
'the card drawn is a spade' and 'the card drawn is an ace' respectively, then
P(E) = 13
1
4
1
and P(F)
52
4
52
13
Also E and F is the event ' the card drawn is the ace of spades' so that
P(E ∩F) = 1
52
Hence
P(E|F) =
1
P(E
F)
1
152
P(F)
4
13
Since P(E) = 1
4 = P (E|F), we can say that the occurrence of event F has not
affected the probability of occurrence of the event E We also have
P(F|E) =
1
P(E
F)
1
52
P(F)
1
P(E)
13
4
Again, P(F) = 1
13 = P(F|E) shows that occurrence of event E has not affected
the probability of occurrence of the event F |
1 | 6908-6911 | 4 Independent Events
Consider the experiment of drawing a card from a deck of 52 playing cards, in which
the elementary events are assumed to be equally likely If E and F denote the events
'the card drawn is a spade' and 'the card drawn is an ace' respectively, then
P(E) = 13
1
4
1
and P(F)
52
4
52
13
Also E and F is the event ' the card drawn is the ace of spades' so that
P(E ∩F) = 1
52
Hence
P(E|F) =
1
P(E
F)
1
152
P(F)
4
13
Since P(E) = 1
4 = P (E|F), we can say that the occurrence of event F has not
affected the probability of occurrence of the event E We also have
P(F|E) =
1
P(E
F)
1
52
P(F)
1
P(E)
13
4
Again, P(F) = 1
13 = P(F|E) shows that occurrence of event E has not affected
the probability of occurrence of the event F Thus, E and F are two events such that the probability of occurrence of one of
them is not affected by occurrence of the other |
1 | 6909-6912 | If E and F denote the events
'the card drawn is a spade' and 'the card drawn is an ace' respectively, then
P(E) = 13
1
4
1
and P(F)
52
4
52
13
Also E and F is the event ' the card drawn is the ace of spades' so that
P(E ∩F) = 1
52
Hence
P(E|F) =
1
P(E
F)
1
152
P(F)
4
13
Since P(E) = 1
4 = P (E|F), we can say that the occurrence of event F has not
affected the probability of occurrence of the event E We also have
P(F|E) =
1
P(E
F)
1
52
P(F)
1
P(E)
13
4
Again, P(F) = 1
13 = P(F|E) shows that occurrence of event E has not affected
the probability of occurrence of the event F Thus, E and F are two events such that the probability of occurrence of one of
them is not affected by occurrence of the other Such events are called independent events |
1 | 6910-6913 | We also have
P(F|E) =
1
P(E
F)
1
52
P(F)
1
P(E)
13
4
Again, P(F) = 1
13 = P(F|E) shows that occurrence of event E has not affected
the probability of occurrence of the event F Thus, E and F are two events such that the probability of occurrence of one of
them is not affected by occurrence of the other Such events are called independent events © NCERT
not to be republished
PROBABILITY 543
Definition 2 Two events E and F are said to be independent, if
P(F|E) = P (F) provided P (E) ≠ 0
and
P (E|F) = P (E) provided P (F) ≠ 0
Thus, in this definition we need to have P (E) ≠ 0 and P(F) ≠ 0
Now, by the multiplication rule of probability, we have
P(E ∩ F) = P(E) |
1 | 6911-6914 | Thus, E and F are two events such that the probability of occurrence of one of
them is not affected by occurrence of the other Such events are called independent events © NCERT
not to be republished
PROBABILITY 543
Definition 2 Two events E and F are said to be independent, if
P(F|E) = P (F) provided P (E) ≠ 0
and
P (E|F) = P (E) provided P (F) ≠ 0
Thus, in this definition we need to have P (E) ≠ 0 and P(F) ≠ 0
Now, by the multiplication rule of probability, we have
P(E ∩ F) = P(E) P (F|E) |
1 | 6912-6915 | Such events are called independent events © NCERT
not to be republished
PROBABILITY 543
Definition 2 Two events E and F are said to be independent, if
P(F|E) = P (F) provided P (E) ≠ 0
and
P (E|F) = P (E) provided P (F) ≠ 0
Thus, in this definition we need to have P (E) ≠ 0 and P(F) ≠ 0
Now, by the multiplication rule of probability, we have
P(E ∩ F) = P(E) P (F|E) (1)
If E and F are independent, then (1) becomes
P(E ∩ F) = P(E) |
1 | 6913-6916 | © NCERT
not to be republished
PROBABILITY 543
Definition 2 Two events E and F are said to be independent, if
P(F|E) = P (F) provided P (E) ≠ 0
and
P (E|F) = P (E) provided P (F) ≠ 0
Thus, in this definition we need to have P (E) ≠ 0 and P(F) ≠ 0
Now, by the multiplication rule of probability, we have
P(E ∩ F) = P(E) P (F|E) (1)
If E and F are independent, then (1) becomes
P(E ∩ F) = P(E) P(F) |
1 | 6914-6917 | P (F|E) (1)
If E and F are independent, then (1) becomes
P(E ∩ F) = P(E) P(F) (2)
Thus, using (2), the independence of two events is also defined as follows:
Definition 3 Let E and F be two events associated with the same random experiment,
then E and F are said to be independent if
P(E ∩ F) = P(E) |
1 | 6915-6918 | (1)
If E and F are independent, then (1) becomes
P(E ∩ F) = P(E) P(F) (2)
Thus, using (2), the independence of two events is also defined as follows:
Definition 3 Let E and F be two events associated with the same random experiment,
then E and F are said to be independent if
P(E ∩ F) = P(E) P (F)
Remarks
(i)
Two events E and F are said to be dependent if they are not independent, i |
1 | 6916-6919 | P(F) (2)
Thus, using (2), the independence of two events is also defined as follows:
Definition 3 Let E and F be two events associated with the same random experiment,
then E and F are said to be independent if
P(E ∩ F) = P(E) P (F)
Remarks
(i)
Two events E and F are said to be dependent if they are not independent, i e |
1 | 6917-6920 | (2)
Thus, using (2), the independence of two events is also defined as follows:
Definition 3 Let E and F be two events associated with the same random experiment,
then E and F are said to be independent if
P(E ∩ F) = P(E) P (F)
Remarks
(i)
Two events E and F are said to be dependent if they are not independent, i e if
P(E ∩ F ) ≠ P(E) |
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