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Miscellaneous Exercise on Chapter 13
1 A and B are two events such that P (A) ≠ 0 Find P(B|A), if
(i) A is a subset of B
(ii) A ∩ B = φ
2 A couple has two children,
(i) Find the probability that both children are males, if it is known that at least
one of the children is male |
1 | 7719-7722 | A and B are two events such that P (A) ≠ 0 Find P(B|A), if
(i) A is a subset of B
(ii) A ∩ B = φ
2 A couple has two children,
(i) Find the probability that both children are males, if it is known that at least
one of the children is male (ii) Find the probability that both children are females, if it is known that the
elder child is a female |
1 | 7720-7723 | Find P(B|A), if
(i) A is a subset of B
(ii) A ∩ B = φ
2 A couple has two children,
(i) Find the probability that both children are males, if it is known that at least
one of the children is male (ii) Find the probability that both children are females, if it is known that the
elder child is a female 3 |
1 | 7721-7724 | A couple has two children,
(i) Find the probability that both children are males, if it is known that at least
one of the children is male (ii) Find the probability that both children are females, if it is known that the
elder child is a female 3 Suppose that 5% of men and 0 |
1 | 7722-7725 | (ii) Find the probability that both children are females, if it is known that the
elder child is a female 3 Suppose that 5% of men and 0 25% of women have grey hair |
1 | 7723-7726 | 3 Suppose that 5% of men and 0 25% of women have grey hair A grey haired
person is selected at random |
1 | 7724-7727 | Suppose that 5% of men and 0 25% of women have grey hair A grey haired
person is selected at random What is the probability of this person being male |
1 | 7725-7728 | 25% of women have grey hair A grey haired
person is selected at random What is the probability of this person being male Assume that there are equal number of males and females |
1 | 7726-7729 | A grey haired
person is selected at random What is the probability of this person being male Assume that there are equal number of males and females 4 |
1 | 7727-7730 | What is the probability of this person being male Assume that there are equal number of males and females 4 Suppose that 90% of people are right-handed |
1 | 7728-7731 | Assume that there are equal number of males and females 4 Suppose that 90% of people are right-handed What is the probability that
at most 6 of a random sample of 10 people are right-handed |
1 | 7729-7732 | 4 Suppose that 90% of people are right-handed What is the probability that
at most 6 of a random sample of 10 people are right-handed © NCERT
not to be republished
PROBABILITY 583
5 |
1 | 7730-7733 | Suppose that 90% of people are right-handed What is the probability that
at most 6 of a random sample of 10 people are right-handed © NCERT
not to be republished
PROBABILITY 583
5 An urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15
bear a mark 'Y' |
1 | 7731-7734 | What is the probability that
at most 6 of a random sample of 10 people are right-handed © NCERT
not to be republished
PROBABILITY 583
5 An urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15
bear a mark 'Y' A ball is drawn at random from the urn, its mark is noted down
and it is replaced |
1 | 7732-7735 | © NCERT
not to be republished
PROBABILITY 583
5 An urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15
bear a mark 'Y' A ball is drawn at random from the urn, its mark is noted down
and it is replaced If 6 balls are drawn in this way, find the probability that
(i) all will bear 'X' mark |
1 | 7733-7736 | An urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15
bear a mark 'Y' A ball is drawn at random from the urn, its mark is noted down
and it is replaced If 6 balls are drawn in this way, find the probability that
(i) all will bear 'X' mark (ii) not more than 2 will bear 'Y' mark |
1 | 7734-7737 | A ball is drawn at random from the urn, its mark is noted down
and it is replaced If 6 balls are drawn in this way, find the probability that
(i) all will bear 'X' mark (ii) not more than 2 will bear 'Y' mark (iii) at least one ball will bear 'Y' mark |
1 | 7735-7738 | If 6 balls are drawn in this way, find the probability that
(i) all will bear 'X' mark (ii) not more than 2 will bear 'Y' mark (iii) at least one ball will bear 'Y' mark (iv) the number of balls with 'X' mark and 'Y' mark will be equal |
1 | 7736-7739 | (ii) not more than 2 will bear 'Y' mark (iii) at least one ball will bear 'Y' mark (iv) the number of balls with 'X' mark and 'Y' mark will be equal 6 |
1 | 7737-7740 | (iii) at least one ball will bear 'Y' mark (iv) the number of balls with 'X' mark and 'Y' mark will be equal 6 In a hurdle race, a player has to cross 10 hurdles |
1 | 7738-7741 | (iv) the number of balls with 'X' mark and 'Y' mark will be equal 6 In a hurdle race, a player has to cross 10 hurdles The probability that he will
clear each hurdle is 5
6 |
1 | 7739-7742 | 6 In a hurdle race, a player has to cross 10 hurdles The probability that he will
clear each hurdle is 5
6 What is the probability that he will knock down fewer
than 2 hurdles |
1 | 7740-7743 | In a hurdle race, a player has to cross 10 hurdles The probability that he will
clear each hurdle is 5
6 What is the probability that he will knock down fewer
than 2 hurdles 7 |
1 | 7741-7744 | The probability that he will
clear each hurdle is 5
6 What is the probability that he will knock down fewer
than 2 hurdles 7 A die is thrown again and again until three sixes are obtained |
1 | 7742-7745 | What is the probability that he will knock down fewer
than 2 hurdles 7 A die is thrown again and again until three sixes are obtained Find the probabil-
ity of obtaining the third six in the sixth throw of the die |
1 | 7743-7746 | 7 A die is thrown again and again until three sixes are obtained Find the probabil-
ity of obtaining the third six in the sixth throw of the die 8 |
1 | 7744-7747 | A die is thrown again and again until three sixes are obtained Find the probabil-
ity of obtaining the third six in the sixth throw of the die 8 If a leap year is selected at random, what is the chance that it will contain 53
tuesdays |
1 | 7745-7748 | Find the probabil-
ity of obtaining the third six in the sixth throw of the die 8 If a leap year is selected at random, what is the chance that it will contain 53
tuesdays 9 |
1 | 7746-7749 | 8 If a leap year is selected at random, what is the chance that it will contain 53
tuesdays 9 An experiment succeeds twice as often as it fails |
1 | 7747-7750 | If a leap year is selected at random, what is the chance that it will contain 53
tuesdays 9 An experiment succeeds twice as often as it fails Find the probability that in the
next six trials, there will be atleast 4 successes |
1 | 7748-7751 | 9 An experiment succeeds twice as often as it fails Find the probability that in the
next six trials, there will be atleast 4 successes 10 |
1 | 7749-7752 | An experiment succeeds twice as often as it fails Find the probability that in the
next six trials, there will be atleast 4 successes 10 How many times must a man toss a fair coin so that the probability of having
at least one head is more than 90% |
1 | 7750-7753 | Find the probability that in the
next six trials, there will be atleast 4 successes 10 How many times must a man toss a fair coin so that the probability of having
at least one head is more than 90% 11 |
1 | 7751-7754 | 10 How many times must a man toss a fair coin so that the probability of having
at least one head is more than 90% 11 In a game, a man wins a rupee for a six and loses a rupee for any other number
when a fair die is thrown |
1 | 7752-7755 | How many times must a man toss a fair coin so that the probability of having
at least one head is more than 90% 11 In a game, a man wins a rupee for a six and loses a rupee for any other number
when a fair die is thrown The man decided to throw a die thrice but to quit as
and when he gets a six |
1 | 7753-7756 | 11 In a game, a man wins a rupee for a six and loses a rupee for any other number
when a fair die is thrown The man decided to throw a die thrice but to quit as
and when he gets a six Find the expected value of the amount he wins / loses |
1 | 7754-7757 | In a game, a man wins a rupee for a six and loses a rupee for any other number
when a fair die is thrown The man decided to throw a die thrice but to quit as
and when he gets a six Find the expected value of the amount he wins / loses 12 |
1 | 7755-7758 | The man decided to throw a die thrice but to quit as
and when he gets a six Find the expected value of the amount he wins / loses 12 Suppose we have four boxes A,B,C and D containing coloured marbles as given
below:
Box
Marble colour
Red
White
Black
A
1
6
3
B
6
2
2
C
8
1
1
D
0
6
4
One of the boxes has been selected at random and a single marble is drawn from
it |
1 | 7756-7759 | Find the expected value of the amount he wins / loses 12 Suppose we have four boxes A,B,C and D containing coloured marbles as given
below:
Box
Marble colour
Red
White
Black
A
1
6
3
B
6
2
2
C
8
1
1
D
0
6
4
One of the boxes has been selected at random and a single marble is drawn from
it If the marble is red, what is the probability that it was drawn from box A |
1 | 7757-7760 | 12 Suppose we have four boxes A,B,C and D containing coloured marbles as given
below:
Box
Marble colour
Red
White
Black
A
1
6
3
B
6
2
2
C
8
1
1
D
0
6
4
One of the boxes has been selected at random and a single marble is drawn from
it If the marble is red, what is the probability that it was drawn from box A , box B |
1 | 7758-7761 | Suppose we have four boxes A,B,C and D containing coloured marbles as given
below:
Box
Marble colour
Red
White
Black
A
1
6
3
B
6
2
2
C
8
1
1
D
0
6
4
One of the boxes has been selected at random and a single marble is drawn from
it If the marble is red, what is the probability that it was drawn from box A , box B ,
box C |
1 | 7759-7762 | If the marble is red, what is the probability that it was drawn from box A , box B ,
box C © NCERT
not to be republished
584
MATHEMATICS
13 |
1 | 7760-7763 | , box B ,
box C © NCERT
not to be republished
584
MATHEMATICS
13 Assume that the chances of a patient having a heart attack is 40% |
1 | 7761-7764 | ,
box C © NCERT
not to be republished
584
MATHEMATICS
13 Assume that the chances of a patient having a heart attack is 40% It is also
assumed that a meditation and yoga course reduce the risk of heart attack by
30% and prescription of certain drug reduces its chances by 25% |
1 | 7762-7765 | © NCERT
not to be republished
584
MATHEMATICS
13 Assume that the chances of a patient having a heart attack is 40% It is also
assumed that a meditation and yoga course reduce the risk of heart attack by
30% and prescription of certain drug reduces its chances by 25% At a time a
patient can choose any one of the two options with equal probabilities |
1 | 7763-7766 | Assume that the chances of a patient having a heart attack is 40% It is also
assumed that a meditation and yoga course reduce the risk of heart attack by
30% and prescription of certain drug reduces its chances by 25% At a time a
patient can choose any one of the two options with equal probabilities It is given
that after going through one of the two options the patient selected at random
suffers a heart attack |
1 | 7764-7767 | It is also
assumed that a meditation and yoga course reduce the risk of heart attack by
30% and prescription of certain drug reduces its chances by 25% At a time a
patient can choose any one of the two options with equal probabilities It is given
that after going through one of the two options the patient selected at random
suffers a heart attack Find the probability that the patient followed a course of
meditation and yoga |
1 | 7765-7768 | At a time a
patient can choose any one of the two options with equal probabilities It is given
that after going through one of the two options the patient selected at random
suffers a heart attack Find the probability that the patient followed a course of
meditation and yoga 14 |
1 | 7766-7769 | It is given
that after going through one of the two options the patient selected at random
suffers a heart attack Find the probability that the patient followed a course of
meditation and yoga 14 If each element of a second order determinant is either zero or one, what is the
probability that the value of the determinant is positive |
1 | 7767-7770 | Find the probability that the patient followed a course of
meditation and yoga 14 If each element of a second order determinant is either zero or one, what is the
probability that the value of the determinant is positive (Assume that the indi-
vidual entries of the determinant are chosen independently, each value being
assumed with probability 1
2 ) |
1 | 7768-7771 | 14 If each element of a second order determinant is either zero or one, what is the
probability that the value of the determinant is positive (Assume that the indi-
vidual entries of the determinant are chosen independently, each value being
assumed with probability 1
2 ) 15 |
1 | 7769-7772 | If each element of a second order determinant is either zero or one, what is the
probability that the value of the determinant is positive (Assume that the indi-
vidual entries of the determinant are chosen independently, each value being
assumed with probability 1
2 ) 15 An electronic assembly consists of two subsystems, say, A and B |
1 | 7770-7773 | (Assume that the indi-
vidual entries of the determinant are chosen independently, each value being
assumed with probability 1
2 ) 15 An electronic assembly consists of two subsystems, say, A and B From previ-
ous testing procedures, the following probabilities are assumed to be known:
P(A fails) = 0 |
1 | 7771-7774 | 15 An electronic assembly consists of two subsystems, say, A and B From previ-
ous testing procedures, the following probabilities are assumed to be known:
P(A fails) = 0 2
P(B fails alone) = 0 |
1 | 7772-7775 | An electronic assembly consists of two subsystems, say, A and B From previ-
ous testing procedures, the following probabilities are assumed to be known:
P(A fails) = 0 2
P(B fails alone) = 0 15
P(A and B fail) = 0 |
1 | 7773-7776 | From previ-
ous testing procedures, the following probabilities are assumed to be known:
P(A fails) = 0 2
P(B fails alone) = 0 15
P(A and B fail) = 0 15
Evaluate the following probabilities
(i) P(A fails|B has failed)
(ii) P(A fails alone)
16 |
1 | 7774-7777 | 2
P(B fails alone) = 0 15
P(A and B fail) = 0 15
Evaluate the following probabilities
(i) P(A fails|B has failed)
(ii) P(A fails alone)
16 Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls |
1 | 7775-7778 | 15
P(A and B fail) = 0 15
Evaluate the following probabilities
(i) P(A fails|B has failed)
(ii) P(A fails alone)
16 Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II |
1 | 7776-7779 | 15
Evaluate the following probabilities
(i) P(A fails|B has failed)
(ii) P(A fails alone)
16 Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II The ball so drawn is found to be red in colour |
1 | 7777-7780 | Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II The ball so drawn is found to be red in colour Find the probability that the
transferred ball is black |
1 | 7778-7781 | One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II The ball so drawn is found to be red in colour Find the probability that the
transferred ball is black Choose the correct answer in each of the following:
17 |
1 | 7779-7782 | The ball so drawn is found to be red in colour Find the probability that the
transferred ball is black Choose the correct answer in each of the following:
17 If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then
(A) A ⊂ B
(B) B ⊂ A
(C) B = φ
(D) A = φ
18 |
1 | 7780-7783 | Find the probability that the
transferred ball is black Choose the correct answer in each of the following:
17 If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then
(A) A ⊂ B
(B) B ⊂ A
(C) B = φ
(D) A = φ
18 If P(A|B) > P(A), then which of the following is correct :
(A) P(B|A) < P(B)
(B) P(A ∩ B) < P(A) |
1 | 7781-7784 | Choose the correct answer in each of the following:
17 If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then
(A) A ⊂ B
(B) B ⊂ A
(C) B = φ
(D) A = φ
18 If P(A|B) > P(A), then which of the following is correct :
(A) P(B|A) < P(B)
(B) P(A ∩ B) < P(A) P(B)
(C) P(B|A) > P(B)
(D) P(B|A) = P(B)
19 |
1 | 7782-7785 | If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then
(A) A ⊂ B
(B) B ⊂ A
(C) B = φ
(D) A = φ
18 If P(A|B) > P(A), then which of the following is correct :
(A) P(B|A) < P(B)
(B) P(A ∩ B) < P(A) P(B)
(C) P(B|A) > P(B)
(D) P(B|A) = P(B)
19 If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then
(A) P(B|A) = 1
(B) P(A|B) = 1
(C) P(B|A) = 0
(D) P(A|B) = 0
© NCERT
not to be republished
PROBABILITY 585
Summary
The salient features of the chapter are –
� The conditional probability of an event E, given the occurrence of the event F
is given by
P(E
F)
P(E | F)
P(F)
∩
=
, P(F) ≠ 0
� 0 ≤ P (E|F) ≤ 1,
P (E′|F) = 1 – P (E|F)
P ((E ∪ F)|G) = P (E|G) + P (F|G) – P ((E ∩ F)|G)
� P (E ∩ F) = P (E) P (F|E), P (E) ≠ 0
P (E ∩ F) = P (F) P (E|F), P (F) ≠ 0
� If E and F are independent, then
P (E ∩ F) = P (E) P (F)
P (E|F) = P (E), P (F) ≠ 0
P (F|E) = P (F), P(E) ≠ 0
� Theorem of total probability
Let {E1, E2, |
1 | 7783-7786 | If P(A|B) > P(A), then which of the following is correct :
(A) P(B|A) < P(B)
(B) P(A ∩ B) < P(A) P(B)
(C) P(B|A) > P(B)
(D) P(B|A) = P(B)
19 If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then
(A) P(B|A) = 1
(B) P(A|B) = 1
(C) P(B|A) = 0
(D) P(A|B) = 0
© NCERT
not to be republished
PROBABILITY 585
Summary
The salient features of the chapter are –
� The conditional probability of an event E, given the occurrence of the event F
is given by
P(E
F)
P(E | F)
P(F)
∩
=
, P(F) ≠ 0
� 0 ≤ P (E|F) ≤ 1,
P (E′|F) = 1 – P (E|F)
P ((E ∪ F)|G) = P (E|G) + P (F|G) – P ((E ∩ F)|G)
� P (E ∩ F) = P (E) P (F|E), P (E) ≠ 0
P (E ∩ F) = P (F) P (E|F), P (F) ≠ 0
� If E and F are independent, then
P (E ∩ F) = P (E) P (F)
P (E|F) = P (E), P (F) ≠ 0
P (F|E) = P (F), P(E) ≠ 0
� Theorem of total probability
Let {E1, E2, ,En) be a partition of a sample space and suppose that each of
E1, E2, |
1 | 7784-7787 | P(B)
(C) P(B|A) > P(B)
(D) P(B|A) = P(B)
19 If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then
(A) P(B|A) = 1
(B) P(A|B) = 1
(C) P(B|A) = 0
(D) P(A|B) = 0
© NCERT
not to be republished
PROBABILITY 585
Summary
The salient features of the chapter are –
� The conditional probability of an event E, given the occurrence of the event F
is given by
P(E
F)
P(E | F)
P(F)
∩
=
, P(F) ≠ 0
� 0 ≤ P (E|F) ≤ 1,
P (E′|F) = 1 – P (E|F)
P ((E ∪ F)|G) = P (E|G) + P (F|G) – P ((E ∩ F)|G)
� P (E ∩ F) = P (E) P (F|E), P (E) ≠ 0
P (E ∩ F) = P (F) P (E|F), P (F) ≠ 0
� If E and F are independent, then
P (E ∩ F) = P (E) P (F)
P (E|F) = P (E), P (F) ≠ 0
P (F|E) = P (F), P(E) ≠ 0
� Theorem of total probability
Let {E1, E2, ,En) be a partition of a sample space and suppose that each of
E1, E2, , En has nonzero probability |
1 | 7785-7788 | If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then
(A) P(B|A) = 1
(B) P(A|B) = 1
(C) P(B|A) = 0
(D) P(A|B) = 0
© NCERT
not to be republished
PROBABILITY 585
Summary
The salient features of the chapter are –
� The conditional probability of an event E, given the occurrence of the event F
is given by
P(E
F)
P(E | F)
P(F)
∩
=
, P(F) ≠ 0
� 0 ≤ P (E|F) ≤ 1,
P (E′|F) = 1 – P (E|F)
P ((E ∪ F)|G) = P (E|G) + P (F|G) – P ((E ∩ F)|G)
� P (E ∩ F) = P (E) P (F|E), P (E) ≠ 0
P (E ∩ F) = P (F) P (E|F), P (F) ≠ 0
� If E and F are independent, then
P (E ∩ F) = P (E) P (F)
P (E|F) = P (E), P (F) ≠ 0
P (F|E) = P (F), P(E) ≠ 0
� Theorem of total probability
Let {E1, E2, ,En) be a partition of a sample space and suppose that each of
E1, E2, , En has nonzero probability Let A be any event associated with S,
then
P(A) = P(E1) P (A|E1) + P (E2) P (A|E2) + |
1 | 7786-7789 | ,En) be a partition of a sample space and suppose that each of
E1, E2, , En has nonzero probability 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)
� Bayes' theorem If E1, E2, |
1 | 7787-7790 | , En has nonzero probability 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)
� Bayes' theorem If E1, E2, , En are events which constitute a partition of
sample space S, i |
1 | 7788-7791 | 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)
� Bayes' theorem If E1, E2, , En are events which constitute a partition of
sample space S, i e |
1 | 7789-7792 | + P (En) P(A|En)
� Bayes' theorem If E1, E2, , En are events which constitute a partition of
sample space S, i e E1, E2, |
1 | 7790-7793 | , En are events which constitute a partition of
sample space S, i e E1, E2, , En are pairwise disjoint and E1 4 E2 4 |
1 | 7791-7794 | e E1, E2, , En are pairwise disjoint and E1 4 E2 4 4 En = S
and A be any event with nonzero probability, then
i
i
1
P(E )P(A|E )
P(E | A)
P(E )P(A|E )
i
n
j
j
j
� A random variable is a real valued function whose domain is the sample
space of a random experiment |
1 | 7792-7795 | E1, E2, , En are pairwise disjoint and E1 4 E2 4 4 En = S
and A be any event with nonzero probability, then
i
i
1
P(E )P(A|E )
P(E | A)
P(E )P(A|E )
i
n
j
j
j
� A random variable is a real valued function whose domain is the sample
space of a random experiment � The probability distribution of a random variable X is the system of numbers
X
:
x1
x2 |
1 | 7793-7796 | , En are pairwise disjoint and E1 4 E2 4 4 En = S
and A be any event with nonzero probability, then
i
i
1
P(E )P(A|E )
P(E | A)
P(E )P(A|E )
i
n
j
j
j
� A random variable is a real valued function whose domain is the sample
space of a random experiment � The probability distribution of a random variable X is the system of numbers
X
:
x1
x2 xn
P(X)
:
p 1
p 2 |
1 | 7794-7797 | 4 En = S
and A be any event with nonzero probability, then
i
i
1
P(E )P(A|E )
P(E | A)
P(E )P(A|E )
i
n
j
j
j
� A random variable is a real valued function whose domain is the sample
space of a random experiment � The probability distribution of a random variable X is the system of numbers
X
:
x1
x2 xn
P(X)
:
p 1
p 2 p n
where,
1
0,
1,
1, 2, |
1 | 7795-7798 | � The probability distribution of a random variable X is the system of numbers
X
:
x1
x2 xn
P(X)
:
p 1
p 2 p n
where,
1
0,
1,
1, 2, ,
n
i
i
i
p
p
i
n
=
>
=
=
∑
© NCERT
not to be republished
586
MATHEMATICS
� Let X be a random variable whose possible values x1, x2, x3, |
1 | 7796-7799 | xn
P(X)
:
p 1
p 2 p n
where,
1
0,
1,
1, 2, ,
n
i
i
i
p
p
i
n
=
>
=
=
∑
© NCERT
not to be republished
586
MATHEMATICS
� Let X be a random variable whose possible values x1, x2, x3, , xn occur with
probabilities p1, p2, p3, |
1 | 7797-7800 | p n
where,
1
0,
1,
1, 2, ,
n
i
i
i
p
p
i
n
=
>
=
=
∑
© NCERT
not to be republished
586
MATHEMATICS
� Let X be a random variable whose possible values x1, x2, x3, , xn occur with
probabilities p1, p2, p3, pn respectively |
1 | 7798-7801 | ,
n
i
i
i
p
p
i
n
=
>
=
=
∑
© NCERT
not to be republished
586
MATHEMATICS
� Let X be a random variable whose possible values x1, x2, x3, , xn occur with
probabilities p1, p2, p3, pn respectively The mean of X, denoted by μ, is
the number
1
n
i
i
i
x p |
1 | 7799-7802 | , xn occur with
probabilities p1, p2, p3, pn respectively The mean of X, denoted by μ, is
the number
1
n
i
i
i
x p The mean of a random variable X is also called the expectation of X, denoted
by E (X) |
1 | 7800-7803 | pn respectively The mean of X, denoted by μ, is
the number
1
n
i
i
i
x p The mean of a random variable X is also called the expectation of X, denoted
by E (X) � Let X be a random variable whose possible values x1, x2, |
1 | 7801-7804 | The mean of X, denoted by μ, is
the number
1
n
i
i
i
x p The mean of a random variable X is also called the expectation of X, denoted
by E (X) � Let X be a random variable whose possible values x1, x2, , xn occur with
probabilities p(x1), p(x2), |
1 | 7802-7805 | The mean of a random variable X is also called the expectation of X, denoted
by E (X) � Let X be a random variable whose possible values x1, x2, , xn occur with
probabilities p(x1), p(x2), , p(xn) respectively |
1 | 7803-7806 | � Let X be a random variable whose possible values x1, x2, , xn occur with
probabilities p(x1), p(x2), , p(xn) respectively Let μ = E(X) be the mean of X |
1 | 7804-7807 | , xn occur with
probabilities p(x1), p(x2), , p(xn) respectively Let μ = E(X) be the mean of X The variance of X, denoted by Var (X) or
σx
2, is defined as
2
2
1
Var(X)=
(
μ)
(
)
n
x
i
i
i
x
p x
or equivalently σx
2 = E (X – μ)2
The non-negative number
2
1
Va r(X) =
(
μ)
(
)
n
x
i
i
i
x
p x
is called the standard deviation of the random variable X |
1 | 7805-7808 | , p(xn) respectively Let μ = E(X) be the mean of X The variance of X, denoted by Var (X) or
σx
2, is defined as
2
2
1
Var(X)=
(
μ)
(
)
n
x
i
i
i
x
p x
or equivalently σx
2 = E (X – μ)2
The non-negative number
2
1
Va r(X) =
(
μ)
(
)
n
x
i
i
i
x
p x
is called the standard deviation of the random variable X � Var (X) = E (X2) – [E(X)]2
� Trials of a random experiment are called Bernoulli trials, if they satisfy the
following conditions :
(i) There should be a finite number of trials |
1 | 7806-7809 | Let μ = E(X) be the mean of X The variance of X, denoted by Var (X) or
σx
2, is defined as
2
2
1
Var(X)=
(
μ)
(
)
n
x
i
i
i
x
p x
or equivalently σx
2 = E (X – μ)2
The non-negative number
2
1
Va r(X) =
(
μ)
(
)
n
x
i
i
i
x
p x
is called the standard deviation of the random variable X � Var (X) = E (X2) – [E(X)]2
� Trials of a random experiment are called Bernoulli trials, if they satisfy the
following conditions :
(i) There should be a finite number of trials (ii) The trials should be independent |
1 | 7807-7810 | The variance of X, denoted by Var (X) or
σx
2, is defined as
2
2
1
Var(X)=
(
μ)
(
)
n
x
i
i
i
x
p x
or equivalently σx
2 = E (X – μ)2
The non-negative number
2
1
Va r(X) =
(
μ)
(
)
n
x
i
i
i
x
p x
is called the standard deviation of the random variable X � Var (X) = E (X2) – [E(X)]2
� Trials of a random experiment are called Bernoulli trials, if they satisfy the
following conditions :
(i) There should be a finite number of trials (ii) The trials should be independent (iii) Each trial has exactly two outcomes : success or failure |
1 | 7808-7811 | � Var (X) = E (X2) – [E(X)]2
� Trials of a random experiment are called Bernoulli trials, if they satisfy the
following conditions :
(i) There should be a finite number of trials (ii) The trials should be independent (iii) Each trial has exactly two outcomes : success or failure (iv) The probability of success remains the same in each trial |
1 | 7809-7812 | (ii) The trials should be independent (iii) Each trial has exactly two outcomes : success or failure (iv) The probability of success remains the same in each trial For Binomial distribution B (n, p), P (X = x) = nCx q n–x px, x = 0, 1, |
1 | 7810-7813 | (iii) Each trial has exactly two outcomes : success or failure (iv) The probability of success remains the same in each trial For Binomial distribution B (n, p), P (X = x) = nCx q n–x px, x = 0, 1, , n
(q = 1 – p)
Historical Note
The earliest indication on measurement of chances in game of dice appeared
in 1477 in a commentary on Dante's Divine Comedy |
1 | 7811-7814 | (iv) The probability of success remains the same in each trial For Binomial distribution B (n, p), P (X = x) = nCx q n–x px, x = 0, 1, , n
(q = 1 – p)
Historical Note
The earliest indication on measurement of chances in game of dice appeared
in 1477 in a commentary on Dante's Divine Comedy A treatise on gambling
named liber de Ludo Alcae, by Geronimo Carden (1501-1576) was published
posthumously in 1663 |
1 | 7812-7815 | For Binomial distribution B (n, p), P (X = x) = nCx q n–x px, x = 0, 1, , n
(q = 1 – p)
Historical Note
The earliest indication on measurement of chances in game of dice appeared
in 1477 in a commentary on Dante's Divine Comedy A treatise on gambling
named liber de Ludo Alcae, by Geronimo Carden (1501-1576) was published
posthumously in 1663 In this treatise, he gives the number of favourable cases
for each event when two dice are thrown |
1 | 7813-7816 | , n
(q = 1 – p)
Historical Note
The earliest indication on measurement of chances in game of dice appeared
in 1477 in a commentary on Dante's Divine Comedy A treatise on gambling
named liber de Ludo Alcae, by Geronimo Carden (1501-1576) was published
posthumously in 1663 In this treatise, he gives the number of favourable cases
for each event when two dice are thrown © NCERT
not to be republished
PROBABILITY 587
Galileo (1564-1642) gave casual remarks concerning the correct evaluation
of chance in a game of three dice |
1 | 7814-7817 | A treatise on gambling
named liber de Ludo Alcae, by Geronimo Carden (1501-1576) was published
posthumously in 1663 In this treatise, he gives the number of favourable cases
for each event when two dice are thrown © NCERT
not to be republished
PROBABILITY 587
Galileo (1564-1642) gave casual remarks concerning the correct evaluation
of chance in a game of three dice Galileo analysed that when three dice are
thrown, the sum of the number that appear is more likely to be 10 than the sum 9,
because the number of cases favourable to 10 are more than the number of
cases for the appearance of number 9 |
1 | 7815-7818 | In this treatise, he gives the number of favourable cases
for each event when two dice are thrown © NCERT
not to be republished
PROBABILITY 587
Galileo (1564-1642) gave casual remarks concerning the correct evaluation
of chance in a game of three dice Galileo analysed that when three dice are
thrown, the sum of the number that appear is more likely to be 10 than the sum 9,
because the number of cases favourable to 10 are more than the number of
cases for the appearance of number 9 Apart from these early contributions, it is generally acknowledged that the
true origin of the science of probability lies in the correspondence between two
great men of the seventeenth century, Pascal (1623-1662) and Pierre de Fermat
(1601-1665) |
1 | 7816-7819 | © NCERT
not to be republished
PROBABILITY 587
Galileo (1564-1642) gave casual remarks concerning the correct evaluation
of chance in a game of three dice Galileo analysed that when three dice are
thrown, the sum of the number that appear is more likely to be 10 than the sum 9,
because the number of cases favourable to 10 are more than the number of
cases for the appearance of number 9 Apart from these early contributions, it is generally acknowledged that the
true origin of the science of probability lies in the correspondence between two
great men of the seventeenth century, Pascal (1623-1662) and Pierre de Fermat
(1601-1665) A French gambler, Chevalier de Metre asked Pascal to explain
some seeming contradiction between his theoretical reasoning and the
observation gathered from gambling |
1 | 7817-7820 | Galileo analysed that when three dice are
thrown, the sum of the number that appear is more likely to be 10 than the sum 9,
because the number of cases favourable to 10 are more than the number of
cases for the appearance of number 9 Apart from these early contributions, it is generally acknowledged that the
true origin of the science of probability lies in the correspondence between two
great men of the seventeenth century, Pascal (1623-1662) and Pierre de Fermat
(1601-1665) A French gambler, Chevalier de Metre asked Pascal to explain
some seeming contradiction between his theoretical reasoning and the
observation gathered from gambling In a series of letters written around 1654,
Pascal and Fermat laid the first foundation of science of probability |
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