Mathematics Chapter 17 Introduction To Mathematical Modelling
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    NCERT Solution For Class 9 About 2.html

    Introduction To Mathematical Modelling Here is the CBSE About 2.html Chapter 17 for Class 9 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 9 About 2.html Introduction To Mathematical Modelling Chapter 17 NCERT Solutions for Class 9 About 2.html Introduction To Mathematical Modelling Chapter 17 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 9 About 2.html.

    Question 1
    CBSEENMA9003987

    In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.

    Solution

    Let E be the event of hitting the boundary.

    Then,
    straight P left parenthesis straight E right parenthesis equals fraction numerator his space the space boundary over denominator Total end fraction equals 6 over 30 equals 1 half equals 0.2

    ∴ Probability of not hitting the boundary
    = 1 – Probability of hitting the boundary
    = 1 – P(E) = 1 – 0.2 = 0.8.


    Question 2
    CBSEENMA9003988

    1500 families with 2 children were selected randomly, and the following data were recorded:

    Number of girls in a family

    2

    1

    0

    Number of families

    475

    814

    211

    Compute the probability of a family, chosen at random, having

    (0 2 girls (ii) 1 girl (iii) No girl.

    Also check whether the sum of these probabilities is 1.

    Solution

    Total number of families
    = 475 + 814 + 211 = 1500
    (i) Probability of a family, chosen at random,
    having 2 girls = 475 over 1500 equals 19 over 60
    (ii) Probability of a family, chosen at random,
    having 1 girl  equals 814 over 1500 equals 407 over 750
    (iii) Probability of a family, chosen at random,
    having no girl  equals 211 over 1500
    Sum of these probabilities
    equals 19 over 60 plus 407 over 750 plus 211 over 1500
equals fraction numerator 475 plus 814 plus 211 over denominator 1500 end fraction equals 1500 over 1500 equals 1
    Hence, the sum is checked.

    Question 3
    CBSEENMA9003989

    In a particular section of Class IX, 40 students were asked about the months of their birth, the following graph was prepared for the data so obtained. Find the probability that a student of the class was bom in August.


    Solution

    Total number of students bom in the year =3+4+2+2+5+1+2+6+3+4+4+4=40
    Number of students bom in August = 6
    ∴ Probability that a student of the class was
    born in August = 6 over 40 equals 3 over 20.

    Question 5
    CBSEENMA9003991

    An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:

    Monthly income

    Vehicles per family

    (in र)

    0

    1

    2

    Above 2

    Less than 7000

    10

    160

    25

    0

    7000-10000

    0

    305

    27

    2

    10000-13000

    1

    535

    29

    1

    13000–16000

    2

    469

    59

    25

    16000 or more

    1

    579

    82

    88

    Suppose a family is chosen. Find the probability that the family chosen is
    (i) earning  र 10000–13000 per month and owning exactly 2 vehicles.
    (ii) earning र 16000 or more per month and owning exactly l vehicle.
    (iii) earning less than र 7000 per month and does not own any vehicle.
    (iv) earning र 13000–16000 per month and owning more than 2 vehicles.
    (v) owning not more than I vehicle.

    Solution

    Total number of families selected = 2400
    (i) Number of families earning र10000–13000 per month and owning exactly 2 vehicles = 29
    ∴ Probability that the family chosen is earning र10000–13000 per month and owning
    exactly 2 vehicles = 29 over 2400.

    (ii) Number of families earning र16000 or more per month and owning exactly 1 vehicle = 579

    ∴ Probability that the family chosen is earning र16000 or more per month and owning
    exactly 1 vehicle = 579 over 2400 equals 193 over 800.

    (iii) Number of families earning less than र 7000 per month and does not own any vehicle =10
    Probability that the family chosen is earning less than र7000 per month and does not own any vehicle = 10 over 2400 equals 1 over 240
    (iv) Number of families earning र13000–16000 per month and owning more than 2 vehicles = 25

    ∴ Probability that the family chosen is earning र13000–16000 per month and owning
    more than 2 vehicles = 25 over 2400 equals 1 over 96.
    (v) Number of families owning not more than 1 vehicle

    = Number of families owning 0 vehicle + Number of families owning 1 vehicle = (10 + 0+ 1 + 2 + 1) +(160 + 305 + 535 + 469 +579) = 14 + 2048 = 2062
    ∴ Probability that the family chosen owns not more than 1 vehicle = 2062 over 2400 equals 1031 over 1200



    Question 6
    CBSEENMA9003992

    Marks (out of 100)

    Number of students

    0-20

    7

    20-30

    10

    30-40

    10

    40-50

    20

    50-60

    20

    60-70

    15

    70–above

    8

    Total

    90

    (i)  Find the probability that a student obtained less than 20% in the mathematics test.
    (ii)  Find the probability that a student obtained marks 60 or above.

    Solution

    Total number of students = 90

    (i) Number of students obtaining less than 20% in the mathematics test = 7
    ∴ Probability that a student obtained less than 20% in mathematics test = 7 over 90.
    (ii) Number of students obtaining marks 60 or above = 15 + 8 = 23
    Probability that a student obtained marks 60 or = 23 over 90.

    Question 7
    CBSEENMA9003993

    To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table:

    Opinion

    Number of students

    like

    dislike

    135

    65

    Find the probability that a student chosen at random
    (i)  likes statistics,
    (ii)  does not like it.

    Solution

    Total number of students = 200

    (i) Number of students who like statistics = 135 Probability that a student chosen at random like statistics = 135 over 200 equals 27 over 40

    (ii) Number of students who do not like statistics = 65
    Probabi lity that a student chosen at random does not like it = 65 over 200 equals 13 over 40.

    Aliter: Probability that a student chosen at andom does like statistics

    = 1 – probability that a student chosen at random likes statistics
    equals 1 minus 27 over 40 equals 13 over 40.

    Question 8
    CBSEENMA9003994

    Refer to Q.2, Exercise 14.2. What is the empirical probability that an engineer lives: (i) less than 7 km from her place of work? (ii) more than or equal to 7 km from her place of work? (iii) within 1 half km from her place of work?

    Solution

    Total number of female engineers = 40
    (i) Number of female engineers whose distance (in km) from their residence to their place of work is less than 7 km = 9.
    ∴ Probability that an engineer lives less than 7 km from her plae of work = 9 over 40.

    (ii) Number of female engineers whose distance (in km) from their residence to their place of work is more than or equal to 7 km = 31.

    ∴Probability that an engineer lives more than or equal to 7 km from her place of residence  equals 31 over 40.

    Aliter: Probability that an engineer lives more than or equal to 7 km from her place of residence = 1 – probability that an engineer lives less than 7 km from her place of work equals 1 minus 9 over 40 equals 31 over 40

    (iii) Number of female engineers whose distance (in km) from their residence to their place of work is within 1 half space space km space equals space 0 comma
    therefore  Probablity  that an engineer lives within 1 half space km from her place of work = 0 over 40 equals 0.

    Question 11
    CBSEENMA9003997

    Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg):

    4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00

    Find the probability that any of these bags chosen at random contains more than 5 kg of flour.

    Solution

    Total number of bags of wheat flour =11.
    Number of bags of wheat flour containing more than 5 kg of flour = 7.
    Probability that any of the bags, chosen at
    random, contains more than 5  kg of flour = 7 over 11

    Question 12
    CBSEENMA9003998

    In Q.5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 - 0.16 on any of these days

    Solution

    Total number of days = 30

    Number of days on which the concentration of sulphur dioxide is in the interval 0.12 – 0.16 = 2.

    ∴ Probability that the concentration of sulphur dioxide is in the interval 0.12–0.16 on any of   these days equals 2 over 30 equals 1 over 15

    Question 13
    CBSEENMA9003999

    In Q.1, Exercise 14.2, you were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB. 

    Solution

    Total number of students = 30
    Number of students having blood groups AB = 3
    ∴ Probability that a student of this class, selected at random, has blood group
    AB equals 3 over 30 equals 1 over 10 equals 0.1
    Frequency Distribution Table


    Question 14
    CBSEENMA9004000

    A and B are the only two outcomes of an event. Probability P(A) = 0.72 then what will be the probability P(B) and why?

    Solution

    P(A) + P(B) = 1
             [ because Sum of the probablities of all the outcomes of an event is 1]
    rightwards double arrow space space space space space space 0.72 space plus space straight P left parenthesis straight B right parenthesis space equals space 1
rightwards double arrow space space space space space straight P left parenthesis straight B right parenthesis space equals space 1 minus 0.72 equals 0.28
            

    Question 15
    CBSEENMA9004001

    A survey of 500 families was conducted to know their opinion about a particular detergent powder. If 375 families liked the detergent powder and the remaining families disliked it, find the probability that a family chosen at random

    (i)  likes the detergent powder
    (ii)  does not like it.

    Solution

    Total number of families = 500

    (i) Number of families who like the detergent powder = 375
    ∴ Probability that a family chosen at random likes the detergent powder 
    equals 375 over 500 equals 3 over 4

    (ii) Number of families who dislike the detergent powder = 500 – 375 = 125
    ∴ Probability that a family chosen at random does not like the detergent powder
    equals 125 over 500 equals 1 fourth


    Question 16
    CBSEENMA9004002

    1500 families with 2 children were released randomly and the following data was recorded:

    No. of girls

    No. of families

    0

    1

    2

    211

    814

    475

    If a family is chosen at random, find the probability that it has

    (i)  at most one girl
    (ii)  at least one girl 

    Solution

    Total number of families = 1500

    (i) at most one girl means 0 girl or 1 girl.
    ∴ Number of families which have at most one girl
    = 211 +814= 1015
    ∴Probability that it has at most one girl
    equals 1015 over 1500 equals 203 over 300

    (ii) at least one girl means 1 girls or 2 girls.

    ∴ Number of families which have at least one girl
    = 814 + 475 = 1289
    ∴ Probability that it has at least one girl
    equals 1289 over 1500.

    Question 17
    CBSEENMA9004003

    Out of the past 250 consecutive days, its weather forecasts were correct 175 times.

    (i)  What is the probability that on a given day it was correct?
    (ii)  What is the probability that it was not correct on a given day?

    Solution

    Total number of days = 250
    (i) Number of days on which the weather forecasts were correct = 175
    ∴ Probability that on a given day it was correct
    equals 175 over 250 equals 7 over 10
    (ii) Probability that it was not correct on a given day equals 1 minus 7 over 10 equals 3 over 10

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    Question 18
    CBSEENMA9004004

    A die is rolled 25 times and outcomes are recorded as under:

    Outcomes

    Frequency

    1

    9

    2

    4

    3

    5

    4

    6

    5

    1

    6

    0

    It is thrown one more time. Find the probability of getting

    (a)  an even number
    (b)  a multiple of 3
    (c)  a prime number.

    Solution

    Total number of times a die is rolled = 25
    (a) Even numbers are 2,4, 6.
    ∴ Probability of getting an even number
    equals fraction numerator 4 plus 6 plus 0 over denominator 25 end fraction equals 10 over 25 equals 2 over 5

    (b) Multiples of 3 are 3, 6.
    ∴ Probability of getting a multiple of 3
    equals fraction numerator 5 plus 0 over denominator 25 end fraction equals 5 over 25 equals 1 fifth

    (c) 2, 3, 5 are prime numbers.
    ∴ Probability of getting a prime number
    equals fraction numerator 4 plus 5 plus 1 over denominator 25 end fraction equals 10 over 25 equals 2 over 5.

    Question 19
    CBSEENMA9004005

    A bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls. If one ball is drawn at random, find the probability that it is

    (i)  black
    (ii)  not green.

    Solution

    In the bag,
    number of red balls = 5
    number of white balls = 8
    number of green balls = 4
    number of black balls = 7
    ∴ Total number of balls in the bag = 5 + 8 + 4 + 7 = 24
    (i) Number of black balls = 7
    ∴ Probability that the ball drawn is black
    equals 7 over 24

    (ii) Number of balls that are not green
    = 5 + 8 + 7 = 20
    ∴ Probability that the ball drawn is not green
    equals 20 over 24 equals 5 over 6

    Question 20
    CBSEENMA9004006

    Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box. Find the probability that the number on the card is
    (a)  a number less than 14
    (b)  a number which is a perfect square
    (c)  a prime number less than 20.

    Solution

    Total number of cards in the box = 100 (a) Numbers less than 14 are
    2, 3,4, 5. 6, 7, 8, 9, 10, 11,12, 13 Their number =12
    ∴ Probability that the number on the card is a number less than 14
    equals 12 over 100 equals 3 over 25

    (b) Perfect square numbers are
    4, 9, 16,25, 36,49, 64,81, 100 Their number = 9
    ∴ Probability that the number on the card is a number which is a perfect square
    equals 9 over 100

    (c) Prime numbers less than 20 are 2, 3, 5,7, 11, 13, 17, 19 Their number = 8
    ∴ Probability that the number in the card is a prime number less than 20
    equals 8 over 100 equals 2 over 25.

    Question 21
    CBSEENMA9004007

    On a particular day, the number of vehicles passing through a crossing is given below:

    Vehicle

    Frequency

    Two-wheeler

    57

    Three-wheeler

    33

    Four-wheeler

    30

    A particular vehicle is chosen at random. What is the probability that it is not a four-wheeler?

    Solution

    Number of two wheelers = 57
    Number of three wheelers = 33
    Number of four wheelers = 30
    ∴ Total number of vehicles = 57 + 33 + 30 = 120
    Number of vehicles that is not a four-wheeler
    = 57 + 33 = 90
    ∴ Probability that the vehicle chosen at random is not a four-wheeler
    equals 90 over 120
equals 3 over 4

    Question 22
    CBSEENMA9004008

    30 plants were planted in each school out of 12 schools. After a month the number of plants that survived are given below:

    School

    Number of plants survived

    1

    22

    2

    15

    3

    12

    4

    24

    5

    27

    6

    10

    7

    13

    8

    22

    9

    17

    10

    9

    11

    20

    12

    25

    What is the probability of survival of
    (i)  more than 20 plants in a school?
    (ii)  less than 10 plants in a school?
    (iii) exactly 22 plants in a school?

    Solution

    Total number of schools =12
    (i) Number of schools in which more than 20 plants survived = 5
    ∴ Probability of survival of more than 20 plants in a school equals 5 over 12

    (ii) Number of schools in which less than 10 plants survived = 1

    ∴ Probability of survival of less than 10 plants in a school = 1 over 12

    (iii) Number of schools in which exactly 22 plants survived = 2

    Probability of survival of exactly 22 plants in a school = 2 over 12 equals 1 over 6.

    Question 23
    CBSEENMA9004009

    The percentages of marks obtained by a student in examination are given below:

    Examination

    subjects

    % marks

    I

    58

    11

    64

    III

    76

    IV

    62

    V

    85

    Find the probability that the student gets

    (i)   a first class i.e. at least 60% marks
    (ii)  a distinction i.e. 75% or above
    (ii)  marks between 70% and 80%.

    Solution

    Total number of subjects = 5

    (i) Number of subjects in which the student gets a first class = 4

    ∴ Probability that the students gets a first class equals 4 over 5

    (ii) Number of subjects in which the student gets a distinction = 2

    ∴ Probability that the student gets a distinction = 2 over 5

    (iii) Number of subjects in which the student gets marks between 70% and 80% = 2

    ∴ Probability that the students gets marks between 70% and 80% = 2 over 5

    Question 24
    CBSEENMA9004010

    At a hospital, a doctor compiled the following data about 400 patients whom he could cure of hepatitis:

    Time for cure

    No. of patients

    < 1 month

    210

    1–2 months

    105

    2-3 months

    60

    > 3 months

    25

    Another case of hepatitis is reported. What is the probability that this patient will be cured in

    (i)   less than 2 months?
    (ii)  1 month or more but not more than 3 months?

    Solution

    Total number of patients = 400

    (i) Number of patients who were cured in less than 2 months = 210+ 105 = 315
    ∴ Probability that the patient will be cured in less than 2 months = 315 over 400 equals 63 over 80

    (ii) Number of patients who were cured in 1 month or more but not more than 3 months

    = 105 + 60 = 165
    ∴ Probability that the patient will be cured in
    1 month or more but not more than 3 months  equals 165 over 400 equals 33 over 80.

    Question 26
    CBSEENMA9004012

    On a busy road, following data was observed about cars passing through it and number of occupants

    No. of occupants

    No. of cars

    1

    29

    2

    26

    3

    23

    4

    17

    5

    5

    Suppose another car passes by. Find the chance that it has
    (i)  exactly 5 occupants
    (ii)  more than 2 occupants (iii) less than 5 occupants.

    Solution

    otal number of cars = 29 + 26 + 23 + 17 + 5= 100

    (i) Number of cars having exactly 5 occupants = 5

    ∴ Probability that it has exactly 5 occupants = 5 over 100 equals 1 over 20

    (ii) Number of cars having more than 2 occupants = 23 + 17 + 5 = 45

    ∴ Probability that it has more than 2 occupants equals 45 over 100 equals 9 over 20

    (iii) Number of cars having less than 5 occupants = 29 + 26 + 23 + 17 = 95

    ∴ Probability that it has less than 5 occupants equals 95 over 100 equals 19 over 20.

    Question 27
    CBSEENMA9004013

    The king, queen and jack of clubs are removed from a deck of 52 cards and then well shuffled. One card is selected at random from the remaining cards. Find the probability of getting

    (a)  a heart
    (b)  a king
    (c)  the 10 of hearts.

    Solution

    Total number of cards in the deck when king, queen and jack of clubs are removed = 52–3 = 49 (a) Number of cards which are ‘a heart’ = 13
    therefore  Probablity of getting a heart = 13 over 49
    (b) Number of cards which are ‘a king’ = 3
    therefore  Probablity of getting a king = 3 over 49
    (c) Number of cards which are ‘the 10 of heart’ = 1
    therefore   Probablity of getting the 10 of heart's = 1 over 49

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    Question 47
    CBSEENMA9004033

    A die is thrown. Find the probability of getting an odd number. 

    Solution

    Solution not provided.
    Ans.   1 half

    Question 56
    CBSEENMA9004042

    The minimum probablity of an event is 

    • 0

    • 1

    • 1 half
    • - 1

    Solution

    A.

    0

    Question 57
    CBSEENMA9004043
    Question 58
    CBSEENMA9004044
    Question 59
    CBSEENMA9004045

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    CBSEENMA9004076
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    CBSEENMA9004078
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    CBSEENMA9004088

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