NCERT Solutions for Class 9 Maths Chapter 15 Probability

 

NCERT solutions for class 9 maths chapter 15 Probability: Predictions are common in our life. Whether it will rain today, whether India will win the match etc. Predictions are based on experiments and observation. Solutions of NCERT for class 9 maths chapter 15 Probability is covering the solutions to these kinds of real-life predictions. Suppose you are doing an experiment of tossing a coin 20 times then each toss is known as a trial and the possible outcomes of a toss are head and tail. If in a toss a tail has occurred we call that an event 'tail has occurred' and vice versa. In these CBSE NCERT solutions for class 9 maths chapter 15 Probability is designed to provide you step by step solutions to all such problems. In this chapter, there is only one exercise that consists of a total of 13 questions including some activities. NCERT solutions for class 9 maths chapter 15 Probability is covering every practice problem to help you while preparing for the final examinations. Apart from probability, you can access the free NCERT solutions by the given link.

\\The \ empirical \ probability \ P(E) \ of \ an \ event \ E \ happening,\ \\P(E)=\frac{Number \ of \ trials \ in \ which \ the \ event \ happened}{The \ total \ number \ of \ trials}

 

 

NCERT solutions for class 9 maths chapter 15 Probability Excercise: 15.1

Q1 In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she batswoman. Find the probability that she did not hit a boundary.

Answer:

From the above question, the data of interest is,

Total Number of balls batswoman played = 30

Number of times batswoman hits a boundary =6

Therefore, we can say,

Number of times batswoman could not hit a boundary =24

 P(she did not hit a boundary) 

= \frac{Number\: of \:times\: batswoman \:cannot \:hit \:a \:boundary }{Total \:Number \:of \:balls \:batswoman \:played }

      =24/30 =0.80

Ans:0.80

Q2 (i) 1500 families with 2 children were selected randomly, and the following data were recorded: Compute the probability of a family, chosen at random, having 2 girls

Number of girls in a family 2 1 0
Number of families 475 814 211

Answer:

From the above question, the data that we can take is,

Total Number of families= 475+814+211 = 1500

Number of families having 2 girls in the family =475

We know:

Empirical (or experimental) probability P(E) of this event can be written as

P(Family, chosen at random has 2 girls) =

\frac{Familes\:with \:2 \:girls}{Total \:number \:of \:Families}

= 475/1500 = 19/60

Q2 (ii) 1500 families with 2 children were selected randomly, and the following data were recorded: Compute the probability of a family, chosen at random, having 1 girl

Number of girls in a family 2 1 0
Number of families 475 814 211

Answer:

From the above question, the data that we can take is,

Total Number of families= 475+814+211 = 1500

Number of families having 1 girl in the family =814

We know:

Empirical (or experimental) probability P(E) of this event can be written as

P(Family, chosen at random has 1 girl) = 

\frac{Families\:with\: 1\: girl}{Total\:Number\: of \:families}

= 814/1500 = 407/1500

Q2 (iii) 1500 families with 2 children were selected randomly, and the following data were recorded: Compute the probability of a family, chosen at random, having no girl

Number of girls in a family 2 1 0
Number of families 475 814 211

Answer:

From the above question, the data that we can take is,

Total Number of families= 475+814+211 = 1500

Number of families having no girls in the family =211

We know:

Empirical (or experimental) probability P(E) of this event can be written as

P(Family, chosen at random has no girls) =

 \frac{Families\:with\:no\:girl}{Total Number of families}

= 211/1500

Ans:= 211/1500

Q3 Refer to Example 5, Section 14.4, Chapter 14. Find the probability that a student of the the class was born in August.

Answer:

Total no. of students = 40 

Total no. of students who all are born in August  =6

P( a student of the class was born in August) =6/40

=3/20

Ans: 3/20

Q4 Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

Outcome

3 Heads

2 Heads

1 Head

0 Head

Frequencies

23 

72

77

28

Answer

Total number of times coins tossed = 200

Total number of possible outcomes = 72

Required probability = 72/200 => 9/25

Q5 (i) An organization 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:

Suppose a family is chosen. Find the probability that the family chosen is earning ` Rs. 10000 – Rs.13000 per month and owning exactly 2 vehicles.

Monthly income (in Rs) Vehicles per family
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

 

Answer:

Although it is given that,

Total no. of families= 2400

Let us find this by adding all the cases

= 10+0+1+2+1+160+305+535+469+579+25+27+29+59+82+0+2+1+25+88 =2400 

Earning Rs. 10000 – Rs.13000 per month and owning exactly 2 vehicles =29

P(Earning Rs. 10000–Rs.13000 per month and owning exactly 2 vehicles)= P1

P1 = 29/2400

Q5 (ii) An organization 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 (in Rs) Vehicles per family
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 earning Rs.16000 or more per month and owning exactly 1 vehicle.

Answer:

Although it is given that,

Total no. of families= 2400

Let us find by this adding all the cases

= 10+0+1+2+1+160+305+535+469+579+25+27+29+59+82+0+2+1+25+88 =2400 

Total no. of families earning Rs.16000 or more per month and owning exactly 1 vehicle = 579

P(earning Rs.16000 or more per month and owning exactly 1 vehicle )= P2

P2 = 579/2400

Q5 (iii) An organization 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 (in Rs) Vehicles per family
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 chose is earning less than Rs. 7000 per month and does not own any vehicle.

Answer:

Although it is given that,

Total no. of families= 2400

Let us find by this adding all the cases

= 10+0+1+2+1+160+305+535+469+579+25+27+29+59+82+0+2+1+25+88 =2400 

Total no. of families, those are earning less than Rs. 7000 per month and not having any vehicle = 10

P(earning less than Rs. 7000 per month and does not have any vehicle)= P3

P3 = 10/2400

=1/240

Ans: 1/240

Q5 (iv) An organization 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 (in Rs) Vehicles per family
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 chose is earning `Rs.13000 – 16000 per month and owning more than 2 vehicles.

Answer:

Although it is given that,

Total no. of families= 2400

Let us find by this adding all the cases

= 10+0+1+2+1+160+305+535+469+579+25+27+29+59+82+0+2+1+25+88 =2400 

Total number of families,those are earning Rs.13000 – 16000 per month and owning more than 2 vehicles = 25

P(earning Rs.13000 – 16000 per month and owning more than 2 vehicles)= P4

P4 = 25/2400

=1/96

Ans:1/96

Q5 (v) An organization 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 (in Rs) Vehicles per family
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 owning not more than 1 vehicle.

Answer:

Although it is given that,

Total no. of families= 2400

Let us find by this adding all the cases

= 10+0+1+2+1+160+305+535+469+579+25+27+29+59+82+0+2+1+25+88 =2400 

Total number of families owning not more than 1 vehicle= 10+0+1+2+1+160+305+535+469+579 = 2062

Let, P(families owning not more than 1 vehicle)= P5

P5 = 2062/2400

Q6 (i) Refer to Table 14.7, Chapter 14. Find the probability that a student obtained less than 20% in the mathematics test.

Answer:

Marks Number of students
0-20 2
20-30 10
30-40 10
40-50 20
50-60 20
60-70 15
70- above 8
Total 90

From the above question, the data of our interest is:

Total no. of students =90 

Total no. of students who obtained less than 20% in the mathematics test= 7

P(student obtained less than 20% in the mathematics test) = 7/90 

Ans: 7/90

Q6 (ii) Refer to Table 14.7, Chapter 14. Find the probability that a student obtained marks 60 or above.

Answer:

Marks Number of students
0-20 2
20-30 10
30-40 10
40-50 20
50-60 20
60-70 15
70- above 8
Total 90

From the above question, the data of our interest is:

Total no. of students =90 

Total no. of students who obtained marks 60 or above = 15+8 =23

P(a student obtains marks 60 or above) = 23/90 

Ans: 23/90 

Q7 (i) 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  135
dislike 65

Find the probability that a student is chosen at random likes statistics,

Answer:

From the above question, the data of our interest is:

Total no. of students =135+65 =200

Total no. of students who like statistics = 135

P(students like statistics )= 135/200 =27/40 

Ans: 27/40

Q7 (ii) 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  135
dislike 65

Find the probability that a student chosen at random does not like it.

Answer:

From the above question, the data of our interest is:

Total no. of students =135+65=200

Total no. of students who do not like it.= 65

P(a student does not like it) = 65/200

= 13/40 

Ans: 13/40

Q8 (i) Refer to Q.2, Exercise 14.2. What is the empirical probability that an engineer lives less than 7 km from her place of work?

Answer:

The distance (in km) of 40 engineers from their residene to their place of work were found as follows:

5 3 10 20 25 11 13 7 12 31
19 10 12 17 18 11 32 17 16 2
7 9 7 8 3 5 12 15 18 3
12 14 2 9 6 15 15 7 6 12

Total no. of engineers = 40

Total no. of engineers who are living less than 7 km from their workplace = 9 

Therefore we can say,

P(engineers who are living less than 7 km from their workplace)= 

\frac{Total\: no.\: of\: engineers\: who\: are\: living\: less\: than\: 7 km \:from their\: workplace }{Total\:no.\:of\:engineers}

= 9/40

Ans: 9/40

Q8 (ii) Refer to Q.2, Exercise 14.2. What is the empirical probability that an engineer lives more than or equal to 7 km from her place of work?

Answer:

The distance (in km) of 40 engineers from their residene to their place of work were found as follows:

5 3 10 20 25 11 13 7 12 31
19 10 12 17 18 11 32 17 16 2
7 9 7 8 3 5 12 15 18 3
12 14 2 9 6 15 15 7 6 12

Total no. of engineers = 40

Total no. of engineers who are living less than 7 km from their workplace = 31

Therefore we can say,

P(engineers who are living more than or equal to 7 km from their workplace)=31/40

Ans:  31/40

Q8 (iii) Refer to Q.2, Exercise 14.2. What is the empirical probability that an engineer lives within 1/2 km from her place of work?

Answer:

The distance (in km) of 40 engineers from their residene to their place of work were found as follows:

5 3 10 20 25 11 13 7 12 31
19 10 12 17 18 11 32 17 16 2
7 9 7 8 3 5 12 15 18 3
12 14 2 9 6 15 15 7 6 12

Hey, don't you think its too simple:

Well, there is no such engineer whose distance between residence and place of work is less than 1/2 km

Therefore,

engineer whose distance between residence and place of work is less than 1/2 km =0 

P(engineer whose distance between residence and place of work is less than 1/2 km) = 0

Ans: 0

Q9 Activity : Note the frequency of two-wheelers, three-wheelers and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.

Answer:

This activity can be taken as a general problem:

Assumption:

Let the frequency of two-wheelers = x

Let the frequency of three-wheelers = y

Let the frequency of  four-wheelers = z

Total no. of vehicles= x+y+z

therefore,

P(anyone vehicle out of the total vehicles I have observed is a two-wheeler) = 

\frac{x}{x+y+z}

 

Q10 Activity: Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.

Answer:

This activity can be taken as a general problem: 

Well, we know the divisibility by 3 is when the sum of all the digits is divisible by 3

So,

The student will write the number between 100-999

There are 900 3-digit numbers, which are 100, 101, 102, 103, ..., 999.

The first 3-digit numbers that are exactly divisible by 3 is 102, 105, ..... 999

total numbers which are divisible by 3 = 300

P(the sum of all the digits is divisible by 3) = 300/900

=1/3

Ans:1/3

Q11 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.

Answer:

From the above question, the data of our interest is:

Total no. of bags = 11

Total no. of bags that contain more than 5 kg flour = 7

P(Bag contain more than 5 kg flour) = 7/11

Ans: 7/11 

Q12 In Q.5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulfur dioxide in the air in parts per million of a certain city for30 days. Using this table, find the probability of the concentration of sulfur dioxide in the interval 0.12 - 0.16 on any of these days.

Answer:

The data is representing the concentration of sulfur dioxide in the air in parts per million (ppm) of a city. The data obtained for a month of 30 days is as follows:

0.03 0.08 0.08 0.09 0.04 0.17
0.16 0.05 0.02 0.06 0.18 0.20
0.11 0.08 0.12 0.13 0.22 0.07
0.08 0.01 0.10 0.06 0.09 0.18
0.11 0.07 0.05 0.07 0.01 0.04

Total no. of days =30

No. of days in which Conc. of sulfur dioxide in the interval 0.12 - 0.16 = 2

P(Conc. of sulphur dioxide in the interval 0.12 - 0.16) =2/30

Ans: 1/15

Q13 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.

Answer:

The below-written data is representing the blood groups of 30 students study in class VIII.
A, B, O, O, B, A, O, O, AB, O, A, O, B, A, O
A, AB, O, A, A, B, A, B, O, O, O, AB, B, A, O

Total no. of students = 30 

Total no. of students of this class who has blood group AB =3

P(Student of this class has blood group AB)= 3/30 => 1/10

Ans: 1/10

NCERT solutions for class 9 maths chapter wise

Chapter No.

Chapter Name

Chapter 1

NCERT solutions for class 9 maths chapter 1 Number Systems

Chapter 2

CBSE NCERT solutions for class 9 maths chapter 2 Polynomials

Chapter 3

Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry

Chapter 4

NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables

Chapter 5

CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry

Chapter 6

Solutions of NCERT class 9 maths chapter 6 Lines And Angles

Chapter 7

NCERT solutions for class 9 maths chapter 7 Triangles

Chapter 8

CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals

Chapter 9

Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles

Chapter 10

NCERT solutions for class 9 maths chapter 10 Circles

Chapter 11

CBSE NCERT solutions for class 9 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 9 maths chapter 12 Heron’s Formula

Chapter 13

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes

Chapter 14

CBSE NCERT solutions for class 9 maths chapter 14 Statistics

Chapter 15

NCERT solutions for class 9 maths chapter 15 Probability

NCERT solutions for class 9 subject wise

CBSE NCERT Solutions for Class 9 Maths

Solutions of NCERT Solutions for Class 9 Science

How to use NCERT solutions for class 9 maths chapter 15 Probability?

  • Try to understand the concept of this chapter using the theory given in the NCERT textbook.
  • Connect the concepts to real-life problems.
  • Go through some examples to understand the way of solving a problem.
  • Practice the questions given in the practice exercises.
  • During the practice, you can use NCERT solutions for class 9 maths chapter 15 Probability as an assistant. 

 

 

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