JEE Main Important Physics formulas
ApplyAs per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
NCERT Solutions for Class 9 Maths Chapter 15 Probability
Edited By Ramraj Saini | Updated on May 08, 2023 11:36 AM IST
NCERT Solutions for Class 9 Maths Chapter 15 Probability
Probability Class 9 Questions And Answers are provided here. These NCERT solutions are prepared by subject matter experts considering latest CBSE syllabus 2023 which are very helpful for exams as these cover most concepts in details and simple language. 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 NCERT solutions for class 9 maths chapter 15 Probability is designed to provide you step by step solutions to all such problems.
This Story also Contains
- NCERT Solutions for Class 9 Maths Chapter 15 Probability
- Probability Class 9 Questions And Answers PDF Free Download
- Probability Class 9 Solutions - Important Formulae And Points
- Probability Class 9 NCERT Solutions (Intext Questions and Exercise)
- Highlights of NCERT Solutions for Class 9 Maths Chapter 15 Probability
- NCERT solutions for class 9 maths - Chapter Wise
- NCERT solutions for class 9 - Subject Wise
In this NCERT Book chapter probability class 9, there is only one exercise that consists of a total of 13 questions including some activities in. Class 9 maths chapter 15 NCERT solutions is covering every practice problem to help you while preparing for the final examinations. Apart from the probability chapter 15 maths class 9. Here you will get NCERT solutions for class 9 Maths also.
Also, Read
Probability Class 9 Questions And Answers PDF Free Download
Probability Class 9 Solutions - Important Formulae And Points
Probability (P): Probability is a measure that quantifies the likelihood of an event occurring. It is typically expressed as a ratio.
Probability (P(E)) = Number of Favourable Outcomes / Total Number of Outcomes
The probability of any event lies between 0 and 1, where 0 represents an impossible event, and 1 represents a certain event.
Trial: A trial is defined as a set of observations of an event in which one or more outcomes are observed. It's essentially one attempt or occurrence of an experiment.
Event: An event is defined as a collection of observations performed to observe an experiment. It represents the specific outcomes or results that are of interest in the context of the experiment or situation.
Free download NCERT Solutions for Class 9 Maths Chapter 15 Probability for CBSE Exam.
Probability Class 9 NCERT Solutions (Intext Questions and Exercise)
Class 9 maths chapter 15 question answer - Exercise 15.1
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)
=24/30 =0.80
Ans:0.80
| 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) =
= 475/1500 = 19/60
| 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) =
= 814/1500 = 407/1500
| 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) =
= 211/1500
Ans:= 211/1500
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
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
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)= P 1
P 1 = 29/2400
| 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
| 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
| 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
| 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
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
| 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
| 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
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)=
= 9/40
Ans: 9/40
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
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
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) =
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
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
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
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
Highlights of NCERT Solutions for Class 9 Maths Chapter 15 Probability
The Class 9 maths chapter 15 NCERT solutions, which focuses on Probability, has several noteworthy features.
- The information contained within these solutions is authentic, straightforward, and easy to comprehend.
- They provide answers to all the questions found in the exercise at the conclusion of the chapter, as well as alternative methods for solving examples presented throughout.
- These solutions were created by Careers360 Mathematics experts who conducted extensive research on each topic.
- As a result, students can depend on these solutions to prepare for their Class 9 Maths exams, as they include shortcut methods, tips, and tricks that simplify the process of solving complex questions.
if students are interested in class 9 maths ch 15 question answer can access the following link.
Probability Class 9 Exercise 15.1
NCERT solutions for class 9 maths - Chapter Wise
| Chapter No. | Chapter Name |
| Chapter 1 | Number Systems |
| Chapter 2 | Polynomials |
| Chapter 3 | Coordinate Geometry |
| Chapter 4 | Linear Equations In Two Variables |
| Chapter 5 | Introduction to Euclid's Geometry |
| Chapter 6 | Lines And Angles |
| Chapter 7 | Triangles |
| Chapter 8 | Quadrilaterals |
| Chapter 9 | Areas of Parallelograms and Triangles |
| Chapter 10 | Circles |
| Chapter 11 | Constructions |
| Chapter 12 | Heron’s Formula |
| Chapter 13 | Surface Area and Volumes |
| Chapter 14 | Statistics |
| Chapter 15 | NCERT solutions for class 9 maths chapter 15 Probability |
NCERT solutions for class 9 - Subject Wise
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.
Also Check NCERT Books and NCERT Syllabus here:
Frequently Asked Questions (FAQs)
1. What are the important topics in Maths chapter 15 class 9 Probability?
Basic probability, independent events, and conditional probability are the important topics of this chapter. students should prioritize important topics, and scheduling time according to priority will give aces to aspirants and help to excel in exams. For ease you can study probability class 9 pdf both online and offline mode.
2. How many chapters are there in CBSE class 9 maths ?
There are 15 chapters in starting from numbers systems to probability in the CBSE class 9 maths. students can find a list in NCERT syllabus as well as in NCERT Book.
3. What is the meaning of probability according to ch 15 maths class 9?
As per the NCERT Solutions for Class 9 Maths Chapter 15, Probability refers to the likelihood of an event occurring and is considered a mathematical discipline that deals with such random events. The probability of an event is expressed as a value ranging from zero to one. The study of probability in mathematics allows us to predict the likelihood of various events occurring.
4. Where can I find the complete probability class 9 NCERT solutions ?
Here you will get the detailed NCERT solutions for class 9 maths by clicking on the link. you can practice these class 9 probability NCERT solutions to get indepth understanding of the concepts that are essential for exams.
Also Read
Articles
Dec 16, 2024
Upcoming School Exams
Application Date:21 November,2024 - 20 December,2024
Application Date:21 November,2024 - 20 December,2024
Late Fee Application Date:11 December,2024 - 20 December,2024
Get answers from students and experts
Popular Questions
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
In the reaction,
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
| Option 1)
0.02 |
Option 2)
3.125 × 10-2 |
| Option 3)
1.25 × 10-2 |
Option 4)
2.5 × 10-2 |
Colleges After 12th
Popular Course After 12th
Applications for Admissions are open.
JEE Main Important Chemistry formulas
ApplyAs per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
TOEFL ® Registrations 2024
ApplyAccepted by more than 11,000 universities in over 150 countries worldwide
Pearson | PTE
ApplyRegister now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 31st DEC'24! Trusted by 3,500+ universities globally
JEE Main high scoring chapters and topics
ApplyAs per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE
JEE Main Important Mathematics Formulas
ApplyAs per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters
News and Notifications
Back to top