NCERT Solutions for Class 9 Maths Chapter 15 Probability (Ex 15.1) Exercise 15.1
Class 9 is a crucial academic year in the educational career of students. This academic session familiarises them with a comprehensive learning schedule and sets a foundation for the Class 10 board examinations. As a result, students must perform well in the Class 9 examinations. Moreover, the knowledge of the subjects that students gain in Classes 9 and 10 helps them decide which subjects they should opt for in Class 11 and their higher education. A good understanding of the curriculum of the subjects in Class 9 is also very essential for learners who wish to score well in any competitive examinations. Many students can find Mathematics intimidating since it introduces them to a wide range of complicated concepts that are difficult to understand. FTo facilitate the students’ understanding of Class 9 Maths Chapter 15 Exercise 15.1, Extramarks provides them with NCERT Solutions For Class 9 Maths Chapter 15 Exercise 15.1In addition, they must practise many questions along with these questions to keep up with all the concepts of the subject. Students in Class 9 face a lot of challenges in their curriculum as the academic year introduces them to several new concepts. Extramarks offers NCERT Solutions for Class 9 Maths Chapter 15 Exercise 15.1 to assist students with their mathematics preparation.
NCERT Solutions for Class 9 Maths Chapter 15 Probability (Ex 15.1) Exercise 15.1
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Access NCERT Solutions for Class 9 Mathematics Chapter 15 – Probability
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Numerous topics are included in the curriculum of Class 9 Mathematics, such as numbers, shapes, formulas, the spaces in which these are contained, quantities and their changes, and much more. Mathematical phenomenons are developed in Science in order to make predictions based on experimental laws. Due to the fact that Mathematics is independent of any experiment, the accuracy of such predictions is solely a function of the model’s suitability. There are many fields where Mathematics is essential, such as Medicine, Finance, Engineering, Computer Science, etc. In addition to the students of Class 9, every student should be familiar with the subject because it has many reallife applications, such as knowledge of weights and currency. In academics and in real life, mathematics concepts can be applied to solve a wide variety of problems. Young learners are able to structure their reasoning and problemsolving skills when they learn the subject thoroughly. Practising mathematical problems is also a great brain exercise.
NCERT Solutions for Class 9 Maths Chapter 15 Probability Exercise 15.1
Class 9 Mathematics introduces students to a wide range of concepts, some of which may be challenging. Extramarks provides students with the NCERT Solutions For Class 9 Maths Chapter 15 Exercise 15.1, so they can better understand the concepts and resolve their doubts regarding the chapter. Examination preparation is easier with these solutions. After thoroughly going over the NCERT Solutions For Class 9 Maths Chapter 15 Exercise 15.1, students can solve any complicated problems that they can encounter during the examinations. Students have always been the main focus of Extramarks. Students can obtain good grades in examinations by using their resources. The NCERT Solutions For Class 9 Maths Chapter 15 Exercise 15.1 have been compiled by the educational experts at Extramarks. Students can easily understand these solutions because of the simple language in which they are curated. These solutions are detailed in a stepbystep manner, making them easy to understand for students. The NCERT Solutions For Class 9 Maths Chapter 15 Exercise 15.1 can be easily downloaded on a variety of devices. Additionally, students can enhance their mathematical and problemsolving skills by practising these solutions.
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Chapter wise NCERT Solutions for Class 9 Maths
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The government of India controls and manages CBSE, a national board of secondary education for public and private schools in India. A number of schools in India and abroad are affiliated with it. IAS officer Nidhi Chibber is the current chairperson of CBSE. Central Board of Secondary Education’s major objectives are:
 Without compromising the quality of their education, all children should receive a stressfree, childcentred education.
 Establishing norms for the implementation of various academic activities, including quality issues. Coordinate academic activities and supervise other agencies involved in the implementation of the Board’s educational and training programs.
 Organising examinations and raising academic standards.
 Establishing examination conditions and conducting public examinations at the end of classes 10 and 12. A certificate of qualification should be awarded to successful candidates from affiliated schools after the examinations.
 There should be a variety of programs aimed at building the capacity and empowerment of teachers in order to update their professional competencies.
 Integrating psychological, pedagogical, and social principles into academic excellence.
All CBSEaffiliated schools follow the NCERT curriculum, especially for Classes 9 to12. NCERT also prepares and publishes model textbooks, supplementary materials, newsletters, journals, educational kits, and multimedia digital materials related to school education, and conducts, promotes, and coordinates research. The organisation was established by the Government of India in 1961 to assist and advise the Central and State Governments on school education policy and programs. NCERT collaborates in networks with state education departments, universities, NGOs, and other organisations to develop and disseminate innovative educational practices. As a coordinating body, it also works to achieve the goal of educating all children up to the age of fourteen. Research, development, training, extension, publication, and dissemination of school education activities are the major activities of the NCERT. Further, the NCERT collaborates with international organisations, visits foreign delegations, and trains educators from developing countries.
NCERT Solutions for Class 9
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Practice is the key to success in the subject of Mathematics. There are so many operations, calculations, concepts, properties, and more in the subject that students may find it difficult to comprehend. As a result, Mathematics can pose a challenge to students in Class 9. Therefore, Extramarks provides students with NCERT Solutions For Class 9 Maths Chapter 15 Exercise 15.1 in order to deeply understand the concepts of the Chapter. But, for the preparation of the examinations, the NCERT textbooks are not enough, as students are required to practise a number of questions to stay on top of these concepts. A thorough review of the NCERT Solutions For Class 9 Maths Chapter 15 Exercise 15.1 is the first and foremost step in preparing the chapter for the examinations. Therefore, Extramarks provides them with the NCERT solutions of all the chapters of Class 9 Mathematics, along with the NCERT Solutions For Class 9 Maths Chapter 15 Exercise 15.1. Students can easily find the important questions of the chapter on the Extramarks website. It helps students practise the concepts again so that they do not forget any properties or formulas and can handle any complicated problem they may encounter in the examinations. Along with the NCERT Solutions For Class 9 Maths Chapter 15 Exercise 15.1, students can also find solved sample papers and past years’ papers on Extramarks. The purpose of this is to provide students with a model for solving their own question papers. Extramarks recommends students carefully review the NCERT Solutions For Class 9 Maths Chapter 15 Exercise 15.1 prior to their examinations.
CBSE Study Materials for Class 9
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CBSE Study Materials
The NCERT Solutions For Class 9 Maths Chapter 15 Exercise 15.1 are a beneficial learning tool for the students. To facilitate a smooth learning process for students, these solutions have been prepared according to the most recent examination pattern. Students should thoroughly review Chapter 15 Probability since it is an important chapter. In order to score well in Mathematics, students must understand the basic concepts of the curriculum. Therefore, Extramarks provides them with the NCERT Solutions For Class 9 Maths Chapter 15 Exercise 15.1. Each topic in Class 9 Chapter 15 Mathematics must be thoroughly understood. Students need to memorise formulasby heart in order to structure their basic answers. With Extramarks NCERT Solutions For Class 9 Maths Chapter 15 Exercise 15.1, students can gain a deeper understanding of concepts through stepbystep explanations and detailed answers. The solutions have been compiled by Extramarks experts, making them a reliable resource. The NCERT Solutions For Class 9 Maths Chapter 15 Exercise 15.1 are provided on the Extramarks website, making them a trustworthy resource for the preparation of school examinations.
Q.1 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.
Ans
Number of hits a boundary = 6
Total number of balls played= 30
Number of ball she did not hit = 30 – 6
= 24
Probability of not hitting boundary = (24/30)
= (4/5)
= 0.8
Thus, the probability that she did not hit a boundary.
Q.2 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
(i) 2 girls (ii) 1 girl (iii) No girl
Also check whether the sum of these probabilities is 1.
Ans
$\begin{array}{l}\text{Number of families}=\text{15}00\\ \left(\text{i}\right)\text{\hspace{0.17em}}\mathrm{}\text{P}\left(\text{2 girls}\right)\text{\hspace{0.17em}}=\frac{\mathrm{Number}\text{of families having two girls}}{\mathrm{Total}\text{families}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{475}{1500}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{19}{60}\\ \left(\text{ii}\right)\text{P}\left(\text{1 girl}\right)=\frac{\mathrm{Number}\text{of families having one girl}}{\mathrm{Total}\text{families}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{}\frac{814}{1500}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{407}{750}\\ \left(\text{iii}\right)\text{P}\left(\text{No girl}\right)=\frac{\mathrm{Number}\text{of families having one girl}}{\mathrm{Total}\text{families}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\text{}\frac{211}{1500}\\ \mathrm{}\text{The sum of probabilities}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{475}{1500}\text{}+\text{}\frac{814}{1500}+\frac{211}{1500}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1500}{1500}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{1}\end{array}$
Q.3 In a particular section of Class IX, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained:
Find the probability that a student of the class was born in August.
Ans
$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of student in class IX}=\text{40}\\ \text{Number of students born in August}=6\\ \mathrm{P}\left(\mathrm{A}\text{student born in August}\right)=\frac{\text{Number of students born in August}}{\mathrm{Number}\text{of student in class IX}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{6}{40}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{20}\\ \text{Thus, the probability of a student born in August is}\frac{3}{20}.\end{array}$
Q.4 Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:
Outcome  3 heads  2 heads  1 head  No head 
Frequency  23  72  77  28 
If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Ans
$\begin{array}{l}\mathrm{Total}\text{\hspace{0.17em}\hspace{0.17em}n}\mathrm{umber}\text{of tossing of two coins}=\text{200}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Frequency of coming up two heads}=72\\ \mathrm{P}\left(2\text{heads coming up}\right)=\frac{\text{Frequency of coming up two heads}}{\mathrm{Total}\text{\hspace{0.17em}\hspace{0.17em}n}\mathrm{umber}\text{of tossing of two coins}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{72}{200}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{9}{25}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=0.36\\ \mathrm{Thus},\text{the probability of coming up two heads is 0.36.}\end{array}$
Q.5 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 Rs)  Vehicles per family


Less than 7000
7000 – 10000 10000 – 13000 13000 – 16000 16000 or more 

Suppose a family is chosen. Find the probability that the family chosen is
(i) earning Rs 10000 – 13000 per month and owning exactly 2 vehicles.
(ii) earning Rs 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than Rs 7000 per month and does not own any vehicle.
(iv) earning Rs 13000 – 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Ans
$\begin{array}{l}\mathrm{Total}\text{number of families}=2400\\ \left(\mathrm{i}\right)\text{ Number of families earning Rs 1}0000\text{}\u2013\text{13}000\text{per month}\\ \text{and}\mathrm{}\text{owning exactly 2 vehicles}=29\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{exactly}\text{two vehicles}\right)=\frac{29}{2400}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}Number of families earning Rs 16}000\text{or more per month}\\ \text{and}\mathrm{}\text{owning exactly 1 vehicle}=579\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{exactly}\text{two vehicles}\right)=\frac{579}{2400}\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}Number of families earning less than Rs 7}000\text{per month}\\ \text{and}\mathrm{}\text{does not own any vehicle}=10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\text{does not own any vehicle}\right)=\frac{10}{2400}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{240}\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}Number of families earning Rs 13}000\u2013\text{16}000\text{per month}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and}\mathrm{}\text{more than 2 vehicles}=25\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\text{more than 2 vehicles}\right)=\frac{25}{2400}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{96}\\ \left(\mathrm{v}\right)\text{\hspace{0.17em}Number of families owning not more than 1 vehicle}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=10+1+2+1+160+305+535\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}+469+579\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2062\\ \text{\hspace{0.17em}}\mathrm{P}\left(\text{owning not more than 1 vehicle}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2062}{2400}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1031}{1200}\end{array}$
Q.6 A teacher wanted to analyse the performance of two sections of students in a mathematics test of 100 marks. Looking at their performances, she found that a few students got under 20 marks and a few got 70 marks or above. So she decided to group them into intervals of varying sizes as follows:
0 – 20, 20 – 30, . . ., 60 – 70, 70 – 100. Then she formed the following table:
Marks  Number of students 
0 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – above 
7 10 10 20 20 15 8 
(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.
Ans
$\begin{array}{l}\mathrm{Total}\text{number of students in a class}=7+10+10+20+20+15+\text{8}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=9\text{0}\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}Number of students obtained marks less than 20\%}=7\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\text{a student obtained less than 2}0\mathrm{\%}\text{marks}\right)=\frac{7}{90}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}Number of students obtained marks 6}0\text{or above}=15+8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=23\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\text{a student obtained marks 6}0\text{or above}\right)=\frac{23}{90}\end{array}$
Q.7 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.
Ans
$\begin{array}{c}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Total}\text{number of students}=135+65\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=200\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of students who likes statistics}=135\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{who}\text{likes statistics}\right)=\frac{135}{200}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{27}{40}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}\mathrm{Number}\text{of students who dislike statistics}=65\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{who}\text{likes statistics}\right)=\frac{65}{200}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{13}{40}\end{array}$
Q.8 The distance (in km) of 40 engineers from their residence 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.
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 (1/2) km from her place of work?
Ans
$\begin{array}{l}\text{\hspace{0.17em}}\mathrm{Total}\text{number of engineers}=\text{40}\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}}\mathrm{Number}\text{of engineers living in less than 7 km from her place of work}=9\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\text{engineers living in less than 7 km\hspace{0.17em}from her place}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{9}{40}\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}}\mathrm{Number}\text{of engineers living in more than or equal to 7 km from her place of work}=31\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\text{engineers living in more than or equal to 7 km\hspace{0.17em}from her place}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{31}{40}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}\mathrm{Number}\text{of engineers living within}\frac{1}{2}\text{\hspace{0.17em}\hspace{0.17em}km from her place of work}=0\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{P}\left(\text{engineers living within\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}\hspace{0.17em}km\hspace{0.17em}from her place}\right)\\ \text{\hspace{0.17em} \hspace{0.17em}}=\frac{0}{40}\\ \text{\hspace{0.17em}\hspace{0.17em}}=0\end{array}$
Q.9 Activity : Note the frequency of twowheelers, threewheelers and fourwheelers 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 twowheeler.
Ans
Let us consider the data obtained from this activity took place any time are given below:
Types of vehicle  Number of vehicle 
Two – Wheeler  40 
Three – Wheeler  35 
Four – Wheeler  25 
Total vehicles  100 
$\begin{array}{l}\mathrm{Total}\text{number of vehicles}=\text{100}\\ \text{\hspace{0.17em}Number of twowheelers}=\text{40}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}P}(\mathrm{Two}\mathrm{wheeler})=\frac{\text{Number of twowheelers}}{\mathrm{Total}\text{number of vehicles}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{40}{100}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{2}{5}\end{array}$
Q.10 Activity : Ask all the students in your class to write a 3digit 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.
Ans
Let there are 30 students in a class and the 3digits numbers written by them are given below:
100, 101, 271, 284, 290, 319, 152, 128, 342, 505, 504, 405, 519, 203, 901, 982, 802, 107, 308, 310, 412, 516, 418, 712, 672, 503, 615, 725, 528, 817
$\begin{array}{l}\text{The 3}\text{digits numbers which are divisible by 3 are}:\\ \text{342,\hspace{0.17em}5}0\text{4,\hspace{0.17em}4}0\text{5,\hspace{0.17em}519,\hspace{0.17em}516,\hspace{0.17em}612,\hspace{0.17em}615,\hspace{0.17em}}\mathrm{}\text{528}\\ \text{Total numbers divisible by 3}=\text{8}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}P}\left(\text{Number divisible by 3}\right)=\frac{\text{Total numbers divisible by 3}}{\mathrm{Total}\text{written}\mathrm{Numbers}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{8}{30}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{4}{15}\\ \text{Thus, the probability of a selected number to be divisible by 3}\\ \text{is}\frac{4}{15}.\end{array}$
Q.11 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.
Ans
$\begin{array}{l}\mathrm{Number}\text{of flour bags = 11}\\ \text{Number of flour bags contains more than 5 kg of flour}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=7\\ \mathrm{P}\left(\mathrm{bag}\text{contains more than 5 kg flour}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{\text{bags containing more than 5 kg flour}}{\mathrm{Total}\text{number of flour}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{7}{11}\\ \text{Thus},\text{the required probability of a selected bag having more than 5 kg flour is}(\text{7}/\text{11}).\end{array}$
Q.12 Frequency distribution table of the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days is given below.
Concentration of SO_{2} (in ppm)  Number of days 
0.00 – 0.04  4 
0.04 – 0.08  9 
0.08 – 0.12  9 
0.12 – 0.16  2 
0.16 – 0.20  4 
0.20 – 0.24  2 
Total  30 
Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 – 0.16 on any of these days.
Ans
$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Total}\text{number of days}=\text{30}\\ \text{Number of days having the concentration of sulphur dioxide in}\\ \text{the interval}0.\text{12}0.\text{16}=2\\ \mathrm{P}(\mathrm{day}\text{having}0.\text{12}0.{\text{16 concentration of SO}}_{\text{2}})\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{2}{30}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{15}\end{array}$
Q.13 Frequency distribution table of the blood groups of 30 students of a class is given below.
Number of students  
A  9 
B  6 
AB  3 
O  12 
Total Students  30 
Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
Ans
$\begin{array}{l}\mathrm{Total}\text{number of students}=\text{30}\\ \text{Number of students having blood group AB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{Blood}\text{group AB}\right)=\frac{\begin{array}{l}\mathrm{Number}\text{of students having blood}\\ \text{group AB}\end{array}}{\mathrm{Total}\text{number of students}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{3}{30}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0.1\\ \text{Thus, the required probability is 0.1.}\end{array}$
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