# NCERT Solutions for Class 9 Mathematics Chapter 15 – Probability

It is remarkable that a science, which began with the consideration of games of chance, should be elevated to the rank of the most important subject of human knowledge. —Pierre Simon Laplace

Mathematics has  a wide range of applications in real life. It encourages analytical thinking to identify the problem and find the solution,strong brain muscles, encourages critical thinking ability, problem-solving ability  and much more. Higher education becomes relatively easy once students have a conceptual understanding of the mathematical concepts  covered in lower grades. Hence, Mathematics is  one of the core subjects in f almost all the competitive examinations.

Probability is the study of predictions and assumptions. To master the skill of making accurate  predictions , one needs to learn the art of  predicting things using certain principles. Its synonyms are –  possibility, prospect, odds, likely, unlikelihood, chance, likelihood, presumption, feasibility, credibility and improbability. The random experiments of probability, the process of solving different applications using probability and the role of probability in day-to-day life are some of the essential topics covered in this chapter.

The students need to have the proper knowledge of this chapter. As a result, in the NCERT Solutions for Class 9 Mathematics Chapter 15, we have covered every topic in detail with a proper analysis of each concept. It is designed by the subject matter experts  who meticulously follow the CBSE curriculum and guidelines to provide concise and accurate notes to the students.Needless to say, they completely understand what is legitimate as per the board’s standards  and hence it is reliable and trustworthy. Students can take note of  the to-be-focused points from the chapter and use them in  their mathematical calculations. Thus, they can strengthen their  conceptual understanding of every topic covered in this chapter.

Extramarks believes and encourages 360° learning among students., i.e. learning based on the students’ overall development.Extramarks understands how important it is to give step-by-step explanations for making the concepts easy for students and even clarify their doubts via live classes. Moreover, it has practice questions to  step up your performance and thereby prepare well for the examinations to get excellent results.

Extramarks’ website is a one stop solution to all your queries and problems. You can access all the NCERT-related resources like  NCERT Solutions, NCERT Exemplar etc., from the Extramarks’ website. Moreover, these practice questions will  step up your preparation significantly and help you to get excellent results in the examination. You needn’t waste your precious time looking  for any kind of assistance in other subjects because Extramarks has solutions for all classes from 1 to 12.  Extramarks is known for providing authentic, reliable and to the point study material and students have complete faith and trust in our services.

## Key Topics Covered In Class 9 Mathematics Chapter 15

Chapter 15 Class 9 Mathematics is about Probability and its applications.

When you toss a coin, you have two chances to get a head or a tail. Thus, there is a half possibility of getting a head and a half possibility of getting a tail. This is called probability.

Probability is the study of all the possible outcomes you will get for an event. It is merely an assumptions-based study which helps in predictions in various disciplines of life. It explains how to study an event and how to draw possible outcomes from a certain event.

It encourages students to develop analytical skills and  hence, they will be able to approach problems in a smarter way. The NCERT Class 9 Mathematics Chapter 15 is covered in the NCERT Solutions for Class 9 Mathematics Chapter 15 and is available on the Extramarks’ website.

Students can learn basic probability formulas on determining empirical probability and defining its limiting value. Moreover, NCERT solutions provide tips to  help students to  quickly learn every topic covered in the chapter. Moreover, it has practice questions to  step up your performance and  prepare well for the examinations to get excellent results.

### Introduction

In this chapter, we will learn about probability, its meaning and the chances of an outcome. Probability is used in many fields, for example, physical sciences, commerce, biological science, medical science, weather forecast etc. All the fundamentals of probability are covered in this chapter.

We have covered all the concepts and topics in our study material, students can visit  our Extramarks’ website   to access NCERT Solutions for Class 9 Mathematics Chapter 15.

### Probability – an experimental approach

The concept of probability was given by the French philosopher Chevalier De Mere in 1654. He did some experiments  and observed their  outcomes.. Most of his experiments were on coins and dice problems.

You will learn about the experiments in probability with the activities given below,

• Activity1 :  It is based on  an experiment on coins. Students can toss two coins and draw conclusions according to its observation.
• Activity2 : In this activity, a class is divided into groups of 2-3 students. Each group has to toss coins individually and draw conclusions according to their  observations.
• Activity3 : It is based on the experiments of dice problems. Students need to toss dice  and draw conclusions according to its observation.

There is an empirical formula available on probability which is given below:

[ P(E) = numbers of the trials in which the events happened / the total number of trials ]

In this chapter, there are some experiments which are available in a very detailed manner on our Extramarks’ website.  Students can refer to NCERT Solutions for Class 9 Mathematics Chapter 15.

Summary

In this chapter, we have learnt about the probability

• The chapter  states the history of probability and the scientist who gave the concept of probability.
•   There are many applications of probability in the following areas: :
• Physical Sciences
• Commerce
• Biological Science
• Medical Science
• Weather Forecast
• Students can  perform  experiments on probability with the help of coins.
• In this chapter, we learnt about the empirical formula of probability.

[ P(E) = numbers of the trials in which the events happened / the total number of trials ]

In this chapter, we will get complete information regarding probability and perform some experiments on the dice and  coins.. To get further details and practice the exercises related to this topic, students can access the NCERT Solutions for Class 9 Mathematics Chapter 15 from the Extramarks’ website.

#### NCERT Solutions for Class 9 Mathematics Chapter 15 Exercise &  Solutions.

One can find all the NCERT Solutions to the exercises covered in the chapter in the NCERT Solutions for Class 9 Mathematics Chapter 15. The subject matter experts give the solutions. All the solutions are in a step-by-step format,  for making the concepts easy for students and even clarify their doubts via live classes.  The highlighted points from the chapter, which will help students to learn every topic covered in the chapter quickly. Moreover, it has practice questions to step up your performance and thereby prepare well for the examinations to get excellent results.

All the solutions are given point-wise for the students to quickly understand and solve mathematical problems with ease. Students can confidently rely on our learning solutions for their studies and exam preparation. It’s important to go beyond the NCERT books by including solutions as an integral  part of your  study schedule. This encourages the students to master the topic and increases their confidence in achieving a high grade. Students swear by Extramarks because of the absolute trust and confidence it has built over the years.

You can avail of NCERT Solutions for Class 9 Mathematics Chapter 15 from the Extramarks website.

Click on the  links below to access  exercise-specific questions and solutions of  NCERT Solutions for Class 9 Mathematics Chapter 15:

• Chapter 15: Exercise 15.1

Along with Class 9 Mathematics Solutions, you may  explore NCERT Solutions on our Extramarks’ website for all primary and secondary classes.

• NCERT Solutions Class 1
• NCERT Solutions Class 2
• NCERT Solutions Class 3
• NCERT Solutions Class 4
• NCERT Solutions Class 5
• NCERT Solutions Class 6
• NCERT Solutions Class 7
• NCERT Solutions Class 8
• NCERT Solutions Class 9
• NCERT Solutions Class 10
• NCERT Solutions Class 11
• NCERT Solutions Class 12

#### NCERT Exemplar Class 9 Mathematics

NCERT Exemplar Class 9 Mathematics book will help to step up your preparation by testing your problem solving and analytical skills. . It has a compilation of all the advanced level questions designed as per NCERT textbooks based on the latest CBSE syllabus. The   questions follow  the competitive examinations pattern. This will definitely  help students in solving all types of tricky and difficult problems   confidently.

It covers complex theories that confidently prepare students to face the upcoming examinations. It  has plenty of examples for helping students to leverage their performance. It is a complete source of information for CBSE students preparing for their 9th standard examinations. The students think logically about a problem after referring to the NCERT Solutions and NCERT Exemplar.

They have  basic as well as advanced level questions. . It contains extra questions from the NCERT Class 9 Mathematics textbook. Students can refer to NCERT Exemplar for Class 9 Mathematics for more practice. Since the difficulty level is graded, hence its important to go through NCERT syllabus first, then move to solutions and exemplar respectively.They can be rest assured  that nothing remains untouched and every example solution, exercise  has been covered in the chapter and hence they are  confident of their preparation and to ace the exam with excellent results.

#### Key Features for NCERT Solutions for Class 9 Mathematics Chapter 15

Consistency during preparation is quite necessary for the students to excel. Hence, NCERT Solutions for Class 9 Mathematics Chapter 15 helps students to practice these questions on a regular basis.  . Some of  the key features are:

• You can find all the resources you need for revision  in our NCERT Solutions for Class 9 Mathematics Chapter 15.
• It has easy-to-learn and understandable key points that help students learn fast.
• After completing the NCERT Solutions for Class 9 Mathematics Chapter 15, students will be relieved  to find  all the related concepts they studied in previous classes  in the higher classes too.

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:

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

 0 1 2 Above 2
Less than 7000

7000 – 10000

10000 – 13000

13000 – 16000

16000 or more

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

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{}–\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–\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 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.

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

$\begin{array}{l}\mathrm{Total}\text{number of vehicles}=\text{100}\\ \text{\hspace{0.17em}Number of two-wheelers}=\text{40}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}P}\left(\mathrm{Two}-\mathrm{wheeler}\right)=\frac{\text{Number of two-wheelers}}{\mathrm{Total}\text{number of vehicles}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{40}{100}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{2}{5}\end{array}$

Q.10 Activity : Ask all the students in your class to write a 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.

Ans.

Let there are 30 students in a class and the 3-digits 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.

$\begin{array}{l}\mathrm{Number}\text{of flour bags = 11}\\ \text{Number of flour bags contains more than 5 kg of flour}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=7\\ \mathrm{P}\left(\mathrm{bag}\text{contains more than 5 kg flour}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{\text{bags containing more than 5 kg flour}}{\mathrm{Total}\text{number of flour}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{7}{11}\\ \text{Thus},\text{the required probability of a selected bag having more than 5 kg flour is}\left(\text{7}/\text{11}\right).\end{array}$

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 SO2 (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.

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Total}\text{number of days}=\text{30}\\ \text{Number of days having the concentration of sulphur dioxide in}\\ \text{the interval}0.\text{12}-0.\text{16}=2\\ \mathrm{P}\left(\mathrm{day}\text{having}0.\text{12}-0.{\text{16 concentration of SO}}_{\text{2}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{2}{30}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{15}\end{array}$

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}$