NCERT Solutions for Class 9 Maths Chapter 15 Probability

Question

You will learn about NCERT Solutions Class 9 Maths Chapter 15 Probability in this chapter. To make it easier for students to understand, we designed all of the solutions in a very intuitive manner, using equations, graphs, diagrams, flow charts, shortcuts, tips, and tricks. These NCERT solutions are created by mentors who are subject experts. Our solutions are the best way to clear basic concepts as well as prepare for competitive exams such as Olympiad, NTSE, JEE Mains, JEE Advance, and others. This chapter is critical for preparing for the CBSE term II examination because it carries a significant amount of weightage in the exam.

In this NCERT Solutions Class, 9 Maths Chapter 15 Probability can learn about different topics related to Probability. Our solutions also provide in-text question solutions and practical-based questions for better understanding and learning. All the important topics mentioned in the chapter are:

• Empirical or experimental probabilities.
• Theoretical or classical probabilities P(E) = number of outcomes with E/ Total possible outcomes. Probability of complementary events P(E).
• Probability of sure and impossible events.
• Probability, when die, is thrown twice.
• Probability of card questions.
• Uses of probability using distance and area.
• In this chapter, we will also learn about various new terms such as Fair, unbiased, bias, random toss, interference, and equally likely outcomes.
• Probability of a sure event is 1.
• Probability of an impossible event is 0.
• An event that has only one outcome is known as an elementary event.
• Sum of all the probabilities of all elementary events of an experiment is 1.
• Probability of any event will never be less than 0 or greater than 1.

In our NCERT Solutions Class 9 Maths, we have prepared all the solutions of the particular chapter in the pointwise method. All the tough concepts are explained in the pointwise method. One must go refer to our solution for their studies, it certainly helps students secure good marks in their exams. Get started now to ace your examination.

Answer 1

NCERT Solutions for Class 9 Maths Chapter 15 Probability

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.

Solution:

According to the question,

Total number of balls = 30

Numbers of boundary = 6

Number of time batswoman didn’t hit boundary = 30 – 6 = 24

Probability she did not hit a boundary = \(\frac{24}{30}\) = \(\frac 45\)

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.

Solution:

Here, total number of families = 1500

(i) ∵ Number of families having 2 girls = 475

∴ Probability of selecting a family having 2 girls = \(\frac{475}{1500} = \frac{19}{60}\)

(ii) ∵ Number of families having 1 girl = 814

∴ Probability of selecting a family having 1 girl = \(\frac{814}{1500} = \frac{407}{750}\)

(iii) Number of families having no girl = 211

Probability of selecting a family having no girl = \(\frac{211}{1500}\)

Now, the sum of the obtained probabilities

\(= \frac{19}{60} + \frac{407}{750} + \frac{211}{1500} \)

\(= \frac{475 + 814 + 211}{1500} \)

\(= \frac{1500}{1500} = 1\)

i.e., Sum of the above probabilities is 1.

3. In a particular section of class IX, 40 students were asked about the month 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.

Solution:

From the graph, we have

Total number of students born in various months = 40

Number of students born in August = 6

∴ Probability of a student of the Class-IX who was born in August \(= \frac 6{40}= \frac3{20}\)

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.

Solution:

Total number of times the three coins are tossed = 200

Number of outcomes in which 2 heads coming up = 72

∴ Probability of 2 heads coming up = \(\frac{72}{200}\) = \(\frac9{25}\)

∴ Thus, the required probability = \(\frac9{25}\)

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.

Suppose a family is chosen. Find the probability that the family chosen is

(i) earning ₹ 10000-13000 per month and owning exactly 2 vehicles.

(ii) earning ₹16000 or more per month and owning exactly 1 vehicle.

(iii) earning less than ₹ 7000 per month and does not own any vehicle.

(iv) earning ₹13000-16000 per month and owning more than 2 vehicles.

(v) owning not more than 1 vehicle.

Solution:

Here, total number of families = 2400

(i) ∵ Number of families earning Rs. 10000 – Rs. 13000 per month and owning exactly 2 vehicles = 29

∴ Probability of a family earning Rs. 10000 – Rs. 13000 per month and owning exactly 2 vehicles = \(\frac{29}{2400}\)

(ii) ∵ Number of families earning Rs. 16000 or more per month and owning exactly 1 vehicle = 579

∴ Probability of a family earning Rs. 16000 or more per month and owning exactly 1 vehicle = \(\frac{579}{2400}\)

(iii) ∵ Number of families earning less than Rs. 7000 per month and do not own any vehicle = 10

∴ Probability of a family earning less than Rs. 7000 per month and does not own any vehicle = \(\frac{10}{2400}\) = \(\frac{1}{240}\)

(iv) ∵ Number of families earning Rs. 13000 – Rs. 16000 per month and owning more than 2 vehicles = 25

∴ Probability of a family earning Rs. 13000 – Rs. 16000 per month and owning more than 2 vehicles = \(\frac{25}{2400} = \frac1{96}\)

(v) ∵ Number of families owning not more than 1 vehicle

= [Number of families having no vehicle] + [Number of families having only 1 vehicle]

= [10 + 1 + 2 + 1] + [160 + 305 + 535 + 469 + 579] = 14 + 2048 = 2062

∴ Probability of a family owning not more than 1 vehicle = \(\frac{2062}{2400} = \frac{1031}{1200}\)

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

(i) Find the probability that a student obtained less than 20% in the mathematics test.

(ii) Find the probability that a student obtained marks 60 or above.

Solution:

Total number of students = 90

(i) From the given table, number of students who obtained less than 20% marks = 7

Probability of a student obtaining less than 20% marks = \(\frac7{90}\)

(ii) From the given table, number of students who obtained marks 60 or above = [Number of students in class-interval 60 – 70] + [Number of students in the class interval 70 – above]

= 15 + 8 = 23

∴ Probability of a student who obtained 23 marks 60 or above = \(\frac{23}{90}\)

Answer 2

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

Find the probability that a student chosen at random

(i) likes statistics,

(ii) does not like it.

Solution:

Total number of students whose opinion is obtained = 200

(i) ∵ Number of students who like statistics = 135

∴ Probability of selecting a student who likes statistics = \(\frac{135}{200} = \frac{27}{40}\)

(ii) ∵ Number of students who do not like statistics = 65

∴ Probability of selecting a student who does not like statistics = \(\frac{65}{200} = \frac{13}{40}\)

8. The distance (in km) of 40 engineers from their residence to their place of work were found as follows

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?

Solution:

Here, total number of engineers = 40

(i) ∵ Number of engineers who are living less than 7 km from their work place = 9

∴ Probability of an engineer who is living less than 7 km from her place of work = \(\frac9{40}\)

(ii) ∵ Number of engineers living at a distance more than or equal to 7 km from their work place = 31

∴ Probability of an engineer who is living at distance more than or equal to 7 km from her place of work = \(\frac{31}{40}\)

(iii) ∵ The number of engineers living within 1/2 km from their work place = 0

∴ Probability of an engineer who is living within 1/2 km from her place of work = \(\frac0{40}\) = 0

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

4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00

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

Solution:

Here, total number of bags = 11

∵ Number of bags having more than 5 kg of flour = 7

∴ Probability of a bag having more than 5 kg of flour = \(\frac7{11}\)

10. A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows

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

Solution:

Here, total number of days = 30

∵ The number of days on which the sulphur dioxide concentration is in the interval 0.12 – 0.16 = 2

∴ Probability of a day on which sulphur dioxide is in the interval 0.12 – 0.16 = \(\frac2{30} = \frac1{15}\)

11. The blood groups of 30 students of class VIII are recorded as follows

A, B, 0, 0, AB, 0, A, 0, B, A, 0, B, A, 0, 0, A, AB, 0, A, A, 0, 0, AB, B, A, B, 0

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

Solution:

Here, total number of students = 30

∵ Number of students having blood group AB = 3

∴ Probability of a student whose blood group is AB = \(\frac3{30} = \frac1{10}\)