NCERT Solutions for Class 9 Maths Chapter 15 - Exercise

Exercise – 15.1

1. In a cricket math, 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 times a batswoman strikes a boundary in a game \[ = 6\]

Number of balls played in total \[ = 30\]

Therefore, Number of times that the batswoman does not hit a boundary \[ = 30 - 6 = 24\]Probability (she did not hit a boundary) = Number of times when she does not hit boundary /total number of balls played

\[ = \dfrac{{24}}{{30}} = \dfrac{4}{5}\]

2. \[1500\] families with \[2\] children were selected randomly, and the following data were recorded:

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: Total number of families \[ = 475 + 814 + 211 = 1500\]

i. Number of families with 2 girls \[ = 475\]

Probability (a family with two daughters, picked at random) =      

Number of families having two girls child/ total number of families

\[ \Rightarrow \dfrac{{475}}{{1500}} = \dfrac{{19}}{{60}}\]

ii. Number of families with 1 girl child \[ = 814\]

Probability (a family with one daughter, picked at random) =
Number of families having one girl child /total number of families

\[ \Rightarrow \dfrac{{814}}{{1500}} = \dfrac{{407}}{{750}}\]

iii. Number of families with no girl child \[ = 211\]

Probability (a family with no daughter, picked at random) =
Number of families having one girl child /total number of families

\[ \Rightarrow \dfrac{{211}}{{1500}}\]

Sum of all probabilities \[ = \dfrac{{19}}{{60}} + \dfrac{{407}}{{750}} + \dfrac{{211}}{{1500}}\]       (taking LCM of denominator)

\[ \Rightarrow \dfrac{{475 + 814 + 211}}{{1500}}\]

\[ \Rightarrow \dfrac{{1500}}{{1500}} = 1\]

Hence, the sum of all the probabilities is 1.
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 class was born in August.

Ans:

Number of students born in August \[ = 6\]

Total number of students\[ = 40\]

Probability (student born in August) =
Number of students born in August  Total number of students

\[ \Rightarrow \dfrac{6}{{40}} = \dfrac{3}{{20}}\]

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.

Ans: Number of times two head appears \[ = 72\]

Number of times the coins were tossed in total \[ = 200\]

Probability (two heads will appears) =
Number of times two heads appears  /Number of times the coins were tossed in total

\[ \Rightarrow \dfrac{{72}}{{200}} = \dfrac{9}{{25}}\]

5. An organization selected \[2400\] families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:

Suppose a family is chosen, find the probability that the family chosen is

i. Earning Rs.\[1000 - 13000\]per month and owning exactly \[2\]vehicles.

ii. Earning Rs. \[16000\]or more per month and owning exactly \[1\] vehicles.

iii. Earning less than Rs \[7000\]per month and owning more than \[2\]vehicles.

iv. Earning Rs. \[13000 - 16000\]per month and owning more than \[2\]vehicles

v. Owning not more than \[1\] vehicle.

Ans: Total number of families surveyed \[ = 10 + 160 + 25 + 0 + 0 + 305 + 27 + 2 + 1 + 535 + 29 + 1 + 2 + 469 + 59 + 25 + 1 + 579 + 82 + 88 = 2400\]

i. Number of families earning Rs. \[10000 - 13000\] per month and owning exactly 2 vehicles \[ = 29\]

Therefore, Probability \[ = \dfrac{{29}}{{2400}}\]

ii. Number of families earning less than Rs. \[ = \dfrac{{10}}{{2400}} = \dfrac{1}{{240}}\] per month and owning exactly 1 vehicles \[ = 579\]

Therefore, Probability \[ = \dfrac{{579}}{{2400}}\]

iii. Number of families earning Rs. \[7000\] per month and does not own any vehicles \[ = 10\]

Therefore, Probability \[ = \dfrac{{10}}{{2400}} = \dfrac{1}{{240}}\]

iv. Number of families earning Rs. \[13000 - 16000\] per month and owning more than 2 vehicles \[40\]

Therefore, Probability \[ = \dfrac{{25}}{{2400}}\] \[ = \dfrac{1}{{96}}\]

v. Number of families owning not more than 1 vehicle \[ = 10 + 160 + 0 + 305 + 1 + 535 + 2 + 469 + 1 + 579 = 2062\]

Therefore, Probability \[ = \dfrac{{2062}}{{2400}} = \dfrac{{1031}}{{1200}}\]

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.

Ans: Number of total students \[ = 90\]

Number of students with less than \[20\]% marks in the test \[ = 7\]

Therefore, Probability \[ = \dfrac{7}{{90}}\]

Number of students securing marks \[60\]or above \[ = 15 + 8 = 23\]

Therefore, Probability \[ = \dfrac{{23}}{{90}}\]

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

Ans: Total number of students \[ = 135 + 65\]

i) Number of students who likes statistics \[ = 135\]

Probability (students liking statistics) \[ = \dfrac{{135}}{{200}}\] \[ = \dfrac{{27}}{{40}}\]

ii) Number of students who do not like statistics \[ = 65\]

Probability (students not liking statistics) \[ = \dfrac{{65}}{{200}} = \dfrac{{13}}{{40}}\]

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 ½ km from her place of work? 

Ans:

i. Total number of engineers \[ = 40\]

Number of engineers who live less than 7 km from their workplace \[ = 9\]

Therefore, Probability (engineers who live less than 7 km from their workplace) \[ = \dfrac{9}{{40}}\]

ii. Number of engineers who live more than or equal to 7 km from their workplace \[ = 40 - 9\] \[ = 31\]

Therefore, Probability (engineers who live more than or equal to 7 km from their workplace) \[ = \dfrac{{31}}{{40}}\]

iii. Number of engineers  who live within ½ km from her place of work \[ = 0\]

Therefore, Probability (engineers who live within ½ km from her place of work) \[ = 0\]

9. Activity

Note the frequency of two – wheeler, three-wheeler and four-wheelers going past during a time – interval in front of 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 these values

Total number of vehicles \[ = 75 + 25 + 50 = 150\]

Therefore, Probability (two wheeler observation) \[ = \dfrac{{75}}{{150}} = \dfrac{1}{2}\]

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 him/her is divisible by\[3\]? Remember that a number is divisible by\[3\], if the sum of it’s digits is divisible by\[3\].

Ans:

Let the total number of students in class \[ = 20\]

Let’s numbers written by students are:

123 347 267 521 346 159 410 217 313 351

205 414 630 176 452 946 911 716 363 615

We know that, for a number to be divisible by \[3,\]the sum of it’s digits must be divisible by \[3\]

So, numbers divisible by \[3\] are

\[ \Rightarrow 123,267,159,351,414,630,363,615\]

Probability (numbers divisible by\[3\]) \[ = \dfrac{8}{{20}} = \dfrac{2}{5}\]

11. Eleven bags of wheat flour, each marked \[5\] kg, actually contained the following weights of flour (in kg): 

\[ \Rightarrow 123,267,159,351,414,630,363,615\] Find the probability that any of these bags chosen at random contains more than 5 kg of flour.

Ans: Total number of bags \[ = 11\]

Number of bags having more than 5 kg of flour \[ = 7\]

Therefore, Probability \[ = \dfrac{7}{{11}}\]

12. The below frequency distribution table represents 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.

Ans: Number days for which the concentration of sulphur dioxide was in the interval of \[0.12 - 0.16 = 2\]

Total number of days \[ = 30\]

Therefore, Probability \[ = \dfrac{2}{{30}}\]

13. The below frequency distribution table represents 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.

Ans: Total number of students \[ = 30\]

Number of students having blood group AB \[ = 3\]

Therefore, Probability (students having blood group AB) \[ = \dfrac{3}{{30}} = \dfrac{1}{{10}}\]

NCERT Solutions for Class 9 Maths Chapter 15 Probability Exercise 15.1

Opting for the NCERT solutions for Ex 15.1 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 15.1 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Chapter wise NCERT Solutions for Class 9 Maths

• Chapter 1 - Number Systems

• Chapter 2 - Polynomials

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introduction to Euclids Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelograms and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Heron's Formula

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 9 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 9 Maths Chapter 15 Exercise 15.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 9 Maths Chapter 15 Exercise 15.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

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