NCERT Solutions for Class 9 Maths Chapter 15 Probability Statistics are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 15 Probability.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 15 |

Chapter Name |
Probability |

Exercise |
Ex 15.1 |

Number of Questions Solved |
13 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 15 Probability

Ex 15.1 Class 9 Maths Question 1.

In a cricket match, a bats woman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.

Solution:

Given, a bats woman play 30 balls, therefore total number of trials = 30 and number of events of hitting the boundary = 6

Now, number of balls in which she is not hitting the boundary

= 30 – 6 = 24

∴ The probability that she did not hit a boundary = \(\cfrac { 24 }{ 30 } \) = \(\cfrac { 4 }{ 5 } \)

Ex 15.1 Class 9 Maths Question 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:

Let E_{0}, E_{1} and E_{2} be the event of getting no girl, 1 girl and 2 girls.

Ex 15.1 Class 9 Maths Question 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.

Solution:

Total number of students in class IX, n(S) = 40

Number of students born in the month of August, n(E) = 6

Probability, that the students of the class was born in August

= \(\cfrac { n(E) }{ n(S) } \) = \(\cfrac { 6 }{ 40 } \) = \(\cfrac { 3 }{ 20 } \)

Ex 15.1 Class 9 Maths Question 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.

Solution:

In tossing of three coins, getting two heads comes out 72 times. i.e., n(E) = 72

The total number of tossed three coins n(s) = 200

∴ Probability of 2 heads coming up = \(\cfrac { n(E) }{ n(S) } \) = \(\cfrac { 72 }{ 200 } \) = \(\cfrac { 9 }{ 25 } \)

Ex 15.1 Class 9 Maths Question 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 :

Solution:

Montly income(in Rs) |
Vehicles per Family |
|||

Less Than 7000 |
010 |
1160 |
225 |
Above 20 |

7000-10000 | 0 | 305 | 27 | 2 |

10000 -13000 | 1 | 535 | 29 | 1 |

13000-16000 | 2 | 469 | 59 | 25 |

16000 or more | 1 | 579 | 82 | 88 |

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

(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.

Solution:

Total number of families selected by the organisation, n(S) = 2400

(i) The number of families earning Rs. 10000 – 13000 per month and owning exactly 2 vehicles,

n(E_{1}) = 29

Ex 15.1 Class 9 Maths Question 6.

**Table**

Marks |
Number of students |

0-20 | 7 |

20-30 | 10 |

30-40 | 10 |

40-50 | 20 |

50 – 60 | 20 |

60-70 | 15 |

70 – above | 8 |

Total |
90 |

(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:

**(i)** Total number of students in a class, n(S) = 90

The number of students less than 20% lies in the interval 0 – 20,

i.e., n(E) = 7

∴ The probability, that a student obtained less than 20% in the Mathematics

test = \(\cfrac { n(E) }{ n(S) } \) = \(\cfrac { 23 }{ 90 } \)

**(ii)** The number of students obtained marks 60 or above lies in the marks interval 60 – 70 and 70-above

i.e., n(F) =15+8 = 23

∴ The probability that a student obtained marks 60 or above = \(\cfrac { n(E) }{ n(S) } \) = \(\cfrac { 23 }{ 90} \)

Ex 15.1 Class 9 Maths Question 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 | 135 |

Dislike | 65 |

Find the probability that a student chosen at random

(i) likes statistics

(ii) does not like it.

Solution:

Total number of students, n(S) = 200

**(i)** The number of students who like Statistics n(E) = 135

∴ The probability, that the student like Statistics

\(\cfrac { n(F) }{ n(S) } \) = \(\cfrac { 135 }{ 200} \) = \(\cfrac { 27 }{ 40} \)

**(ii)** The number of student Who does not like Statistics, n (E) = 65

∴ The probability, that the student does not like Statistics = \(\cfrac { n(F) }{ n(S) } \) = \(\cfrac { 65 }{ 200} \) = \(\cfrac { 13 }{ 40} \)

Ex 15.1 Class 9 Maths Question 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 \(\cfrac { 1 }{ 2 } \) km from her place of work?

Solution:

Total number of engineers lives, n(S) = 40

**(i)** The number of engineers whose residence is less than 7 km from their place, n(E) = 9

∴ The probability that an engineer lives less than 7 km from their place of work

= \(\cfrac { n(E) }{ n(S) } \) = \(\cfrac { 9 }{ 40} \)

**(ii)** The number of engineers whose residence is more than or equal to 7 km from their place of work, n(F) = 40 – 9 = 31

∴ The probability, that an engineer lives more than or equal to 7 km from their place of work

= \(\cfrac { n(F) }{ n(S) } \) = \(\cfrac { 31 }{ 40} \)

**(iii)** The number of engineers whose residence within \(\cfrac { 1 }{ 2} \) km from their place of work, i.e., n(G) = 0

∴ The probability, that an engineer lives within \(\cfrac { 1 }{ 2} \) km from their place

= \(\cfrac { n(G) }{ n(S) } \) = \(\cfrac { 0 }{ 40} \) =0

Ex 15.1 Class 9 Maths Question 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?

Solution:

After observing in front of the school gate in time interval 6 : 30 to 7 : 30 am respective frequencies of different types of vehicles are :

Types of Vehicles |
Frequency |

Two wheelers | 550 |

Three wheelers | 250 |

Four wheelers | 80 |

∴ Total number of vehicle, n(S) = 550 + 250 + 80 = 880

Number of two-wheelers, n(E) = 550

robability of observing two-wheelers = \(\cfrac { n(E) }{ n(S) } \) = \(\cfrac { 550 }{ 880} \) = \(\cfrac { 5 }{ 8} \)

Ex 15.1 Class 9 Maths Question 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.

Solution:

Suppose, there are 40 students in a class.

∴ The probability of selecting any of the student = \(\cfrac { 40 }{ 40 } \) = 1

A three digit number start from 100 to 999

Total number of three digit numbers = 999 – 99 = 900

∴ Multiple of 3 in three digit numbers = (102, 105,…, 999)

∴ Number of multiples of 3 in three digit number = \(\cfrac { 900 }{ 3 } \) = 300

i.e., n(E) = 300

∴ The probability that the number written by her/him is divisible by 3

\(\cfrac { n(E) }{ n(S) } \) = \(\cfrac { 300 }{ 900 } \) = \(\cfrac { 1 }{ 3 } \)

Ex 15.1 Class 9 Maths Question 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.

Solution:

The total number of wheat flour bags, n(S) = 11

Bags, which contains more than 5 kg of flour, (E)

= {5.05, 5.08, 5.03, 5.06, 5.08, 5.04, 5.07,}

∴ n(E)=7

∴ Required probability = \(\cfrac { n(E) }{ n(S) } \) = \(\cfrac { 7 }{ 11 } \)

Ex 15.1 Class 9 Maths Question 12.**
**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 :

0.03 | 0.08 | 0.08 | 0.09 | 0.04 | 0.17 |

0.16 | 0.05 | 0.02 | 0.06 | 0.18 | 0.20 |

0.11 | 0.08 | 0.12 | 0.13 | 0.22 | 0.07 |

0.08 | 0.01 | 0.10 | 0.06 | 0.09 | 0.18 . |

0.11 | 0.07 | 0.05 | 0.07 | 0.01 | 0.04 |

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:

Now, we prepare a frequency distribution table

Interval |
Frequency |

0.01-0.04 | 5 |

0.04-0.08 | 11 |

0.08-0.12 | 7 |

0.12-0.16 | 2 |

0.16-0.20 | 4 |

0.20-0.24 | 1 |

Total |
30 |

The total number of days for data, to prepare sulphur dioxide, n(S) = 30

The frequency of the sulphur dioxide in the interval 0.12-0.16, n(E) = 2

∴ Required probability = \(\cfrac { n(E) }{ n(S) } \) = \(\cfrac { 2 }{ 30 } \) = \(\cfrac { 1 }{ 15 } \)

Ex 15.1 Class 9 Maths Question 13.

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

A,B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB,

B,A, O, B, A, B, O

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 randon, has blood group AB.

Blood group |
Frequency |

A | 9 |

B | 6 |

O | 12 |

AB | 3 |

Total |
30 |

The total number of students in class VIII, n(S) = 30

The number of students who have blood group AB, n(E) = 3

∴ The probability that a student has a blood group

AB = \(\cfrac { n(E) }{ n(S) } \) = \(\cfrac { 3 }{ 30 } \) \(\cfrac { 1 }{ 10 } \)

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