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 E0, E1 and E2 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 | 0 10 |
1 160 |
2 25 |
Above 2 0 |
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(E1) = 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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