NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 15 |

Chapter Name |
Probability |

Exercise |
Ex 15.1 |

Number of Questions Solved |
2 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1

**Question 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:**

Since, batswoman plays 30 balls, therefore total number of trials is n(S) = 30.

Let E be the event of hitting the boundary.

∴ n(E) = 6

The number of balls not hitting the target

n(E’) = 30-6=24

The probability that she does not hit a boundary = \(\frac { n(E’) }{ n(S) }\) = \(\frac { 24 }{ n(30) }\) = \(\frac { 4 }{ 5 }\)

**Question 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.**

**Solution:**

**Question 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:**

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

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

Probability, that the students of the class was born in August = \(\frac { n(E) }{ n(S) }\) = \(\frac { 6 }{ 40 }\) = \(\frac { 3 }{ 20 }\)

**Question 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:**

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 = \(\frac { n(E) }{ n(S) }\) = \(\frac { 72 }{ 200 }\) = \(\frac { 9 }{ 25 }\)

**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.**

**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:**

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

(i) The number of families earning ₹ 10000-13000 per month and owing exactly 2 vehicles, n(E_{1}) = 29

(ii) The number of families earning ₹ 16000 or more per month and owing exactly 1 vehicle, n(E_{2}) = 579

(iii) The number of families earning less than ₹ 7000 per month and does not own any vehicle, n(E_{3}) = 10

(iv) The number of families earning ₹ 13000-16000 per month and owing more than 2 vehicles, n(E_{4}) = 25

(v) The number of families owing not more than 1 vehicle,

n(E_{5}) = (10 + 1 + 2 + 1) + (160 + 305 + 535 + 469 + 579)

=14 + 2048 = 2062

**Question 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:**

(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 = \(\frac { n(E) }{ n(S) }\) = \(\frac { 7 }{ 90 }\)

(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 = \(\frac { n(E) }{ n(S) }\) = \(\frac { 23 }{ 90 }\)

**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**

**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

(ii) The number of students who does not like Statistics, n(F) = 65

∴ The probability, that the student does not like Statistics

**Question 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 \(\frac { 1 }{ 2 }\) 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

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

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

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

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

**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 .

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

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

∴ Probability of observing two-wheelers = \(\frac { n(E) }{ n(S) }\) = \(\frac { 550 }{ 880 }\) = \(\frac { 5 }{ 8 }\)

**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 digit is divisible by 3.**

**Solution:**

Suppose, there are 40 students in a class.

∴ The probability of selecting any of the student = \(\frac { 40 }{ 40 }\) = 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 = \(\frac { 900 }{ 3 }\) = 300

i.e., n(E) = 300

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

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

**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 =\(\frac { n(E) }{ n(S) }\) = \(\frac { 7 }{ 11 }\)

**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**

**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

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

**Question 13.**

**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:**

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

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