# NCERT Solutions For Class 9 Maths Chapter 15

## NCERT Solutions Class 9 Maths Probability

### Ncert Solutions For Class 9 Maths Chapter 15 PDF Free Download

The NCERT Solutions For Class 9 Maths Chapter 15 are intended for students of class 9 looking to improve their knowledge in the subject and test their knowledge as well. This chapter is one of the most important chapter given in the syllabus of mathematics and questions are expected in various examinations from this chapter. The chapter will prove to be helpful and will act as the fundamental pillar in building greater concepts on probability in higher grades. There are various questions given in this chapter for getting clarity in this subject.

### NCERT Solutions For Class 9 Maths Chapter 15 Exercises

You will be facing various interesting questions such as what is the probability of a wicket keeper not dropping a catch in an ODI match. If a set of data is given regarding a number of families, can you find out if a person is chosen randomly, will it be a girl ?, learn how to solve this type of questions here in this chapter. For your good understanding of this chapter, you will find such questions in NCERT solutions in class 9 Maths and we have provided table containing values out which you have to find out the probability such as if birth month of  students are given, what are the chances of kid picked in random to be born in the month of april.

What is the chance of getting heads, when a coin is tossed ?, learn how to find out the answers to that kind of question here in this chapter. You will be finding out income of employees in an organization and even the marks of about 100 students in class 6 and what are the chances of a student scoring more than 50 ?. These type of questions will be easier to solve by using the method given below.

We have even more questions such as preparing a frequency table with the given values in the question and if learnt the technique, will be very beneficial to you in the future.

1. In an ODI match, a wicket keeper drops a catch 6 times out of 30 catches he gets. Find the probability of the wicketkeeper not dropping a catch.

Sol:         Let E be the event of dropping a catch

P(E)=  Probability of the wicketkeeper dropping a catch = $\frac{6}{30}$ = $\frac{1}{5}$

Thus, Probability of not dropping a catch is 1-P(E) = $1-\frac{1}{5}$ = $\frac{4}{5}$

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

 No of girls in a family 2 1 0 No of families 500 300 200

What is the probability that a family, chosen at random, has

i. 2 girls

ii. 1 girl

iii. No girl

Also check whether the sum of these probabilities is 1.

Sol:         Total number of families = $500+300+200$= 1000

i. P(Probability of 2 girls) = $\frac{500}{1000}$ = $\frac{1}{2}$

ii. P(Probability of 1 girl) = $\frac{300}{1000}$ = $\frac{3}{10}$

iii. P(Probability of No girl) = $\frac{200}{1000}$ = $\frac{1}{5}$

Sum of all Probabilities = $P(Probability of 2 girls) + P(Probability of 1 girl)+ P(Probability of No girl)$  = $\frac{1}{2}+\frac{3}{10}+\frac{1}{5}$ = 1

3. From the following table, Find the probability of a student selected at random being born in April.

 Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec No of students 5 6 4 2 8 1 5 4 4 3 7 1

Sol:     Total number of students:  $5+6+4+2+8+1+5+4+4+3+7+1$ = 50

P( Student born in April) =$\frac{2}{50}$ = $\frac{1}{25}$

4. Three coins are tossed simultaneously 210 times with the following frequencies of different outcomes:

If the three coins are simultaneously tossed again, What is the probability of not getting even a single head?

Sol:         Number of times the coins were tossed = 210

P( Not getting even a single head) = $\frac{73}{210}$

5. An organization selected 1900 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:

Vehicles per family

 Income level 0 1 2 More than 2 Less than 6000 10 60 25 0 6000 to 11000 0 205 27 2 11000 to 14000 1 435 29 1 14000 to 17000 2 369 59 25 17000 or more 1 479 82 88

Find the probability that a family chosen at random is

(i) Earning 11000 – 14000 per month and owning exactly 1 vehicle.

(ii) Earning 17000 or more per month and owning exactly 2 vehicles.

(iii) Earning less than 6000 per month and does not own any vehicle.

(iv) Earning 6000-11000 per month and owning more than 2 vehicles

(v) Owning more than 2 vehicles

Sol:       Total number of families: 1900

• P( Earning 11000-14000 and owning 1 vehicle) = $\frac{435}{1900}$
• P( Earning 17000 or more and owning 2 vehicles) = $\frac{82}{1900}$
• P( Earning less than 6000 and doesn’t own any vehicle) = $\frac{10}{1900}$
• P( Earning 6000-11000 and owning more than 2 vehicles) = $\frac{2}{1900}$
• P( Owning more than 2 vehicles) = $\frac{116}{1900}$

6. In a science test, the marks of 100 students of class VI are listed in the following table:

 Marks (Out of 100) No of students 0-20 18 20-30 17 30-40 15 40-50 8 50-60 14 60-70 21 70-100 7 Total 100

i. What is the probability that a student selected at random has scored less than 40?

ii. What is the probability that a student selected at random has scored more than 50?

Sol:         Total number of students = 100

Number of students having scored less than 40% = $18+17+15$ = 50

Number of students having scored more than 50% = $14+21+7$ = 42

i. P(Less than 40%) = $\frac{50}{100}$ = $\frac{1}{2}$

ii. P(More than 50%) = $\frac{42}{100}$ = $\frac{21}{50}$

7. The distance (in km) of 20 doctors from their residence to their place of work were found as follows:

5, 6, 4, 7, 2, 9, 1, 6, 4, 3, 5, 32, 4, 6, 21, 15, 4,15,18,5

What is the probability that a doctor  lives:

(i) Less than 6 km from her place of work?

(ii)  More than or equal to 6 km from her place of work?

(iii) Within 0.5 km from her place of work?

Sol:  Total number of doctors = 20

Number of doctors with travelling distance below 6km = 10

Number of doctors with travelling distance more than or equal to 6 km = 10

Number of doctors with travelling distance below 0.5 km = 0

(i) P( Less than 6 km) = $\frac{10}{20}$ =$\frac{1}{2}$

(ii) P( More than or equal to 6) = $\frac{10}{20}$ = $\frac{1}{2}$

(iii) P(Less than 0.5 km) = $\frac{0}{20}$ = 0

8. A survey of 100 students was conducted to know the opinion of the students about the subject statistics which is recorded in the following table.

 Opinion Number of students Like 37 Dislike 63

What is the probability that a student selected at random

i. Likes the subject?

ii. Dislikes the subject?

Sol:       Total number of students = 100

Number of students liking the subject = 37

Number of students disliking the subject = 63

i. P(liking the subject) = $\frac{37}{100}$

ii. P(Disliking the subject) = $\frac{63}{100}$

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

5.08 ,4.95, 5.00, 4.96, 5.08, 4.98, 5.04, 5.07 and 5.00

What is the probability that any of these bags chosen at random contains less than 5 kg of flour?

Sol:         Total number of wheat bags = 9

Number of wheat bags weighing less than 5 kg = 3

P( Weighing less than 5 kg) = $\frac{3}{9}$ = $\frac{1}{3}$

10. 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.

The blood groups of 10 students of Class VIII are recorded as follows:
O,AB,A,B,O,A,AB,AB,O and AB

Represent this data in the form of a frequency distribution table. Use this table to determine the probability that a student of this class, selected random, has blood group O.

Sol:     Frequency Distribution Table:

 Blood Group Frequency A 2 B 1 AB 4 O 3

Total number of students = 10

Number of students having blood group O = 3

P(Blood group O) = $\frac{3}{10}$

Thus, use our downloadable NCERT Solutions For Class 9 Maths Chapter 15 pdf to learn all the concepts of the chapter probability from anywhere in the world. The solutions given here are very easily understandable so that students does not face any difficulties regarding any of the answers.