Download NCERT Solutions for Class 9 Maths Chapter 15 Exercise 15.1 – Probability. This Exercise contains 13 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 9 for Maths NCERT solutions for other Chapters, you can click the link at the end of this Note.

### NCERT Solutions for Class 9 Maths Chapter 15 Exercise 15.1 – Probability

**NCERT Solutions for Class 9 Maths Chapter 15 Exercise 15.1 – Probability**

**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 :**

Total number of balls she played = 30

Number of balls she hits a boundary = 6

Therefore,

Number of balls she doesn’t hit a boundary

= 30 – 6

= 24

P(she did not hit a boundary) = 24/30

= 4/5

Hence, the probability of not hitting a boundary is 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

(ii) 1 girl

(iii) No girl

**Also check whether the sum of these probabilities is 1.**

** ****Solution :**

(i) Total number of families = 1500

Number of families having 2 girls = 475

P(2 girls) = 475/1500

= 19/60

Hence, the probability of a family having 2 girls is 19/60

(ii)Total number of families = 1500

Number of families having 1 girl child = 814

P(1 girls) = (814/1500)

= 407/750

Hence, the probability of a family having 1 girl is 407/750

(iii) Total number of families = 1500

Number of families having no girl = 211

P(no girls) = (211/1500)

Hence, the probability of a family having no girl is (211/1500).

Now, Sum of the probabilities obtained in parts (i), (ii) and (iii)

= (19/60) + (407/750) + (211/1500)

= (475+814+211)/1500

= (1500/1500)

= 1

Thus, we get the sum of their probabilities is 1.

**3. Refer to Example 5, Section 14.4, Chapter 14(NCERT Class IX Mathematics). Find the ****probability that a student of the class was born in August.**

** ****Solution :**

Given below is the Example 5, Section 14.4, Chapter 14(NCERT Class IX Mathematics):

**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:**

Here,

Total number of students = 40

Number of students born in August = 6

P(Student was born in August) = 6/40

= 3/20

Hence, the probability that a student of the class was born in august is 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.**

** ****Solution :**

Number of total tosses = 200

Frequency of getting 2 heads = 72

P(2 heads) = 72/200

= 9/25

Hence, the probability of coming up 2 heads = 9/25

**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 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 :**

(i) Total number of families = 2400

Number of families earning Rs 10000-13000 per month and owning exactly 2 vehicles = 29

Probability of choosing such a family = 29/2400

Hence, Probability of families earning Rs 10000-13000 per month and owning exactly 2 vehicles is 29/2400.

(ii) Total number of families = 2400

Number of families earning Rs 16000 or more per month and owning exactly 1 vehicle = 579

Probability of choosing such a family = (579/2400)

Hence, Probability of families earning Rs 16000 or more per month and owning exactly 1 vehicle is (579/2400)

(iii) Total number of families = 2400

Number of families earning less than Rs 7000 per month and does not own any vehicle = 10

Probability of choosing such a family = (10/2400)

= (1/240)

Hence, Probability of families earning less than Rs 7000 per month and does not own any vehicle is 1/240.

(iv) Total number of families = 2400

Number of families earning Rs 13000-16000 per month and owning more than 2 vehicles = 25

Therefore, Probability of choosing such a family = (25/2400)

= (1/96)

Hence, Probability of families earning Rs 13000-16000 per month and owning more than 2 vehicles = 1/96

(v) Total number of families = 2400

Number of families not having more than one vehicle

= Number of families with no vehicles + Number of families with 1 vehicle

= (10 + 0 + 1 + 2 + 1) + (160 + 305 + 535 + 469 + 579)

= 14 + 2048

= 2062

Therefore, Probability of choosing such a family = (2062/2400)

= (1031/1200)

Hence, Probability of families not having more than one vehicle = (1031/1200)

**6. Refer to Table 14.7, Chapter 14.**

**(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 :**

Table 14.7, Chapter 14 is as follows :

** **

(i) Total number students = 90

Number of students who obtained less than 20% marks = 7

P(Student obtained less than 20% marks) = 7/90

Hence, probability that a student obtained less than 20% marks = 7/90

(ii) Total number students = 90

Number of students who obtained marks 60 or above = 15 + 8 = 23

P(Student obtained marks 60 or above) = 23/90

Hence, probability that a student obtained marks 60 or above = 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.**

**Solution :**

(i) Total number students = 200

Number of students who like statistics = 135

P(Student likes statistics) = 135/200

= 27/40

Hence, probability that a student likes statistics = 27/40

(ii) Total number students = 200

Number of students who doesn’t like statistics = 65

P(Student dislikes statistics) = 65/200

= 13/40

Hence, probability that a student doesn’t like statistics = 13/40

**8. Refer to Q.2, Exercise 14.2. 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?**

** ****Solution :**

Q.2, Exercise 14.2 is as follows :

The distance (in km) of 40 engineers from their residence to their place of work were found as follows:

Rearranging the data in ascending order :

2, 2, 3, 3, 3, 5, 5, 6, 6, 7, 7, 7, 7, 8, 9, 9, 10, 10, 11, 11, 12, 12, 12, 12, 12, 13, 14, 15, 15, 15, 16, 17, 17, 18, 18, 19, 20, 25, 31, 32

(i) Total number engineers = 40

Number of engineers who live less than 7 km from her place of work = 9

P(an engineer lives less than 7 km from her place of work) = 9/40

Hence, probability that the engineer lives less than 7 km from her place of work = 9/40

(ii) Total number engineers = 40

Number of engineers who live more than or equal to 7 km from her place of work = 31

P(an engineer lives more than or equal to 7 km from her place of work) = 31/40

Hence, Probability that the engineer lives more than or equal to 7 km from her place of work = 31/40

(iii) Total number engineers = 40

Number of engineers who live within ½ km from her place of work = 0

P(an engineer lives within ½ km from her place of work) = 0/40 = 0

Hence, Probability that the engineer lives within ½ km from her place of work = 0.

**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 :**

The frequency of two wheelers, three wheelers and four wheelers going past during 9:00 AM to 10:00 AM is as follows :

Total number of vehicles = 30

Total no. of 2 wheelers = 15

P(2 wheelers) = 15/30

= 1/2

Hence, the probability of 2 wheelers is 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 ****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 :**

Total three digit numbers are :

100, 101, …………………, 998, 999

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

Three digit numbers divisible by 3 are :

102, 105, 108, …………………, 996, 999

Number of three digit numbers divisible by 3 = 300

P(the number is divisible by 3) = 300/900

= 1/3

Thus, probability that the number written by him/her is divisible by 3 is 1/3

**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 : **

Rearranging the given data in ascending order :

4.97, 4.98, 5.00, 5.00, 5.03, 5.04, 5.05, 5.06, 5.07, 5.08, 5.08

Total number of bags = 11

Number of bags containing more than 5 kg of flour = 7

P(the bag contains more than 5 kg of flour) = 7/11

Hence, the probability that the chosen bag contains more than 5 kg flour is 7/11

**12. In Q.5, Exercise 14.2, 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 :**

Q.5, Exercise 14.2 is as follows :

**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:**

The grouped frequency distribution table for the given data is as follows :

Total number of days = 30

Number of days having concentration of sulphur dioxide in the interval 0.12-0.16 = 2

P(the concentration of sulphur dioxide in the interval 0.12 – 0.16) = 2/30

= 1/15

Hence, the probability of the concentration of sulphur dioxide in the interval 0.12 – 0.16 on any of the day is 1/15

** 13. ****In Q.1, Exercise 14.2, 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 :**

Q.1, Exercise 14.2 is as follows :

**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.**

The grouped frequency distribution table for the given data is as follows :

Total number of students = 30

Number of students having blood group AB = 3

P(selected student has blood group AB) = 3/30

= 1/10

Hence, the probability that a student of this class, selected at random, has blood group AB is 1/10

**NCERT Solutions for Class 9 Maths Chapter 15 Exercise 15.1 – Probability, has been designed by the NCERT to test the knowledge of the student on the topic – 2 Probability – an Experimental Approach**

**Download NCERT Solutions for Class 9 Maths Chapter 15 Exercise 15.1 – Probability**

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