Skip to content
Related Articles

Related Articles

Improve Article

Class 9 RD Sharma Solutions – Chapter 25 Probability – Exercise 25.1 | Set 2

  • Last Updated : 19 Jan, 2021

Question 11. Given below is the frequency distribution table regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days.

Conc. of SO2

0.00-0.04

0.04-0.08

0.08-0.12

0.12-0.16

0.16-0.20

0.20-0.24

No of days

4

8

9

2

4

3

Find the probability of concentration of sulphur dioxide in the interval 0.12-0.16 on any of these days.

Solution:

Total number of days = 30

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

Probability of concentration of SO2 in the internal 0.12 – 0.16 = 



= Favorable Outcome / Total outcome 

= 2/30 = 0.06

Question 12. An organization 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:

 Vehicles per family

Monthly Income(in Rs)

1

2

Above 2

Less than 7000

10

 160 

25 

 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.

(vi) owning at least one vehicle.

Solution:

Total numbers of families selected by the organization to Survey= 2400.         -(According to Question)

(i) Let E1 be the event of selecting of family earning ₹(10000 -13000) 

per month and owning exactly two vehicles.

Numbers of families earning ₹10000 –13000

per month and owning exactly 2 vehicles = 29



Required probability P(E1) = 29/2400

(ii) Let E2 be the event of selecting of family earning ₹16000 or

more per month and owning exactly 1 vehicle.

Number of families earning ₹16000 or 

more per month and owning exactly 1 vehicle = 579

Required probability,P(E2) = 579/2400

(iii) Let E3 be the event of selecting of family earning than ₹ 7000 per month and

doesn’t own any vehicle.

Number of families earning but ₹7000 per month and 

doesn’t own any vehicle = 10

Required probability, P(E3)= 10/2400 = 1/240

(iv) Let E4 be the event of selecting a family earning ₹(13000 -16000) per month and 

owning quite 2 vehicles.

Number of families earning ₹13000-16000 per month and

owning quite 2 vehicles = 25

Required probability, P(E4) = 25/2400 = 1/96

(v) Let E5 be the event of selecting a family owning less than 1 vehicle.

Number of families owning less than 1 vehicle i.e. the number 

of families owning 0 vehicle and 1 vehicle = 10+160+0+305+1+535+2+469+1+579 = 2069

Required probability, P(E5) = 2062/2400 = 1031/1200



(vi) Let P(E6) is the probability that the family of owning atleast one vehicle 

P(E6) = (160+305+535+469+579+25+27+29+29+82+0+2+1+25+88)/2400 

= 2356/2400 = 589/600

Question 13. The following table gives the lifetimes of 400 neon lamps :

Lifetime

300-400

400-500

500-600

600-700

700-800

800-900

900-1000

Number of lambs

14

56

60

86

74

62

48

A bulb is selected in random. Find the probability that the lifetime of the selected bulb is

(i) less than 400?

(ii) Between 300 to 800?

(iii) At least 700hours?

Solution:

(i) The probability that the lifetime of the selected bulb is less than 400

= Favorable outcomes / Total outcome

= 14/400 = 7/400

(ii) The probability that the lifetime of the selected bulb is between 300 – 800 hours

= Favorable outcomes / Total outcome

= (14 +56 +60 +86 +74) / 400



 = 29/40

(iii) The probability that the lifetime of the selected bulb is at least 700 hours

= Favorable outcomes / Total outcome

= (74 +62+ 48)/400 = 23/50

Question 14. Given below is the frequency distribution of wages (in Rs) of 30 workers in a certain factory:

Wages(in Rs)

110-130

130-150

150-170

170-190

190-210

210-230

230-250

No. of workers

3

4

5

6

5

4

3

A worker is selected at random. Find the probability that 

(i) Less than Rs150

(ii) Atleast Rs210

(iii)More than or equal to Rs150 but less than Rs210.

Solution:

(i) The probability that his wages are less than Rs 150 = 

= Favorable outcomes / Total outcome

=(3 + 4) / 30 = 7 / 30

(ii)The probability that his wages are at least Rs 210 

= Favorable outcomes / Total outcome

= (3 + 4) / 30 = 7 / 30

(iii) The probability that his wages are more than or equal to 150 but less than Rs 200

= Favorable outcomes / Total outcome

= (5 + 6 + 5) / 30 = 16 / 30 = 8 / 15




My Personal Notes arrow_drop_up
Recommended Articles
Page :