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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

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

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