Class 9 RD Sharma Solutions – Chapter 25 Probability – Exercise 25.1 | Set 2
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) |
0 |
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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