RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1

Other Exercises

Question 1.
A coin is tossed 1000 times with the following frequencies
Head : 455, Tail : 545.
Compute the probability for each event.
Solution:
Total number of events (m) 1000
(i) Possible events (m) 455
∴Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 455 }{ 1000 }$$
= $$\frac { 91 }{ 200 }$$ = 0.455
(ii) Possible events (m) = 545
∴Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 545 }{ 1000 }$$ = $$\frac { 109 }{ 200 }$$ = 0.545

Question 2.
Two coins are tossed simultaneously 500 times with the following frequencies of different
outcomes:
One tail : 290 times
Find the probability of occurrence of each of these events.
Solution:
Two coins are tossed together simultaneously 500 times
∴ Total outcomes (n) 500
(i) 2 heads coming (m) = 95 times
∴Probability P(A) = $$\frac { m }{ n }$$
= $$\frac { No. of possible events }{ Total number of events }$$
= $$\frac { 95 }{ 500 }$$ = $$\frac { 19 }{ 100 }$$ = 0.19
(ii) One tail (m) = 290 times
∴Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 290 }{ 500 }$$ = $$\frac { 580 }{ 1000 }$$ = $$\frac { 58 }{ 100 }$$ = 0.58
(iii) No head (m) = 115 times
∴Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 115 }{ 500 }$$ = $$\frac { 23 }{ 100 }$$ = 0.23

Question 3.
Three coins are tossed simultaneously 1oo times with the following frequencies of different outcomes:

If the three coins are simultaneously tossed again, compute the probability of:
(iii) at least one head coming up.
(iv) getting more heads than tails.
(v) getting more tails than heads.
Solution:
Three coins are tossed simultaneously 100 times
Total out comes (n) = 100

(i) Probability of 2 heads coming up (m) = 36
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 36 }{ 100 }$$ = 0.36
(ii) Probability of 3 heads (m) = 12
ProbabilityP(A)= $$\frac { m }{ n }$$ = $$\frac { 12 }{ 100 }$$ = 0.12
(iii) Probability of at least one head coming up (m) = 38 + 36 + 12 = 86
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 86 }{ 100 }$$ = 0.86
(iv) Probability of getting more heads than tails (m) = 36 + 12 = 48
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 48 }{ 100 }$$ = 0.48
(v) Getting more tails than heads (m) = 14 + 38 = 52
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 52 }{ 100 }$$ = 0.52

Question 4.
1500 families with 2 children were selected randomly and the following data were recorded:

If a family is chosen at random, compute the probability that it has:
(i) No girl
(ii) 1 girl
(iii) 2 girls
(iv) at most one girl
(v) more girls than boys
Solution:
Total number of families (n) = 1500

(i) Probability of a family having no girls (m) = 211
∴Probability P(A)= $$\frac { m }{ n }$$ = $$\frac { 211 }{ 1500 }$$ = 0.1406
(ii) Probability of a family having one girl (in) = 814
∴Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 814 }{ 1500 }$$ = 0.5426
(iii) Probability of a family having 2 girls (m) = 475
∴Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 475 }{ 1500 }$$ = 0.3166
(iv) Probability of a family having at the most one girls
∴m = 814 + 211 = 1025
∴Probability P(A) =$$\frac { m }{ n }$$ = $$\frac { 1025 }{ 1500 }$$ = 0.6833
(v) Probability of a family having more girls than boys (m) = 475
∴Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 475 }{ 1500 }$$ = 0.3166

Question 5.
In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that on a ball played:
(i) he hits boundary
(ii) he does not hit a boundary.
Solution:
Total balls played (n) 30
No. of boundaries = 6
(i) When the batsman hits the boundary = 6
∴m = 6
∴Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 6 }{ 30 }$$ = $$\frac { 1 }{ 5 }$$ = 0.2
(ii) When the batsman does not hit the boundary (m) = 30 – 6 = 24
∴Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 24 }{ 30 }$$ = $$\frac { 4 }{ 5 }$$ = 0.8

Question 6.
The percentage of marks obtained by a student in monthly unit tests are given below:

Find the probability that the student gets:
(i) more than 70% marks
(ii) less than 70% marks
(iii) a distinction.
Solution:
Percentage of marks obtain in

(i) Probability of getting more than 70% marks (m) = In unit test II, III, V = 3
Total unit test (n) = 5
∴Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 3 }{ 5 }$$ = 0.6
(ii) Getting less then 70% marks = units test I and IV
∴m = 2
Total unit test (n) = 5
∴Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 2 }{ 5 }$$ = 0.4
(iii) Getting a distinction = In test V (76 of marks)
∴m = 1
Total unit test (n) = 5
∴Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 1 }{ 5 }$$ = 0.2

Question 7.
To know the opinion of the students about Mathematics, 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 Mathematics
(ii) does not like it.
Solution:
Total number of students (n) = 200

(i) Probability of students who like mathematics (m) = 135
∴Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 135 }{ 200 }$$ = 0.675
(ii) Probability of students who dislike mathematics (m) = 65
∴Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 65 }{ 200 }$$ = 0.325

Question 8.
The blood groups of 30 students of class IX 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,
A student is selected at random from the class from blood donation. Find the probability that the blood group of the student chosen is:
(i) A (ii) B (iii) AB (iv) O
Solution:
Total number of students of IX class = 30
No. of students of different blood groups
A AB B O
9 3 6 12

Question 9.
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:
Number of total bags (n) = 11
No. of bags having weight more than 5 kg (m) = 7
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 7 }{ 11 }$$

Question 10.
Following table shows the birth month of 40 students of class IX.

Find the probability that a student was born in August.
Solution:
Total number of students (n) = 40

Number of students who born in Aug. (m) = 6
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 6 }{ 40 }$$ = $$\frac { 3 }{ 20 }$$

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.

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 (n) = 30

Probability of cone, of S02 of the interval 0.12-0.16 (m) = 2
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 2 }{ 30 }$$ = $$\frac { 1 }{ 15 }$$

Question 12.
A company selected 2400 families at random and survey them to determine a relationship between income level and the number of vehicles in a home. The information gathered is listed in the table below:

If a family is chosen, find the probability that the family is:
(i)earning Rs 10000-13000 per month and owning exactly 2 vehicles.
(ii)earning Rs 16000 or more per month and owning exactly I 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 vehicle.
(v)owning not more than 1 vehicle.
(vi)owning at least one vehicle.
Solution:
Total number of families (n) = 2400

(i) Number of families earning income Rs 10000-13000 and owning exactly 2 vehicles (m) = 29
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 29 }{ 2400 }$$
(ii) Number of families earning income Rs 16000 or more having one vehicle (m) = 579
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 579 }{ 2400 }$$
(iii) Number of families earning income less than Rs 7000 having no own vehicle (m) = 10
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 10 }{ 2400 }$$ = $$\frac { 1 }{ 240 }$$
(iv) Number of families having X13000 to X16000 having more than two vehicles (m) = 25
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 25 }{ 2400 }$$ = $$\frac { 1 }{ 96 }$$
(v) Number of families owning not more than one vehicle (m)
= 10 + 1 + 2 + 1 + 160 + 305 + 533 + 469 + 579 = 2062
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 2062 }{ 2400 }$$ = $$\frac { 1031 }{ 1200 }$$
(vi) Number of families owning at least one vechile (m) = 2048 + 192 + 110 = 2356
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 2356 }{ 2400 }$$ = $$\frac { 589 }{ 600 }$$

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

A bulb is selected at random. Find the probability that the life time of the selected bulb is: (i) less than 400 (ii) between 300 to 800 hours (iii) at least 700 hours.
Solution:
Total number of neon lamps (n) = 400

A bulb is chosen:
(i)No. of bulbs having life time less than 400 hours (m) = 14
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 14 }{ 400 }$$ = $$\frac { 7 }{ 200 }$$
(ii)No. of bulbs having life time between 300 to 800 hours (m) = 14 + 56 + 60 + 86 + 74 = 290
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 290 }{ 400 }$$ = $$\frac { 29 }{ 40 }$$
(iii)No. of bulbs having life time at least 700 hours (m) = 74 + 62 + 48 = 184
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 184 }{ 400 }$$ = $$\frac { 23 }{ 50 }$$

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

A worker is selected at random. Find the probability that his wages are:
(i) less than Rs 150
(ii) at least Rs 210
(iii) more than or equal to 150 but less than Rs 210.
Solution:
Number of total workers (n) = 30

A worker is selected.
(i)No. of workers having less than Rs 150 (m) = 3 + 4 = 7
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 7 }{ 30 }$$
(ii)No. of workers having at least Rs 210 (m) = 4 + 3 = 7
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 7 }{ 30 }$$
(iii)No. of workers having more than or equal to Rs 150 but less than Rs 210 = 5 + 6 + 5 = 16
Probability P(A) = $$\frac { m }{ n }$$ = $$\frac { 16 }{ 30 }$$ = $$\frac { 8 }{ 15 }$$

Hope given RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 are helpful to complete your math homework.

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