R.D. Sharma Solutions Class 9th: Ch 25 Probability Exercise 25.1

Chapter 25 Probability R.D. Sharma Solutions for Class 9th Exercise 25.1

Exercise 25.1

1. A coin is tossed 1000 times with the following frequencies:
Head: 455, Tail: 545
Compute the probability for each event.

Solution

The coin is tossed 1000 times. So, the total number of trials is 1000.
Let A be the event of getting a head and B be the event of getting a tail.
The number of times A happens is 455 and the number of times B happens is 545.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Therefore, we have
P(A) = 455/1000 = 0.455
P(B) = 545/1000 = 0.545

2. Two coins are tossed simultaneously 500 times with the following frequencies of different outcomes:
Two heads: 95 times
One tail: 290 times
No head: 115 times
Find the probability of occurrence of each of these events.

Solution

The total number of trials is 500.
Let A be the event of getting two heads, B be the event of getting one tail and C be the event of getting no head.
The number of times A happens is 95, the number of times B happens is 290 and the number of times C happens is 115.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Therefore, we have
P(A) = 95/500 = 0.19
P(B) = 290/500 = 0.58
P(C) = 115/500 = 0.23

3. Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes:
Outcome No head One headTwo Head Three Head
Frequency 14 3836 12

If the three coins are simultaneously tossed again, compute the probability of:
(i) 2 heads coming up.
(ii) 3 heads coming up.
(iii) at least one head coming up.
(iv) getting more heads than tails.
(v) getting more tails than heads.

Solution

The total number of trials is 100.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) m/n

(i) Let A be the event of getting two heads.
The number of times A happens is 36.
Therefore,we have
P(A) = 36/100 = 0.36

(ii) Let B be the event of getting three heads
The number of times B happens is 12.
Therefore, we have.
P(B) = 12/100 = 0.12

(iii) Let C be the event of getting at least one head.
The number of times C happens is 38+36+12 = 86
Therefore, we have
P(C) = 86/100 = 0.86

(iv) Let D be the event of getting more heads than tails.
The number of times D happens is 36+12=48.
Therefore, we have
P(D) = 48/100 = 0.48

(v) Let E be the event of getting more tails than heads .
The number of times E happens is 14 + 38 = 52.
Therefore, We have
P(E) = 52/100 = 0.52.

4. 1500 families with 2 children were selected randomly and the following data were recorded:
Number of girls in a family 0 12
Number of families 211 814475

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

The total number of trials is 1500.
Remember the empirical or experimental or observed frequency approach to probability. If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) m/n

(i) Let A be the event of getting two heads.
The number of times A happens is 211.
Therefore,we have
P(A) = 211/1500 = 0.1406

(ii) Let B be the event of getting three heads
The number of times B happens is 814.
Therefor, we have.
P(B) = 814/1500 = 0.5426

(iii) Let C be the event of having two girls.
The number of times C happens is 475.
Therefore, we have
P(C) = 475/1500 = 0.3166

(iv) Let D be the event of having at most one girl.
The number of times D happens is 211+814 = 1025.
Therefore, we have
P(D) = 1025/1500 = 0.6833

(v) Let E be the event of having more girls than boys.
The number of times E happens is 475.
Therefore, we have
P(E) = 475/1500 = 0.3166

5. In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays.
(i) he hits boundary
(ii) he does not hit a boundary.

Solution

The total number of trials is 30.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted byand is given by
P(A) = m/n

(i) Let A be the event of hitting boundary.
The number of times A happens is 6.
Therefore, we have
P(A) = 6/30 = 0.2

(ii) Let B be the event of does not hitting boundary.
The number of times B happens is 30-6=24.
Therefore, we have
P(B) = 24/30 = 0.8

6. The percentage of marks obtained by a student in monthly unit tests are given below:
Unit Test I IIIIIIVV
Percentage of marks obtained 69 71736876
Find the probability that the student gets:
(i) More than 70% marks
(ii) Less than 70% marks
(iii) A distinction

Solution

The total number of trials is 5.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n

(i) Let A be the event of getting more than 70% marks.
The number of times A happens is 3.
Therefore, we have
P(A) = 3/5 = 0.6

(ii) Let B be the event of getting less than 70% marks.
The number of times B happens is 2.
Therefore, we have
P(B) = 2/5 = 0.4

(iii) Let C be the event of getting a distinction.
The number of times C happens is 1.
Therefore, we have
P(C) = 1/5 = 0.2

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:
Opinion LikeDislike
Number of student 13565
Find the probability that a student chosen at random (i) likes Mathematics (ii) does not like it.

Solution

The total number of trials is 200.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n

(i) Let A be the event of liking mathematics.
The number of times A happens is 135.
Therefore, we have
P(A) = 135/200 = 0.675

(ii) Let B be the event of disliking mathematics .
The number of times A happens is 65.
Therefore, we have
P(B) 65/200 = 0.325.

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, Fin the probability that the blood group of the student chosen is:
(i) A
(ii) B
(iii) AB
(iv) O

Solution

The total number of trials is 30.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n

(i) Let A1 be the event that the blood group of a chosen student is A.
The number of times A1 happens is 9.
Therefore, we have
P(A1) = 9/30 = 0.3

(ii) Let A2 be the event that the blood group of a chosen student is B.
The number of times A2 happens is 6.
Therefore, we have
P(A2) = 6/30 = 0.2

(iii) Let A3 be the event that the blood group of a chosen student is AB.
The number of times A3 happens is 3.
Therefore, we have
P(A3) = 3/30 = 0.1

(iv) Let A4 be the event that the blood group of a chosen student is O.
The number of times A4 happens is 12.
Therefore, we have

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

The total number of trials is 11.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Let A1 be the event that the actual weight of a chosen bag contain more than 5 Kg of flour.
The number of times A1 happens is 7.
Therefore, we have P(A1) = 7/11

10. Following table shows the birth month of 40 students of class IX.
Jan. FebMarchApril MayJuneJuly Aug.Sept.Oct. Nov.Dec.
3 422 512 534 44
Find the probability that a student was born in August.

Solution

The total number of trials is 40.
Remember the empirical or experimental or observed frequency approach to probability. If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Let A1 be the event that the birth month of a chosen student is august.
The number of times A1 happens is 5.
Therefore, we have
P(A1) = 5/40 = 1/8

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 0.00-0.040.04-0.080.08-0.12 0.12-0.160.16-0.200.20-0.24
No. of Days 489 243
Find the probability of concentration of sulphur dioxide in the interval 0.12-0.16 on any of these days.

Solution

The total number of trials is 30.
Remember the empirical or experimental or observed frequency approach to probability. If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Let A1 be the event that the concentration of sulphur dioxide in a day is 0.12-0.16 parts per million. The number of times A1 happens is 2.
Therefore, we have
P(A1) = 2/30 = 1/15

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 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 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 vehicle.
(v) owning not more than 1 vehicle
(vi) owning at least one vehicle.

Solution

The total number of trials is 2400.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n

(i) Let A1 be the event that a chosen family earns Rs 10000-13000 per month and owns exactly 2 vehicles.
The number of times A1 happens is 29.
Therefore, we have P(A1) = 29/2400

(ii) Let A2 be the event that a chosen family earns Rs 16000 or more per month and owns exactly 1 vehicle.
The number of times A2 happens is 579.
Therefore, we have P(A2) = 579/2400.

(iii) Let A3 be the event that a chosen family earns less than Rs 7000 per month and does not owns any vehicles.
The number of times A3 happens is 10.
Therefore, we have
P(A3) = 10/2400 = 1/240

(iv) Let A4 be the event that a chosen family earns Rs 13000-16000 per month and owns more than 2 vehicles.
The number of times A4 happens is 25.
Therefore, we have
P(A4) = 25/2400 = 1/96


(v) Let A5 be the event that a chosen family owns not more than 1 vehicle (may be 0 or 1). In this case the number of vehicles is independent of the income of the family.
The number of times A5 happens is
(10+0+1+2+1) + (160+305+535+469+579) = 2062.
Therefore, we have
P(A5) = 2062/2400 = 1031/1200

(vi) Let A6 be the event that a chosen family owns atleast 1 vehicle (may be 1 or 2 or above 2). In this case the number of vehicles is independent of the income of the family.
The number of times A6 happens is
(160+305+535+469+579) + (25+27+29+29+82) + (0+2+1+25+88) = 2356
Therefore, we have
P(A6) = 2356/2400 = 589/600

13. The following table gives the life time of 400 neon lamps:
Life time(in hours) 300-400400-500500-600 600-700700-800800-900900-1000
Number of lamps: 145660 86746248
A bulb is selected of random, Find the probability that the the life time of the selected bulb is:
(i) less than 400
(ii) between 300 to 800 hours
(iii) at least 700 hours.

Solution

The total number of trials is 400.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n

(i) Let A1 be the event that the lifetime of a chosen bulb is less than 400 hours.
The number of times A1 happens is 14.
Therefore, we have
P(A1) = 14/400 = 7/200

(ii) Let A2 be the event that the lifetime of a chosen bulb is in between 300 to 800 hours.
The number of times A2 happens is 14+56+60+86+74 = 290.
Therefore, we have
P(A2) = 290/400 = 29/40

(iii) Let A3 be the event that the lifetime of a chosen bulb is atleast 700 hours.
The number of times A3 happens is 74+62+48=184.
Therefore, we have
P(A3) = 184/400 = 23/50

14. Given below is the frequency distribution of wages (in Rs) of 30 workers in a certain factory:
Wages (in Rs) 110-130130-150150-170 170-190190-210210-230230-250
Number of workers 345 6543
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

The total number of trials is 30.

Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n

(i) Let A1 be the event that the wages of a worker is less than Rs 150. The number of times A1 happens is 3 + 4 = 7
Therefore, we have P(A1) = 7/30.

(ii) Let A2 be the event that the wages of a worker is atleast Rs 210.
The number of times A2 happens is 4+3 = 7 .
Therefore, we have P(A2) = 7/30.

(iii) Let A3 be the event that the wages of a worker is more than or equal to Rs 150 but less than Rs 210.
The number of times A3 happens is 5+6+5 = 16.
Therefore, we have
P(A3) = 16/30 = 8/15

GET OUR ANDROID APP

Get Offline Ncert Books, Ebooks and Videos Ask your doubts from our experts Get Ebooks for every chapter Play quiz while you study

Download our app for FREE

Study Rankers Android App Learn more

Study Rankers App Promo