## RD Sharma Class 9 solutions Chapter 25 Probability

### RD Sharma Class 9 Solution Chapter 25 Probability Ex 25.1

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:

Two heads : 95 times

One tail : 290 times

No head: 115 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:

(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:

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:

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) = \(\frac { m }{ n } \) = \(\frac { 16 }{ 30 } \) = \(\frac { 8 }{ 15 } \)

### Probability Class 9 RD Sharma Solutions VSAQS

Question 1.

Define a trial.

Solution:

When we perform an experiment, it is called a trial of the experiment.

Question 2.

Define an elementary event.

Solution:

An outcome of a trial of an experiment is called an elementary event.

Question 3.

Define an event.

Solution:

An event association to a random experiment is said to occur in a trial.

Question 4.

Define probability of an event.

Solution:

In n trials of a random experiment if an event A happens m times, then probability of happening

of A is given by P(A) = \(\frac { m }{ n } \)

Question 5.

A bag contains 4 white balls and some red balls. If the probability of drawing a white ball from the bag is \(\frac { 2 }{ 5 } \), find the number of red balls in the bag

Solution:

No. of white balls = 4

Let number of red balls = x

Then total number of balls (n) = 4 white + x red = (4 + x) balls

Question 6.

A die is thrown 100 times. If the probability of getting an even number is \(\frac { 2 }{ 5 } \). How many times an odd number is obtained?

Solution:

Total number of a die is thrown = 100

Let an even number comes x times, then probability of an even number = \(\frac { x }{ 100 } \)

Question 7.

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes

Find the probability of getting at most two heads.

Solution:

Total number of three coins are tossed (n) = 200

Getting at the most 2 heads (m) = 72 + 77 + 28 = 177

Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 177 }{ 200 } \)

Question 8.

In the Q. No. 7, what is the probability of getting at least two heads?

Solution:

Total number of possible events = 200

No. of events getting at the least = 2 heads (m) = 23 + 72 = 95

Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 95 }{ 200 } \) = \(\frac { 19 }{ 40 } \)

### RD Sharma Class 9 PDF Chapter 25 Probability MCQS

Mark the correct alternative in each of the following:

Question 1.

The probability of an impossible event is

(a) 1

(b) 0

(c) less than 0

(d) greater than 1

Solution:

The probability of an impossible event is 0 (b)

Question 2.

The probability on a certain event is

(a) 0

(b) 1

(c) greater than 1

(d) less than 1

Solution:

The probability of a certain event is 1 (b)

Question 3.

The probability of an event of a trial is

(a) 1

(b) 0

(c) less than 1

(d) more than 1

Solution:

The probability of an even of a trial is less than 1 (c)

Question 4.

Which of the following cannot be the probability of an event?

(a) \(\frac { 1 }{ 3 } \)

(b) \(\frac { 3 }{ 5 } \)

(c) \(\frac { 5 }{ 3 } \)

(d) 1

Solution:

The probability of an event is less than 1

\(\frac { 5 }{ 3 } \) i.e .\(1\frac { 2 }{ 3 } \) is not the probability

Question 5.

Two coins are tossed simultaneously. The probability of getting atmost one head is

(a) \(\frac { 1 }{ 4 } \)

(b) \(\frac { 3 }{ 4 } \)

(c) \(\frac { 1 }{ 2 } \)

(d) \(\frac { 1 }{ 4 } \)

Solution:

Total number of possible events (n) = 2 + 2 = 4

Number of events coming at the most 1 head (m) 2 times + 1 times = 3

Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 3 }{ 4 } \) (b)

Question 6.

A coin is tossed 1000 times, if the probability of getting a tail is 3/8, how many times head is obtained?

(a) 525

(b) 375

(c) 625

(d) 725

Solution:

No. of times a coin is tossed (n) = 1000

Probability of getting a tail = \(\frac { 3 }{ 8 } \)

Let No. of tail come = x

Probability P(A) = \(\frac { m }{ n } =\frac { x }{ 1000 } \)

\(\frac { x }{ 1000 } \) = \(\frac { 3 }{ 8 } \)

=> \(\frac { x }{ 1000 } =\frac { 3 }{ 8 } \) => \(\frac { 3X1000 }{ 8 } =3X125\)

=> x = 375

∴ No. of heads are obtained = 1000 – 375 = 625 (c)

Question 7.

A dice is rolled 600 times and the occurrence of the outcomes 1, 2, 3, 4, 5 and 6 are given below:

The probability of getting a prime number is

(a)\(\frac { 1 }{ 3 } \)

(b)\(\frac { 2 }{ 3 } \)

(c)\(\frac { 49 }{ 60 } \)

(d)\(\frac { 39 }{ 125 } \)

Solution:

Total number of times a dice is rolled (n) = 600

Now total number of times getting a prime number i.e. 2, 3, 5, (m) = 30 + 120 + 50 = 200

Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 200 }{ 600 } \) = \(\frac { 1 }{ 3 } \) (a)

Question 8.

The percentage of attendance of different classes in a year in a school is given below:

What is the probability that the class attendance is more than 75%?

(a) \(\frac { 1 }{ 6 } \)

(b) \(\frac { 1 }{ 3 } \)

(c) \(\frac { 5 }{ 6 } \)

(d) \(\frac { 1 }{ 2 } \)

Solution:

Percentage of attendance of different classes

Total attendance more than 75% (m) VIII,VII and VI = 3 classes

and total number of classes (n) = 6

Probability P(A) = \(\frac { 3 }{ 6 } \) = \(\frac { 1 }{ 2 } \).

Question 9.

A bag contains 50 coins and each coin is marked from 51 to 100. One coin is picked at random.

The probability that the number on the coin is not a prime number, is

(a) \(\frac { 1 }{ 5 } \)

(b) \(\frac { 3 }{ 5 } \)

(c) \(\frac { 2 }{ 5 } \)

(d) \(\frac { 4 }{ 5 } \)

Solution:

Total number of coins (n) = 50

Prime numbers between 51 to 100 are 53, 59, 6, 67, 71, 73, 79, 83, 89, 97 = 10

Numbers which are not primes (m) = 50 – 10 = 40

Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 40 }{ 50 } \) = \(\frac { 4 }{ 5 } \)(d)

Question 10.

In a football match, Ronaldo makes 4 pals from 10 penalty kids. The probability of converting a penalty kick into a goal by Ronaldo,is

(a) \(\frac { 1 }{ 4 } \)

(b) \(\frac { 1 }{ 6 } \)

(c) \(\frac { 1 }{ 3 } \)

(d) \(\frac { 2 }{ 5 } \)

Solution:

No. of penalty kicks (n) = 10

No. of goal scored (m) = 4

Probability of converting a penally Into goals P(A) = \(\frac { 4 }{ 10 } \) = \(\frac { 2 }{ 5 } \)(d)

RD Sharma Class 9 solutions Chapter 25 Probability

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RD Sharma Class 9 Solutions