In this chapter, we provide RD Sharma Solutions for Chapter 25 Probability Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 25 Probability Maths pdf, free RD Sharma Solutions for Chapter 25 Probability Maths book pdf download. Now you will get step by step solution to each question.

## 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 .(1frac { 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

**All Chapter RD Sharma Solutions For Class 9 Maths**

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