RD Sharma Solution Class 9 Maths Chapter 25 – Probability (Updated for 2024)

RD Sharma Solution Class 9 Maths Chapter 25 – Probability: RD Sharma Solution Class 9 Maths Chapter 25 is one of the student’s favorite chapters as it does not require those heavy calculations and you can easily understand its concept with proper guidance. But, it doesn’t make the chapter less important, this unit will ask the most scoring questions in your mathematics exams. To know more about the RD Sharma Solution Class 9 Maths Chapter 25, read the whole blog.

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RD Sharma Class 9 Solutions Chapter 25

Exercise-wise RD Sharma solutions class 9 Maths Chapter 25 – Probability

RD Sharma Solutions Class 9 Maths Chapter 25 Exercise 25.1

Access answers of RD Sharma Solutions Class 9 Maths Chapter 25

Exercise 25.1 Page No: 25.13

Question 1: A coin is tossed 1000 times in the following sequence:

Head: 455, Tail: 545

Compute the probability of each event.

Solution:

The coin is tossed 1000 times, which means, the number of trials is 1000.

Let us consider, the event of getting head and the event of getting tail to be E and F respectively.

Number of favorable outcomes = Number of trials in which the E happens = 455

So, Probability of E = (Number of favorable outcomes) / (Total number of trials)

P(E) = 455/1000 = 0.455

Similarly,

Number of favorable outcomes = Number of trials in which the F happens = 545

Probability of the event getting a tail, P(F) = 545/1000 = 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:

We know that, Probability of any event = (Number of favorable outcomes) / (Total number of trials)

Total number of trials = 95 + 290 + 115 = 500

Now,

P(Getting two heads) = 95/500 = 0.19

P(Getting one tail) = 290/500 = 0.58

P(Getting no head) = 115/500 = 0.23

Question 3: Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes:

Outcome	No head	One head	Two heads	Three heads
Frequency	14	38	36	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:

We know, Probability of an event = (Number of Favorable outcomes) / (Total Number of outcomes)

In this case, the total number of outcomes = 100.

(i) Probability of 2 Heads coming up = 36/100 = 0.36

(ii) Probability of 3 Heads coming up = 12/100 = 0.12

(iii) Probability of at least one head coming up = (38+36+12) / 100 = 86/100 = 0.86

(iv) Probability of getting more Heads than Tails = (36+12)/100 = 48/100 = 0.48

(v) Probability of getting more tails than heads = (14+38) / 100 = 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:

We know, Probability of an event = (Number of Favorable outcomes) / (Total Number of outcomes)

In this case, the total number of outcomes = 211 + 814 + 475 = 1500.

(Here, total numbers of outcomes = total number of families)

(i) Probability of having no girl = 211/1500 = 0.1406

(ii) Probability of having 1 girl = 814/1500 = 0.5426

(iii) Probability of having 2 girls = 475/1500 = 0.3166

(iv) Probability of having at the most one girl = (211+814) /1500 = 1025/1500 = 0.6833

(v) Probability of having more girls than boys = 475/1500 = 0.31

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 a boundary (ii) He does not hit a boundary.

Solution:

Total number of balls played by a player = 30

Number of times he hits a boundary = 6

Number of times he does not hit a boundary = 30 – 6 = 24

We know, Probability of an event = (Number of Favorable outcomes) / (Total Number of outcomes)

Now,

(i) Probability (he hits boundary) = (Number of times he hit a boundary) / (Total number of balls he played)

= 6/30 = 1/5

(ii) Probability that the batsman does not hit a boundary = 24/30 = 4/5

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

Find the probability that the student gets

(i) More than 70% of marks

(ii) Less than 70% of marks

(iii) A distinction

Solution:

Total number of unit tests taken = 5

We know, Probability of an event = (Number of Favorable outcomes) / (Total Number of outcomes)

(i) Number of times student got more than 70% = 3

Probability (Getting more than 70%) = 3/5 = 0.6

(ii) Number of times student got less than 70% = 2

Probability (Getting less than 70%) = 2/5 = 0.4

(iii) Number of times student got a distinction = 1

[Marks more than 75%]

Probability (Getting a distinction) = 1/5 = 0.2

Question 7: To know the opinion of the students about Mathematics, a survey of 200 students was conducted. The data was recorded in the following table:

Find the probability that students are chosen at random:

(i) Likes Mathematics (ii) Does not like it.

Solution:

Total number of students = 200

Students like mathematics = 135

Students dislike Mathematics = 65

We know, Probability of an event = (Number of Favorable outcomes) / (Total Number of outcomes)

(i) Probability (Student likes mathematics) = 135/200 = 0.675

(ii) Probability (Student does not like mathematics) = 65/200 = 0.325

Important Topics – RD Sharma Solution Class 9 Maths Chapter 25

Here is the list of topics that you should go through before starting a chapter.

• Understanding the concept of Probability
• Calculating probability in coins
• Calculating probability in different daily situations

This is the complete blog on RD Sharma Solutions Class 9 Maths Chapter 25. To know more about the CBSE Class 9 Maths, ask in the comments. 

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