Introduction to RD Sharma Solution Class 12 - Chapter 31 Probability - Exercise 31.1

RD Sharma is a well-renowned book author in the field of mathematics for Indian students. The solutions provided by RD Sharma are a great resource for solving complex mathematical problems in an easy way.

In this article, we will discuss the RD Sharma solution for Class 12 - Chapter 31 Probability - Exercise 31.1. This exercise covers important topics such as probability, conditional probability, Bayes' theorem, and event space.

Chapter 31: Probability

The 31st chapter of RD Sharma Class 12 covers Probability, which is an important branch of mathematics. In this chapter, students learn about the different types of probability such as theoretical probability, empirical probability, and conditional probability. They also learn about the basics of probability, such as events, sample spaces and their various properties.

Exercise 31.1

Exercise 31.1 in RD Sharma for Class 12 is a set of 31 questions which cover different aspects of Probability. Here are some of the topics covered in Exercise 31.1:

• Theoretical Probability
• Sample Space
• Event Space
• Combinations

RD Sharma Solution Class 12 - Chapter 31 Probability - Exercise 31.1

RD Sharma Solution Class 12 - Chapter 31 Probability - Exercise 31.1 provides detailed solutions to all the questions present in the exercise. The solutions are presented in a step-by-step manner, making it easy for students to follow and understand. The RD Sharma solution also includes diagrams and examples wherever needed, making complex topics easy to understand.

To give you an idea of what the RD Sharma solution looks like, here is an example of a solution to a question in Exercise 31.1:

Question

A bag contains 8 red and 5 black balls. If two balls are drawn at random from the bag, without replacement, find the probability that they are both red.

Solution

Total number of balls in the bag = 8 + 5 = 13

When two balls are drawn at random without replacement, the sample space will be:

S = {(1st ball, 2nd ball)}

Total number of possible outcomes = 13C2 = (13 x 12) / (2 x 1) = 78

Number of ways of selecting two red balls = 8C2 = (8 x 7) / (2 x 1) = 28

Therefore, the probability of selecting two red balls is:

P(Two red balls) = Number of ways of selecting two red balls / Total number of possible outcomes = 28 / 78 = 14 / 39

Markdown code for the example solution

#### Question

A bag contains 8 red and 5 black balls. If two balls are drawn at random from the bag, without replacement, find the probability that they are both red.

#### Solution

Total number of balls in the bag = 8 + 5 = 13

When two balls are drawn at random without replacement, the sample space will be:

`S = {(1st ball, 2nd ball)}`

Total number of possible outcomes = `13C2 = (13 x 12) / (2 x 1) = 78`

Number of ways of selecting two red balls = `8C2 = (8 x 7) / (2 x 1) = 28`

Therefore, the probability of selecting two red balls is:

`P(Two red balls) = Number of ways of selecting two red balls / Total number of possible outcomes
                = 28 / 78
                = 14 / 39`

As can be seen from the above example, the RD Sharma solution is presented in a clear and concise way, making it easy for students to follow and understand.

In conclusion, the RD Sharma Solution for Class 12 - Chapter 31 Probability - Exercise 31.1 is a great resource for students who want to improve their understanding of Probability. By providing detailed solutions to all the questions in the exercise, students can improve their problem-solving skills and gain a deeper understanding of the concepts covered in the chapter.