RD Sharma Class 9 Solutions Chapter 25 - Probability (Ex 25.1) Exercise 25.1

RD Sharma Solutions

RD Sharma Class 9 Solutions Chapter 25 - Probability (Ex 25.1) Exercise 25.1

RD Sharma Class 9 Solutions Chapter 25 - Probability (Ex 25.1) Exercise 25.1 - Free PDF

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RD Sharma Class 9 Solutions Chapter 25 - Probability (Ex 25.1)

Chapter 25, Probability of Class 9

Probability refers to the chance of occurrence of an event, that is, how likely it is for a particular event to occur. It always ranges between 0 to 1 where 0 means that the event is impossible to occur and 1 means the event will surely occur.
Any process that can be repeated indefinitely and have a set of possible outcomes is known as an experiment.
Each individual process of an experiment is known as a trial, that is, a single event that is performed to determine the outcome.
The probability of an event to happen is calculated as the number of favourable outcomes divided by the total number of outcomes.
The favourable event is the one whose outcome is the same as the expected outcome.
The sum of the probability of occurrence of an unfavourable event and the probability of occurrence of a favourable event in an experiment is always equal to 1.

FAQs on RD Sharma Class 9 Solutions Chapter 25 - Probability (Ex 25.1) Exercise 25.1

1. What will be the solution for question 1 RD Sharma Class 9 Chapter 25, Probability, exercise 25.1?

In question 1 of RD Sharma Class 9 Chapter 25, Probability, exercise 25.1 it is given that the coin is tossed 1000 times, that is, the total number of trials is 1000. Now, we will have to determine the probability of getting ahead which will be equal to the number of times the head appears divided by the total number of trials, therefore, this will be 455/1000 which is equal to 0.455. Similarly, the probability of getting a tail will be 545 (the number of times the tail occurs) divided by 1000 (number of trials). Thus, the probability of occurrence of the tail will be 0.545. Now, when we add the total probability, that is, 0.455 and 0.545, it came to one. Thus, we can say that these two are the only opposite outcomes.

2. How can I solve question 9 of Class 9 RD Sharma Chapter 25, Probability, exercise 25.1?

According to question 9 of RD Sharma Class 9 Chapter 25, Probability, exercise 25.1, it is given that eleven bags of wheat flour are marked 5 kg each but the actual weight of these bags is 4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07 and 5.00kgs, and now we need to find the probability of choosing a bag that contains more than 5kg flour. Therefore, the probability of choosing a bag that contains more than 5kg flour will be equal to the number of bags having more than 5kg, that is, 7 divided by the total number of bags, that is, 11. Therefore, 7/11 will be the answer.

3. What will be the solution to question 10 of Class 9 RD Sharma Chapter 25, Probability, exercise 25.1?

Question 10 of Class 9 RD Sharma Chapter 25, Probability, exercise 25.1is based on the concept similar to question 10 of the same exercise. In this question, the students are provided with data of the number of students of a class born in a particular month. Now, we need to find the probability of a student being born in august if we choose any student from the class randomly. Therefore, the probability of the chosen student to be born in August will be the number of students of the class born in august divided by the total strength of the class. Therefore, the answer will be 6/40 that is simplified to 3/20.

4. How can I solve question 11 of Class 9 RD Sharma Chapter 25, Probability, exercise 25.1?

Question 11 of Class 9 RD Sharma Chapter 25, Probability, exercise 25.1 provides the students with a frequency distribution table regarding the concentration of sulfur dioxide in parts per million exposed in the air of a city for 30 days and we are required to calculate the probability of concentration of sulfur dioxide in the range 0.12 - 0.16. Therefore, the probability of concentration of sulfur dioxide in the internal 0.12 - 0.16 will be equal to the number of days in which the concentration of air in the city has reached that range divided by the total number of days. Therefore, the answer to this question will be 2/30, that is, 0.06.

5. How can question 2 of Class 9 RD Sharma Chapter 25, Probability, exercise 25.1 be solved?

In question 2 of Class 9 RD Sharma Chapter 25, Probability, exercise 25.1, the students need to calculate the probability of getting two heads, one tail, and no head. The total number of trials in the question is 500, therefore the probability of getting two heads will be 95/500, that is, 0.19. The probability of getting one tail will be 290/500, that is, 0.58 and the probability of getting no head will be 115/500, which is equal to 0.23.