RS Aggarwal Solutions Class 9 Maths Chapter 19: RS Aggarwal Solutions for Class 9 Maths Chapter 19 - Probability provide a detailed guide to understanding the fundamental concepts of probability. This chapter introduces students to the concept of probability, which is essential in various real-life situations and mathematical problems.
With these solutions, students can learn about the basic principles of probability, including sample space, events, and probability of events. The step-by-step solutions provide in this chapter help students develop a clear understanding of different probability scenarios and how to calculate probabilities using various methods. Practicing with these solutions enables students to strengthen their problem-solving skills and prepare effectively for exams. Overall, RS Aggarwal Solutions for Class 9 Maths Chapter 19 - Probability is as an invaluable resource for students looking to master the concept of probability.RS Aggarwal Solutions Class 9 Maths Chapter 19 PDF
Question 1.
Solution:
Number of trials = 500 times Let E be the no. of events in each case, then ∴No. of heads (E1) = 285 times and no. of tails (E2) = 215 times ∴ Probability in each case will be ∴(i)P(E1) =285/500 = 0.57 (ii) P(E2) = 43/100 = 0.43Question 2.
Solution:
No. of trials = 400 Let E be the no. of events in each case, then No. of 2 heads (E1) = 112 No. of one head (E2) = 160 times and no. of O. head (E3) = 128 times ∴ Probability in each case will be: ∴ (i)P(E1) = 112/400 = 0.28 (ii)P(E2) = 160/400= 0.40 (iii) P(E3) = 128/400= 0.32 Ans.Question 3.
Solution:
Number of total trials = 200 Let E be the no. of events in each case, then No. of three heads (E1) = 39 times No. of two heads (E2) = 58 times No. of one head (E3) = 67 times and no. of no head (E4) = 36 times ∴ Probability in each case will be . (i) P(E1) =39/200 = 0.195 (ii) P(E3) =58/200 = 0.335 (iii) P(E4) = 67/200= 0.18 (iv) P(E2) = 36/200= 0.29Question 5.
Solution:
No. of ladies on whom survey was made = 200. Let E be the no. of events in each case. No. of ladies who like coffee (E1) = 142 No. of ladies who like coffee (E2) = 58 Probability of (1) P(E1) = 142/200 = 0.71 (ii) P(E2) =58/200 = 0.29 Ans.Question 6.
Solution:
Total number of tests = 6 No. of test in which the students get more than 60% mark = 2 Probability will he P(E) = 1/3Question 7.
Solution:
No. of vehicles of various types = 240 No. of vehicles of two wheelers = 64. Probability will be P(E) = 7/20= 0.35Question 11.
Solution:
Total number of patients of various age group getting medical treatment = 360 Let E be the number of events, then (i) No. of patient which are 30 years or more but less than 40 years = 60. P(E) = 60/360 (ii) 50 years or more but less than 70 years = 50 + 30 = 80 P(E) = 80/360 (iii) Less than 10 years = zero P(E) = 0/360= 0 (iv) 10 years or more 90 + 50 + 60 + 80 + 50 + 30 = 360CBSE Class 9 Maths Syllabus | CBSE Class 9 Science Syllabus |
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