Class 9: Maths Chapter 15 solutions. Complete Class 9 Maths Chapter 15 Notes.
RS Aggarwal Solutions for Class 9 Maths Chapter 15–Probability
RS Aggarwal 9th Maths Chapter 15, Class 9 Maths Chapter 15 solutions
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) = 285500 = 57100 = 0.57
(ii) P(E2) = 215500 = 43100 = 0.43
Question 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) = 112400 = 28100 = 0.28
(ii)P(E2) = 160400 = 40100= 0.40
(iii) P(E3) = 128400 = 32100 = 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) = 39200 = 0.195
(ii) P(E3) = 67200 = 0.335
(iii) P(E4) = 36200 = 18100 = 0.18
(iv) P(E2) = 58200 = 29100 = 0.29
Question 4.
Solution:
Solution No. of trials = 300 times
Let E be the no. of events in each case, then
No. of outcome of 1(E1) = 60
No. of outcome of 2(E2) = 72
No. of outcome of 3(E3) = 54
No. of outcome of 4(E4) 42
No. of outcome of 5(E5) = 39
No. of outcome of 6(E6) = 33
The probability of
(i) P(E3) = 54300 = 18100 = 0.18
(ii) P(E6) = 33100 = 11100= 0.11
(iii) P(E5) = 39300 = 13100 = 0.13
(iv) P(E1) = 60300 = 20100= 0.20 Ans.
Question 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) = 142200 = 71100 = 0.71
(ii) P(E2) = 58200 = 29100 = 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) = 26 = 13Ans.
Question 7.
Solution:
No. of vehicles of various types = 240
No. of vehicles of two wheelers = 64.
Probability will be P(E) = 84240 = 720 = 0.35 Ans.
Question 8.
Solution:
No. of phone numbers are one page = 200
Let E be the number of events in each case,
Then (i) P(E5) = 24200 = 12100 = 0.12
(ii) P(E8) = 16200 = 8100 = 0.08 Ans.
Question 9.
Solution:
No. of students whose blood group is checked = 40
Let E be the no. of events in each case,
Then (i) P(E0) = 1440 = 720 = 0.35
(ii) P(EAB) = 640 = 320 = 0.15 Ans.
Question 10.
Solution:
No. of total students = 30.
Let E be the number of elements, this probability will be of interval 21 – 30
P(E) = 630 = 15 = 0.2 Ans.
Question 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) = 60360 = 16
(ii) 50 years or more but less than 70 years = 50 + 30 = 80
P(E) = 80360 = 29
(iii) Less than 10 years = zero
P(E) = 0360 = 0
(iv) 10 years or more 90 + 50 + 60 + 80 + 50 + 30 = 360
RS Aggarwal Solutions for Class 9 Maths Chapter 15: Download PDF
RS Aggarwal Solutions for Class 9 Maths Chapter 15–Probability
Download PDF: RS Aggarwal Solutions for Class 9 Maths Chapter 15–Probability PDF
Chapterwise RS Aggarwal Solutions for Class 9 Maths :
- Chapter 1–Real Numbers
- Chapter 2–Polynomials
- Chapter 3–Introduction to Euclid’s Geometry
- Chapter 4–Lines and Triangles
- Chapter 5–Congruence of Triangles and Inequalities in a Triangle
- Chapter 6–Coordinate Geometry
- Chapter 7–Areas
- Chapter 8–Linear Equations in Two Variables
- Chapter 9–Quadrilaterals and Parallelograms
- Chapter 10–Area
- Chapter 11–Circle
- Chapter 12–Geometrical Constructions
- Chapter 13–Volume and Surface Area
- Chapter 14–Statistics
- Chapter 15–Probability
About RS Aggarwal Class 9 Book
Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal's book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.
He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.
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