Ex 15.1 NCERT Solutions Class 9 Chapter 15
The solutions to exercise 15.1 (from class 9th's maths NCERT book) are given below. The exercise has 13 questions which have been explained thoroughly with proper reasoning.
Answer: E = Event when she didn’t hit the boundary
She hit the boundary 6 times out of 30 balls, so she failed to hit the boundary 24 times out of 30 balls
P(E) = 24/30
P(E) = 4/5 or 0.80
(i) No of families having 2 girls = 475
Total families = 1500
P( 2 girls) = No of families having 2 girls/ Total families
P( 2 girls) = 475/1500
P( 2 girls) = 95/300 => 19/60
(ii) No of families having 1 girl = 814
Total families = 1500
P(1 girl) = No of families having 1 girl/ Total families
P(1 girl) = 814/1500 => 407/750
(iii) No of families having no girls = 211
Total families = 1500
P( no girl) = No of families having no girls/ Total families
P( no girl) = 211/1500
Yes, they all add up to give 1
(475/1500) + (814/1500) + (211/1500) = 1500/1500 => 1
Here is the question
Total students = 40
Number of students born in August = 6
P(that a student of the class is born in August)
= No of students born in August/Total students
= 6/40
= 3/20
Total times the coins are tossed = 200
P (2 heads) = 72/200 => 9/25
(i) P(earning Rs 10000 – 13000 per month and owning exactly 2 vehicles) = 29/2400
(ii) P(earning Rs 16000 or more per month and owning exactly 1 vehicle) = 579/2400
(iii) P(earning less than Rs 7000 per month and does not own any vehicle) = 10/2400 => 10/240
(iv) P(earning Rs 13000 – 16000 per month and owning more than 2 vehicles) = 25/2400 = 1/96
(v) P(owning not more than 1 vehicle) = 2062/2400 => 1031/1200
[how?; Add the numbers given in 0 & 1 columns; more than one i.e. 2 & above 2 columns will not be considered]
Since the marks of mathematics are awarded on 100, they all can be treated as a percent
(i) P (a student obtained less than 20% in the mathematics) = 7/90 [ 7 students out of 90 got less than 20]
(ii) P( a student obtained marks 60 or above) = 23/90 (In the above-60 category we will count both 15 and 8)
(i) P( likes statistics) = 135/200 = 27/40
(ii) P (does not like statistics) = 65/200 = 13/40
Here is Q2 from 14.2 Exercise
Originally the table should be
P(less than 7 km from her place of work) = 9/40
P(more than or equal to 7 km from her place of work) = 31/40
P(within 1/2 km from her place of work) = 0/40 = 0 (0.5 is not given in the data)
Let’s assume that the frequency of two-wheelers, three-wheelers, and four-wheelers every morning in front of a school is 100, 200 and 300
Total vehicles = 100+200+300 = 600
P (Two-wheeler) = 200/600 => ⅓
Suppose the numbers given by 20 students are
101, 112, 115, 212, 123, 127, 124, 111, 219, 615, 413, 516, 723, 221, 321,333, 234, 444, 323, 999
Total numbers = 20
Numbers divisible by 3 = 11/20
Bags containing more than 5 kg flour = 7
Total bags = 11
P( A bag has more than 5 kg flour) = 7/11
Here is the question: the data is given for 30 days
As per the question, we will make a frequency distribution table with class size 0.04
P( concentration of sulphur dioxide in the interval 0.12 - 0.16) = 2/30 = 1/15
Here is the data
Ungrouped frequency distribution table
P( probability that a student of this class has blood group AB) = 3/30 => 1/10
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