﻿ Class 9 NCERT Math Solution
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Exercise - 15.1

Question-1 :-  In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.

Solution :-
```  Let E be the event that she doesn't hit a boundry.
Total number of balls played = 30
Number of times the batswoman hits a boundary = 6
Number of times that the batswoman does not hit a boundary = 30 − 6 = 24
P(E) = 24/30 = 4/5
```

Question-2 :-  1500 families with 2 children were selected randomly, and the following data were recorded: Compute the probability of a family, chosen at random, having
(i) 2 girls
(ii) 1 girl
(iii) No girl
Also check whether the sum of these probabilities is 1.

Solution :-
```  Let E, F and G are be the event that 2 girls, 1 girl and no girl respectively.
Total number of families = 475 + 814 + 211 = 1500

(i) 2 girls
P(E) = 475/1500 = 19/60

(ii) 1 girl
P(F) = 814/1500 = 407/750

(iii) No girl
P(G) = 211/1500

Verification :
Sum of all Probabilities
= P(E) + P(F) + P(G)
= 19/60 + 407/750 + 211/1500
= (475 + 814 + 211)/1500
= 1500/1500
= 1
```

Question-3 :-  In a particular section of Class IX, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained: Find the probability that a student of the class was born in August.

Solution :-
```  Let E be the event that students born in the month of August.
Number of studen ts born in the month of August = 6
Total number of students = 40
P(E) = 6/40 = 3/20
```

Question-4 :-  Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes: If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.

Solution :-
```  Let E be the event that 2heads will comes up.
Number of times 2 heads come up = 72
Total number of times the coins were tossed = 200
P(E) = 72/200 = 9/25
```

Question-5 :-  An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below: Suppose a family is chosen. Find the probability that the family chosen is
(i) earning ₹ 10000 – 13000 per month and owning exactly 2 vehicles.
(ii) earning ₹ 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than ₹ 7000 per month and does not own any vehicle.
(iv) earning ₹ 13000 – 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.

Solution :-
```  Number of total families surveyed = 10 + 160 + 25 + 0 + 0 + 305 + 27 + 2 + 1 + 535 + 29 + 1 + 2 + 469 + 59 + 25 + 1 + 579 + 82 + 88 = 2400

(i) Let E be the event that earning ₹ 10000 – 13000 per month and owning exactly 2 vehicles.
Number of families earning ₹ 10000 – 13000 per month and owning exactly 2 vehicles = 29
P(E) = 29/2400

(ii) Let F be the event that earning ₹ 16000 or more per month and owning exactly 1 vehicle.
Number of families earning ₹ 16000 or more per month and owning exactly 1 vehicle = 579
P(E) = 579/2400

(iii) Let G be the event that earning less than ₹ 7000 per month and does not own any  vehicle.
Number of families earning less than ₹ 7000 per month and does not own any vehicle = 10
P(G) = 10/2400 = 1/240

(iv) Let H be the event that earning ₹ 13000 – 16000 per month and owning more than 2 vehicles.
Number of families earning ₹ 13000 – 16000 per month and owning more than 2 vehicles = 25
P(H) = 25/2400 = 1/96

(v) Let I be the event that owning not more than 1 vehicle.
Number of families owning not more than 1 vehicle = 10 + 160 + 0 + 305 + 1 + 535 + 2 + 469 + 1 + 579 = 2062
P(I) = 2062/2400 = 1031/1200
```

Question-6 :-  A teacher wanted to analyse the performance of two sections of students in a mathematics test of 100 marks. Looking at their performances, she found that a few students got under 20 marks and a few got 70 marks or above. So she decided to group them into intervals of varying sizes as follows: 0 - 20, 20 - 30, . . ., 60 - 70, 70 - 100. Then she formed the following table: (i) Find the probability that a student obtained less than 20% in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.

Solution :-
```  Total number of students = 90

(i) Let E be the event that a student obtained less than 20% in the mathematics test.
Number of students getting less than 20 % marks in the test = 7
P(E) = 7/90

(ii) Let F be the event that a student obtained marks 60 or above.
Number of students obtaining marks 60 or above = 15 + 8 = 23
P(F) = 23/90
```

Question-7 :- To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table: Find the probability that a student chosen at random
(i) likes statistics,
(ii) does not like it

Solution :-
```  Total number of students = 135 + 65 = 200

(i) Let E be the event that likes statistics.
Number of students liking statistics = 135
P(E) = 135/200 = 27/40

(ii) Let F be the event that doesn't like statistics.
Number of students who do not like statistics = 65
P(F) = 65/200 = 13/40
```

Question-8 :-  The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
5  3  10  20  25  11  13  7  12  31 19  10  12  17  18  11  32  17  16  2 7  9  7  8  3  5  12  15  18  3 12  14  2  9  6  15  15  7  6  12
What is the empirical probability that an engineer lives:
(i) less than 7 km from her place of work?
(ii) more than or equal to 7 km from her place of work?
(iii) within 1/2 km from her place of work?

Solution :-
```  Total number of engineers = 40
(i) Let E be the event that less than 7 km from her place of work.
Number of engineers living less than 7 km from their place of work = 9
P(E) = 9/40

(ii) Let F be the event that more than or equal to 7 km from her place of work.
Number of engineers living more than or equal to 7 km from their place of work = 40 − 9 = 31
P(F) = 31/40

(iii) Let G be the event that within 1/2 km from her place of work.
Number of engineers living within 1/2 km from her place of work = 0
P(G) = 0/40 = 0
```

Question-9 :-  Activity : Note the frequency of two-wheelers, three-wheelers and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.

Solution :-
```  Do Activity Yourself.
```

Question-10 :-  Activity : Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.

Solution :-
```  Do Activity Yourself.
```

Question-11 :-  Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg):
4.97  5.05  5.08  5.03  5.00  5.06  5.08  4.98  5.04  5.07  5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.

Solution :-
```  Let E be the event that any of these bags chosen at random contains more than 5 kg of flour.
Number of total bags = 11
Number of bags containing more than 5 kg of flour = 7
P(E) = 7/11
```

Question-12 :-  A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:
0.03  0.08  0.08  0.09  0.04  0.17  0.16  0.05  0.02  0.06  0.18  0.20  0.11  0.08  0.12  0.13  0.22  0.07  0.08  0.01  0.10  0.06  0.09  0.18  0.11  0.07  0.05  0.07  0.01  0.04
you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 - 0.16 on any of these days.

Solution :-
```  Let E be the event that the concentration of sulphur dioxide was in the interval of 0.12 − 0.16.
Total number of days = 30
Number days for which the concentration of sulphur dioxide was in the interval of 0.12 − 0.16 = 2
P(E) = 2/30 = 1/15
```

Question-13 :-  The blood groups of 30 students of Class VIII are recorded as follows:
A,  B,  O,  O,  AB,  O,  A,  O,  B,  A,  O,  B,  A,  O,  O,  A,  AB,  O,  A,  A,  O,  O,  AB,  B,  A,  O,  B,  A,  B,  O.
you were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.

Solution :-
```  Let E be the event that students having blood group AB.
Number of students having blood group AB = 3
Total number of students = 30
P(E) = 3/30 = 1/10
```
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