# JAC Class 9 Maths Solutions Chapter 15 Probability Ex 15.1

Jharkhand Board JAC Class 9 Maths Solutions Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 15 Probability Ex 15.1

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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:
Total number of balls = 30.
Number of boundaries = 6.
Number of times she didn’t hit boundary = 30 – 6 = 24
Probability she did not hit a boundary
= $$\frac{24}{30}$$ = $$\frac{4}{5}$$

Question 2.
1500 families with 2 children were selected randomly, and the following data were recorded :

 Number of girls in a family 2 1 0 Number of families 475 814 211

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:
Total number of families = 1500
(i) Number of families having 2 girls = 475

Probability = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}$$
= $$\frac{475}{1500}$$ = $$\frac{19}{60}$$

(ii) Number of families having 1 girl = 814
Probability = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}$$
= $$\frac{814}{1500}$$ = $$\frac{407}{750}$$

(iii) Number of families having no girl = 211
Probability = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}$$
= $$\frac{211}{1500}$$

Sum of the probability
= $$\frac{19}{60}+\frac{407}{750}+\frac{211}{1500}$$
= $$\frac{475+814+211}{1500}$$
= $$\frac{1500}{1500}$$ = 1

Question 3.
In a particular section of class IX, 40 students were asked about the month of their birth and the following graph was prepared for the data so obtained :
Img 1
Solution :
Total number of students = 40
Number of students born in August = 6
Required probability = $$\frac{6}{40}$$ = $$\frac{3}{20}$$

Question 4.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes :

 Outcome 3 heads 2 heads 1 head No head Frequency 23 72 77 28

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Solution :
Number of times 2 heads coming up = 72
Total number of times the coins were tossed = 200
Required probability = $$\frac{72}{200}$$ = $$\frac{9}{25}$$

Question 5.
An organisation selected 2400 families at random and surveyed them to determine their income level and the number of vehicles in a family. The information gathered is listed in the table below :
Img 2
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:
The total number of families = 2400
(i) Number of families earning ₹ 10000 – 13000 per month and owning exactly 2 vehicles = 29
Required probability = $$\frac{29}{2400}$$

(ii) Number of families earning ₹ 16000 or more per month and owning exactly 1 vehicle = 579
Required probability = $$\frac{579}{2400}$$

(iii) Number of families earning less than ₹ 7000 per month and does not own any vehicle = 10
Required probability = $$\frac{10}{2400}$$ = $$\frac{1}{240}$$

(iv) Number of families earning ₹ 13000 – 16000 per month and owning more than 2 vehicles = 25
Required probability = $$\frac{25}{2400}$$ = $$\frac{1}{96}$$

(v) Number of families owning not more than 1 vehicle
10 + 0 + 1 + 2 + 1 + 160 + 305 + 535 + 469 + 579 = 2062
Required probability = $$\frac{2062}{2400}$$ = $$\frac{1031}{1200}$$

Question 6.
Following table shows the performance of two sections of students in Mathematics test of 100 marks.
(i) Find the probability that a student obtained less than 20% marks in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.

 Marks Number of students 0 – 20 7 20 – 30 10 30 – 40 10 40 – 50 20 50 – 60 20 60 – 70 15 70 – above 8 Total 90

Solution:
Total number of students = 90
(i) Number of students obtained less than 20% in mathematics test = 7
Required probability = $$\frac{7}{90}$$

(ii) Number of students obtained marks 60 or above = 15 + 8 = 23
Required probability = $$\frac{23}{90}$$

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 :

 Opinion Number of students like 135 dislike 65

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) Number of students who likes statistics = 135
Required probability = $$\frac{135}{200}$$ = $$\frac{27}{40}$$

(ii) Number of students who dislikes statistics = 65
Required probability = $$\frac{65}{200}$$ = $$\frac{13}{40}$$

Question 8.
The distance (in km) of 40 engineers from their residence to their place of work were found as follows: 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 $$\frac{1}{2}$$ km from her place of work?
Solution:
The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
5, 3, 10, 2, 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.
Total number of engineers = 40.
(i) Number of engineers living less than 7 km from her place of work = 9.
Required probability = $$\frac{9}{40}$$.

(ii) Number of engineers living more than or equal to 7 km from her place of work = 40 – 9 = 31.
Required probability = $$\frac{31}{40}$$
(iii) Number of engineers living within $$\frac{1}{2}$$ km from her place of work = 0.
Required probability = $$\frac{0}{40}$$ = 0.

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Question 9.
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:
Total number of bags = 11
Numbers of bags containing more than 5 kg of flour = 7.
Required probability = $$\frac{7}{11}$$

Question 10.
A study was conducted to find out 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.
The data obtained for 30 days is as follows:
Img 3
Solution :
Total number of days data recorded = 30.
Numbers of days in which sulphur dioxide is in the interval 0.12 – 0.16 = 2.
Required probability = $$\frac{2}{30}$$ = $$\frac{1}{15}$$ .

Question 11.
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.
Represent this data in the form of a frequency distribution table. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
Solution :
Total numbers of students = 30.
Numbers of students having blood group AB = 3.
Required probability = $$\frac{3}{30}$$ = $$\frac{1}{10}$$.