RBSE Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1

​Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 15 Probability Ex 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.
Answer:
Since in a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays, i.e. she missed the boundary 30 - 6 = 24 times out of 30 balls.
∴ The probability that a batswoman does not hit a boundary
= \(\frac{\text { Number of not hitting a boundary }}{\text { Total number of trials }(i . e . \text { balls) }}\)
= \(\frac{24}{30}\),i.e. \(\frac{4}{5}\).

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.
Answer:
Let E₀, E₁, and E₂ be the events of getting no girl, 1 girl and 2 girls,
(i) ∴ P(E₂) = Probability of a family having 2 girls
= \(\frac{\text { Number of families having } 2 \text { girls }}{\text { Total number of families }}\)
= \(\frac{475}{1500}=\frac{19}{60}\)

(ii) ∴ P(E₁) = Probability of a family having 1 girl
= \(\frac{\text { Number of families having } 1 \text { girl }}{\text { Total number of families }}\)
= \(\frac{814}{1500}=\frac{407}{750}\)

(iii) ∴ P(E₀) = Probability of a family having no girl
= \(\frac{\text { Number of families having no girl }}{\text { Total number of families }}\)
= \(\frac{211}{1500}\)
∴ Sum of probabilities = P(E₀) + P(E₁) + P(E₂)
= \(\frac{211}{1500}+\frac{407}{750}+\frac{19}{60}\)
= \(\frac{211+814+475}{1500}=\frac{1500}{1500}\) = 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 to represent the data :

Find the probability that a student of the class was born in August.
Answer:
Clearly from the bar graph, six students were born in the month of August out of 40 students of a particular section of class IX.
Probability that a student of the class was born in August
= \(\frac{\text { Number of students born in August }}{\text { Total number of students }}\)
= \(\frac{6}{40}=\frac{3}{20}\)

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.
Answer:
Since, three coins are tossed 200 times, so the total number of trials is 200.
Probability of getting 2 heads = \(\frac{\text { Number of outcomes having } 2 \text { heads }}{\text { Total number of trials }}\)
= \(\frac{72}{200}=\frac{9}{25}\)

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.
Answer:
The total number of families = 2400
(i) Number of families earning ₹ 10000 - 13000 per month and owning exactly 2 vehicles = 29.
P (Families earning ₹ 10000 - 13000 per month and owning exactly 2 vehicles)
= \(\frac{29}{2400}\)

(ii) Number of families earning ₹ 16000 or more per month and owning exactly 1 vehicle = 579.
∴ P (Families earning ₹ 16000 or more per month and owning exactly 1 vehicle)
= \(\frac{579}{2400}=\frac{193}{800}\)

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

(iv) Number of families earning ₹ 13000 - 16000 per month and owning more than 2 vehicles = 25
∴ P (Families earning ₹ 13000 - 16000 per month and owning more than two
= \(\frac{25}{2400}=\frac{1}{96}\)

(v) Number of families owning not more than 1 vehicle
= Families having no vehicle + Families having 1 vehicle
= (10 + 0 + 1 + 2 + 1) + (160 + 305 + 535 + 469 + 579)
= 14 + 2048 = 2062
P (Families Owning not more than 1 vehicle) = \(\frac{2062}{2400}=\frac{1031}{1200}\)

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:

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

(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.
Answer:
Total number of students in mathematics test is 90.
(i) Clearly, from the given table, the number of students who obtained less than 20% marks in the mathematics test = 7.
P (a student obtaining less than 20% marks) = \(\frac{7}{90}\)

(ii) Clearly, from the given table, number of students who obtained marks 60 or above = (students in 60-70) + (students 70 and above 70)
= 15 + 8 = 23
∴ P (a student obtaining marks 60 and above) = \(\frac{23}{90}\)

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.
Answer:
The total number of students = 200
(i) P (a student likes statistics) = \(\frac{\text { Number of students who like statistics }}{\text { Total number of students }}\)
= \( \frac{135}{200}=\frac{27}{40}\)

(ii) P (a student does not like statistics) = \(\frac{\text { Number of students who do not like statistics }}{\text { Total number of students }}\)
= \(\frac{65}{200}=\frac{13}{40}\)

Question 8.
The distances (in km) of 40 female 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?
Answer:
Total number of female engineers = 40
(i) Number of female engineers living at a distance less than 7 km from their place of work = 9
Probability that a female engineer lives at a distance less than 7 km from her place of work = \(\frac{9}{40}\)

(ii) Number of female engineers living more than or equal to 7 km away from her place of work = 30
Probability that a female engineer lives more than or equal to 7 km away from 31 her place of work = \(\frac{31}{40}\)

(iii) Since, there is no engineer living at a distance less than \(\frac{1}{2}\)km from her place of work.
∴ Probability that an engineer lives within\(\frac{1}{2}\)km from her place of work = \(\frac{0}{40}\) = 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.
Answer:
After observing in front of the school gate from time interval 6:30 to 7:30 am. The respective frequencies of different types of vehicles are :

Types of vehicles	Frequency
Two-wheelers	550
Three-wheelers	250
Four-wheelers	80

∴ Total number of vehicles = 550 + 250 + 80 = 880
Number of two-wheelers = 550
Hence, probability of observing two-wheelers
= \(\frac{\text { Number of two }-\text { wheelers }}{\text { Total number of vehicles }}=\frac{550}{880}=\frac{5}{8}\)

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.
Answer:
Suppose there are 40 students in a class.
The probability of selecting any of the student = \(\frac{40}{40}\) = 1
A three-digit number start from 100 to 999.
Total number of three-digit numbers = 999 - 99 = 900
Now, multiples of 3 in three-digit numbers = {102, 105, ..., 999}
Then, total number of multiples of 3 in three-digit numbers = \(\frac{900}{3}\) = 300
∴ The probability that the number written by her/him is divisible by 3 = \(\frac{300}{900}=\frac{1}{3}\).

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.
Answer:
Total number of wheat flour bags =11
Number of bags having more than 5 kg = 7
P (a bag contains more than 5 kg) = \(\frac{7}{11}\).

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:

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.
Answer:

Interval	Frequency
0.00 - 0.04	4
0.04 - 0.08	9
0.08 - 0.12	9
0.12 - 0.16	2
0.16 - 0.20	4
0.20 - 0.24	2
Total	30

Total number of days = 30

Concentration of sulphur dioxide in the interval 0.12 - 0.16 on any day = 2
Required probability = \(\frac{2}{30}=\frac{1}{15}\)

Question 13.
The blood groups of 30 students of class IX are recorded as follows :

A student is selected at random from the class from blood donation. Find the probability that the blood group of the student chosen is :
(i) A
(ii) B
(iii) AB
(iv) O
Answer:
Total number of students = 30
(i) Number of students having blood group A = 9
∴ Probability that a student of this class has a blood group A = \(\frac{9}{30}=\frac{3}{10}\) = 0.3

Similarly, (ii) P (a student having blood group B) = \(\frac{6}{30}=\frac{1}{5}\) = 0.2

(iii) P (a student having blood group AB) = \(\frac{3}{30}=\frac{1}{10}\) = 0.1

(iv) P (a student having blood group O) = \(\frac{12}{30}=\frac{4}{10}\) = 0.4.