The Mathematics of Maybe Introduction to Probability Class 9 Extra Questions Maths Chapter 7

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Class 9 The Mathematics of Maybe Introduction to Probability Extra Questions

Extra Questions of The Mathematics of Maybe Introduction to Probability

Class 9 Maths Chapter 7 Extra Questions – The Mathematics of Maybe Introduction to Probability Extra Questions Class 9

The Mathematics of Maybe Introduction to Probability Class 9 Short Question Answer

Question 1.
If we throw a die, then the upper face shows 1, 2, 3, 4, 5 or 6. Suppose we throw a die 150 times and get 2 for 75 times. What is the experimental probability of getting a ‘2’?
Solution:
Let E be the event of getting 2.
∴ P(E) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of trials }}\)
= \(\frac{75}{150}=\frac{1}{2}\)
= 0.5

Question 2.
A coin is tossed 200 times and is found that a tails comes up for 120 times. Find the experimental probability of getting a tails.
Solution:
Number of trials = 200
Number of favourable outcomes = Number of getting a tails = 120
Let the probability of getting a tails is P(E).
∴ P(E) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of trials }}\)
= \(\frac{120}{200}=\frac{6}{10}\)
= 0.6

Question 3.
How many times the coin was tossed, if the probability of getting a heads is 0.4 and it appeared up for 24 times?
Solution:
Number of favourable outcomes = 24
Let the total number of trials is n.
Here, P(E) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of trials }}\)
⇒ 0.4 = \(\frac{1}{2}\)
⇒ \(\frac{4}{10}=\frac{24}{n} \quad \Rightarrow \frac{n}{24}=\frac{10}{4}\)
⇒ n = \(\frac{10}{4}\) × 24 = 60

Question 4.
In a cricket match, if the probability P(E) of hitting the boundary is 0.3, then find the probability of not-hitting the boundary.
Solution:
Probability of hitting the boundary P(E) = 0.3
∴ Probability of not-hitting the boundary = 1 – P(E)= 1 – 0.3 = 0.7

Question 5.
In a sample study of 640 people, it was found that 500 people go for yoga regularly in the morning. If a person is selected at random, then what is the probability that the person does not go for yoga exercise?
Solution:
Since, the sample study is made on 640 people.
Number of people going for yoga regularly = 500
Number of people who do not go for yoga = 640 – 500 = 140
⇒ Total possible outcomes = 640
Number of favourable outcomes = 140
∴ Probability of people who do not go for yoga
= \(\frac{140}{640}=\frac{7}{32}\) = 0.21875

Question 6.
A die is thrown 270 times and the outcomes are recorded as in the following table:

If a die is thrown at random, find the probability of getting:
(i) 1 or 2
(ii) a prime number
(ii) a number greater than 3
Solution:
Here, total number of trials = 270
(i) Number of events of getting 1 or 2 = 36 + 45 = 81
∴ P(getting 1 or 2)
= \(=\frac{\text { Number of favourable outcomes }}{\text { Total number of trials }}\)
= \(\frac{81}{270}=\frac{3}{10}\)

(ii) ∵ Number of events of getting a prime number (2, 3 or 5) = 45
∴ P(getting 2, 3 or 5)
= \(=\frac{\text { Number of favourable outcomes }}{\text { Total number of trials }}\)
= \(\frac{153}{270}=\frac{17}{30}\)

(iii) Number of events of getting 4, 5 or 6 = 18 + 75 + 63 = 156
∴ P(getting 4, 5 or 6)
= \(=\frac{\text { Number of favourable outcomes }}{\text { Total number of trials }}\)
= \(\frac{156}{270}=\frac{26}{45}\)

Question 7.
In class IX. there are 40 students of whom 25 are girls and 15 are boys. They plan to spread awareness about water-management in different localities. Their class teacher has to select onc student for the group representative. She writen the name of each student on a separate card, put them in a bag and stir thoroughly. She then, draws one card at random from the bag.
(i) What is the probability that the name written on the card is the name of a girl student?
(ii) What is the probability that the name written on the card is the name of a boy student?
Solution:
Number of girls = 25
Number of boys = 15
Total number of students = 25 + 15 = 40
(i) Total number of outcomes = 40
Favourable number of outcomes = 25
⇒ Probability that the representative of the group is a girl-student = \(\frac{25}{40}=\frac{5}{8}\)

(ii) Total number of outcomes = 40
Favourable number of outcomes = 15
⇒ Probability that the representative of the group is a boy-student= \(\frac{15}{40}=\frac{3}{8}\)

The Mathematics of Maybe Introduction to Probability Class 9 Long Question Answer

Question 1.
A rectangular lawn measures 50 m by 30 m. In the centre of the lawn, there is a circular pond with a diameter of 14 m. During a storm, a lightning bolt hits the lawn at random. What is the probability that the lightning hits the pond? (Use π = \(\frac{22}{7}\))
Solution:
Step 1: Calculate total area
Area of rectangle = Length × Width
50 m × 30 m
= 1500 m²

Step 2: Calculate favorable area (Pond)
Diameter 14m, so Radius (r) = 7 m.
Area of Circle = πr² = \(\frac{22}{7}\) × 7 × 7
Area of Pond = 22 × 7
= 154 m²

Step 3: Calculate Probability Area of pond
P = \(\frac{\text { Area of pond }}{\text { Total area of lawn }}\)
= \(\frac{154}{1500}=\frac{77}{750}\)

Question 2.
A bag-I contains four cards numbered 1, 3, 5 and 7 respectively. Another bag-11 contains three cards numbered 2, 4 and 6 respectively. A card is drawn at random from each bag. Find the probability that the sum of two cards drawn is 9.
Solution:

Number of all possible outcomes = 12
Number of favourable outcomes = 3
⇒ P(sum = 9) = \(\frac{3}{12}=\frac{1}{4}\)

Question 3.
The ages of 400 workers in a factory are recorded in the table below:

A worker is selected at random. Find the probability that the worker is:
(i) At least 41 years old.
(ii) Under 31 years old.
(iii) Not older than 40 years.
Solution:
Total Workers = 400
(i) P(At least 41 years old):
Workers in categories (41 – 50) and (51 – 60)
= 80 + 50 = 130
Required probability
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac{130}{400}=\frac{13}{40}\)
= 0.325

(ii) P(Under 31 years old):
Workers in category (18 – 30) = 120
Required probability
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac{120}{400}=\frac{3}{10}\)
= 0.3

(iii) P(Not older than 40):
This means 40 or younger (Categories 18 – 30 and 31 – 40)
Number of workers = 120 + 150 = 270
Required probability
\(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac{270}{400}=\frac{27}{40}\)
= 0.675

The Mathematics of Maybe Introduction to Probability Class 9 Case Based Questions

Question 1.
A travel agency surveyed 2,000 people to find out their preferred mode of travel for long-distance trips. The results are categorized by age group in the table below:

If a person is chosen at random from this survey, find the probability that the person:
(i) Prefers to travel by Aeroplane.
(ii) Is over 50 years old and prefers to travel by Train.
(iii) Is under 30 years old and does not prefer a Private Car.
(iv) Prefers to travel by Bus.
(v) Is in the 30-50 age group and prefers either a Train or a Bus.
Solution:
Total people surveyed = 2000
(i) Number of people preferring Aeroplane = (300+ 150 + 20) = 470
Required probability
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac{470}{2000}\)
= 0.235

(ii) Number of people who are over 50 and prefers train = 180
Required probability
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac{180}{2000}\)
= 0.09

(iii) Number of people who are under 30 and do not prefer private Car = (250 + 300 + 100) = 650 Required probability
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac{650}{2000}\)
= 0.325

(iv) Number of people who prefer Bus
= (100+ 100+ 100) = 300
Required probability .
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac{300}{2000}\)
= 0.15

(v) Number of people under age 30-50 and prefer Train/Bus = (200 + 100) = 300
Required probability
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac{300}{200}\)
= 0.15

Question 2.
Define the sample space for each of the following experiments and calculate the required probabilities:
(i) A fair coin is tossed twice. What is the probability of obtaining at most one head?
(ii) A set of eight cards is numbered 1 to 8 and placed in a container. If one card is drawn at random, what is the probability that the number on the card is a multiple of 3?
(iii) A standard six-sided die is rolled once. What is the probability of obtaining a prime number?
(iv) A box contains 4 yellow pens, 3 green pens, and 3 black pens. If a pen is selected at random, what is the probability that the selected pen is neither yellow nor green?
(v) Three coins are tossed simultaneously. What is the probability of obtaining no heads (all tails)?
Solution:
(i) Sample Space S = {HH, HT, TH, TT}
Total outcomes = 4
Favourable outcomes (0 or 1 head) = {HT, TH, TT}
= 3 outcomes
Required probability
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac{3}{4}\)
= 0.75

(ii) Sample Space S = {1, 2, 3, 4, 5, 6, 7, 8}
Total outcomes = 8
Favourable outcomes (multiples of 3) = {3, 6} = 2 outcomes
Required probability
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac{2}{8}=\frac{1}{4}\)
= 0.75

(iii) Sample Space S = {1, 2, 3, 4, 5, 6}
Total outcomes = 6
Favourable outcomes (prime numbers)
= {2, 3, 5} = 3 outcomes
Required probability
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac{3}{6}=\frac{1}{2}\)
= 0.5

(iv) Total pens = 4 + 3 + 3 = 10
Event: neither yellow nor green = black pens only = 3
Required probability
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac{3}{4}\)
= 0.3

(v) Sample Space S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Total outcomes = 8
Favourable outcome (no heads) = {TTT} = 1 outcome
Required probability
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac{1}{8}\)
= 0.125

The Mathematics of Maybe Introduction to Probability Class 9 Competency Based Questions

Question 1.
A bag contains 24 marbles, some blue and some green. If the probability of drawing a green marble is \(\frac{2}{3}\), then the number of blue marbles is
(a) 6
(b) 8
(c) 12
(d) 16
Solution:
Total marbles = 24
Then the number of green marbles = \(\frac{2}{3}\) x 24 = 16
So, the number of blue marbles = 24 – 16 = 8
Hence, (b) is the correct answer.

Question 2.
A die is rolled thrice, and the number of 6s is recorded. Which of the following lists is a sample space for this experiment?
(a) {1, 2, 3}
(b) {0, 1, 2}
(c) {0, 1, 2, 3, 4}
(d) {0, 1, 2, 3}
Solution:
When a die is rolled three times, the number of 6s that can appear in the rolls can range from 0 (no 6s) to 3 (all three rolls showing 6s). Therefore, the sample space for the number of 6s recorded is {0, 1, 2, 3}.
Hence, (d) is the correct answer.

Question 3.
Ansh has 3 shirts of different colours (yellow, red and blue) and 2 types of pants (jeans and shorts). List all the possible combinations of outfits consisting of one shirt and one pair of pants. Display your answer in a table format.
Solution:
Here’s the list of all possible combinations of outfits consisting of one shirt and one pair of pants, displayed in a table format:

Shirts	Pants	Combination of outfits
Yellow	Jeans	Yellow shirt, Jeans
	Shorts	Yellow shirt, Shorts
Red	Jeans	Red shirt, Jeans
	Shorts	Red shirt, Shorts
Blue	Jeans	Blue shirt, Jeans
	Shorts	Blue shirt, Shorts

Thus, the number of all possible combinations of outfits = 3 × 2 = 6.

Question 4.
A student chooses a snack combination consisting of one main course and one drink. The main course options are Patties (P), Burger (B), or Sandwich (S). The drink options are Juice (J) or Milk (M).
(i) Draw a tree diagram to show the sample space.
(ii) If the student chooses randomly, what is the probability they do NOT choose a Sandwich?
(iii) What is the probability they choose a Burger and Juice?
Solution:
(i) Tree Diagram:

Sample Space (S): {(P, J), (P, M), (B, J), (B, M), (S, J), (S, M)}
Total outcomes = 6

(ii) P(Not a Sandwich):
Outcomes without Sandwich: {(P, J), (P, M), (B, J), (B, M)}
Number of favorable outcomes = 4
Required probability = \(\frac{4}{6}=\frac{2}{3}\)

(iii) P(Burger and Juice):
Favorable outcome: {(B, J)}
Required probability
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}=\frac{1}{6}\)

Question 5.
A die is thrown 500 times with the following frequencies for each outcome:

Find the probability of: (i) an even number, (ii) a number less than 3.
Solution:
Total trials = 500.
(i) Even outcomes = {2, 4, 6}.
Total frequency = 75 + 75 + 95 = 245.
P(Even) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac{245}{500}\)
= 0.49

(ii) Outcomes less tahn 3 = {1, 2}
Total frequency = 80 + 75 = 155.
P(Less tahn 3) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac{155}{500}\)
= 0.31

Question 6.
Two fair 6-sided dice are rolled simultaneously. Calculate the probability that:
(i) The sum of the numbers is a multiple of 4.
(ii) The product of the numbers is a perfect square.
Solution:
Total outcomes for two dice = 6 × 6 = 36
(z) Sum is a multiple of 4 (Sum = 4, 8, or 12):
Favourable outcomes = {(1, 3), (2, 2), (3, 1), (2, 6), (3,5), (4, 4), (5,3), (6,2), (6, 6)}
Total favourable outcomes = 9
Required probability
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}=\frac{9}{36}=\frac{1}{4}\)

(ii) Product is a perfect square (1, 4, 9, 16, 25, 36):
Favourable outcomes = (1, 1), (1, 4), (2, 2), (4, 1), (3, 3), (4, 4), (5, 5), (6, 6)
Total favourable = 8
Required probability
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}=\frac{8}{36}=\frac{2}{9}\)

The Mathematics of Maybe Introduction to Probability Class 9 Extra Questions for Practice

Multiple Choice Questions

Question 1.
A bag contains 3 red, 5 black, and 2 white balls. A ball is drawn at random. What is the probability that the ball drawn is not white?
(a) \(\frac{1}{5}\)
(b) \(\frac{4}{5}\)
(c) \(\frac{3}{10}\)
(d) \(\frac{1}{2}\)

Question 2.
In a survey of 200 students, 120 like Mathematics while 80 dislike it. If a student is chosen at random, what is the probability that the student dislikes Mathematics?
(a) 0.4
(b) 0.6
(c) 0.8
(d) 1.2

Question 3.
Which of the following cannot be the probability of an event?
(a) 15%
(b) 0.7
(c) \(\frac{2}{3}\)
(d) -1.5

Assertion Reason Questions

Two statements are given, one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the options (a), (b), (c), and (d) given below.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.

Question 1.
Assertion (A): The probability of getting an even number in a single throw of a fair die is —.
Reason (R): For a fair die. the outcomes {2, 4, 6} are the only possible outcomes.

Question 2.
Assertion (A): If P(E) = 0.35, then P(not E) = 0.65.
Reason (R): For any event E, the sum of the probability of an event and its complement is always 1.

Short Answer Type Questions

Question 1.
A coin is tossed 500 times with the following outcomes:

Compute the probability for each event.

Question 2.
Two coins are tossed simultaneously 200 times.
The outcomes are recorded as follows: 2 Heads-58; 1 Head-82; 0 Heads-60
Find the probability of getting at least 1 head.

Question 3.
Define Sample Space. Write the sample space for tossing three coins simultaneously.

Question 4.
A die is thrown 1000 times with the following frequencies for outcomes 1, 2, 3, 4, 5, and 6:

Find the probability of getting (a) a prime number (b) neither prime nor even number

Question 5.
Over the past 200 working days, the number of defective parts produced by a machine is given below:

Determine the probability that tomorrow’s output will have more than 4 defective parts.

Question 6.
Explain the difference between Experimental Probability and Theoretical Probability using the example of tossing a coin.

Long Answer Type Questions

Question 1.
A bag contains 3 red, 2 blue and 4 green pens. Two pens are drawn without replacement. What is the probability that both are of (i) same colour (ii) different colours?

Question 2.
A company selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The data is as follows:

Suppose a family is chosen. Find the probability that the family chosen is:
(i) Earning ₹30000-39000 per month and owning exactly 2 vehicles.
(ii) Earning ₹50000 or more per month and owning exactly 1 vehicle.
(iii) Earning less than ₹21000 per month and does not own any vehicle.
(iv) Earning ₹39000-50000 and owning more than 2 vehicles.
(v) Owning not more than 1 vehicle.

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