Chapter 7 - The mathematics of maybe:Introduction to probability

NCERT Solutions for Class 9 Maths Chapter 7 The mathematics of maybe:Introduction to probability

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1.0Download NCERT Solutions for Class 9 Maths Chapter 7 – The Mathematics of Maybe: Introduction to Probability

Chapter 7: The Mathematics of Maybe: Introduction to Probability helps students understand how mathematics can be used to analyse uncertain events and predict possible outcomes. The chapter builds analytical thinking through practical observations and experiments.Download the NCERT Solutions for Class 9 Maths Chapter 7 PDF, featuring detailed step-by-step solutions for every textbook exercise.

NCERT Solutions Class 9 Maths Chapter 7

2.0Learning Outcomes Class 9 Maths Chapter 7: The Mathematics of Maybe: Introduction to Probability

Understand the concept of probability and measure the likelihood of an event on a scale from 0 to 1.
Distinguish between subjective and objective probability in different situations.
Identify random experiments, sample spaces, and events.
Calculate experimental probability using observations from repeated trials.
Find theoretical probability by identifying favourable and total possible outcomes.
Compare experimental and theoretical probability and understand the Law of Large Numbers.
Represent possible outcomes using tree diagrams and sample spaces.
Apply probability concepts to solve real-life and textbook problems involving chance and uncertainty.

3.0Detailed Class 9 Maths Chapter 7 – The Mathematics of Maybe: Introduction to Probability Solutions

Exercise: 7.1

1.Rank the following events on a scale from 0 (Impossible) to 1 (Certain). Label each event: Impossible, less likely, equally likely (even chance), more likely, certain. Give reasons why you gave each event its ranking. (i) The next Monday will come after Sunday. (ii) It will snow in Mumbai in July. (iii) An elephant will walk through your classroom today. (iv) You will greet at least one friend at school tomorrow.

Sol. (i) Certain (Probability = 1) Days of week always follow a fixed order Sunday is always followed by Monday. This will definitely happen so its probability is 1 . (ii) Impossible (Probability = 0) Mumbai has a tropical climate. In July It experiences the monsoon season with hot and humid weather. Snowfall is not possible there. So, this event cannot happen at all, its probability. (iii) Impossible (Probability = 0) Under normal situations, it is not possible for an elephant to walk into school classroom. This event practically cannot happen so its probability is 0 (or nearby equal to 0 ). (iv) More likely (Probability is close to 1 , but not exactly 1).

In school there are many friends. It is very likely that you will greet at least one friend. The chances are very high so this event is "more likely". As it is not "certain" because there could be rare situations (like you are absent or school is closed) but normally it will happen.

Exercise: 7.2

1.A teacher mixes a large bag of sweets of different colours and randomly selects a sample of 30 sweets. She counts the number of sweets of each colour: 10 red sweets | 8 green sweets | 7 yellow sweets | 5 blue sweets (i) Calculate the probability that a randomly picked sweet from the sample is green. (ii) If there are 600 sweets in total in the large bag, estimate how many are likely to be yellow, based on the sample results.

Sol. (i) Let Probability of picking sweet from the sample is green denoted by P(Green).

$P (Green) = \frac{Number of green sweets}{Number of total sweets} = \frac{8}{30} = \frac{4}{15} \approx 0.267$

(ii) Number of yellow sweets $= 7$ total sweets in sample $= 30$ $P ($ Yellow $) = \frac{7}{30}$ Estimated number of yellow sweets in $600 = \frac{7}{30} \times 600 = 140$

Thus 140 sweets are likely to be yellow.

2. A survey is conducted at a school where a random sample of 40 students is asked about their favourite club. The responses are: 14 students: Science Club | 11 students: Arts Club | 9 students: Sports Club | 6 students: Debate Club

Assume there are 800 students in the whole school. (i) What is the probability that a randomly chosen student from the sample prefers the Arts Club? (ii) Using the sample results, estimate how many students in the whole school are likely to prefer the Sports Club.

Sol. (i) Number of students who prefer arts club = 11 total students in sample $= 40$ $P ($ arts-club $) = \frac{11}{40} = 0.275$ (ii) Number of students who are in sports club $= 9$ $P ($ sports club $) = \frac{9}{40}$ Estimated number in whole school $= \frac{9}{40} \times 800 = 180$ . approximately 180 students in the whole school prefer the sports club.

3. Toss a coin 20 times and record the result each time (heads or tails). (i) How many times did you get heads? (ii) How many times did you get tails? (iii) Calculate the experimental probability of getting heads. (iv) If you toss the coin once more, what is the probability of getting tails?

Sol. Let a coin tossed 20 times out of which 11 times we get heads & 9 times tails. (i) 11 times (ii) 9 times (iii) P (heads) $= \frac{Number of heads}{Total number of tosses}$

$= \frac{11}{20} = 0.55$

(iv) This is a new independent toss. The theoretical probability of getting tails on any single fair coin toss is always

4.Toss a paper cup into the air 100 times. After each toss record whether the cup lands on its bottom, upside down on its top or on its side Figure. Assign probabilities to the outcomes by using experimental probability.

Sol. I got following observations from the experiment. Bottom = 30, top = 20, side = 50 $P ($ Bottom $) = \frac{30}{100} = 0.3$

What is the probability of getting an even number when rolling a fair 6 -sided die? Sol. Total possible outcomes $S = 1, 2, 3, 4, 5, 6$ Even numbers $= 2, 4, 6$ $P ($ even number $) = \frac{no. of favourable outcomes}{no. of total outcomes} = \frac{3}{6} = \frac{1}{2} = 0.5$

Suppose you roll a 6 -sided die 12 times and get a ' 3 ' three times. (i) What is the experimental probability of rolling a ' 3 '? (ii) What is the theoretical probability of rolling a ' 3 '? (iii) Why might these probabilities be different? What would you expect to happen if you roll the die 60,600, or 6000 times?

Sol. (i) Experimental Probability

$= \frac{Number of times 3 comes}{Total rolls} = \frac{3}{12} = \frac{1}{4}$

(ii) Theoretical Probability A die has 6 numbers and only one number is $3 = \frac{1}{6}$ (iii) They are different because the die was rolled only 12 times.

If we roll the die $60, 600, 6000$ times, the experimental probability will become closer to $\frac{1}{6}$ .

Exercise: 7.3

1.When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?

Sol. Die outcomes are $= 1, 2, 3, 4, 5, 6$ Sample space $= {1, 2, 3, 4, 5, 6}$ Total number of possible outcomes $= 6$

2. For the following experiments write down the sample space $S$ . (i) Rolling a die and tossing a coin together. (ii) Choosing a random integer between -5 and +5 . (iii) A box containing 5 green and 7 red balls. One ball is drawn at random.

Sol. (i) Rolling a die and tossing a coin together

Die outcomes $= 1, 2, 3, 4, 5, 6$ Coin outcomes = H(Heads), T(Tails) Each outcome is a pair = (die number, coin result)

There S(sample - space). $(1, H), (1, T), (2, H), (2, T), (3, H)$ , $(3, T), (4, H), (4, T), (5, H), (5, T), (6, H)$ , (6, T)

Hence $n (S) = 12$ (ii) Choosing random integer between -5 and 5

Integer between -5 & 5 (including both endpoints) $- 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5$ Therefore, sample space $S = - 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5$ $n (S) = 11$ (iii) A box contain 5 green and 7 red balls.

One ball is chosen at random. Case (1) If balls are identical S = Green ball, Red ball $n (S) = 2$ Case (2) If balls are not identical. If balls are individually distinguishable, 'S' would have 12 elements $G_{1}, G_{2}, G_{3}, G_{4}$ , $G_{5}, R_{1}, R_{2}, R_{3}, R_{4}, R_{5}, R_{6}, R_{7}$ but for probability purpose the colour based sample space is standard at this level.

3. In a village fair, there are 3 popular snacks available: Samosa, Pakora, and Bhaji. For drinks, villagers can choose either Chai or Lassi. (i) List the sample space of all possible snack and drink combinations a person could choose at the fair. (ii) List the event 'Selecting Samosa as a snack.'

Sol. (i) $S =$ (samosa, chai), (samosa, lassi), (pakora, chai), (pakora, lassi), (bhaji, chai), (bhaji, lassi) $n (S) = 6$ (ii) $E =$ (samosa, chai), (samosa, lassi)

The event has 2 outcomes out of 6 total outcomes. $P ($ samosa $) = \frac{2}{6} = \frac{1}{3}$

Exercise: 7.4

1.There are two fruit baskets A and B . Basket A has one apple and two oranges. Basket B has one banana and one mango. You randomly pick one fruit from each basket. (i) Draw a tree diagram showing all possible pairs of fruits. (ii) List the sample space. (iii) What is the probability of picking one apple and one banana?

Sol.

(i) Possible pairs formed =

6 =

possible outcomes (ii)

{(A, B), (A, M), (O_{1}, B), (O_{1}, M), (O_{2}, B)

(O_{2}, M)} =

sample space (iii) Probability of picking 1 apple and 1 Banana

= \frac{1}{6}

2. Let us say that you have a box containing 3 red pens, 4 black pens and 2 green pens. You pick a pen (without looking) from the box and put it back. Then your friend does the same. (i) What are the possible outcomes of the pen colours? Can you draw a tree diagram representing the possible outcomes? (ii) Can you use the tree diagram to guess the probability that both you and your friend pick pens of the same colour?

Sol. (i) Possible outcomes and tree diagram Total pens $= 3 ($ red $) + 4 ($ black $) +$ 2 (green) $= 9$ Pens Because the pen is replaced before your friend's turn the total number of pens and the probability for each colour remain exactly the same for both picks. The probability are: $P ($ Red $) = \frac{3}{9} = \frac{1}{3}$ $P ($ Black $) = \frac{4}{9}$ $P ($ Green $) = \frac{2}{9}$ Possible outcomes (sample space) There are $9 \times 9 = 81$ total possible combinations, but they fall into 9 distinct colour pairings. $S = {(R, R), (R, B), (R, G), (B, R), (B, B)$ , $(B, G), (G, R), (G, B), (G, G)}$

(ii) Probability of picking the same colour.

To find the probability of both people picking the same colour, you calculate the individual probabilities of both picking red, both picking black, both green and then add those values together. (a) Probability of both Red

$P (R, R) = \frac{3}{9} \times \frac{3}{9} = \frac{1}{9}$

(b) Probability of both Black

$\Rightarrow P (B, B) = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}$

$\Rightarrow P (G, G) = \frac{2}{9} \times \frac{2}{9} = \frac{4}{81}$

Total Probability of same colour

$P (same colour) = \frac{1}{9} + \frac{16}{81} + \frac{4}{81} = \frac{29}{81}$

End of Chapter Exercise

1. Fill in the blanks. (i) The probability of an impossible event is $____$ . (ii) The set of all possible outcomes of a random experiment is called the $____$ . (iii) The probability of an event that is certain to happen is $____$ . (iv) Tossing a fair coin has a probability of $____$ for getting heads.

Sol. (i) zero (ii) sample-space (iii) 1 (iv) $\frac{1}{2}$

2. In a survey of 50 students, 15 students said they liked football. The number of students who like football is 15 , and the $____$ (frequency/relative frequency) is $____$ (fill in the fraction or decimal).

Sol. The number of students who like football is 15, and the relative frequency is $\frac{15}{50} = 0.3$

3. Which of the following experiments have equally likely outcomes? Explain. (i) A driver attempts to start a car. The car starts or does not start. (ii) Tossing a fair coin once. (iii) Rolling a fair 6 -sided die. (iv) Choosing a marble randomly from a bag that contains 3 red marbles and 7 blue marbles. (v) A baby is born. It is a boy or a girl.

Sol. (i) Not equally likely. The car starting depends on the condition of the car, fuel, battery etc. The two outcomes (starts/does not start) are not equally likely under normal conditions. (ii) Yes equally likely Heads and tails each have probability $\frac{1}{2}$ , fair coin gives equal chance to both outcomes. (iii) Yes, equally likely. Each of the 6 faces ( 1 to 6 ) has an equal probability $= \frac{1}{6}$ (iv) Not equally likely.

$P (red) = \frac{3}{10} P (Blue) = \frac{7}{10}$

The two colours have different probabilities. (v) Yes equally likely. Biologically, the probability of a boy or girl is approximately $\frac{1}{2}$ each so these can be considered roughly equally likely.

4. Write the sample space and calculate the probability based on the given information. (i) Two coins are tossed at the same time. What is the probability of getting at least one head? (ii) Ten identical cards numbered 1 to 10 are placed in a box. One card is drawn at random. What is the probability of drawing a card with an even number? (iii) A die is rolled once. What is the probability of getting a number greater than 4 ? (iv) A bag contains 3 red balls, 2 blue balls, and 1 green ball. One ball is picked at random. What is the probability that it is not red? (v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?

Sol. (i) $S = {HH, HT, TH, TT}$ $n (S) = 4$ Event ' E ' $=$ at least one head $= {HH, HT, TH}$ $n (E) = 3$ $P ($ at least one head $) = \frac{n ( E )}{n ( S )} = \frac{3}{4}$ (ii) $S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}$ $n (S) = 10$ Even numbers $= {2, 4, 6, 8, 10}$ $n (E) = 5$ Hence, P (even number) $= \frac{n ( E )}{n ( S )} = \frac{1}{2}$ (iii) $S = {1, 2, 3, 4, 5, 6}$ $n (S) = 6$ Number greater than $4 = {5, 6}$ $n (E) = 2$ $P ($ number greater than 4 $)$ $= \frac{n ( E )}{n ( S )} = \frac{2}{6} = \frac{1}{3}$ (iv) Total balls $= 3 + 2 + 1 = 6$

Not red $=$ blue + green $= 2 + 1 = 3$ $P ($ not red $) = \frac{3}{6} = \frac{1}{2}$ (v) $S = {HHH, HHT, HTH, THH, HTT, THT$ , TTH, TTT} $n (S) = 8$ Exactly two heads $= {HHT, HTH$ , THH} $n (E) = 3$ $P ($ exactly two heads $) = \frac{n ( E )}{n ( S )} = \frac{3}{8}$ 5. A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?

Sol. $S = {$ strawberry, lemon, mint $}$ $n (S) = 3$ n(E) = 1 (strawberry) $P ($ strawberry $) = \frac{1}{3}$

6. A child has 2 shirts (one red and one blue) and 3 types of pants (jeans, khakis, 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.

Sol. Possible combinations of outfits consisting of one shirt and one pair of pants. (1) Red shirt + Jeans (2) Red shirt + Khakis (3) Red shirt + shorts (4) Blue shirt + Jeans (5) Blue shirt + Khakis (6) Blue shirt + shorts

So total combinations $= 6$ of outfits

7. A tyre company records distances before replacement in 1000 cases.

Distance $(Km)$	Number of cases
Less than $4000$	20
$4001$ to $9000$	210
$9001$ to $14000$	325
More than $14000$	445

Find the probability that a randomly chosen type lasts: (i) Less than 4000 km . (ii) Between 4000 and 14000 km . (iii) More than 14000 km .

Sol. (i) $P ($ Less than 4000 Km $) = \frac{20}{1000} = \frac{1}{50}$ (ii) P (Between 4000 and 14000 km )

$= \frac{535}{1000} = \frac{107}{200}$

(iii) P(More than 14000 km )

$= \frac{445}{1000} = \frac{89}{200}$

The letters of the word 'PEACE' are placed on cards. Leela draws a card without looking.

$P E A C E$

(i) What is the probability that it is a $P, E$ or C? (ii) What is the probability that it is not an E?

Sol. Total number of cards in PEACE $= 5$ Cards with $P, E$ or $C = 4$ $P = 1, E = 2, C = 1$ (i) Therefore,

$P (P, E or C) = \frac{4}{5}$

(ii) $P (not E) = P (\overline{E}) = \frac{3}{5}$

9. A game of chance consists of spinning an arrow see Figure which comes to rest pointing at one of the numbers $1, 2, 3, 4$ , 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at

(i) 8 ? (ii) An odd number? (iii) A number greater than 2 ? (iv) A number less than 9 ? (v) A multiple of 3 ?

Sol. (i) $P (8) = \frac{1}{8} =$ Probability of getting 8 (ii) $P ($ odd number $}) = \frac{4}{8}$ (iii) Number greater than 2 are $= {3, 4, 5$ , 6, 7, 8} $P ($ number greater than 2 $) = \frac{6}{8} = \frac{3}{4}$ (iv) $P ($ number less than 9 $) = \frac{8}{8} = 1$ (v) $P ($ multiple of 3 $) = \frac{2}{8} = \frac{1}{4}$

10. A basket contains 4 red balls and 5 blue balls. One ball is drawn and laid aside, and a second ball is drawn. Draw a tree diagram to represent the possible outcomes and probabilities. Use the tree diagram to answer the following questions. (i) What is the probability of drawing a red ball and then a blue ball? (ii) What is the probability of drawing 2 blue balls?

Sol. Basket contains total balls $= 9$ Red balls $(R) = 4$ Blue balls ( B ) $= 5$ One ball is drawn not replaced then second ball is drawn.

Possible outcomes with Probabilities (i) Red and Red $(R, R) = \frac{4}{9} \times \frac{3}{8} = \frac{12}{72} = \frac{1}{6}$ (ii) Red then blue $(R, B) = \frac{4}{9} \times \frac{5}{8} = \frac{5}{18}$ (iii) Blue and Red ( $B, R$ ) $= \frac{5}{9} \times \frac{4}{8} = \frac{5}{18}$ (iv) Blue and blue (b, b) $= \frac{5}{9} \times \frac{4}{8} = \frac{5}{18}$ (i) Probability of Drawing a Red ball and then a blue ball

$P (R, B) = \frac{4}{9} \times \frac{5}{8} = \frac{20}{72} = \frac{5}{18}$

(ii) Probability of drawing 2 Blue balls

11.I throw a pair of 6 -sided dice. Write down an event that has a probability of 0 and an outcome that has a probability of 1 .

Sol. Event with probability 0 (Impossible) Getting a sum of 1 (least sum when 2 dice are thrown is $1 + 1 = 2$ ) is impossible. Outcome with Probability = 1 (Certain) Getting a sum that is less than 13 (maximum sum is $6 + 6 = 12$ , which is always less than 13 , so this is certain).

12. Write the sample space and calculate the probability based on the given information. (i) Two dice are rolled. What is the probability that the sum is a prime number greater than 5 ? (ii) A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours? (iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total? (iv) A four-digit number is formed using the digits $1, 2, 3$ , and 4 with no repetition. What is the probability that the number is even? (v) A student takes a multiple-choice test with 3 questions, each having 4 options (A, B, C, D), with only one correct answer. What is the probability that the student guesses and gets exactly 2 answers correct? Sol. (i) When 2 dice are rolled total outcomes $= 6^{2} = 36$ ∴ Sum can be from 2 to 12 . but ATQ (According to Question) sum required is prime number greater than 5 . When sum is 7 outcomes are: $= (1, 6) (6, 1) (2, 5) (5, 2) (3, 4) (4, 3) = 6$ outcomes.

When sum is 11 : $= (6, 5) (5, 6) = 2$ outcomes. Total favourable outcomes $= 8$ Required Probability $= \frac{8}{36} = \frac{2}{9}$ (ii) Total balls $= 4 + 3 + 2 = 9$ balls. ∴ balls are of different colours. 1st ball can be chosen in 9 ways second ball have 8 ways total ways $= 72$ Now Favourable outcomes:

I	II
Red	Green	$4 \times 3 = 12$
Green	Red	$3 \times 4 = 12$
Red	blue	$4 \times 2 = 8$
blue	Red	$2 \times 4 = 8$
Green	Blue	$3 \times 2 = 6$
Blue	Green	$2 \times 3 = 6$

Total favourable outcomes $= 24 + 16 + 12 = 52$ $P ($ different colours $) = \frac{52}{72} = \frac{13}{18}$ (iii) Favourable outcomes $= 2$ {HHT, HTH}

Total outcomes for 3 coins $= 2^{3} = 8$ Probability $= \frac{2}{8} = \frac{1}{4}$ (iv) Total arrangement of 4 different digits $= 4 \times 3 \times 2 \times 1 = 24$

Condition for even number: A number is even if its last digit is even. even digits here: 2 & 4.

Case 1:

When last digit = 2 No. of ways to arrange then $= 3 \times 2 \times 1 = 6$

Case 2:

When last digit = 4 No. of ways $= 3 \times 2 \times 1 = 6$ Now total ways: (favourable outcomes) $= 6 + 6 = 12$ $P ($ even number $) = \frac{12}{24} = \frac{1}{2}$ (v) Each question has 4 option and only 1 is correct.

So, for each question: Probability of correct answer $= \frac{1}{4}$ Probability of wrong answer $} = \frac{3}{4}$ Total question = 3 We need Probability of exactly 2 correct answers. possible case for exactly 2 correct. There are 3 ways this can happen: ${(C, C, W), (C, W, C), (W, C, C)}$

Probability of each case: Each case has 2 correct answers $= \frac{1}{4} \times \frac{1}{4}$ 1 wrong answer $= \frac{3}{4}$ So, Probability of one case $= \frac{1}{4} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{64}$ Since there are 3 such cases, so $P ($ exactly 2 correct $) = 3 \times \frac{3}{64} = \frac{9}{64}$

13. A box contains 4 balls numbered 1 to 4. Record a sample space using a tree diagram for the following experiments: (i) A ball is drawn, and the number is recorded. Then the ball is returned, and a second ball is drawn and recorded. (ii) A ball is drawn and recorded. Without replacing the first ball, the experimenter draws and records a second ball. (iii) What are the sizes of these two sample spaces? Sol. (i) sample space (all ordered pairs) $S = (1, 1), (1, 2), (1, 3), (1, 4), (2, 1)$ , $(2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3)$ , $(3, 4), (4, 1), (4, 2), (4, 3), (4, 4)$ $n (S) = 16$ (ii) sample space (when both are different) $S_{1} = (1, 2), (1, 3), (1, 4), (2, 1), (2, 3)$ , $(2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2)$ , $(4, 3)$ $n (S_{1}) = 12$ with replacement $n (S) = 16$ without replacement $n (S_{1}) = 12$

14. List the elements of a sample space for the simultaneous tossing of a coin and drawing a card from a set of 6 cards numbered 1 through 6.

Sol. Coin outcomes (H, T) Card outcomes 1, 2, 3, 4, 5, 6 $S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)$ , (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)} $n (S) = 12$

15. Three coins are tossed, and the number of heads is recorded. Which of the following lists is a sample space for this experiment? Why do the other lists fail to qualify as a sample space? (i) ${1, 2, 3}$ (ii) ${0, 1, 2}$ (iii) ${0, 1, 2, 3, 4}$ (iv) ${0, 1, 2, 3}$

Sol. When 3 coins are tossed, possible number of heads $= 0, 1, 2$ & 3 (i) ${1, 2, 3}$ -Not a valid sample space. It does not include 0 heads (all tails = TTT) (ii) ${0, 1, 2}$ -Not a valid sample space. It does not include 3 heads (HHH) (iii) ${0, 1, 2, 3, 4}$ -Not a valid sample space. The value 4 is not possible since only 3 coins are tossed. Maximum heads $= 3$ (iv) ${0, 1, 2, 3}$ -Yes this is the correct sample space. It includes all possible values: 0 heads (TTT), 1 head, 2 head, 3 heads (HHH)

16. Suppose you drop a dye at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with a diameter of 1 m ?

Sol. Area of rectangle $= 3 \times 2 = 6 m^{2}$ Diameter of circle $= 1 m$ So, radius $= 0.5 m$ . Area of circle $= π r^{2}$ $= π (0.5)^{2} = 0.25 π = \frac{π}{4}$ $P ($ lands inside circle $) = \frac{Area of circle}{area of rectangle}$ $= \frac{π}{24}$

4.0Important Key Concepts of Class 9 Maths Chapter 7

Topic	What Students Learn
Probability Scale	Ranking events from 0 (impossible) to 1 (certain), and reasoning through everyday examples to place them correctly
Random Experiments	Identifying experiments whose outcomes cannot be predicted with certainty, such as tossing a coin or rolling a die
Experimental Probability	Calculating probability from observed data or repeated trials, using the ratio of favourable outcomes to total trials
Theoretical Probability	Calculating probability using the ratio of favourable outcomes to all possible outcomes in an idealised scenario
Sample Space	Identifying and listing all possible outcomes of a random experiment
Events	Understanding events as subsets of the sample space, and calculating their probability
Tree Diagrams	Using tree diagrams to visualise and count outcomes of compound events, such as picking two items in sequence
Real-Life Applications	Applying probability to practical situations like surveys, sampling, and estimating outcomes in larger populations

5.0Exercise-wise NCERT Class 9 Maths Solutions

Chapter 7 builds probability step by step — from ranking events by likelihood to calculating experimental and theoretical probability using sample spaces and tree diagrams.

Exercise	Topics Covered
Exercise 7.1	Ranking events from impossible to certain on the probability scale, with reasoning based on real-life examples
Exercise 7.2	Calculating experimental probability from survey data and sample results, and estimating outcomes for larger populations
Exercise 7.3	Identifying the sample space for random experiments and calculating theoretical probability
Exercise 7.4	Using tree diagrams to represent compound events and calculate the probability of combined outcomes
End-of-Chapter Exercises	Mixed practice combining probability ranking, experimental and theoretical probability, sample spaces, and tree diagrams

6.0Quick Revision on Class 9 Maths Chapter 7 - The Mathematics of maybe : Introduction to Probability

7.0Related Study Materials Class 9 Maths

Enhance your preparation with ALLEN’s Class 9 Maths study material as per the latest NCERT syllabus. Along with NCERT Solutions, check updated syllabus, NCERT Books, chapter-wise notes, revision notes, sample papers and previous year question papers to strengthen your concepts, revise efficiently and prepare confidently for school and CBSE examinations.

CBSE Class 9 Maths Syllabus

Class 9 Maths Revision Notes

NCERT Textbook for Class 9 Maths

CBSE Sample Papers for Class 9 Maths

8.0Advantages of Chapter 7 Maths Class 9 NCERT Solutions

Recognises Number Patterns: Helps students identify and extend different numerical sequences.
Explains Arithmetic Progressions: Develops a clear understanding of arithmetic progressions and the common difference.
Finds Terms Systematically: Teaches how to determine any term in a sequence using the correct approach.
Strengthens Pattern-Based Reasoning: Improves logical thinking by analysing relationships between consecutive terms.
Simplifies Sequence Problems: Provides step-by-step solutions for sequence and progression-based questions.
Builds a Foundation for Higher Maths: Prepares students for advanced topics involving sequences and series.

NCERT Solutions for Class 9 Maths Chapter 7 The mathematics of maybe:Introduction to probability

1.0Download NCERT Solutions for Class 9 Maths Chapter 7 – The Mathematics of Maybe: Introduction to Probability

NCERT Solutions Class 9 Maths Chapter 7

2.0Learning Outcomes Class 9 Maths Chapter 7: The Mathematics of Maybe: Introduction to Probability

Understand the concept of probability and measure the likelihood of an event on a scale from 0 to 1.
Distinguish between subjective and objective probability in different situations.
Identify random experiments, sample spaces, and events.
Calculate experimental probability using observations from repeated trials.
Find theoretical probability by identifying favourable and total possible outcomes.
Compare experimental and theoretical probability and understand the Law of Large Numbers.
Represent possible outcomes using tree diagrams and sample spaces.
Apply probability concepts to solve real-life and textbook problems involving chance and uncertainty.