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

Chapter 7 of Ganita Manjari, “The Mathematics of Maybe: Introduction to Probability,” takes students beyond guesswork into the mathematics of chance. Starting from everyday situations — a coin toss before a cricket match, a bag of mixed sweets, a spinning arrow at a village fair — the chapter builds up the probability scale, experimental and theoretical probability, sample spaces and events, and tree diagrams for multi-step experiments. Every idea is grounded in a worked example before students attempt the Exercise Sets, so that the formula P(E) = Number of favourable outcomes ÷ Number of all possible outcomes is always tied to a concrete situation rather than memorised in isolation. This page provides complete NCERT solutions for Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability, with detailed answers to every Exercise Set question from the Ganita Manjari textbook.

Last Updated: September 23, 2026

7.1 Probability Scale — Exercise Set 7.1 (Page 158)

The chapter opens by asking why a coin toss is considered a fair way to decide which cricket team bats first. A fair coin has exactly two equally likely outcomes — Heads and Tails — each with probability 1/2, the result cannot be predicted in advance, and every toss is independent of the ones before it, so neither team gets an advantage. This idea of randomness — all possible outcomes are known, but no single trial can be predicted — is the foundation of the probability scale: every event is ranked on a scale from 0 (impossible) to 1 (certain), with “less likely,” “equally likely (even chance),” and “more likely” describing everything in between.

Q1. 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.

Answer:

(i) Rank = 1, Label: Certain. The days of the week always follow the same fixed order, and Monday always immediately follows Sunday, so this is guaranteed to happen.

(ii) Rank = 0, Label: Impossible. Mumbai has a tropical climate and is in the middle of its monsoon season in July; temperatures never drop anywhere near freezing, so snowfall cannot occur.

(iii) Rank = 0, Label: Impossible. Classrooms are enclosed, structured spaces in urban or semi-urban school buildings that large wild animals have no access to, so this event practically cannot occur.

(iv) Rank ≈ 0.9, Label: More likely. School is a social setting where interaction with classmates is the norm, so the chance of greeting at least one friend is very high — but it is not “certain” because rare situations (being absent, school being closed) could prevent it.

7.2 Experimental and Theoretical Probability — Exercise Set 7.2 (Page 164)

Once the probability scale is established, the chapter introduces two objective ways to measure likelihood. Experimental probability is calculated from real trials or survey data: Number of times an event occurs ÷ Total number of trials. Theoretical probability instead uses logical reasoning, assuming all outcomes are equally likely: Number of favourable outcomes ÷ Number of all possible outcomes. A “Think and Reflect” box makes the point sharply: if a fair die shows a 4 eight times in a row, the probability of rolling a 4 on the ninth try is still exactly 1/6 ≈ 0.16, because each roll is an independent event and the die has no memory of past results. Believing a streak “must end” is called the Gambler’s Fallacy — in the short run random streaks can and do happen, but in the long run (the Law of Large Numbers) each outcome settles down to its theoretical probability as the number of trials grows.

Q1. 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.

Answer: Total sweets in sample = 10 + 8 + 7 + 5 = 30.

(i) P(green) = 8/30 = 4/15 ≈ 0.267 or 26.7%.

(ii) P(yellow) = 7/30. Estimated yellow sweets in the bag = (7/30) × 600 = 140 sweets.

Q2. 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.

Answer: Total students in sample = 14 + 11 + 9 + 6 = 40.

(i) P(Arts Club) = 11/40 = 0.275 or 27.5%.

(ii) P(Sports Club) = 9/40. Estimated students in the school = (9/40) × 800 = 180 students.

Q3. 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?

Answer: This is a hands-on activity, so your actual counts of heads and tails will vary each time you perform it — that variation is itself the point of the exercise. As an illustration, suppose a trial gives 11 heads and 9 tails out of 20 tosses:

(i) Heads = 11 (ii) Tails = 9

(iii) Experimental P(heads) = 11/20 = 0.55.

(iv) Each toss is an independent event, so the theoretical probability of tails on the next toss is always P(tails) = 1/2, regardless of what happened in the previous 20 tosses.

Q4. 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. Assign probabilities to the outcomes by using experimental probability — Answer: This is a hands-on activity — actual counts will vary with the cup used. As an…

Answer: This is a hands-on activity — actual counts will vary with the cup used. As an illustration, suppose out of 100 tosses the cup lands on its bottom 35 times, on its top 15 times, and on its side 50 times:

P(bottom) = 35/100 = 0.35, P(top) = 15/100 = 0.15, P(side) = 50/100 = 0.50. Since a paper cup is an irregular shape, its outcomes are not equally likely, so experimental probability (not theoretical probability) must be used to estimate them.

Q5. What is the probability of getting an even number when rolling a fair 6-sided die? — Answer: Sample space S = {1, 2, 3, 4, 5, 6}, so total outcomes = 6. Even numbers = {2,…

Answer: Sample space S = {1, 2, 3, 4, 5, 6}, so total outcomes = 6. Even numbers = {2, 4, 6}, so favourable outcomes = 3. P(even number) = 3/6 = 1/2 = 0.5 or 50%.

Q6. 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 rolled the die 60, 600, or 6000 times?

Answer:

(i) Experimental P(3) = 3/12 = 1/4 = 0.25 or 25%.

(ii) Theoretical P(3) = 1/6 ≈ 0.167 or 16.7% (only one favourable face out of six equally likely faces).

(iii) The two differ because experimental probability is based on a small, limited sample of only 12 trials, where chance variation has a large effect, while theoretical probability assumes an ideal, perfectly fair die. By the Law of Large Numbers, as the number of trials increases — 60, then 600, then 6000 rolls — the experimental probability of rolling a 3 will drift closer and closer to the theoretical value of 1/6, with the 6000-roll estimate expected to be the closest of all.

7.3 Sample Spaces and Events — Exercise Set 7.3 (Page 168)

To calculate probability precisely, every outcome of an experiment must first be listed as a sample space S, and the number of elements in it is written n(S). An event is then any subset of the sample space that satisfies a given condition, written E, with n(E) favourable outcomes; then P(E) = n(E)/n(S). A “Think and Reflect” box illustrates why the level of detail in a sample space matters: forecasting whether it will rain tomorrow might use the simple sample space {Rain, No Rain}, but if the amount of rainfall matters, the sample space must be expanded to {No Rain, Drizzle, Light Rain, Heavy Rain} — a sample space must always be detailed enough to match what the question is actually asking.

Q1. When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space? — Answer: Sample space S = {1, 2, 3, 4, 5, 6}. Each face gives a distinct outcome, so the…

Answer: Sample space S = {1, 2, 3, 4, 5, 6}. Each face gives a distinct outcome, so the total number of possible outcomes, n(S) = 6.

Q2. 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.

Answer:

(i) Pairing each of the 6 die faces with each of the 2 coin outcomes: S = {(1,H), (1,T), (2,H), (2,T), (3,H), (3,T), (4,H), (4,T), (5,H), (5,T), (6,H), (6,T)}. n(S) = 6 × 2 = 12.

(ii) Taking “between -5 and +5” as inclusive of both endpoints: S = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}. n(S) = 11.

(iii) By colour, since only two outcome-types are possible when drawing one ball: S = {Green, Red}, n(S) = 2, with 5 favourable outcomes for Green and 7 for Red out of 12 total balls — note that Green and Red are not equally likely here, since P(Green) = 5/12 and P(Red) = 7/12. If each ball is treated as a separate, distinguishable object instead, the sample space becomes S = {G1, G2, G3, G4, G5, R1, R2, R3, R4, R5, R6, R7}, with n(S) = 12 equally likely outcomes.

Q3. 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.’

Answer:

(i) Pairing each of the 3 snacks with each of the 2 drinks: S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}. n(S) = 6.

(ii) E = {(Samosa, Chai), (Samosa, Lassi)}. n(E) = 2, so P(Samosa) = 2/6 = 1/3.

7.4 Tree Diagrams — Exercise Set 7.4 (Page 172)

For experiments with two or more sequential steps, a tree diagram is the most reliable way to map out every possible outcome without missing any. Branches spread out from a single starting point for each outcome of the first step, and further branches spread from each of those for the second step; every complete path from start to tip is one outcome in the sample space. A “Think and Reflect” box builds toward this: when two coins are tossed, the sample space is S = {HH, HT, TH, TT}, so the probability of getting exactly one head and one tail is P(HT or TH) = 2/4 = 1/2 = 50%.

Q1. 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?

Answer:

(i) Basket A has 3 individual fruits (Apple, Orange1, Orange2), each branching into the 2 fruits of Basket B (Banana, Mango), giving 3 × 2 = 6 complete paths.

(ii) Sample space S = {(Apple, Banana), (Apple, Mango), (Orange1, Banana), (Orange1, Mango), (Orange2, Banana), (Orange2, Mango)}. n(S) = 6.

(iii) Favourable outcome for one apple and one banana = {(Apple, Banana)}, so n(E) = 1. P(Apple and Banana) = 1/6 ≈ 0.167 or 16.7%. (This uses the fact that Basket A actually contains 3 separate, equally likely fruits, not 2 fruit “types” — since there are two identical oranges, picking “an orange” is twice as likely as picking “the apple.”)

Q2. 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?

Answer:

(i) Total pens = 3 + 4 + 2 = 9. Since the pen is replaced, both picks have the same probabilities: P(Red) = 3/9, P(Black) = 4/9, P(Green) = 2/9. The tree diagram has 3 first-pick branches (R, B, G), each splitting into the same 3 second-pick branches, giving 9 possible colour-pairs: (R,R), (R,B), (R,G), (B,R), (B,B), (B,G), (G,R), (G,B), (G,G).

(ii) The matching-colour paths are (R,R), (B,B), and (G,G). Because the three colours are not equally represented in the box, each path must be weighted by its own probability rather than treated as 3-out-of-9 equally likely outcomes: P(R,R) = (3/9)×(3/9) = 9/81, P(B,B) = (4/9)×(4/9) = 16/81, P(G,G) = (2/9)×(2/9) = 4/81. Adding these: P(same colour) = (9 + 16 + 4)/81 = 29/81 ≈ 0.358 or 35.8%.

End-of-Chapter Exercises (Page 173)

Q1. 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.

Answer: (i) 0 (ii) sample space (iii) 1 (iv) 1/2 (or 0.5).

Q2. 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) — Answer: The number of students who like football is a frequency of 15, and the relative…

Answer: The number of students who like football is a frequency of 15, and the relative frequency is 15/50 = 3/10 = 0.3.

Q3. Which of the following experiments has 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.

Answer:

(i) Not equally likely — whether a car starts depends on battery health, fuel, and engine condition, not a 50/50 random process.

(ii) Equally likely — a fair coin gives Heads and Tails each a probability of 1/2.

(iii) Equally likely — each of the 6 faces of a fair die has probability 1/6.

(iv) Not equally likely — P(red) = 3/10 and P(blue) = 7/10, which are different.

(v) Equally likely (as an idealised model) — the chance of a boy or a girl is treated as approximately 1/2 each in a large population.

Q4. 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?

Answer:

(i) S = {HH, HT, TH, TT}, n(S) = 4. At least one head = {HH, HT, TH}, n(E) = 3. P = 3/4 = 0.75.

(ii) S = {1,…,10}, n(S) = 10. Even numbers = {2,4,6,8,10}, n(E) = 5. P = 5/10 = 1/2 = 0.5.

(iii) S = {1,2,3,4,5,6}, n(S) = 6. Numbers greater than 4 = {5,6}, n(E) = 2. P = 2/6 = 1/3 ≈ 0.333.

(iv) Total balls = 3 + 2 + 1 = 6. Not red = 2 blue + 1 green = 3. P = 3/6 = 1/2 = 0.5.

(v) S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, n(S) = 8. Exactly two heads = {HHT, HTH, THH}, n(E) = 3. P = 3/8 = 0.375.

Q5. A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy? — Answer: S = {strawberry, lemon, mint}, n(S) = 3. P(strawberry) = 1/3 ≈ 0.333.

Answer: S = {strawberry, lemon, mint}, n(S) = 3. P(strawberry) = 1/3 ≈ 0.333.

Q6. 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 — Answer:

Answer:

Total possible outfits = 2 × 3 = 6.

Q7. A tyre company records distances before replacement in 1000 cases: less than 4000 km — 20 cases; 4001 to 9000 km — 210 cases; 9001 to 14000 km — 325 cases; more than 14000 km — 445 cases. Find the probability that a randomly chosen tyre lasts:
(i) Less than 4000 km.
(ii) Between 4000 and 14000 km.
(iii) More than 14000 km.

Answer: This is experimental probability, based on 1000 actual recorded cases. Total cases = 1000.

(i) P(less than 4000 km) = 20/1000 = 0.02 or 2%.

(ii) Favourable cases = 210 + 325 = 535. P(between 4000 and 14000 km) = 535/1000 = 0.535 or 53.5%.

(iii) P(more than 14000 km) = 445/1000 = 0.445 or 44.5%.

Q8. The letters of the word ‘PEACE’ are placed on cards. Leela draws a card without looking.
(i) What is the probability that it is a P, E, or C?
(ii) What is the probability that it is not an E?

Answer: The letters are P, E, A, C, E, so S has 5 cards, n(S) = 5.

(i) Favourable letters (P, E, C) counting each card = {P, E, C, E}, n(E) = 4. P = 4/5 = 0.8 or 80%.

(ii) Cards that are not E = {P, A, C}, n(E) = 3. P(not E) = 3/5 = 0.6 or 60%.

Q9. A game of chance consists of spinning an arrow, 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?

Answer: S = {1,2,3,4,5,6,7,8}, n(S) = 8.

(i) P(8) = 1/8 = 0.125 or 12.5%.

(ii) Odd numbers = {1,3,5,7}, n(E) = 4. P(odd) = 4/8 = 1/2 = 0.5 or 50%.

(iii) Numbers greater than 2 = {3,4,5,6,7,8}, n(E) = 6. P = 6/8 = 3/4 = 0.75 or 75%.

(iv) All numbers 1–8 are less than 9, n(E) = 8. P = 8/8 = 1 (a certain event).

(v) Multiples of 3 = {3,6}, n(E) = 2. P = 2/8 = 1/4 = 0.25 or 25%.

Q10. 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?

Answer: Total balls = 4 red + 5 blue = 9. Since the first ball is laid aside (not replaced), the second draw is from the remaining 8 balls.

First draw: P(R) = 4/9, P(B) = 5/9. If the first ball drawn is red, the second draw is from 3 red + 5 blue = 8 balls, so P(B | first red) = 5/8. If the first ball drawn is blue, the second draw is from 4 red + 4 blue = 8 balls, so P(B | first blue) = 4/8.

(i) P(Red then Blue) = (4/9) × (5/8) = 20/72 = 5/18 ≈ 0.278.

(ii) P(Blue then Blue) = (5/9) × (4/8) = 20/72 = 5/18 ≈ 0.278.

Q11. 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 — Answer: Event with probability 0 (impossible): getting a sum of 1, since the smallest…

Answer: Event with probability 0 (impossible): getting a sum of 1, since the smallest possible sum with two dice is 1 + 1 = 2. Outcome with probability 1 (certain): getting a sum that lies between 2 and 12 inclusive, since every possible sum of two 6-sided dice falls within this range.

Q12. 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?

Answer:

(i) Rolling two dice gives n(S) = 36 equally likely ordered pairs. Prime numbers greater than 5 and at most 12 are 7 and 11. Sum = 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) — 6 outcomes. Sum = 11: (5,6),(6,5) — 2 outcomes. Total favourable = 8. P = 8/36 = 2/9 ≈ 0.222.

(ii) Total balls = 4 + 3 + 2 = 9. Drawing 2 without replacement: total ordered outcomes = 9 × 8 = 72. P(same colour) = (4/9 × 3/8) + (3/9 × 2/8) + (2/9 × 1/8) = 12/72 + 6/72 + 2/72 = 20/72 = 5/18. So P(different colours) = 1 − 5/18 = 13/18 ≈ 0.722.

(iii) S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, n(S) = 8. First coin heads AND exactly 2 heads total = {HHT, HTH}, n(E) = 2. P = 2/8 = 1/4 = 0.25.

(iv) Total 4-digit arrangements of 1,2,3,4 without repetition = 4! = 24. A number is even only if its last digit is 2 or 4: for each choice of last digit, the remaining 3 digits can be arranged in 3! = 6 ways, giving 6 + 6 = 12 favourable numbers. P(even) = 12/24 = 1/2 = 0.5.

(v) For each question, P(correct) = 1/4, P(wrong) = 3/4. Exactly 2 correct out of 3 can happen in 3 ways (C(3,2) = 3): CCW, CWC, WCC. Each such path has probability (1/4)×(1/4)×(3/4) = 3/64. Total P(exactly 2 correct) = 3 × 3/64 = 9/64 ≈ 0.141 or 14.1%.

Q13. 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?

Answer:

(i) With replacement, both draws have the same 4 options: S₁ = {(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)}.

(ii) Without replacement, the second draw excludes the ball already drawn: S₂ = {(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2),(3,4),(4,1),(4,2),(4,3)}.

(iii) n(S₁) = 4 × 4 = 16 (with replacement). n(S₂) = 4 × 3 = 12 (without replacement).

Q14. List the elements of a sample space for the simultaneous tossing of a coin and drawing of a card from a set of 6 cards numbered 1 through 6 — Answer: Coin outcomes: {H, T}. Card outcomes: {1,2,3,4,5,6}. S =…

Answer: 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) = 2 × 6 = 12.

Q15. 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}

Answer: With 3 coins, the number of heads can be 0, 1, 2, or 3, so the correct sample space is (iv) {0, 1, 2, 3}.

(i) {1, 2, 3} is incomplete — it omits 0 heads (all tails, TTT), which is possible.

(ii) {0, 1, 2} is incomplete — it omits 3 heads (HHH), which is possible.

(iii) {0, 1, 2, 3, 4} is incorrect — 4 heads is impossible since only 3 coins are tossed.

Q16. Suppose you drop a dye at random on a rectangular region measuring 3 m by 2 m. What is the probability that it will land inside a circle marked on the region with a diameter of 1 m? — Answer: This is geometric probability, where probability is the ratio of favourable area…

Answer: This is geometric probability, where probability is the ratio of favourable area to total area.

Area of rectangle = 3 × 2 = 6 m². Circle has diameter 1 m, so radius = 0.5 m, and area = π × (0.5)² = 0.25π m² = π/4 m².

P(landing inside the circle) = Area of circle ÷ Area of rectangle = (π/4)/6 = π/24 ≈ 0.131 or about 13.1%.

Practice more: Extra Questions for Class 9 Mathematics Chapter 7

Quick revision: Revision Notes for Class 9 Mathematics Chapter 7

Frequently Asked Questions

What is the formula for probability used in this chapter?
Probability of an event E, P(E) = number of outcomes favourable to E / total number of equally likely outcomes — this classical (theoretical) definition is the basis for every problem in this introductory chapter.

Can probability ever be negative or greater than 1?
No — probability always lies between 0 and 1 (inclusive): 0 means an impossible event, 1 means a certain event, and every other value represents a degree of likelihood in between.

Written by Satish

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