Class 9 Maths Ganita Manjari Chapter 7 Solutions (NCERT 2026–27) – The Mathematics of Maybe: Introduction to Probability

These Class 9 Maths Ganita Manjari Chapter 7 solutions cover The Mathematics of Maybe: Introduction to Probability from the new NCF-2023 textbook (2026–27). Every exercise is solved step by step — experimental and theoretical probability, sample spaces, events and tree diagrams — so you can revise the whole chapter quickly.

Class: 9 Subject: Mathematics Book: Ganita Manjari (Part 1) Chapter: 7 Exercises: 7.1–7.4, End-of-Chapter Session: 2026–27
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Chapter 7 Overview

Chapter 7 of Ganita Manjari, The Mathematics of Maybe: Introduction to Probability, introduces probability as a measure of likelihood on a scale from 0 (impossible) to 1 (certain). It distinguishes experimental probability (from actual trials) and theoretical probability (assuming equally likely outcomes), explains sample spaces and events, and uses tree diagrams for multi-step experiments. The Class 9 Maths Ganita Manjari Chapter 7 solutions below work through every exercise step by step.

Key Concepts

Probability scale: every probability lies between 0 (impossible) and 1 (certain); 0 ≤ P(E) ≤ 1.

Experimental probability = (number of times the event occurred) / (total number of trials).

Theoretical probability = (number of favourable outcomes) / (number of possible outcomes), when outcomes are equally likely.

Sample space (S): the set of all possible outcomes; its size is n(S). An event is a subset of the sample space.

Tree diagram: a branching diagram that lists all outcomes of a multi-step experiment.

Law of Large Numbers: as trials increase, experimental probability approaches theoretical probability.

Key Formulas

Theoretical: P(E) = favourable outcomes ÷ total outcomes.

Experimental: P(E) = times event occurred ÷ total trials.

P(impossible) = 0; P(certain) = 1; P(not E) = 1 − P(E).

Estimate for a population = sample probability × population size.

Exercise Set 7.1

1. Rank these events from 0 (Impossible) to 1 (Certain) and label each (Impossible / less likely / equally likely / more likely / certain) with reasons: (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.

SOLUTION (i) Certain (1) — Monday always follows Sunday. (ii) Impossible / almost 0 — Mumbai does not get snow, especially in July. (iii) Impossible (0) — an elephant in a classroom does not happen. (iv) More likely / close to certain — you almost always meet a friend at school.

Exercise Set 7.2

1. From a sample of 30 sweets — 10 red, 8 green, 7 yellow, 5 blue: (i) Probability that a randomly picked sweet is green. (ii) If the bag has 600 sweets in total, estimate how many are yellow.

SOLUTION (i) P(green) = 8/30 = 4/15 ≈ 0.27. (ii) Yellow estimate = (7/30) × 600 = 140 sweets.

2. From 40 students — 14 Science, 11 Arts, 9 Sports, 6 Debate (school has 800 students): (i) Probability a student prefers the Arts Club. (ii) Estimate how many in the whole school prefer the Sports Club.

SOLUTION (i) P(Arts) = 11/40 = 0.275. (ii) Sports estimate = (9/40) × 800 = 180 students.

3. Toss a coin 20 times and record each result. (i) heads count (ii) tails count (iii) experimental probability of heads (iv) probability of tails on the next toss.

SOLUTION (i)–(iii) Record your own data; experimental P(heads) = (heads observed) / 20. (iv) The next toss is independent and the coin is fair, so P(tails) = 1/2.

4. Toss a paper cup 100 times; record bottom / top / side and assign experimental probabilities.

SOLUTION For each landing position, experimental probability = (number of times it landed that way) / 100. (Outcomes here are not equally likely, so only experimental probability applies.)

5. What is the probability of getting an even number when rolling a fair 6-sided die?

SOLUTIONEven numbers = {2, 4, 6} → P = 3/6 = 1/2.

6. You roll a die 12 times and get a 3 three times. (i) experimental probability of a 3 (ii) theoretical probability of a 3 (iii) why they differ.

SOLUTION (i) Experimental = 3/12 = 1/4. (ii) Theoretical = 1/6. (iii) They differ because of the small number of trials; with 60, 600 or 6000 rolls the experimental value gets closer to 1/6 (Law of Large Numbers).

Exercise Set 7.3

1. When a single 6-sided die is rolled, how many possible outcomes are in the sample space?

SOLUTIONS = {1, 2, 3, 4, 5, 6}, so n(S) = 6.

2. Write the sample space S for: (i) Rolling a die and tossing a coin together. (ii) Choosing a random integer between −5 and +5. (iii) A box of 5 green and 7 red balls; one ball drawn.

SOLUTION (i) S = {1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T}, n(S) = 12. (ii) S = {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}, n(S) = 11. (iii) S = {Green, Red} (by colour).

3. A fair offers 3 snacks (Samosa, Pakora, Bhaji) and 2 drinks (Chai, Lassi). (i) List the sample space of all snack–drink combinations. (ii) List the event “Selecting Samosa as a snack.”

SOLUTION (i) S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)} — 6 outcomes. (ii) E = {(Samosa, Chai), (Samosa, Lassi)}.

Exercise Set 7.4

1. Basket A: 1 apple, 2 oranges; Basket B: 1 banana, 1 mango. One fruit is picked from each. (i) Draw a tree diagram of all pairs. (ii) List the sample space. (iii) Probability of one apple and one banana.

SOLUTION (i)–(ii) Pairs: (apple, banana), (apple, mango), (orange, banana), (orange, mango) — with 2 oranges these split further, giving 3 × 2 = 6 equally likely pairs. (iii) P(apple and banana) = P(apple from A) × P(banana from B) = (1/3) × (1/2) = 1/6.

2. A box has 3 red, 4 black, 2 green pens. You pick one (with replacement); then your friend does the same. (i) Possible colour outcomes / tree diagram. (ii) Probability that both pick the same colour.

SOLUTION Each pick: P(R) = 3/9, P(B) = 4/9, P(G) = 2/9. (ii) P(same) = (3/9)2 + (4/9)2 + (2/9)2 = (9 + 16 + 4)/81 = 29/81 ≈ 0.36.
Think and Reflect When two coins are tossed, what is the probability of getting one head and one tail? Answer. Favourable outcomes are HT and TH out of {HH, HT, TH, TT}, so P = 2/4 = 1/2.

Class 9 Maths Ganita Manjari Chapter 7 Solutions — End-of-Chapter Exercises

1. Fill in the blanks. (i) The probability of an impossible event is ___. (ii) The set of all possible outcomes is called the ___. (iii) The probability of a certain event is ___. (iv) Tossing a fair coin gives probability ___ for heads.

SOLUTION(i) 0; (ii) sample space; (iii) 1; (iv) 1/2.

2. In a survey of 50 students, 15 like football. The frequency is 15, and the ___ (relative frequency) is ___.

SOLUTIONRelative frequency = 15/50 = 3/10 = 0.3.

3. Which experiments have equally likely outcomes? Explain. (i) A car starts or does not start. (ii) Tossing a fair coin. (iii) Rolling a fair die. (iv) A marble from 3 red + 7 blue. (v) A baby is a boy or a girl.

SOLUTION (ii) and (iii) are equally likely (a fair coin and a fair die). (v) boy/girl is taken to be (approximately) equally likely. (i) is not equally likely (depends on the car); (iv) is not equally likely (more blue than red).

4. Write the sample space and find the probability: (i) Two coins tossed — at least one head. (ii) Cards 1–10 — an even number. (iii) A die — a number greater than 4. (iv) 3 red, 2 blue, 1 green ball — not red. (v) Three coins — exactly two heads.

SOLUTION (i) S = {HH, HT, TH, TT}; at least one head = 3/4. (ii) Even cards = {2,4,6,8,10} → 5/10 = 1/2. (iii) {5, 6} → 2/6 = 1/3. (iv) not red = 3 of 6 → 1/2. (v) Exactly two heads = {HHT, HTH, THH} → 3/8.

5. A bag has 3 candies: strawberry, lemon, mint. One is picked. Probability of strawberry?

SOLUTIONP(strawberry) = 1/3.

6. A child has 2 shirts (red, blue) and 3 pants (jeans, khakis, shorts). List all outfit combinations in a table.

SOLUTION 6 outfits: (Red, Jeans), (Red, Khakis), (Red, Shorts), (Blue, Jeans), (Blue, Khakis), (Blue, Shorts).

7. Tyre distances before replacement (1000 cases): <4000 km: 20; 4001–9000: 210; 9001–14000: 325; >14000: 445. Find the probability a tyre lasts: (i) less than 4000 km (ii) between 4000 and 14000 km (iii) more than 14000 km.

SOLUTION (i) 20/1000 = 0.02. (ii) (210 + 325)/1000 = 535/1000 = 0.535. (iii) 445/1000 = 0.445.

8. The letters of “PEACE” are on cards; one is drawn. (i) Probability it is a P, E or C. (ii) Probability it is not an E.

SOLUTION PEACE has 5 letters (P, E, A, C, E). (i) P, E, E, C are favourable → 4/5. (ii) not E = {P, A, C} → 3/5.

*9. A spinner is equally likely to stop at 1–8. Probability it points at: (i) 8 (ii) an odd number (iii) a number greater than 2 (iv) a number less than 9 (v) a multiple of 3.

SOLUTION (i) 1/8. (ii) {1,3,5,7} → 4/8 = 1/2. (iii) {3,4,5,6,7,8} → 6/8 = 3/4. (iv) all 8 → 1. (v) {3, 6} → 2/8 = 1/4.

*10. A basket has 4 red and 5 blue balls. One ball is drawn and set aside, then a second is drawn. Using a tree diagram: (i) Probability of red then blue. (ii) Probability of two blue balls.

SOLUTION (i) (4/9) × (5/8) = 20/72 = 5/18. (ii) (5/9) × (4/8) = 20/72 = 5/18.

*11. Throw a pair of dice. Write an event with probability 0 and an outcome with probability 1.

SOLUTION Probability 0: “the sum is 1” (impossible — minimum sum is 2). Probability 1: “the sum is between 2 and 12” (certain — always true).

*12. Write the sample space and find the probability: (i) Two dice — sum is a prime number greater than 5. (ii) 4 red, 3 green, 2 blue; two drawn without replacement — both different colours. (iii) Three coins — first is heads and exactly two heads in total. (iv) Four-digit number from 1,2,3,4 (no repetition) — the number is even. (v) 3-question MCQ (4 options each), guessing — exactly 2 correct.

SOLUTION (i) Primes > 5 up to 12 are 7 and 11: P = (6 + 2)/36 = 8/36 = 2/9. (ii) P(both different) = 1 − P(same) = 1 − (6 + 3 + 1)/36 = 26/36 = 13/18. (iii) First H and exactly two heads = {HHT, HTH} → 2/8 = 1/4. (iv) Even ⇒ last digit 2 or 4: 2 × 3! = 12 of 24 → 1/2. (v) C(3,2)(1/4)2(3/4) = 3 × 3/64 = 9/64.

*13. A box has 4 balls numbered 1–4. Use a tree diagram for: (i) Draw, record, replace, draw again. (ii) Draw and record, then draw a second without replacing. (iii) The sizes of the two sample spaces.

SOLUTION (i) With replacement: 4 × 4 = 16 outcomes. (ii) Without replacement: 4 × 3 = 12 outcomes. (iii) Sizes are 16 and 12.

*14. List the sample space for tossing a coin and drawing one card from 6 cards numbered 1–6.

SOLUTION S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} — 12 elements.

*15. Three coins are tossed and the number of heads is recorded. Which list is the sample space? (i) {1,2,3} (ii) {0,1,2} (iii) {0,1,2,3,4} (iv) {0,1,2,3}.

SOLUTION The number of heads can be 0, 1, 2 or 3, so the sample space is (iv) {0, 1, 2, 3}. (i) misses 0, (ii) misses 3, and (iii) includes 4, which is impossible.

*16. A dye is dropped at random on a 3 m × 2 m rectangle. What is the probability it lands inside a circle of diameter 1 m?

SOLUTION P = area of circle / area of rectangle = π(0.5)2 / (3 × 2) = 0.25π/6 = π/24 ≈ 0.131.

Common Mistakes to Avoid

Watch out for these

  • Giving a probability greater than 1 or less than 0 — always 0 ≤ P ≤ 1.
  • Swapping favourable and total outcomes in the fraction.
  • The Gambler’s Fallacy — past results do not change an independent event’s probability.
  • Forgetting that “without replacement” reduces the total for the second draw.
  • Writing a sample space with a missing or repeated outcome.
  • Treating non-equally-likely outcomes as equally likely (use experimental probability then).

Practice MCQs & Assertion–Reason

1. The probability of a certain event is:

(a) 0    (b) 1    (c) ½    (d) −1

2. The probability of an impossible event is:

(a) 1    (b) ½    (c) 0    (d) 2

3. The probability P(E) of any event satisfies:

(a) 0 ≤ P(E) ≤ 1    (b) P(E) > 1    (c) P(E) < 0    (d) P(E) = 2

4. The sample space of tossing one coin is:

(a) {H}    (b) {T}    (c) {H, T}    (d) {H, H}

5. The probability of getting an even number on a fair die is:

(a) 1/6    (b) 1/3    (c) 1/2    (d) 2/3

6. The probability of getting heads on a fair coin is:

(a) 1    (b) 0    (c) 1/2    (d) 1/4

7. On a fair die, the probability of a number greater than 4 is:

(a) 1/6    (b) 1/3    (c) 1/2    (d) 2/3

8. Tossing two coins, the probability of at least one head is:

(a) 1/4    (b) 1/2    (c) 3/4    (d) 1

9. The number of outcomes when two coins are tossed is:

(a) 2    (b) 3    (c) 4    (d) 6

10. The probability of getting a prime number on a fair die is:

(a) 1/3    (b) 1/2    (c) 2/3    (d) 1/6

Answer key: 1-(b), 2-(c), 3-(a), 4-(c), 5-(c), 6-(c), 7-(b), 8-(c), 9-(c), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: The probability of an impossible event is 0.

Reason: An impossible event has no favourable outcomes.

A-R 2. Assertion: The probability of any event lies between 0 and 1.

Reason: The number of favourable outcomes cannot exceed the total number of outcomes.

A-R 3. Assertion: For a fair coin, P(head) = P(tail) = 1/2.

Reason: The two outcomes are equally likely.

A-R 4. Assertion: After getting heads 5 times in a row, a tail is more likely on the next toss.

Reason: A coin has no memory, so each toss is independent.

A-R 5. Assertion: Experimental probability gets closer to theoretical probability as the number of trials increases.

Reason: This is the Law of Large Numbers.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

  • Probability measures likelihood on a scale 0 (impossible) to 1 (certain).
  • Theoretical P(E) = favourable / total (equally likely outcomes); experimental P(E) = successes / trials.
  • P(not E) = 1 − P(E).
  • The sample space S lists all outcomes; an event is a subset of S.
  • Tree diagrams list outcomes of multi-step experiments.
  • Experimental probability approaches theoretical as trials increase (Law of Large Numbers); independent events have no memory.

How to score full marks in this chapter

Write the sample space clearly first, then count favourable outcomes carefully. State the formula P(E) = favourable/total before substituting, and keep answers as simple fractions. For “without replacement” problems, reduce the total at each step; for multi-step ones, a tree diagram avoids mistakes.

Frequently Asked Questions

What is Class 9 Maths Ganita Manjari Chapter 7 about?

It introduces probability — the 0-to-1 scale, experimental and theoretical probability, sample spaces and events, and tree diagrams.

What is the formula for theoretical probability?

P(E) = number of favourable outcomes ÷ number of possible outcomes, with 0 ≤ P(E) ≤ 1.

What is the difference between experimental and theoretical probability?

Experimental probability is based on actual trial data; theoretical probability assumes equally likely outcomes. They converge as trials increase.

Are these Class 9 Maths Ganita Manjari Chapter 7 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Manjari textbook for 2026–27.

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