NCERT Solutions for Class 9 Maths Chapter 7 The Mathematics of Maybe Introduction to Probability – step-by-step answers to every exercise and in-text question of the chapter, as per the latest NCERT syllabus for 2027. Read them online or download the free question-answer PDF.
Chapter 7 turns the everyday words impossible, likely and certain into a number between 0 and 1. Two honest ways of fixing that number are developed side by side: experimental probability, read off data you actually collected, and theoretical probability, argued from a list of outcomes that are equally likely. Everything else in the chapter — sample space, event, tree diagram — exists to make that list correct, because the counting formula is only as good as the list it counts.
NCERT Solutions for Class 9 Maths Chapter 7 The Mathematics of Maybe Introduction to Probability PDF
All the answers are on this page below. The PDF also explains every answer – why it is right and how to check it – so download it for revision.
| Chapter | Download | |
|---|---|---|
| Solutions PDF (English medium) | The Mathematics of Maybe Introduction to Probability | View PDF |
| Solutions PDF (Hindi medium) | संभावनाओं का गणित — प्रायिकता का परिचय | View PDF |
NCERT Solutions for Class 9 Maths Chapter 7 The Mathematics of Maybe Introduction to Probability – Questions and Answers
All 35 questions of the chapter (textbook pages 155 – 173) are answered below, in the order of the book – in-text questions, activities and the end-of-chapter exercises. Each answer is written the way you would write it in the exam.
In-text Questions — Page 155Section 7.1 What is Probability?
No. Each of the three examples — rain today, the hockey result tomorrow, the monthly lucky draw — is a random event. We know the whole list of results that could occur, but not which one will occur.
Think and Reflect — Page 156Section 7.1.1 What is Randomness?
Because the coin gives both captains exactly the same chance, and gives neither of them any way to influence the result.
Captain A’s chance of winning the toss = 1/2
Captain B’s chance of winning the toss = 1/2
In-text Questions — Page 157Section 7.1.1 What is Randomness?
Rain is random because it depends on too many interacting factors — temperature, humidity, wind patterns, pressure — and it is so sensitive to them that no measurement is ever complete or exact enough to settle the question in advance.
Think and Reflect — Page 157Section 7.1.1 What is Randomness?
Yes. That is randomness:
- Your friend knows the complete sample space, S = {H, T}, before the coin leaves your hand.
- Your friend knows each result carries probability 1/2.
- Your friend cannot know which of the two this particular toss will give.
Exercise Set 7.1 — Page 159Section 7.1.2 The Probability Scale
Order on the scale, from 0 to 1: (ii) and (iii) → (iv) → (i).
| Event | Label | Where it sits | Reason |
|---|---|---|---|
| (i) The next Monday will come after Sunday | Certain | P = 1 | Not a random event at all. The order of the days is fixed by the calendar, so there is no other possible outcome. |
| (ii) It will snow in Mumbai in July | Impossible | P = 0 | Snow needs freezing temperatures (around 0 °C or below). Mumbai is coastal and tropical; July is its monsoon, with temperatures around 25–30 °C. The favourable outcome does not exist. |
| (iii) An elephant will walk through your classroom today | Impossible (or as near 0 as makes no difference) | P ≈ 0 | Classroom doors are not elephant-sized and elephants are not loose in towns. In a forest-edge school with a wild-elephant corridor nearby it would be “extremely unlikely” rather than strictly impossible. |
| (iv) You will greet at least one friend at school tomorrow | More likely | P close to 1 | You have many friends and you greet them every day, so it is far more likely to happen than not. It is not certain, though — tomorrow could be a holiday, or you could be absent. |
Think and Reflect — Page 163Section 7.2.3 Analysing Statistical Data Using Probability
Correct — and worth stating exactly. For a fair die the next roll is a fresh experiment with sample space S = {1, 2, 3, 4, 5, 6}, so
(the book rounds this to ≈ 0.16)
Exercise Set 7.2 — Pages 165–166Section 7.2 Measuring Probability Objectively
Check the sample first: 10 + 8 + 7 + 5 = 30 sweets. ✓
(i) Every sweet in the sample is equally likely to be picked, so count green ones against all of them.
= 4/15 ≈ 0.267 or 26.7%
(ii) The sample gives the yellow proportion; apply it to the whole bag.
Expected yellow in 600 = 600 × 7/30
= 20 × 7
= 140 yellow sweets
Check the sample: 14 + 11 + 9 + 6 = 40 students. ✓
(i)
= 0.275 or 27.5%
(ii) The Sports Club share in the sample is 9 out of 40. Apply that share to all 800 students.
Estimated number in school = 800 × 9/40
= 20 × 9
= 180 students
This one you must actually perform — the answer is your own data. Here is a worked set of results so you can see the method; your counts will differ, and that is the point.
| Toss | 1–5 | 6–10 | 11–15 | 16–20 |
|---|---|---|---|---|
| Result | H T H H T | T H T H H | H T T H T | H H T H T |
(i) Heads = 11 (ii) Tails = 9 (11 + 9 = 20 ✓)
(iii)
= 11/20
= 0.55 or 55%
(iv) This part is not answered from your data. The 21st toss is a fresh, independent trial of a fair coin.
Here the sample space is S = {bottom, top, side}, with n(S) = 3 — but you cannot write 1/3 for each. A paper cup is not symmetric, so the three outcomes are not equally likely. The only honest way to get their probabilities is to toss and count.
Sample result of 100 tosses (your own counts will differ):
| Landing position | Tally (out of 100) | Relative frequency | Experimental probability |
|---|---|---|---|
| Bottom (upright) | 22 | 22/100 | 0.22 |
| Top (upside down) | 18 | 18/100 | 0.18 |
| Side | 60 | 60/100 | 0.60 |
| Total | 100 | 100/100 | 1.00 |
Experimental P(bottom) = 22/100 = 0.22
Experimental P(top) = 18/100 = 0.18
Check: 0.60 + 0.22 + 0.18 = 1 ✓
E = ‘an even number’ = {2, 4, 6}, so n(E) = 3
P(E) = n(E)/n(S) = 3/6
= 1/2 = 0.5 or 50%
(i) From the data:
= 3/12
= 1/4 = 0.25 or 25%
(ii) From the symmetry of a fair die:
Theoretical P(3) = 1/6 = 0.1666… ≈ 0.167 or 16.7%
(iii) They differ because 12 rolls is a very small number of trials. We expect about 12 × 1/6 = 2 threes, and just one extra 3 moves the fraction from 0.167 to 0.25. If the die is rolled 60, 600 or 6000 times, the experimental probability of a ‘3’ will come closer and closer to 1/6 ≈ 0.167. This is the Law of Large Numbers.
Think and Reflect — Page 167Section 7.3.1 Sample Space
The point is that a sample space is not fixed by the weather — it is chosen by the question you are asking. Both lists below are correct sample spaces for tomorrow; they answer different questions.
| Question being studied | Suitable sample space | n(S) |
|---|---|---|
| Should I carry an umbrella? | S = {Rain, No Rain} | 2 |
| Will the school ground be fit for the match? | S = {No Rain, Drizzle, Light Rain, Heavy Rain} | 4 |
Exercise Set 7.3 — Pages 167–168Section 7.3 Elements of Probability: Sample Spaces and Events
n(S) = 6
A die has six faces and exactly one of them lands on top, so there are 6 possible outcomes.
(i) Pair each of the 6 die faces with each coin face:
(1, T), (2, T), (3, T), (4, T), (5, T), (6, T)}
n(S) = 6 × 2 = 12
(ii) Taking “between” to leave out −5 and +5:
n(S) = 9
(If the ends −5 and +5 are included, S = {−5, −4, …, 4, 5} and n(S) = 11.)
(iii) One ball from 5 green and 7 red. Label the balls so that every ball is equally likely:
If only the colour is recorded, S = {Green, Red} — but these two are not equally likely (5/12 and 7/12).
(i) Pair every snack with every drink:
(Pakora, Chai), (Pakora, Lassi),
(Bhaji, Chai), (Bhaji, Lassi)}
n(S) = 3 × 2 = 6
(ii) The event is the set of outcomes in which the snack is a samosa (the first row of the grid):
n(E) = 2
If a villager chooses snack and drink completely at random, then P(E) = 2/6 = 1/3.
Think and Reflect — Page 169Section 7.4 Tree diagrams
Yes. Read it off the tree of Fig. 7.6, whose four paths are equally likely.
E = ‘one head and one tail’ = {HT, TH}, n(E) = 2
P(E) = 2/4 = 1/2 = 0.5 or 50%
Exercise Set 7.4 — Page 169Section 7.4 Tree diagrams
(i) Basket A holds 3 fruits, so the first stage has three branches; label the two oranges separately so that all three are equally likely. Basket B holds 2 fruits, so each branch then splits in two.
(ii) Reading the six leaves:
(Orange 1, Banana), (Orange 1, Mango),
(Orange 2, Banana), (Orange 2, Mango)}
n(S) = 3 × 2 = 6, each with probability 1/6
(iii) Exactly one leaf is favourable:
or, multiplying along that path: 1/3 × 1/2 = 1/6 ✓
The box holds 3 + 4 + 2 = 9 pens. Because the pen is put back, the box is identical for your friend’s turn, so both stages carry the same branch probabilities: red 3/9, black 4/9, green 2/9.
(i) 3 colours × 3 colours = 9 ordered pairs:
(ii) “Same colour” is made of the three paths RR, BB and GG. Multiply along each, then add:
P(BB) = 4/9 × 4/9 = 16/81
P(GG) = 2/9 × 2/9 = 4/81
P(same colour) = (9 + 16 + 4)/81
= 29/81 ≈ 0.358 or about 36%
End-of-Chapter Exercises — Pages 169–173Chapter 7 review (questions marked * are the harder set)
| Part | Answer | Reason |
|---|---|---|
| (i) Probability of an impossible event | 0 | No outcome is favourable, so P = 0/n(S) = 0. |
| (ii) Set of all possible outcomes | sample space (S) | By definition; its size is the sample size n(S). |
| (iii) Probability of a certain event | 1 | Every outcome is favourable, so P = n(S)/n(S) = 1. |
| (iv) Heads on a fair coin | 1/2 (0.5 or 50%) | S = {H, T}, both equally likely, one favourable. |
The blanks are filled by relative frequency and 15/50 = 3/10 = 0.3.
Relative frequency = 15/50 = 3/10 = 0.3 or 30% (a fraction of the total)
(ii), (iii) and (v) have equally likely outcomes; (i) and (iv) do not.
| Experiment | Equally likely? | Explanation |
|---|---|---|
| (i) Car starts / does not start | No | The result depends on the battery, the fuel and the age of the car — physical facts, not symmetry. A well-serviced car starts far more often than not. |
| (ii) Tossing a fair coin once | Yes | The coin is symmetrical and unbiased, so there is no reason to favour either face: P(H) = P(T) = 1/2. |
| (iii) Rolling a fair 6-sided die | Yes | All six faces are identical in shape and weight, so each has probability 1/6. |
| (iv) A marble from 3 red and 7 blue | No (by colour) | P(red) = 3/10 and P(blue) = 7/10. The ten marbles are equally likely at 1/10 each; the two colours are not. |
| (v) A baby is a boy or a girl | Yes (taken as equally likely) | There is no reason to expect one more than the other, so each is taken as 1/2. (Birth records actually show slightly more boys — about 51 in 100 births — so 1/2 is a very close approximation.) |
At least one head = {HH, HT, TH}, n(E) = 3
P = 3/4
Even = {2, 4, 6, 8, 10}, n(E) = 5
P = 5/10 = 1/2
Greater than 4 = {5, 6}, n(E) = 2
P = 2/6 = 1/3
Not red = {B1, B2, G}, n(E) = 3
P = 3/6 = 1/2
Exactly two heads = {HHT, HTH, THH}, n(E) = 3
P = 3/8
E = {strawberry}, n(E) = 1
P(strawberry) = 1/3 ≈ 0.333 or about 33.3%
There is one candy of each flavour, so the 3 outcomes are equally likely.
Every shirt can go with every pair of pants, so there are 2 × 3 = 6 outfits.
| Shirt \ Pants | Jeans | Khakis | Shorts |
|---|---|---|---|
| Red | Red + Jeans | Red + Khakis | Red + Shorts |
| Blue | Blue + Jeans | Blue + Khakis | Blue + Shorts |
(Blue, Jeans), (Blue, Khakis), (Blue, Shorts)}
n(S) = 2 × 3 = 6
Total cases: 20 + 210 + 325 + 445 = 1000 ✓ — so each probability is (cases in that class) ÷ 1000.
| Distance (km) | Less than 4000 | 4001 to 9000 | 9001 to 14000 | More than 14000 |
|---|---|---|---|---|
| Number of cases | 20 | 210 | 325 | 445 |
| Probability | 0.02 | 0.21 | 0.325 | 0.445 |
(i)
(ii) ‘Between 4000 and 14000’ covers the two middle classes together:
P = 535/1000 = 107/200 = 0.535 or 53.5%
(iii)
Check: 0.02 + 0.535 + 0.445 = 1 ✓ — the three answers cover every one of the 1000 cases exactly once.
PEACE has 5 letters, so there are 5 cards: P, E, A, C, E. Each card is equally likely, so n(S) = 5. Note that E appears on two cards.
| Letter | P | E | A | C |
|---|---|---|---|---|
| Number of cards | 1 | 2 | 1 | 1 |
| Probability | 1/5 | 2/5 | 1/5 | 1/5 |
(i) Favourable cards: P (1 card), E (2 cards) and C (1 card).
P(P, E or C) = 4/5 = 0.8 or 80%
(ii) The cards that are not E are P, A and C.
P(not E) = 3/5 = 0.6 or 60%
Check: P(E) + P(not E) = 2/5 + 3/5 = 1 ✓
The wheel of Fig. 7.7 is cut into 8 equal sectors numbered 1 to 8, and the question tells us they are equally likely. So n(S) = 8 with S = {1, 2, 3, 4, 5, 6, 7, 8}, and every answer is (how many sectors qualify) ÷ 8.
| Part | Favourable sectors | Count | Probability |
|---|---|---|---|
| (i) 8 | {8} | 1 | 1/8 = 0.125 |
| (ii) an odd number | {1, 3, 5, 7} | 4 | 4/8 = 1/2 = 0.5 |
| (iii) a number greater than 2 | {3, 4, 5, 6, 7, 8} | 6 | 6/8 = 3/4 = 0.75 |
| (iv) a number less than 9 | {1, 2, 3, 4, 5, 6, 7, 8} | 8 | 8/8 = 1 |
| (v) a multiple of 3 | {3, 6} | 2 | 2/8 = 1/4 = 0.25 |
(ii) P(odd) = 4/8 = 1/2
(iii) P(> 2) = 6/8 = 3/4
(iv) P(< 9) = 8/8 = 1
(v) P(multiple of 3) = 2/8 = 1/4
The basket has 4 + 5 = 9 balls. The first ball is laid aside, so only 8 balls are left for the second draw.
(i) Red then blue. Multiply along that path:
P(blue second, after a red is laid aside) = 5/8 (5 blue still there, 8 balls left)
P(red then blue) = 4/9 × 5/8 = 20/72
= 5/18 ≈ 0.278
(ii) Two blue balls.
P(blue second) = 4/8 (only 4 blue left out of 8)
P(two blue) = 5/9 × 4/8 = 20/72
= 5/18 ≈ 0.278
Throwing two dice gives n(S) = 6 × 6 = 36 equally likely outcomes, and the possible sums run from 2 to 12.
An event with probability 0 — anything that cannot occur:
No pair (a, b) with a, b ≥ 1 adds to 1, so n(E) = 0
P(E) = 0/36 = 0
An event with probability 1 — anything that must occur:
All 36 outcomes satisfy it, so n(F) = 36
P(F) = 36/36 = 1
(i) S = all 36 ordered pairs (first die, second die). Prime sums greater than 5: 7 and 11.
Sum 11: (5,6), (6,5) → 2
P = 8/36 = 2/9
(ii)
Different-colour pairs = 4×3 + 4×2 + 3×2 = 26
P(different colours) = 26/36 = 13/18
(iii)
First H and exactly two H: HHT, HTH
P = 2/8 = 1/4
(iv)
Even ⇒ units digit 2 or 4: 2 × (3 × 2 × 1) = 12
P(even) = 12/24 = 1/2
(v)
Wrong question: 3 ways; its wrong option: 3 ways → 9
P(exactly 2 correct) = 9/64
(i) With replacement. The ball goes back, so each of the 4 first branches splits into 4.
(3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)}
(ii) Without replacement. The first number cannot come again, so each of the 4 branches splits into only 3.
(3,1), (3,2), (3,4), (4,1), (4,2), (4,3)}
(iii) Sizes.
n(S2) = 4 × 3 = 12
Difference = 4, exactly the four repeats (1,1), (2,2), (3,3), (4,4)
The coin gives 2 results and the card gives 6, and the two happen together, so pair every coin result with every card.
(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
n(S) = 2 × 6 = 12
(iv) {0, 1, 2, 3} is the sample space. With three coins the number of heads can be 0, 1, 2 or 3 — and nothing else.
| List | Valid? | What is wrong |
|---|---|---|
| (i) {1, 2, 3} | No | Leaves out 0. TTT is a genuine result of the experiment, and it would have no outcome to belong to. |
| (ii) {0, 1, 2} | No | Leaves out 3. HHH can happen, so the list is incomplete. |
| (iii) {0, 1, 2, 3, 4} | No | Includes 4, which is impossible — only three coins are tossed, so at most three heads can appear. |
| (iv) {0, 1, 2, 3} | Yes | Complete and with no repeats: every toss of three coins gives exactly one of these four counts. |
From Fig. 7.8 the rectangle is 3 m by 2 m and the circle has diameter 1 m, so its radius is 0.5 m. Here the outcomes are points, not a list you can count — so probability is measured by area instead.
Radius of the circle = 1/2 = 0.5 m
Area of the circle = πr² = π × (1/2)² = π/4 m²
P(lands inside the circle) = area of circle / area of rectangle
= (π/4) ÷ 6
= π/24
≈ 3.1416/24 ≈ 0.131 or about 13.1%
Chapter 7 of 8 – Maths
- Previous: Chapter 6 Measuring Space Perimeter and Area
- Next: Chapter 8 Predicting What Comes Next Exploring Sequences and Progressi
- All chapters: NCERT Solutions Class 9 Maths
Also for This Chapter
NCERT Solutions Class 9 Maths Chapter 7 The Mathematics of Maybe Introduction to Probability – Overview
| Class | Class 9 |
| Subject | Maths |
| Chapter | 7 – The Mathematics of Maybe Introduction to Probability |
| Textbook pages | 155 – 173 |
| Questions solved | 35 |
| Session | 2027 |
| Format | Full answers on this page + PDF |
NCERT Solutions Class 9 Maths Chapter 7 – FAQs
Where can I download NCERT Solutions Class 9 Maths Chapter 7 The Mathematics of Maybe Introduction to Probability PDF?
Read every answer on this page, or download the free PDF from the table at the top.
Is this for the 2027 session?
Yes, it follows the latest NCERT Class 9 Maths book for 2027.
How many questions are solved in NCERT Class 9 Maths Chapter 7?
All 35 questions of the chapter (textbook pages 155 – 173) – the in-text questions, activities and the end-of-chapter exercises.
How should I use these solutions?
Read the chapter in the NCERT book first, attempt the questions yourself, then check your answers here.