NCERT Solutions Class 9 Maths Chapter 7 the Mathematics of Maybe Introduction to Probability

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 9 · M AT H S

NCERT Solutions

Chapter 7: The Mathematics of
Maybe: Introduction to
Probability

NCERT Textbook — Ganita Manjari

BOOK PAGES SECTIONS QUESTIONS MEDIUM

155 – 173 12 35 English

Solutions, notes, sample papers & more at 46 pages

Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

CLASS 9 · MATHS · GANITA MANJARI

NCERT Solutions — Chapter 7: The Mathematics of
Maybe: Introduction to Probability
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.

TEXTBOOK BOOK PAGES

Ganita Manjari (Class 9) 155 – 173

SECTIONS QUESTIONS

12 35

MEDIUM

English

In-text Questions — Page 155
Section 7.1 What is Probability?

Q1 Can we predict these outcomes with 100% certainty?

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.

Why it happens: knowing the list is not the same as knowing the entry. For the lucky
draw we can say with certainty that exactly one slip will be pulled out, and that it will
be one of the slips in the box. What no amount of thinking will tell us is whose slip.
That gap between "what could happen" and "what will happen" is exactly what the
word chance names.

What we can do is grade our confidence. The book gives two ways to say it:

in words — impossible, less likely, equally likely, more likely, certain;
as a number on a scale from 0 to 1, which is what the rest of the chapter builds.

Notice that two friends can look at the same sky and give different answers — "the sun is bright,
so no rain" against "it is very hot, so rain later". Both are reading the same evidence differently.
That is a subjective probability, and it is precisely what Section 7.2 replaces with something

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

anyone can check.

Check it yourself: not everything is random. "The next Monday will come after
Sunday" can be predicted with 100% certainty, because it follows from how the week
is defined, not from chance. Certainty is possible — just not for random events.

Think and Reflect — Page 156
Section 7.1.1 What is Randomness?

THINK AND REFLECT

Q1 Such unpredictability can be useful sometimes! For example, in a cricket match, the
fact that a coin is tossed to decide which team will bat first is considered to be a fair
method. Can you explain why?

Because the coin gives both captains exactly the same chance, and gives neither of them any
way to influence the result.

Coin is unbiased ⇒ P(Heads) = P(Tails) = 1/2

Captain A's chance of winning the toss = 1/2

Captain B's chance of winning the toss = 1/2

Why it happens: a fair method must be equal and uncontrollable. The coin delivers
both. It is symmetrical, so there is no physical reason for it to land on one face more
often than the other — that is what the book means by unbiased. And a random toss
means it is allowed to fall freely, so no skill, strength, seniority or home advantage
changes the outcome. Even the choice of who calls does not matter: calling "heads"
and calling "tails" both carry probability 1/2.

Here the unpredictability is the whole point. If the result could be predicted or arranged, the
stronger side would arrange it. Because nobody can, both teams accept the result before it
happens.

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Did you know? The same reasoning is used far beyond cricket — drawing lots to
break a tie in an election, randomly allotting exam-hall seats, or picking one slip out
of a box for the school assembly. In each case fairness comes from every candidate
having the same probability, not from anybody being clever.

In-text Questions — Page 157
Section 7.1.1 What is Randomness?

Q1 Have you wondered what makes an event like rain random?

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.

Why it happens: "random" here does not mean "without a cause". Every raindrop
has a physical cause. It means that the causes are so many, and so finely balanced,
that a tiny difference in today's air — one we could never measure — leads to a
different answer tomorrow. So the exact time and place of rainfall cannot be
predicted with certainty, however good the instruments are.

This is why the sentence "it will rain tomorrow" cannot be given the value 0 or 1 in advance. But
it is not useless either: past records let us count how often rain followed conditions like today's,
and that relative frequency is a genuine, checkable estimate of the probability.

Tip: compare rain with the toss of a coin. A coin has just two outcomes and an
obvious symmetry, so we can reason out P = 1/2 without any data. Rain has no such
symmetry, so its probability must be measured from records rather than argued from
a picture. Both are random; only one can be handled theoretically.

Think and Reflect — Page 157

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Section 7.1.1 What is Randomness?
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THINK AND REFLECT

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a s emyour friend to predict the outcome of a ₹ 1 coin you toss. Do you see that your
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Yes — and the interesting part is where exactly the knowledge stops.

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one trial. Saying P(H) = 1/2 does not half-predict this toss; it says that if the toss were

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Guessing on 20 tosses: expected number right ≈ 20 × 1/2 = 10
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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Section 7.1.2 The Probability Scale

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.

EVENT LABEL WHERE REASON
IT SITS

(i) The next Certain P=1 Not a random event at all. The order of the days
Monday will is fixed by the calendar, so there is no other
come after Sunday possible outcome.

(ii) It will snow in Impossible P=0 Snow needs the air to be at or below 0 °C.
Mumbai in July Mumbai is coastal and tropical; July is its
monsoon, with temperatures around 25–30 °C.
The favourable outcome does not exist.

(iii) An elephant Impossible (or as P≈0 Classroom doors are not elephant-sized and
will walk through near 0 as makes elephants are not loose in towns. In a forest-
your classroom no difference) edge school with a wild-elephant corridor
today nearby it would be "extremely unlikely" rather
than strictly impossible.

(iv) You will greet More likely P close to You have many friends and you greet them
at least one friend 1 every day, so it is far more likely to happen than
at school not. It is not certain, though — tomorrow could
tomorrow be a holiday, or you could be absent.

(iv)
(ii), (iii) (i)

0 1/2 1
impossible less likely even chance more likely certain

The four events placed on the probability scale of Fig. 7.1.

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Why it happens: the scale is doing the same job a number line does for length.
Every event has to land somewhere between 0 and 1, and the further right it sits, the
more confident we are. What decides the position is evidence, not opinion: (i) is
settled by definition, (ii) by climate records, (iii) by what a classroom physically is, (iv)
by your own daily experience.

Tip: keep "impossible" for events with no favourable outcome — P is exactly 0 — and
use "extremely unlikely" for events that are merely rare, like (iii) at a forest-edge
school. A very small probability is still not zero, and the difference matters once you
start calculating.

Think and Reflect — Page 163
Section 7.2.3 Analysing Statistical Data Using Probability

THINK AND REFLECT

Q1 If I have rolled a 4 on a die 8 times in succession, the probability of rolling a 4 again
is still only ≈ 0.16 (assuming the die is fair). Probability does not tell you what will
happen next but predicts what will happen in the long run.

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

P(next roll is a 4) = 1/6 = 0.1666… ≈ 0.167

(the book rounds this to ≈ 0.16)

Why it happens: the die carries no record of the eight rolls that went before. Each
roll is an independent event, so the sample space and the favourable count are the
same as they were on the very first roll. There is no mechanism by which a past
outcome could change the shape of a cube.

The trap is to confuse two different questions:

Before any rolling: P(nine 4s in a row) = (1/6)9 ≈ 0.0000001 — a genuinely tiny number.
After eight 4s have already happened: those eight are now facts, not possibilities. Only the
ninth roll is still uncertain, and for it P = 1/6.

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

The rare thing is the whole run of nine, not the last roll of it. Believing that a 4 has become
"unlikely now" — or, in the other direction, that the die is "hot" — is the Gambler's Fallacy of
page 164.

Check it yourself: there is a sensible worry hiding here, but it is a different one.
Eight 4s in a row is such weak evidence for a fair die that you might reasonably start
doubting the assumption and test whether the die is loaded. That is a question about
the die, not about the ninth roll. As long as the die is fair, 1/6 stands.

Exercise Set 7.2 — Pages 165–166
Section 7.2 Measuring Probability Objectively

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.

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.

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

(ii) The sample gives the yellow proportion; apply it to the whole bag.

P(yellow) = 7/30

Expected yellow in 600 = 600 × 7/30

= 20 × 7

= 140 yellow sweets

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Why it happens: this is statistical probability — the estimate comes from data, not
from symmetry. The sweets are not equally likely by colour (there is no reason a bag
should hold equal numbers), so we cannot argue P(yellow) = 1/4 from "there are four
colours". What is equally likely is the pick of one sweet from the 30 in the sample,
and that is what licenses the fraction 7/30. Scaling to 600 then assumes the sample
is representative of the whole bag.

Tip: 140 is an estimate, not a count. A second sample of 30 might give 6 or 9 yellow
and shift the estimate to 120 or 180. A larger sample would pin it down more tightly
— that is exactly why the chapter recommends bigger, more representative samples.

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.

Check the sample: 14 + 11 + 9 + 6 = 40 students. ✓
(i)

P(Arts Club) = 11/40

= 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.

P(Sports Club) = 9/40 = 0.225

Estimated number in school = 800 × 9/40

= 20 × 9

= 180 students

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

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CLUB SAMPLE COUNT RELATIVE FREQUENCY ESTIMATE OUT OF 800

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Sports 9 9/40 = 0.225 180

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Total 40
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Why it happens: the four relative frequencies add to 1, exactly as the probabilities

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Q3 .cToss a coin 20 times and record the result each time (heads
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experimental probability of getting heads. (iv) If you toss the coin once more, what
is the probability of getting tails?
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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.

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(i) Heads = 11 (ii) Tails = 9 (11 + 9 = 20 ✓)
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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Experimental P(heads) = number of heads / total tosses

= 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.

P(tails on the next toss) = 1/2 = 0.5

Why it happens: parts (iii) and (iv) ask two different questions, and mixing them is
the classic error here. Part (iii) is experimental probability — it summarises 20 trials
that have already happened, so 0.55 is a fact about your data. Part (iv) is theoretical
probability — it is about a trial that has not happened yet, and a fair coin does not
remember that heads led 11–9. Answering (iv) with 9/20 would be the Gambler's
Fallacy in reverse.

Try This: pool the class's results. If 30 students each toss 20 times, you have 600
tosses. The combined proportion of heads will sit far closer to 0.5 than most
individual results did — the Law of Large Numbers, visible in one lesson.

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

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 (See Fig. 7.5).

bottom top side
(upside down)

Fig. 7.5, page 165 — the three ways a paper cup can land (left to right): on its bottom,
upside down on its top, and on its side.

Assign probabilities to the outcomes by using experimental probability.

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.
A typical set of 100 tosses, with the tally turned into probabilities:

LANDING TALLY (OUT OF RELATIVE EXPERIMENTAL
POSITION 100) FREQUENCY 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

Page 11 of 46

Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Experimental P(side) = 60/100 = 0.60

Experimental P(bottom) = 22/100 = 0.22

Experimental P(top) = 18/100 = 0.18
Check: 0.60 + 0.22 + 0.18 = 1 ✓

Why it happens: this experiment is in the chapter precisely to break the habit of
dividing by the number of outcomes. The counting formula P = n(A)/n(S) is valid only
when every outcome is equally likely, and the cup gives no reason to believe that: its
curved wall has far more area to rest on than either rim, so "side" dominates. With
no symmetry to argue from, evidence is the only route — which is the whole
distinction between theoretical and experimental probability.

Check it yourself: record your results in blocks of 20 and plot the running
proportion of "side". It will jump about early on and steady down as the tosses
accumulate. Use a cup of a different shape — a wider, shorter one — and the three
probabilities change, proving they are properties of that particular cup, not of the
number 3.

Q5 What is the probability of getting an even number when rolling a fair 6-sided die?

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

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%

Why it happens: the die is fair, so all six faces are equally likely and the counting
formula applies without further thought. Three of the six faces carry even numbers,
so exactly half the outcomes are favourable. The same argument gives P(odd) = 3/6
= 1/2, and P(even) + P(odd) = 1, as it must, since every roll is one or the other.

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Tip: the answer is 1/2 only because evens and odds are equally represented on this
die. On a die numbered 1, 1, 2, 3, 4, 5 the sample space of faces still has six equally
likely members, but only two show an even number, so P(even) = 2/6 = 1/3. Always
count faces, not the numbers written on them.

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 roll the die 60, 600, or 6000 times?

(i) From the data:

Experimental P(3) = number of 3s / number of rolls
= 3/12

= 1/4 = 0.25 or 25%

(ii) From the symmetry of a fair die:

S = {1, 2, 3, 4, 5, 6}, n(S) = 6; favourable = {3}, n = 1
Theoretical P(3) = 1/6 = 0.1666… ≈ 0.167 or 16.7%

(iii) They differ because 12 rolls is a very short run. In 12 rolls the expected number of 3s is 12 ×
1/6 = 2, and getting 3 instead of 2 is an ordinary fluctuation — one extra 3 shifts the fraction
from 0.167 to 0.25 because the denominator is so small.

NUMBER OF EXPECTED 3S (N ONE EXTRA 3 WHAT TO EXPECT
ROLLS × 1/6) CHANGES P BY

12 2 0.083 Wide swings; 0.25 or 0.08 are
unremarkable

60 10 0.017 Usually within about 0.10 to
0.23

600 100 0.0017 Usually close to 0.15–0.19

6000 1000 0.00017 Very close to 0.167

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

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Why it happens: the experimental probability is a fraction whose denominator is

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Tip: if you rolled 6000 times and still got about 0.25, the sensible conclusion would
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experimental probability is used to test an assumption.
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Section 7.3.1 Sample Space

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whether it will rain or not. However, if we want to include different amounts of
rainfall like drizzle, light rain or heavy rain, we need to expand the sample space to
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{No Rain, Drizzle, Light Rain, Heavy Rain} so that it better matches the level of

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Why it happens: the finer list is built by splitting the single outcome "Rain" into
Drizzle, Light Rain and Heavy Rain. Splitting keeps the list complete and non-
overlapping, so it is still a valid sample space — but notice what it does not do. It
does not make the four outcomes equally likely. Writing P(Heavy Rain) = 1/4 just
because n(S) = 4 would be wrong; these probabilities have to come from rainfall
records, exactly as on page 157.

Tip: "detailed enough" cuts both ways. Too coarse and the sample space cannot
express the event you care about — {Rain, No Rain} has no way to say "heavy rain".
Too fine (rainfall to the nearest millimetre) and you will never collect enough data to
estimate any of the probabilities. Match the detail to the decision.

Exercise Set 7.3 — Pages 167–168
Section 7.3 Elements of Probability: Sample Spaces and Events

Q1 When a single 6-sided die is rolled, what is the total number of possible outcomes in
the sample space?

S = {1, 2, 3, 4, 5, 6}

n(S) = 6

Why it happens: a die has six faces and exactly one of them ends up on top, so
there are six possible results and no more. The list satisfies both requirements of a
sample space — every result appears, and none appears twice — and since the die is
fair, all six are equally likely, each with probability 1/6.

Tip: n(S) counts outcomes, not the answers to some question about them. "Even or
odd" is a way of grouping these six outcomes into two events; it does not turn the
sample space into a two-element set.

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

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.

(i) Die and coin together. Each of the 6 die faces can pair with each of the 2 coin faces.

S = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H),

(1, T), (2, T), (3, T), (4, T), (5, T), (6, T)}

n(S) = 6 × 2 = 12

All 12 are equally likely, each with probability 1/12.
(ii) A random integer between −5 and +5. Read "between" strictly, so the two ends are not
included:

S = {−4, −3, −2, −1, 0, 1, 2, 3, 4}

n(S) = 9

If your teacher intends the endpoints to be included, then S = {−5, −4, …, 4, 5} and n(S) = 11. Say
which reading you are using — that is part of the answer.
(iii) One ball from 5 green and 7 red. Two correct sample spaces, and they are not
interchangeable:

SAMPLE SPACE N(S) EQUALLY LIKELY? USE IT FOR

S = {Green, Red} 2 No — P(G) = 5/12, P(R) = 7/12 Naming what can happen

S = {G1, …, G5, R1, …, R7} 12 Yes — 1/12 each Calculating probabilities

Why it happens: this is the single most important idea in the section. The list
{Green, Red} is a perfectly valid sample space — it is complete and has no repeats —
but its two members are not equally likely, because there are more red balls than
green. So the counting formula must not be used on it: P(green) is 5/12, not 1/2.
Labelling the balls G1…G5, R1…R7 produces 12 outcomes that are equally likely,
because the draw is at random and every ball is as easy to pick as every other. Only
then does n(A)/n(S) give the right answer: 5/12.

Page 16 of 46

Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Tip: whenever you are about to divide by n(S), stop and ask "would I bet the same
amount on each of these?" If the answer is no, refine the list until it is yes.

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.’

(i) Pair every snack with every drink:

S = {(Samosa, Chai), (Samosa, Lassi),

(Pakora, Chai), (Pakora, Lassi),

(Bhaji, Chai), (Bhaji, Lassi)}

n(S) = 3 × 2 = 6

CHAI LASSI

Samosa (Samosa, Chai) (Samosa, Lassi)

Pakora (Pakora, Chai) (Pakora, Lassi)

Bhaji (Bhaji, Chai) (Bhaji, Lassi)

(ii) The event is the set of outcomes in which the snack is a samosa — the shaded first row:

E = {(Samosa, Chai), (Samosa, Lassi)}

n(E) = 2

If a villager chooses snack and drink completely at random, then P(E) = 2/6 = 1/3.

Why it happens: the grid shows why 3 × 2 = 6 and not 3 + 2 = 5. The two choices are
made together, so each snack must be offered every drink; the sample space is one
cell of the grid, not one row or one column. The event 'Samosa as a snack' is an
entire row, which is a reminder that an event is a subset of S and may hold more
than one outcome — here two.

Page 17 of 46

Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Tip: the same 3 × 2 grid is the sample space for a tree diagram with 3 first branches
and 2 second branches — the next section's Fig. 7.6 is exactly this picture drawn
sideways.

Think and Reflect — Page 169
Section 7.4 Tree diagrams

THINK AND REFLECT

Q1 Can you calculate the probability of getting one head and one tail?

Yes. Read it off the tree of Fig. 7.6, whose four paths are equally likely.

S = {HH, HT, TH, TT}, n(S) = 4

E = 'one head and one tail' = {HT, TH}, n(E) = 2

P(E) = 2/4 = 1/2 = 0.5 or 50%

H 1/2 HH 1/4

T 1/2
H 1/2
HT 1/4

T 1/2
H 1/2 TH 1/4

T 1/2
TT 1/4

Four equally likely paths. HT and TH are two different paths — that is why one head and one tail is
twice as likely as two heads.

Page 18 of 46

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

co m
m.
Why it happens: HT and TH are two different paths through the tree — head first

m l a se
then tail, and tail first then head. They give the same description but they are
o
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separate .outcomes, g one. This is why
so the event contains two of the four leaves,anot
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seof each) = 1/2 while P(HH) = 1/4: getting one of each is twice as likely as
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agetting two heads.

. com
The common wrong answer is 1/3, from the list "two heads, one of each, two tails". That list is a
ag
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legitimate sample space — nothing is missing
s em and nothing repeats — but its three members are
not equally likely (1/4, 1/2, 1/4), so g l by 3 is not allowed. The tree protects you from this
adividing
by keeping the two orders apart.

c o m
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Check it yourself: multiply along a path instead of counting leaves. P(HT)

.
= 1/4 and
gla so add them: 1/4
comP(TH) = 1/2 × 1/2 = 1/4; the two paths make up the aevent,
a s em= 1/2. Multiply along a path, add across paths — that rule will carry you
+ 1/4
l
ag through the rest of the chapter.
m a s
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Exercise Set 7.4 — Page 169lase
Section 7.4 Tree diagrams ag

com
m .
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Q1 There are two fruit baskets A and B. Basket A has one apple and two oranges.

o m l a
g of fruits. (ii) List the
.cbasket. (i) Draw a tree diagram showing all possible pairs
Basket B has one banana and one mango. You randomly pick one fruit from each

e m a
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sample space. (iii) What is the probability of picking one apple and one banana?

se m
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asBasket B holds 2 fruits, so each branch then splits in two.
(i) Basket A holds 3 fruits, so the first stage has three branches; label the two oranges separately

a g l
so that all three are equally likely.

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g l as
a

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Banana 1/2 (Apple, Banana) 1/6
Mango 1/2
(Apple, Mango) 1/6
Apple 1/3

Banana 1/2 (Orange 1, Banana) 1/6
Orange 1 1/3
Mango 1/2
(Orange 1, Mango) 1/6
Orange 2 1/3

Banana 1/2 (Orange 2, Banana) 1/6
Mango 1/2
(Orange 2, Mango) 1/6

Basket A has 3 fruits, Basket B has 2, so the tree ends in 3 × 2 = 6 equally likely pairs.

(ii) Reading the six leaves:

S = {(Apple, Banana), (Apple, Mango),

(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:

P(apple and banana) = 1/6 ≈ 0.167

or, multiplying along that path: 1/3 × 1/2 = 1/6 ✓

Why it happens: the two oranges must be kept apart. If you write the sample space
by type — {(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)} —
you get four outcomes, and dividing gives the wrong answer 1/4. Those four are not
equally likely: picking an orange from Basket A has probability 2/3 while picking the
apple has probability 1/3, because there are two oranges and only one apple.
Labelling them Orange 1 and Orange 2 restores equal likelihood, and only then is
counting leaves legitimate.

Page 20 of 46

Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Check it yourself: add the six leaf probabilities: 6 × 1/6 = 1. Now group them by type
— (Apple, ·) totals 2/6 = 1/3 and (Orange, ·) totals 4/6 = 2/3, matching the 1 apple to 2
oranges in the basket. The tree and the basket agree.

Q2 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?

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) Three colours at each stage give 3 × 3 = 9 ordered colour pairs:

S = {RR, RB, RG, BR, BB, BG, GR, GB, GG}

RR 9/81
Red 3/9
Black 4/9
RB 12/81
Green 2/9

RG 6/81
Red 3/9

BR 12/81
Red 3/9
Black 4/9 Black 4/9
BB 16/81
Green 2/9

BG 8/81
Green 2/9

GR 6/81
Red 3/9
Black 4/9
GB 8/81
Green 2/9

GG 4/81

Two picks with replacement. The nine path probabilities add to 81/81 = 1; the three shaded-alike
paths RR, BB, GG give 29/81.

Page 21 of 46

Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

These nine pairs are not equally likely, so their probabilities must be got by multiplying along
each path, not by writing 1/9 each.
(ii) "Same colour" is made of the three paths RR, BB and GG. Multiply along each, then add:

P(RR) = 3/9 × 3/9 = 9/81

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%

COLOUR P(YOU) P(FRIEND) P(BOTH)

Red 3/9 3/9 9/81

Black 4/9 4/9 16/81

Green 2/9 2/9 4/81

Same colour — — 29/81

Different colours — — 52/81

Why it happens: the branches on one path happen one after the other, so their
probabilities multiply; different paths are alternatives, so their probabilities add.
Replacement is what makes the second-stage fractions the same as the first — the
box your friend picks from is exactly the box you picked from. Had you kept the pen,
the second denominator would be 8 and black-then-black would be 4/9 × 3/8.

Check it yourself: all nine path probabilities add to 81/81 = 1, and 29/81 + 52/81 = 1,
so different colours is the more likely result. Black is the single most likely match
(16/81) simply because there are more black pens than any other colour.

End-of-Chapter Exercises — Pages 169–173

Page 22 of 46

Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Chapter 7 review (questions marked * are the harder set)

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.

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.

Why it happens: (i) and (iii) are the two ends of the probability scale, and they
explain why no probability can lie outside it. The favourable count n(A) can never be
smaller than 0 or larger than n(S), so the fraction n(A)/n(S) is trapped between 0 and
1 — which is the statement 0 ≤ P(E) ≤ 1 in the Chapter Summary.

Tip: P(E) = 0 and P(E) = 1 are opposites in a precise sense: if E is impossible then 'not
E' is certain, and P(not E) = 1 − P(E) = 1 − 0 = 1.

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).

The blanks are filled by relative frequency and 15/50 = 3/10 = 0.3.

Frequency of 'likes football' = 15 (a count)

Relative frequency = 15/50 = 3/10 = 0.3 or 30% (a fraction of the total)

Page 23 of 46

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

co m
m.
Why it happens: the sentence deliberately puts the two words side by side.

m l a se
Frequency answers "how many?" and is a whole number that grows as you survey
o
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a g by the total, so it
m
more people.

l a se lies between 0 and 1 and can be compared across surveys of different sizes. It
aisg the relative frequency, not the frequency, that serves as an estimate of probability:
always

P(a randomly chosen student likes football) ≈ 0.3.
co m
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g l as
a
Tip: a second survey of 200 students might give a frequency of 62 — a much bigger
number — yet a relative frequency of 62/200 = 0.31, almost the same. That stability
is exactly why probability is defined as a proportion.
co m
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m l as
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a s em
l
Which of the following experiments have equally likely outcomes? Explain. (i) A
g
Q3
a driver attempts to start a car. The car starts or does not start. (ii) Tossing a fair coin

s
once. (iii) Rolling a fair 6-sided die. (iv) Choosing a marble randomly from a bag that
m a
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contains 3 red marbles and 7 blue marbles. (v) A baby is born. It is a boy or a girl.

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Only (ii) and (iii) have equally likely outcomes. (v) is treated as equally likely in school problems
but is not exactly so.
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EXPERIMENT EQUALLY EXPLANATION
a g l
a s em LIKELY?

agl (i) Car starts / does No The result depends on the battery, the fuel and the age of the
not start car — physical facts, not symmetry. A well-serviced car starts
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far more often than not.

e m
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agl
(ii) Tossing a fair coin Yes
once favour either face: P(H) = P(T) = 1/2.

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All six faces are identical in shape and weight, so each has
.
(iii) Rolling a fair 6- Yes
probability 1/6.
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sided die

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P(red) = 3/10 and P(blue) =g7/10. The ten marbles are equally
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(iv) A marble from 3 No (by colour)

g l a likely at 1/10 each; the two colours are not.

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(v) A baby is a boy or Very nearly, Birth records worldwide show slightly more boys than girls —
m .
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a girl but not exactly roughly 51 boys per 100 births. School problems model it as

.
m symmetry.
1/2 each, which is a good approximation, not a fact of

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Why it happens: two outcomes being the only two outcomes does not make them
equally likely — that is the mistake (i), (iv) and (v) are all testing. Equal likelihood has
to come from somewhere: a physical symmetry, as in the coin and the die, or from
data. In (iv) the symmetry is real but it belongs to the marbles, not to the colours,
which is why relabelling the sample space as {M1, …, M10} makes the counting
formula work again.

Tip: whenever a problem offers you exactly two outcomes, resist the pull of 1/2. Ask
what makes them equal. In (i) nothing does.

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?

(i) Two coins, at least one head.

S = {HH, HT, TH, TT}, n(S) = 4

E = {HH, HT, TH}, n(E) = 3

P(at least one head) = 3/4 = 0.75

Or by the complement: the only way to fail is TT, so P = 1 − 1/4 = 3/4.
(ii) One card from 1 to 10, even number.

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, n(S) = 10

E = {2, 4, 6, 8, 10}, n(E) = 5

P(even) = 5/10 = 1/2 = 0.5

(iii) A die, number greater than 4.

Page 25 of 46

Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

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

E = {5, 6}, n(E) = 2

P(greater than 4) = 2/6 = 1/3 ≈ 0.333

(iv) 3 red, 2 blue, 1 green — not red. Label the balls so that all six are equally likely:

S = {R1, R2, R3, B1, B2, G}, n(S) = 6

E = 'not red' = {B1, B2, G}, n(E) = 3

P(not red) = 3/6 = 1/2 = 0.5

Check with the complement: P(red) = 3/6 = 1/2, so P(not red) = 1 − 1/2 = 1/2. ✓
(v) Three coins, exactly two heads. Each coin doubles the list, so n(S) = 2 × 2 × 2 = 8.

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, n(S) = 8

E = 'exactly two heads' = {HHT, HTH, THH}, n(E) = 3

P(exactly two heads) = 3/8 = 0.375

Why it happens: in every part the working is the same two steps — write a list
whose members are equally likely, then count. The care goes into the list. In (iv) the
colours are not equally likely, so the balls are labelled first; in (v) 'exactly two heads'
has to be split into the three orders in which it can occur, because HHT, HTH and THH
are three different outcomes of the experiment even though they look alike in
words.

Tip: 'at least one' in (i) and 'not red' in (iv) are both easier through the complement,
P(not E) = 1 − P(E). With three coins, 'at least one head' would be 1 − P(TTT) = 1 − 1/8
= 7/8 — one line instead of listing seven outcomes.

Page 26 of 46

Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Q5 A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is
the probability of picking a strawberry candy?

S = {strawberry, lemon, mint}, n(S) = 3

E = {strawberry}, n(E) = 1

P(strawberry) = 1/3 ≈ 0.333 or about 33.3%

Why it happens: here — unlike the bags of coloured balls elsewhere in this exercise
— there is exactly one candy of each kind, so the three outcomes really are equally
likely and 1/3 is honest. The phrase "at random" is what guarantees it: no candy is
easier to reach than another.

Tip: change the bag to 3 strawberry, 1 lemon and 1 mint and the answer becomes
3/5, not 1/3, even though there are still three flavours. Count candies, not names.

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.

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

S = {(Red, Jeans), (Red, Khakis), (Red, Shorts),

(Blue, Jeans), (Blue, Khakis), (Blue, Shorts)}

n(S) = 2 × 3 = 6

Page 27 of 46

Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Why it happens: the two choices are made together, not instead of each other, so
the counts multiply. Each row of the table is one shirt paired with all three pants; two
rows of three cells give six cells. This is the same multiplication that gave 6 outcomes
for the die-and-coin pair in Exercise Set 7.3 and 4 outcomes for two coin tosses in
Fig. 7.6.

Tip: if the child dressed at random, each cell would have probability 1/6, so
P(wearing shorts) = 2/6 = 1/3 — the shorts column. A table is a sample space laid out
in a rectangle.

Q7 A tyre company records distances before replacement in 1000 cases. Distance (km):
Less than 4000 — 20 cases; 4001 to 9000 — 210 cases; 9001 to 14000 — 325 cases;
More than 14000 — 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.

Total cases: 20 + 210 + 325 + 445 = 1000 ✓ — so each probability is (cases in that class) ÷ 1000.

DISTANCE LESS THAN 4001 TO 9001 TO MORE THAN
(KM) 4000 9000 14000 14000

Number of cases 20 210 325 445

Probability 0.02 0.21 0.325 0.445

(i)

P(less than 4000 km) = 20/1000 = 1/50 = 0.02 or 2%

(ii) 'Between 4000 and 14000' covers the two middle classes together:

Favourable cases = 210 + 325 = 535

P = 535/1000 = 107/200 = 0.535 or 53.5%

(iii)

Page 28 of 46

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

co m
e m.
P(more than 14000 km) = 445/1000 = 89/200 = 0.445 or 44.5%
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Check: 0.02 + 0.535 + 0.445 = 1 ✓ — the three answers cover every one of the 1000 cases exactly

ag
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Why it happens: these are experimental probabilities read from real records, not
.
mamong the four distance classes — a tyre is
theoretical ones. There is no symmetry
s e
a g la to fail before 4000 km — so the numbers must
far likelier to last past 14000 km than
come from the data. What licenses the division by 1000 is that one case out of the
thousand is chosen at random, so every recorded case is equally likely to be the one
co m
picked.
e m.
m l as
.co a g
a s emthe printed classes are 'less than 4000' and '4001 to 9000', so a tyre lasting
gl exactly 4000 km falls in no class. The table treats the distances as whole kilometres
Tip:
a
s
and part (ii) asks about the two middle classes, which is how the 535 arises. In your
m a
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own tables, write classes that leave no gaps.

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The letters of the word ‘PEACE’ are placed on cards. Leela draws a card without
co
Q8
looking. (i) What is the probability that it is a P, E or C? (ii) What is the probability
m .
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that it is not an E?
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a PEACE has 5 letters, so there are 5 cards: P, E, A, C, E. Each card is equally likely, so n(S) = 5. Note

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that E appears on two cards.
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LETTER E A C

Number of cards 1 2 1 1

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1/5 1/5
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Probability 2/5

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cards: P (1 card), E (2 cards) and C (1 card).

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P(P, E or C) = 4/5 = 0.8 or 80%

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(ii) The cards that are not

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

n(not E) = 5 − 2 = 3

P(not E) = 3/5 = 0.6 or 60%

Check: P(E) + P(not E) = 2/5 + 3/5 = 1 ✓

Why it happens: the repeated E is the whole point of choosing this word. There are
4 different letters but 5 cards, and it is the cards that are drawn at random and
therefore equally likely. Answering (i) with 3/4 — three chosen letters out of four
distinct letters — ignores that the second E is a real card that Leela can pull out.

Tip: part (ii) is quicker by the complement rule: P(not E) = 1 − P(E) = 1 − 2/5 = 3/5. Also
notice (i) and (ii) are related — 'P, E or C' and 'not E' overlap but are not opposites, so
you cannot get one from the other.

Page 30 of 46

Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Q9 *A game of chance consists of spinning an arrow (see Fig. 7.7.) which comes to rest
pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely
outcomes.

8 1

7 2

6 3

5 4

Fig. 7.7, page 171 — the spinning wheel, cut into eight equal sectors numbered 1 to 8.

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?

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.

Page 31 of 46

Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

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

(i) P(8) = 1/8

(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

Why it happens: the sectors are equal in size, which is the physical reason the eight
outcomes are equally likely — the arrow has as much room to stop in one sector as
in any other. Part (iv) is the interesting one: every number on the wheel is less than 9,
so the event is the whole sample space and P = 1, a certain event. Part (iii) is a
reminder to read the inequality strictly: 'greater than 2' excludes 2 itself, giving 6
sectors, not 7.

Tip: for (v), 9 is a multiple of 3 but is not on this wheel, and 1 and 2 are not multiples
of 3 — only 3 and 6 qualify. Always intersect the description with the sample space
before counting.

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

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?

The basket starts with 4 + 5 = 9 balls. The first ball is laid aside, so only 8 balls remain for the
second draw — and which colour is missing depends on the first result. That is what the second
stage of the tree has to record.

Red 3/8 R then R = 12/72 = 1/6

Blue 5/8
Red 4/9
R then B = 20/72 = 5/18

Blue 5/9
Red 4/8 B then R = 20/72 = 5/18

Blue 4/8
B then B = 20/72 = 5/18

Drawing without replacement: the second-stage fractions have denominator 8, because one ball has
already been laid aside.

(i) Red then blue. Multiply along that path:

P(red first) = 4/9

P(blue second | red gone) = 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.

Page 33 of 46

as e
a g l
Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

co m
e m.
P(blue first) = 5/9
m l as
.co
P(blue second) = 4/8 (only 4 blue left out of 8)
m a g
l a se
g
P(two blue) = 5/9 × 4/8 = 20/72
a= 5/18 ≈ 0.278

co m
e m . ag
PATH WORKING
g l as PROBABILITY

Red, Red 4/9 × 3/8 = 12/72
a 1/6

co m
m.
Red, Blue 4/9 × 5/8 = 20/72 5/18

m as e
.co 5/9 × 4/8 = 20/72
a g l
m
Blue, Red 5/18

l a se
a g Blue, Blue 5/9 × 4/8 = 20/72 5/18

s
(12 + 20 + 20 + 20)/72 = 72/72
a
Total 1
m
m.co agl
l a se
g
Why it happens: the two draws are not independent. Removing a ball changes both
a
the total and the count of that colour, so the second-stage fractions all have

m
denominator 8 and their numerators depend on which branch you came along.

. co
m
Compare Exercise Set 7.4 Q2, where the pen was put back and both stages had

as e
comitself or not. l
denominator 9 — replacement is exactly what decides whether a two-stage tree
. a g
m
ase
repeats

agl
Check it yourself: the four leaf probabilities add to 1, which is the standard test that
se m
com
a tree is complete. Notice that 'red then blue' and 'blue then red' come out equal at
g l a
m . a
ase
5/18 even though the branch fractions differ: 4/9 × 5/8 and 5/9 × 4/8 are the same

agl
product written in a different order.

. com
*I throw a pair of 6-sided dice. Write down an event that e
s m a probability of 0 and
m gl a
has
co an outcome that has a probability of 1.
Q11

. a
a sem
agl c
.

s e m
m a
agl
Throwing two dice gives n(S) = 6 × 6 = 36 equally likely outcomes, and the possible sums run

. co
m
from 2 to 12.

as e
g l
An event with probability 0 — anything that cannot occur:
a

co m
m .
m ase
.co


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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

E = 'the sum of the two dice is 1'

No pair (a, b) with a, b ≥ 1 adds to 1, so n(E) = 0

P(E) = 0/36 = 0

Other correct answers: 'the sum is 13', 'a die shows 7', 'both dice show the same number and
the sum is odd'.
An event with probability 1 — anything that must occur:

F = 'the sum lies between 2 and 12 (both included)'

All 36 outcomes satisfy it, so n(F) = 36

P(F) = 36/36 = 1

Other correct answers: 'each die shows a whole number from 1 to 6', 'the sum is at least 2'.

Why it happens: P = 0 and P = 1 are the two ends of the scale, and they describe
events, not luck. An impossible event has an empty favourable set; a certain event
has the whole sample space as its favourable set. Everything real about two dice lies
strictly between: the commonest sum, 7, has probability only 6/36 = 1/6.

Tip: the question says "an outcome that has a probability of 1", and strictly no single
outcome of this experiment can have probability 1 — each of the 36 pairs has
probability 1/36. What can have probability 1 is the certain event, as written above.
Read it as asking for a result that is bound to happen.

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

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?

(i) Two dice, sum a prime greater than 5. The sample space is all ordered pairs, n(S) = 6 × 6 =
36. Possible sums are 2 to 12; the primes among them are 2, 3, 5, 7, 11, and those greater than
5 are 7 and 11.

Second die

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8
First die

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

All 36 equally likely sums. The 6 rose cells are the sum 7 and the 2 green cells the sum 11 — together
8 of 36.

Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes

Sum 11: (5,6), (6,5) → 2 outcomes

n(E) = 6 + 2 = 8

P = 8/36 = 2/9 ≈ 0.222

(ii) Two balls from 4 red, 3 green, 2 blue — different colours. There are 9 balls; drawing 2
without replacement gives

n(S) = 9C2 = (9 × 8)/2 = 36 equally likely pairs

Same colour: red 4C2 = 6, green 3C2 = 3, blue 2C2 = 1

n(same) = 6 + 3 + 1 = 10, so P(same) = 10/36 = 5/18

P(different) = 1 − 5/18 = 13/18 ≈ 0.722

Without the C notation: count the pairs directly as 4×3 (red-green) + 4×2 (red-blue) + 3×2 (green-
blue) = 12 + 8 + 6 = 26 different-colour pairs, and 26/36 = 13/18. ✓
(iii) Three coins — first is heads AND exactly two heads in all.

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, n(S) = 8

First coin H and exactly two heads: HHT, HTH → n(E) = 2

P = 2/8 = 1/4 = 0.25

Note that THH has exactly two heads but fails the first condition, so it is not counted.
(iv) Four-digit numbers from 1, 2, 3, 4 with no repetition — even.

n(S) = 4! = 4 × 3 × 2 × 1 = 24 numbers

Even ⇒ the units digit must be 2 or 4 → 2 choices

The other three digits can be arranged in 3! = 6 ways

n(E) = 2 × 6 = 12

P(even) = 12/24 = 1/2 = 0.5

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

(v) Three multiple-choice questions, guessing, exactly 2 correct. Each question has 4
options, so

n(S) = 4 × 4 × 4 = 64 equally likely answer-sheets

Choose which 2 of the 3 questions are right: 3 ways (QQ×, Q×Q, ×QQ)

Each right question: 1 way; the wrong one: 3 wrong options

n(E) = 3 × 1 × 1 × 3 = 9
P(exactly 2 correct) = 9/64 ≈ 0.141 or about 14%

Why it happens: every part is the same discipline — build a list of equally likely
outcomes first, then count. In (i) the pairs must be ordered, because (1,6) and (6,1)
are two different throws; treating the 11 sums 2…12 as the sample space would
wrongly give 2/11. In (ii) the pairs are unordered, since two balls come out together;
the complement is used because 'same colour' has three cases and 'different colour'
has three too, but the same-colour count is smaller. In (iv) the condition sits on the
units digit, so it is fixed first and the rest arranged freely. In (v) the position of the
wrong answer matters, which is why the factor 3 appears.

Tip: in (iv) the answer 1/2 has a neat reason. Of the four digits, two are even and two
are odd, and each digit is equally likely to occupy the units place among the 24
arrangements — so exactly half the numbers end in an even digit.

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?

(i) With replacement. The ball goes back, so all four numbers are available again at the second
draw: 4 branches, then 4 more from each.

Page 38 of 46

as e
a g l
Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

co m
e m.
m (1, 1)
l as
.co a g
sem
1
a
agl
2 (1, 2)
3
4 (1, 3)
co m
e m . ag
as
(1, 4)
a g l
1 (2, 1)
m
1
co
m.
2 (2, 2)
m as e
l
3
.co g
a 3)
em
4 (2,
l a s 2

ag (2, 4)

m a s
.co 1 agl
(3, 1)
m
3

l a se
a g 2 (3, 2)
3
4 (3, 3)
co m
.
4
e
s4) m
m l a
.co
(3,
m ag
l a se
g
(4, 1)
a 1
2 (4, 2)
se m
com g l a
. a
3
m
ase
4 (4, 3)
agl (4, 4)

. c om
With replacement: every first number can be followed by any of the four m
s e
com gla
numbers, so n(S) = 4 × 4 =

. a
16.

a s em
agl c
m .
e
S1 = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4),
m a s
m . co
(3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)}
agl
l a se
ag The first ball is kept out, so the number drawn first cannot appear
(ii) Without replacement.
again: each of the 4 branches splits into only 3.

com
m .
m ase
.co


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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

(1, 2)
2
3
(1, 3)
4

(1, 4)

1 (2, 1)
1
3
(2, 3)
4
2
(2, 4)

3 (3, 1)
1
2
(3, 2)
4
4
(3, 4)

(4, 1)
1
2
(4, 2)
3

(4, 3)

Without replacement: the first number cannot repeat, so each branch splits into only 3, giving n(S) = 4
× 3 = 12.

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) Sizes.

Page 40 of 46

Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

n(S1) = 4 × 4 = 16

n(S2) = 4 × 3 = 12

Difference = 4, exactly the four repeats (1,1), (2,2), (3,3), (4,4)

Why it happens: replacement decides how wide the second stage of the tree is. Put
the ball back and the box is unchanged, so the second draw has the same 4 choices
as the first — and the pairs (k, k) are possible. Keep the ball out and one choice has
been used up, so the second draw has 3 choices and no pair can repeat a number.
The four missing outcomes are precisely the four repeats, which is why 16 − 12 = 4.

Tip: the order matters in both lists: (1,2) means 1 first then 2, while (2,1) is the other
way round, and both are separate leaves of the tree. If the question had asked only
which two numbers came out, without order, S2 would shrink to 12 ÷ 2 = 6
unordered pairs.

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.

The coin gives 2 results and the card gives 6, and the two happen together, so pair every coin
result with every card.

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

COIN \ CARD 1 2 3 4 5 6

H (H, 1) (H, 2) (H, 3) (H, 4) (H, 5) (H, 6)

T (T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6)

Page 41 of 46

Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Why it happens: the coin result does not affect the card, and the card does not
affect the coin, so no combination is ruled out and none is favoured: all 12 outcomes
are equally likely at 1/12 each. This is the same list as Exercise Set 7.3 Q2(i), where a
die replaced the six cards — a good sign that the structure, not the props, is what
matters.

Tip: with this sample space you can answer combined questions at once. P(heads
and an even card) = |{(H,2), (H,4), (H,6)}| / 12 = 3/12 = 1/4, which is also 1/2 × 1/2 —
multiplying along a two-stage path.

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}

(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, No Includes 4, which is impossible — only three coins are tossed, so at most
3, 4} three heads can appear.

(iv) {0, 1, 2, Yes Complete and with no repeats: every toss of three coins gives exactly one of
3} these four counts.

Why it happens: a sample space has to satisfy both rules of Section 7.3.1 at once —
nothing left out and nothing that cannot occur. Lists (i) and (ii) break the first rule
and list (iii) breaks the second. Adding an impossible outcome is not harmless: it
would receive probability 0 and would tempt you to divide by 5.

Page 42 of 46

Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Tip: {0, 1, 2, 3} is a correct sample space but its four outcomes are not equally likely,
so P(2 heads) is not 1/4. Going back to the eight equally likely outcomes {HHH, HHT,
HTH, HTT, THH, THT, TTH, TTT} gives P(0) = 1/8, P(1) = 3/8, P(2) = 3/8, P(3) = 1/8 —
which add to 1. Being a valid sample space and being usable with the counting
formula are two different things.

Q16 *Suppose you drop a dye at random on the rectangular region shown in Fig. 7.8.

3m

2m

Fig. 7.8, page 173 — the rectangular region, with the circle inside it (drawn to scale).

What is the probability that it will land inside the circle with a diameter of 1 m?

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.

Page 43 of 46

as e
a g l
Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

co m
e m.
m l as
m .co 3m a g
ase
agl

co m
m . ag
l a se
ag 1 m
2m
co m
em.
m l as
m .co a g
l a se
a g
m a s
m.co agl
l a se
a g
The dye can land anywhere in the 6 m² rectangle; it lands in the circle only over the shaded disc of
area π/4 m².
co m
m .
as e
. coofmthe rectangle = 3 × 2 = 6 m² a g l
m
ase
Area

agl Radius of the circle = 1/2 = 0.5 m

se m
com a
Area of the circle = πr² = π × (1/2)² = π/4 m²
. a g l
m
ase
agl
P(lands inside the circle) = area of circle / area of rectangle

= (π/4) ÷ 6
co m
m .
e
= π/24
m l as
m .co
≈ 3.1416/24 ≈ 0.131 or about 13.1%
a g
l a se
ag
.c
Using π ≈ 22/7 gives (22/7)/24 = 22/168 = 11/84 ≈ 0.131 — the same to three decimal places.

s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Why it happens: "at random" here means every point of the rectangle is as likely to
be hit as every other, and no region is favoured. Under that assumption the chance
of landing in a part of the region is proportional to how much area that part
occupies — so the counting formula n(A)/n(S) becomes area(A)/area(S). The circle
covers less than an eighth of the rectangle, and the answer says so.

Tip: the answer stays π/24 wherever the circle is placed, as long as it lies wholly
inside the rectangle — only the areas matter, not the position. And notice the circle
covers about 13% of the sheet even though its diameter is a third of the rectangle's
length: area shrinks with the square of the length.

Chapter at a glance
Probability measures how likely an event is, on a scale from 0 to 1. For an event E, 0 ≤
P(E) ≤ 1. P(E) = 0 means impossible, P(E) = 1 means certain, P(E) = 1/2 means an even chance.
Nothing outside [0, 1] is a probability.
A random experiment is one you can repeat, whose full list of possible results you know in
advance, but whose result on any single trial you cannot know. The complete list of possible
results is the sample space S; each item in it is an outcome; the number of items is the
sample size n(S). An event is any subset of S — one outcome, several outcomes, or none.
Experimental probability = (number of times the event occurred) ÷ (total number of trials).
It is a relative frequency — it describes the data you have, so different sets of trials give
different values.
Theoretical probability P(A) = (number of favourable outcomes) ÷ (number of possible
outcomes). This division is legitimate only when every outcome in the list is equally
likely. If the outcomes are not equally likely, counting them gives a wrong answer — you
must go back and rewrite the sample space so that its members really are equally likely.
By the Law of Large Numbers, the more trials you perform, the closer the experimental
probability tends to settle to the theoretical probability. It does not say the two must agree
in a short run, and it does not make a run of one result "due" to be corrected — that
mistake is the Gambler's Fallacy. Coins and dice have no memory: independent trials.
A tree diagram lists the outcomes of a multi-step experiment: one complete path from root
to leaf is one outcome. Multiply along a path to get its probability; add the paths that make
up an event. Every tree's leaf probabilities add to 1.

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Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability AglaSem · NCERT Solutions

Quick revision

IDEA IN SYMBOLS / WHAT IT REALLY SAYS WHERE STUDENTS SLIP
FORMULA

Outcome one element of S A single, complete result of Calling a group of results
one trial one outcome

Sample space S = {o1, o2, …, on} Every possible outcome, each Leaving out an outcome, or
listed exactly once listing one twice

Sample size n(S) How many outcomes S Counting types instead of
contains outcomes

Event E⊆S Any selection of outcomes Forgetting that E may hold
from S 0, 1 or many outcomes

Experimental (times event A relative frequency Expecting it to equal the
probability occurred)/(total trials) measured from data theoretical value exactly

Theoretical P(A) = n(A)/n(S) Fraction of the equally likely Using it when the outcomes
probability outcomes that are favourable are not equally likely

Probability scale 0 ≤ P(E) ≤ 1 0 impossible · 1/2 even Writing an answer bigger
chance · 1 certain than 1, or a negative one

Complement P(not E) = 1 − P(E) Every trial either gives E or Adding when you should
does not subtract

Two-stage tree P(path) = product of its The branches on one path Multiplying when you
branches happen one after the other should add, and the reverse

Independence P stays the same whatever A fair die has no memory Gambler's Fallacy: thinking
happened before a 6 is now 'due'

Without second denominator drops The first draw really has Reusing the first-stage
replacement by 1 changed the box fractions at the second
stage

Law of Large experimental → More trials, steadier relative Reading it as a promise
Numbers theoretical frequency about the next trial

Page 46 of 46