The Mathematics of Maybe: Introduction to Probability: Class 9 Ganita Manjari · Chapter 7

Ganita Manjari Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 26 Sep 2026 · Updated 28 Sep 2026

82 min
3 simulations
31 practice questions
32 MCQs

What you will learn in The Mathematics of Maybe: Introduction to Probability

Describe a random experiment and explain why its result cannot be predicted even when every possible result is known
Place events on the probability scale from 0 (impossible) to 1 (certain), writing probabilities as fractions, decimals and percentages
Find the experimental probability (relative frequency) of an event from trials or survey data, and use a sample to estimate counts in a population
Calculate the theoretical probability of an event when all outcomes are equally likely
Explain why experimental and theoretical probabilities differ, how they come together over many trials, and why past results do not change the next one
Write sample spaces and events as sets and count their elements
Draw tree diagrams for experiments with two or three stages, with or without replacement, and use them to find probabilities

Original companion text aligned to NCERT Mathematics Class 9, Chapter 7, pp. 155–173. Not affiliated with or endorsed by NCERT.

7.1 What is Probability?

Will the school bus be late this morning? Will our team win tomorrow's kho-kho final? Will your chit come out when the class draws lots for the one spare seat on the planetarium trip? Nobody can answer these questions with complete confidence. Each has a few possible answers, and we have to wait to see which one actually happens.

Length has the ruler and mass has the balance. Probability (प्रायिकता) is the tool mathematics uses to put a number on chance: how likely something is to happen. With it we can compare one “maybe” with another instead of just shrugging.

Situations like these are random. The possible results are known in advance; the actual result is not. In everyday talk we describe such situations with words such as impossible, unlikely, fifty–fifty, likely and certain.

Where do such judgements come from? Before a cricket match, two friends argue about the result. Ria thinks her team will probably win, because its spinners love this pitch. Imran doubts it, because the other side's openers are in great form. Each is weighing the same facts in a personal way. A likelihood judged like this, from someone's own reading of a situation, is a subjective probability. Two sensible people can disagree about it.

Can we do better than a hunch? This chapter builds two objective methods for measuring chance: one uses data, the other uses careful counting. Before that, we need a clear idea of what random means, and a scale on which to measure chance.

7.1.1 What is Randomness?

Shuffle a box of chits carrying your classmates' names and pull one out. You could list every name that might appear, yet nobody can tell whose name it will be this time. That built-in unpredictability is called randomness.

Definition

Random experiment

An action we can carry out again and again under the same conditions. We know the list of possible results, but not the result of any one try.

ExampleRolling a die: the result will be one of 1 to 6, but we cannot tell which.

Each single try of a random experiment is a trial (अभिप्रयोग), and each possible result is an outcome. Look at Fig. 7.1. The six faces of the die are known before we start. The rolls themselves, though, jump about with no pattern we could use to predict the next one.

4/4Any of the six faces can turn up on the next roll.

Fig. 7.1We can list every possible result of a roll, but we cannot say which one comes next.NCERT p. 156

The top row shows all six faces of a die, from one to six, which we know before rolling. Below is a row of eight slots for actual rolls; the last one holds a question mark. In the first step nothing has been rolled yet. In the second step the first three slots fill with a five, a two and another five. In the third step seven rolls are shown: five, two, five, six, one, one and three, with no pattern to follow. In the last step dashed lines run from every one of the six faces to the question-mark slot, showing that any face can come next.

Toss a fair coin. Neither face has an advantage, so the chances are shared equally. We give heads a probability of $\frac{1}{2}$ and tails a probability of $\frac{1}{2}$ . Attaching numbers like these to the possible results of a random experiment is what the branch of mathematics called probability is about. The numbers describe how likely each result is; they never promise which result you will get.

Even rain behaves like a random experiment. Whether a shower reaches your street tomorrow hinges on how a huge mass of air moves, warms and cools. Nudge the starting conditions very slightly and a computer forecast can end up quite different. That is why a weather service never promises rain: it gives a chance of rain, worked out from its models and from years of records.

7.1.2 The Probability Scale

Every probability is a number from 0 to 1. At one end, 0 belongs to events that cannot happen at all; at the other, 1 belongs to events that are bound to happen. Everything else sits in between, and the nearer its number is to 1, the more likely the event.

The probability scale

0 \leq P (E) \leq 1

where

$P (E)$: the probability of an event $E$

The same probability can be written as a fraction, a decimal or a percentage. Suppose a weather app gives the chance of rain in Chennai tomorrow as 0.8. That is $\frac{4}{5}$ , or 80%, so rain is more likely than not. A probability of 0.5, or 50%, means the event and its opposite are equally likely: an even chance.

Fig. 7.2 turns this into a picture. A jar holds ten marbles, some blue and the rest orange, and we take one without looking. Each time an orange marble is swapped for a blue one, the chance of blue creeps a tenth further along the scale. With ten blue marbles it reaches 1.

2/5Three of ten are blue: less likely, $P = 0.3$ .

Fig. 7.2As more of the ten marbles are blue, the chance of picking a blue one climbs the scale.NCERT Fig. 7.1 · p. 158

A jar holds ten marbles, each either blue or orange; the orange ones have dashed outlines and the blue ones are solid with a shine. Below it a scale runs from 0, marked impossible, through 0.5, marked even chance, to 1, marked certain, with a tick at every tenth. In step one no marble is blue and the pointer sits at 0: impossible. In step two three of the ten marbles are blue and the pointer moves to 0.3: less likely. In step three five are blue and the pointer stands at 0.5: even chance. In step four eight are blue and the pointer is at 0.8: more likely. In step five all ten are blue and the pointer reaches 1: certain.

Table 7.1Some events placed on the probability scaleNCERT p. 158

Event	Place on the scale	Why
Pulling a green ball from a bag of only red balls	impossible (0)	There is no green ball to pull.
Drawing a king from a full pack of 52 playing cards	less likely	Only 4 of the 52 cards are kings.
Getting an even number on a die	even chance (0.5)	Three of the six faces, 2, 4 and 6, are even.
Getting more than 1 on a die	more likely	Five of the six faces are bigger than 1.
Getting a number less than 7 on a die	certain (1)	Every face shows a number below 7.

Quick check

1 of 4

Which of these probabilities describes an event that is more likely than not?

0.35

\frac{1}{2}

0.620

Show answer

Answer (C) 0.62

Only 0.62 is bigger than 0.5. A probability of exactly

\frac{1}{2}

is an even chance, not more likely.

7.2 Measuring Probability Objectively

A subjective guess depends on who is guessing. To make probability a fair measurement, anyone must be able to check how the number was found. There are two main routes.

Count what happened. Toss, roll or spin many times yourself, or use records that someone has already kept. Then see what share of the time the event turned up. That share is the event's experimental probability.
Count what could happen. If the set-up is perfectly balanced, every outcome deserves the same share of the chance. Then the share of outcomes that give the event is its theoretical probability.

Compare

Two routes to a probability

Experimental (from evidence)

Needs trials or recorded data.
The answer can change each time you repeat the experiment.
Works even when the outcomes are not equally likely.

Theoretical (from reasoning)

Needs no experiment at all.
Always gives the same answer.
Works only when the outcomes are equally likely.

7.2.1 Experimental Probability: Performing Observations or Experiments

Before collecting data, it helps to know exactly what we are counting. Write down every result the experiment can give, put the list inside curly brackets, and you have its sample space, usually called $S$ . Each item in the list is one outcome. Fig. 7.3 builds the sample spaces for a coin and a die.

3/3Gathered in curly brackets, the outcomes form the sample space $S$ .

Fig. 7.3Listing every outcome of an experiment gives its sample space

S

.NCERT Fig. 7.2 and 7.3 · p. 160

Two experiments sit side by side: tossing a coin on the left and rolling a die on the right. In the first step we see only the spinning coin, marked with a question mark, and the die. In the second step the possible outcomes drop down below each one: a head and a tail for the coin, and the six faces from one to six for the die. In the third step each group of outcomes is enclosed in a box labelled S and written as a set: S equals H, T for the coin and S equals 1, 2, 3, 4, 5, 6 for the die.

Example 1NCERT Example 1 · p. 160

Listing sample spaces

Write the sample space for each experiment. (a) Spinning a spinner with four equal parts coloured red, yellow, green and blue. (b) Picking one chit from a box of chits numbered 1 to 5. (c) A footballer taking one penalty kick.

Try it yourself first. Then reveal the solution one step at a time.

Now suppose we actually carry out an experiment many times and keep count. The fraction of trials in which our event happened is its experimental probability.

Definition

Experimental probability

The fraction of trials in which an event actually happened. It is also called the relative frequency of the event.

ExampleIf a bottle cap lands top-up in 13 of 40 flips, the experimental probability of top-up is $\frac{13}{40} = 0.325$ .

Experimental probability

experimental probability of E = \frac{trials in which E happened}{total number of trials}

Example 2NCERT Example 2 · p. 160

A die rolled 40 times

Kabir rolls a die 40 times, and a 6 turns up 9 times. Find the experimental probability of rolling a 6 as a fraction, a decimal and a percentage.

Try it yourself first. Then reveal the solution one step at a time.

Notice the words in Kabir's rolls. Another student rolling the same die 40 times will almost surely get a different count. Experimental probability describes the data you actually collected. That makes it valuable in science, sport and business, where we often have plenty of records but no neat formula.

Try thisNCERT Exercise Set 7.2, Q3 · p. 165

Pool your tosses

You’ll need

a coin
a partner or a small group
pencil and paper

Predict first

You toss a coin 20 times. How many heads will you get?

exactly 10, every timesomewhere near 10, but not always exactly 1020, if you toss carefullynothing can be said at all

Toss the coin 20 times and tally the heads and tails.
Work out your relative frequency of heads: number of heads ÷ 20.
Combine your tosses with a partner's, then with the whole group's.
Compare the relative frequencies as the number of tosses grows.

Table 7.2Your results

Tosses	Number of heads	Relative frequency of heads
20 (mine)
40 (with a partner)
Whole group

Your entries are saved automatically.

Can’t do it now? Try the simulation · Fig. 7.4

SimulationCoin-toss lab

A virtual coin that you can toss 1, 10, 100 or 1000 times at a press. A line chart shows the fraction of heads so far against the number of tosses, with a dashed line at the theoretical value 0.5 for comparison.

Fig. 7.4Toss a virtual coin in batches and watch the fraction of heads.

Try this

Toss 10 at a time, five times. How far from 0.5 does the line wander in the first 50 tosses?
Now toss 1000 at a time, a few times. What happens to the wobbles?
Reset and start again. Is the early part of the line the same as last time?

Quick check

1 of 3

A bottle cap is flipped 80 times and lands top-up 28 times. What is the experimental probability of top-up?

Show answer

Answer 0.35

\frac{28}{80} = 0.35

7.2.2 Theoretical Probability

Sometimes we need no experiment at all. A well-made coin, a balanced die and a box of identical, well-mixed chits have something in common: nothing favours one outcome over another. We then call the outcomes equally likely (समप्रायिक).

Definition

Equally likely outcomes

Outcomes that each have the same chance of happening, because nothing in the set-up favours any one of them.

ExampleThe six faces of a fair die.

When the outcomes are equally likely, we find a probability by counting. The outcomes that make an event happen are called favourable to it.

Theoretical probability (equally likely outcomes)

P (E) = \frac{number of outcomes favourable to E}{total number of possible outcomes}

Because it comes from reasoning alone, a theoretical probability is the same for everyone who works it out. It describes a perfectly balanced set-up and tells us what to expect over many trials, not what the next trial will give.

Example 3NCERT Example 3 · p. 161

Three events for one die

A fair die is rolled once. Find the probability of getting (a) a 2, (b) a multiple of 3, (c) a number other than 1.

Try it yourself first. Then reveal the solution one step at a time.

Words work the same way as dice. In Fig. 7.5, each letter of ARUNACHAL, as in Arunachal Pradesh, sits on its own tile, and one tile is picked without looking.

2/33 favourable tiles, so $P (A) = \frac{3}{9} = \frac{1}{3}$ .

Fig. 7.5Count the favourable tiles, then divide by all the tiles.NCERT p. 161

Nine tiles in a row spell A, R, U, N, A, C, H, A, L. In the first step all the tiles look the same and a brace notes that there are 9 equally likely tiles. In the second step the three A tiles rise, turn green and get tick marks; the text says 3 of the 9 tiles show A, so the probability of A is three-ninths, or one-third. In the third step the four vowel tiles, the three As and the U, rise, turn pink and get tick marks, giving a probability of four-ninths for a vowel.

Example 4NCERT Example 4 · p. 161

Picking a letter tile

The 9 letters of ARUNACHAL are written on identical tiles and mixed in a bag. One tile is drawn at random. Find the probability that it shows (a) the letter A, (b) a vowel, (c) the letter Z.

Try it yourself first. Then reveal the solution one step at a time.

Theory says a fair die gives a multiple of 3 with probability $\frac{1}{3}$ . Does real rolling agree? Test it in the lab below, first with a handful of rolls and then with hundreds.

SimulationDie-rolling lab

A virtual die that you can roll 1, 10, 100 or 1000 times at a press. A bar chart shows how often each face has come up, with a line at the theoretical share of one-sixth, and a counter compares how often the event 'a multiple of 3' has happened with its theoretical probability of one-third.

Fig. 7.6Roll a virtual die and compare each face's share with the theoretical

\frac{1}{6}

Try this

Roll 12 times. Did each face come up exactly twice?
Roll 600 times. How close is the share of 3s and 6s to $\frac{1}{3}$ ?

Quick check

1 of 3

A bag holds 4 red, 5 green and 3 white beads of the same size. One bead is taken without looking. What is

P (green)

\frac{5}{7}

\frac{5}{12}

\frac{1}{3}

\frac{7}{12}

Show answer

Answer (B) $\frac{5}{12}$

There are

4 + 5 + 3 = 12

equally likely beads and 5 are green, so

P = \frac{5}{12}

. The tempting

\frac{5}{7}

compares green with the other beads instead of with all of them.

7.2.3 Analysing Statistical Data Using Probability

Many events have no symmetry to reason with. Will a customer who walks into a shop buy an umbrella? Will a new phone battery still hold its charge after three years? Here the honest route is data. A shopkeeper stocks umbrellas using last monsoon's sales. A cricket analyst checks how often a batter gets out to spin. A doctor counts how many patients in a trial got better.

Example 5NCERT Example 5 · p. 162

Planning the canteen's breakfast

A school canteen asks 60 students of Class 9 to name their favourite breakfast. The replies are: poha 21, idli 18, paratha 15 and upma 6. (a) One of these 60 students is picked at random. Find the probability that the student's favourite is idli. (b) The school has 900 students. About how many plates of poha should the canteen plan for on a “favourites day”?

Try it yourself first. Then reveal the solution one step at a time.

3/3Scaled up to 900 students: about $0.35 \times 900 = 315$ choose poha.

Fig. 7.7The sample gives a probability; multiplying by 900 gives an estimate for the whole school.NCERT p. 162

Sixty small shapes stand for the 60 students in a canteen survey, with a legend: 21 circles for poha, 18 squares for idli, 15 diamonds for paratha and 6 triangles for upma. In the second step every shape except the poha circles fades, and the probability of poha is shown as 21 out of 60, which is 0.35. In the third step a long bar for the whole school of 900 students appears, with 35 percent of it shaded to show about 315 students choosing poha and about 585 choosing something else.

The 60 students who answered form a sample (प्रतिदर्श). The 900 students of the school form the population we really care about. Asking everyone is slow and costly, so we study a sample and scale up, as in Fig. 7.7.

Would you trust the canteen's plan if all 60 students came from Class 9? Younger children may like quite different food. A sample should be large enough, and it should include every kind of member of the population: here, students of all ages. Choosing a sample with this kind of care is called sampling.

Experiment and theory in the long run

Toss a perfectly fair coin ten times and you might well see 7 heads. Theory still says $\frac{1}{2}$ , while your data say 0.7. Neither is wrong. The theoretical value describes the ideal coin, and the experimental value describes these particular ten tosses. With so few tosses, chance has plenty of room to push the count around.

What happens as the trials pile up? Fig. 7.8 follows one long run of tosses of a fair coin. Early on, the fraction of heads leaps about. After thousands of tosses it hugs the theoretical value 0.5.

4/4After 10 000 tosses: 0.5009, almost exactly one-half.

Fig. 7.8One long run of a fair coin: the fraction of heads jumps about early, then settles near 0.5.NCERT p. 163

A line chart with the number of coin tosses along the bottom, on a scale where each mark is ten times the one before (1, 10, 100, 1000 and 10 000), and the fraction of tosses that were heads up the side, from 0 to 1. A dashed line marks the theoretical value 0.5. In step one only the first 10 tosses are drawn: the fraction starts at 1 and is 0.80 after 10 tosses. In step two the line reaches 100 tosses at 0.56. In step three it reaches 1000 tosses at 0.523. In step four it reaches 10 000 tosses at 0.5009, lying almost on the dashed line.

LawLaw of large numbers

When a random experiment is repeated a very large number of times, the experimental probability of an event settles closer and closer to its theoretical probability.

The law is about the fraction, not the raw count. In the run shown, 10 000 tosses gave 5009 heads and 4991 tails. The counts differ by 18, yet the fraction of heads, 0.5009, is extremely close to 0.5. That is why the law of large numbers needs many trials.

Let's exploreNCERT End-of-chapter Exercises, Q16 · p. 173

Probability from areas

You’ll need

a sheet with a square of side 18 cm drawn on it
a circle of radius 5 cm drawn in the middle of the square
a handful of rice grains

Predict first

Roughly what fraction of the grains that land on the square should end up inside the circle? Give a decimal.

Hold your hand about 30 cm above the sheet and let a few grains fall at a time, without aiming.
Ignore any grain that bounces off the square.
Count the grains inside the circle and the total on the square, first after about 20 grains and then after 100 or more.

3/3300 grains: 73 inside, about 0.243, close to the area ratio.

Fig. 7.9The fraction of grains inside the circle settles near the ratio of the two areas.NCERT Fig. 7.8 · p. 173

A square board with a green circle in its centre. Grains that land inside the circle are filled green dots and grains outside are hollow rings. In step one 20 grains have fallen and 7 are inside, a fraction of 0.35. In step two there are 100 grains with 28 inside, a fraction of 0.28. In step three there are 300 grains with 73 inside, about 0.243. The side panel compares these fractions with the area of the circle divided by the area of the board, which is about 0.242.

The gambler's fallacy

Watch people at a game of chance and you will hear it: “Five heads already? Tails is bound to come now!” The feeling that a result is overdue after a streak is so widespread that it has a name, the gambler's fallacy. It is called a fallacy because it is simply wrong.

Why is it wrong? Nothing inside a coin records what it did before. Each toss is the same coin spinning through the air under the same conditions. Trials like this, where earlier results have no influence on later ones, are called independent. So after five heads, tails still has probability exactly $\frac{1}{2}$ on the next toss, as Fig. 7.10 shows.

3/3Nothing about the streak changes the coin: tails still has probability $\frac{1}{2}$ .

Fig. 7.10However long the streak, the next toss of a fair coin is still fifty–fifty.NCERT p. 164

A row of five coins all show H for heads, followed by a dashed sixth coin with a question mark for toss 6. In the first step only the streak is shown. In the second step a thought bubble appears, saying that tails must be due now. In the third step the thought is struck through and marked not true, and a bar below splits the chances for toss 6 evenly: heads 0.5 and tails 0.5, the same chances as on the very first toss.

Then why does the fraction of heads settle at 0.5 in the long run? Not because tails “catches up”, but because a streak gets diluted. Five extra heads are swamped by the thousands of later tosses, which split roughly half and half. The streak is never cancelled; it simply stops mattering.

Example 6NCERT Example 6 · p. 164

Waiting for a six in ludo

In ludo, Tara needs a 6 to bring a token out. She has rolled 7 times without getting a 6. Her brother says, “After so many misses, the next roll is almost sure to be a 6.” (a) Is he right? What is the probability of a 6 on her next roll? (b) Over her next 60 rolls, about how many 6s should she expect?

Try it yourself first. Then reveal the solution one step at a time.

Definition

Fair (unbiased)

A coin, die or spinner is fair, or unbiased (अनभिनत), when its shape and weight give every outcome the same chance. A toss or roll is random when nobody steers it: the object tumbles and lands however it happens to land.

ExampleA coin with a blob of gum stuck to one face is biased.

Practice · 6 questions

Practice 7.2: questions with answers

1
NumericalEasyNCERT Exercise Set 7.2, Q1
A bangle seller takes a sample of 40 bangles from a big box: 13 maroon, 12 pink, 9 golden and 6 silver.
1. Estimate the probability that a bangle picked at random from the sample is golden.
  Show answer
  Answer 0.225
  $\frac{9}{40} = 0.225$ .
2. The box holds 1000 bangles. Estimate how many of them are silver.
  Show answer
  Answer 150
  $\frac{6}{40} = 0.15$ , and $0.15 \times 1000 = 150$ silver bangles.
2
NumericalMediumNCERT Exercise Set 7.2, Q2
Fifty students, chosen at random from a school of 1200, were asked how they usually come to school. Their answers: walk 18, school bus 15, bicycle 12, car or scooter 5.
1. Estimate the probability that a student in the sample cycles to school.
  Show answer
  Answer 0.24
  $\frac{12}{50} = 0.24$ .
2. About how many of the school's 1200 students come by school bus?
  Show answer
  Answer 360
  $\frac{15}{50} = 0.3$ , and $0.3 \times 1200 = 360$ students.
3
ActivityEasyNCERT Exercise Set 7.2, Q3
Roll a die 30 times and record each result.
1. How many times did you get a 5?
  Show solution
  This depends on your own rolls. Counts from about 2 to 8 are common. Around 5 is most typical, since $30 \times \frac{1}{6} = 5$ .
2. Find your experimental probability of rolling a 5.
  Show solution
  Divide your count by 30. For example, 4 fives give $\frac{4}{30} \approx 0.133$ .
3. What is the probability of a 5 on your next roll?
  Show answer
  Answer 0.1667
  Each roll is independent of the ones before, so $P (5) = \frac{1}{6} \approx 0.167$ , whatever happened in your 30 rolls.
4
ActivityMediumNCERT Exercise Set 7.2, Q4
Toss an empty matchbox 50 times. Each time, record whether it lands on a large face, a long side or a small end.
1. Use your results to give an experimental probability for each way of landing.
  Show solution
  Divide each count by 50. One class got large face 38, long side 10 and small end 2, giving 0.76, 0.2 and 0.04. Your numbers will differ, but they must add up to 1.
2. Should each way of landing have probability $\frac{1}{3}$ ? Explain.
  Show solution
  No. The three outcomes are not equally likely: a large face is a broad, steady base, while a small end is narrow and wobbly. For objects like this, only experiment can tell us the probabilities.
Quick answerEasyNCERT Exercise Set 7.2, Q5
You roll a fair die once. How likely is a prime number? Choose the probability.
$\frac{1}{3}$ $\frac{1}{2}$ $\frac{2}{3}$ $\frac{1}{6}$
Show answer
Answer (B) $\frac{1}{2}$
The primes on a die are 2, 3 and 5 (1 is not prime), so $P = \frac{3}{6} = \frac{1}{2}$ .
6
NumericalMediumNCERT Exercise Set 7.2, Q6
A fair spinner has four equal sectors marked A, B, C and D. Ravi spins it 16 times, and it stops on C 7 times.
1. What is the experimental probability of C?
  Show answer
  Answer 0.4375
  $\frac{7}{16} = 0.4375$ .
2. What is the theoretical probability of C?
  Show answer
  Answer 0.25
  One favourable sector out of four equally likely ones: $\frac{1}{4} = 0.25$ .
3. Why are the two answers different? What would you expect after 40, 400 and 4000 spins?
  Show solution
  Sixteen spins is a small number of trials, so chance can easily push C above its fair share. With 40 spins the fraction may still be some way off 0.25. After 400 and then 4000 spins, it should settle closer and closer to 0.25, as the law of large numbers says.

7.3 Elements of Probability: Sample Spaces and Events

We now make the language of probability more exact by borrowing an idea you already know: the set.

7.3.1 Sample Space

Think of a sample space as a set in the sense you already know: its members are the outcomes of the experiment. We usually call it $S$ . The number of its members is written $n (S)$ and is called the sample size.Here sample size means the number of outcomes in $S$ . In a survey, the same words mean the number of people asked.

Definition

Sample space

Every outcome that a random experiment can produce, collected into one set, usually called $S$ . The number of its elements is written $n (S)$ .

ExampleFor one toss of a coin, $S = {H, T}$ and $n (S) = 2$ .

A correct sample space follows two rules:

Nothing is missing: every possible outcome is included.
Nothing is repeated: each outcome is listed exactly once.

Table 7.3Some experiments and their sample spacesNCERT p. 166

Experiment	Sample space $S$	$n (S)$
Will the school bus arrive on time tomorrow?	${on time, late}$	2
The result of a Test match for one team	${win, loss, draw, tie}$	4
Spinning a spinner with red, green and blue sectors	${R, G, B}$	3
Tossing a ₹1 coin and a ₹2 coin together	${H H, H T, T H, T T}$	4
Rolling two dice	36 ordered pairs, from $(1, 1)$ to $(6, 6)$	36

Look at the two coins. $H T$ means “₹1 coin heads, ₹2 coin tails”, while $T H$ is the other way round. These are different outcomes, even when the coins look alike: imagine that one of them is painted. Listing only “two heads, two tails, one of each” would squash two outcomes into one, and those three results would not be equally likely.

For two dice, a grid is the tidiest way to list the sample space. Each cell of Fig. 7.11 is one outcome: the first number belongs to the first die and the second number to the second die.

3/4Six cells give 7: $P = \frac{6}{36} = \frac{1}{6}$ .

Fig. 7.11A grid lists all 36 outcomes of two dice. The sums are not equally likely.NCERT p. 166

A grid has the first die's numbers 1 to 6 across the top and the second die's numbers 1 to 6 down the side, making 36 cells. In step one each cell shows an ordered pair such as 3,4 and the side panel says 36 outcomes. In step two each cell shows the sum of its two numbers instead. In step three the six cells with a sum of 7, running diagonally from bottom left to top right, turn green, giving a probability of 6 out of 36, or one-sixth. In step four only the two corner cells, with sums 2 and 12, are highlighted: one cell each, so each sum has probability 1 out of 36.

Quick check

1 of 3

A coin is tossed and a spinner with five equal sectors, numbered 1 to 5, is spun. How many elements does the sample space have?

Show answer

Answer 10

Each of the 2 coin results can pair with each of the 5 spinner results:

2 \times 5 = 10

7.3.2 Events

Often we care about a group of outcomes rather than one: “an even number”, “at least one head”, “a vowel”. Such a group is an event (घटना).

Definition

Event

Any collection of outcomes of a random experiment, that is, a subset of the sample space. The event happens when the outcome is one of its elements.

ExampleFor one roll of a die, the event “an even number” is $E = {2, 4, 6}$ .

Fig. 7.12 shows four events for one roll of a die. Two special events stand out. An event with no outcomes at all, like “a number above 6”, is impossible and has probability 0. The event made of the whole sample space is certain and has probability 1.

2/5A prime number: $A = {2, 3, 5}$ .

Fig. 7.12Events are subsets of the sample space, from the empty set to the whole of

S

.NCERT p. 167

A large rectangle labelled S holds the six faces of a die, from one to six. In the first step the sample space alone is shown, with n of S equal to 6. In the second step event A, a prime number, is circled in green around 2, 3 and 5, and P of A is three-sixths, or one-half. In the third step event B, a multiple of 3, is circled in pink around 3 and 6, with probability one-third. In the fourth step event C, a number above 6, circles nothing: the faces fade, C is the empty set and its probability is 0. In the last step event D, a number below 7, circles all six faces, so D equals S and its probability is 1.

When the outcomes are equally likely, the probability of an event is simply the share of the sample space that it takes up.

Probability of an event

P (E) = \frac{n (E)}{n (S)}

where

$n (E)$: the number of outcomes in the event
$n (S)$: the number of outcomes in the sample space

Table 7.4Events written as subsetsNCERT p. 167

Experiment	Event	As a set
Tossing two coins	exactly one head	${H T, T H}$
Rolling one die	a prime number	${2, 3, 5}$
Picking one vegetable from carrot, tomato, potato and spinach	it grows under the ground	${carrot, potato}$
Rolling two dice	a double	${(1, 1), (2, 2), \dots, (6, 6)}$

Quick check

1 of 4

An event is a of the sample space.

Show answer

Answer subset

An event is a collection of outcomes chosen from

S

, so it is a subset of

S

Practice · 3 questions

Practice 7.3: questions with answers

Quick answerEasyNCERT Exercise Set 7.3, Q1
A spinner has nine equal sectors numbered 1 to 9. How many outcomes are in the sample space for one spin?
Show answer
Answer 9
There is one outcome per sector, so $n (S) = 9$ .
2
Short answerMediumNCERT Exercise Set 7.3, Q2
Write down the sample space for each experiment.
1. A spinner with red, green and blue sectors is spun and a coin is tossed.
  Show solution
  Pair each colour with each face: $S = {R H, R T, G H, G T, B H, B T}$ , so $n (S) = 6$ .
2. An integer $x$ with $- 3 \leq x \leq 3$ is chosen at random.
  Show solution
  $S = {- 3, - 2, - 1, 0, 1, 2, 3}$ . Remember to include 0, which gives $n (S) = 7$ .
3. A bag holds 4 white and 6 black counters, and one counter is drawn.
  Show solution
  If we record only the colour, $S = {white, black}$ , but these two outcomes are not equally likely. To use counting, label the counters $W_{1}, \dots, W_{4}$ and $B_{1}, \dots, B_{6}$ . Then $S$ has 10 equally likely elements.
3
Short answerEasyNCERT Exercise Set 7.3, Q3
At a school fete you choose one sweet (jalebi or rasgulla) and one drink (nimbu pani, chaas or coconut water).
1. List the sample space of all sweet-and-drink choices.
  Show solution
  Write J for jalebi, R for rasgulla and N, C, W for the drinks. Then $S = {J N, J C, J W, R N, R C, R W}$ : 6 combinations.
2. List the event “the drink is chaas”.
  Show solution
  ${J C, R C}$ : 2 of the 6 combinations.

7.4 Tree diagrams

Some experiments happen in stages: toss a coin, then toss it again; roll a die, then spin a spinner. These are multi-step experiments. Listing their outcomes by guesswork is risky, because it is easy to miss one or to count one twice.

A tree diagram (वृक्ष आरेख) does the listing for us. Start from a point and draw one branch for each outcome of the first stage. From the end of every branch, draw branches for each outcome of the next stage, and so on. Follow any route from the starting point out to a tip and you have read off one complete outcome of the whole experiment.

Definition

Tree diagram

A diagram of branching lines that lists the outcomes of a multi-step experiment, stage by stage. Each path from the start to the end of a branch is one outcome.

ExampleRolling a die and then tossing a coin gives a tree with $6 \times 2 = 12$ paths.

5/5Exactly two heads: HHT, HTH and THH, so $P = \frac{3}{8}$ .

Fig. 7.13Each path from the start to a tip is one outcome of three tosses.NCERT Fig. 7.6 · p. 168

A tree diagram grows downward from a start point, and every branch has probability one-half. In step one the first toss branches into H and T. In step two each of these branches again, giving four nodes. In step three each of those branches once more, giving eight tips labelled, from left to right, HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. In step four the eight outcomes are boxed together: n of S is 8 and each outcome has probability one-eighth. In step five the three paths to HHT, HTH and THH are highlighted in green, so the probability of exactly two heads is three-eighths.

Example 7NCERT Example 7 · p. 168

Three tosses of a coin

A fair coin is tossed three times. (a) Use a tree diagram to write the sample space. (b) Find the probability of exactly two heads. (c) Find the probability of at least one tail.

Try it yourself first. Then reveal the solution one step at a time.

Look again at part (c). Counting the opposite event, “no tails”, was quicker than counting “at least one tail” directly. One of the two must happen, and they cannot both happen, so their probabilities add up to 1.

A tree can also carry probabilities. Write each stage's probability on its branch, then multiply along a path: $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$ for every path of the three-toss tree. This agrees with counting 8 equally likely paths.

SimulationTwo-coin lab

Two virtual coins tossed together 1, 10, 100 or 1000 times at a press. A bar chart counts the four outcomes HH, HT, TH and TT, and a counter compares how often 'one head and one tail' has happened with its theoretical probability of one-half.

Fig. 7.14Toss two coins together, again and again, and see how often you get one of each.

Try this

Someone claims that “one of each” has probability $\frac{1}{3}$ , because there are three kinds of result. Toss 1000 times and decide.
Which bars grow at about the same rate? Why?

So far every stage was a fresh toss. What if the first stage changes what is left for the second? Then the branch probabilities change too.

ExampleNCERT End-of-chapter Exercises, Q10 · p. 172

Taking without putting back

A box holds 5 toffees of the same size: 2 mango (M) and 3 orange (O). Priya takes one without looking and eats it. Then she takes a second one. Find the probability that (a) both are orange, (b) she gets one toffee of each flavour.

Try it yourself first. Then reveal the solution one step at a time.

4/4MO and OM together: $\frac{12}{20} = \frac{3}{5}$ .

Fig. 7.15Without putting back, the second-pick fractions change. Multiply along each path.NCERT p. 172

A box on the left holds 2 mango and 3 orange toffees. In step one two branches lead to the first pick: mango with probability two-fifths and orange with probability three-fifths. In step two each first pick branches again for the second pick, with changed fractions because one toffee has been eaten: after mango, mango one-quarter and orange three-quarters; after orange, mango two-quarters and orange two-quarters. In step three the four results MM, MO, OM and OO are labelled with their path probabilities: 2, 6, 6 and 6 out of 20. In step four the rows MO and OM are highlighted, giving 12 out of 20, or three-fifths, for one toffee of each flavour.

Quick check

1 of 4

A coin is tossed and then a die is rolled. How many complete paths does the tree diagram have?

Show answer

Answer 12

The coin gives 2 branches, and each splits into 6 for the die:

2 \times 6 = 12

Practice · 2 questions

Practice 7.4: questions with answers

1
DiagramMediumNCERT Exercise Set 7.4, Q1
Tiffin box A holds 3 idlis and 1 vada. Tiffin box B holds 1 laddoo, 1 barfi and 1 peda. Without looking, you take one item from each box.
1. Draw a tree diagram for the two picks.
  Show solution
  First stage (box A): four branches, idli 1, idli 2, idli 3 and vada. From each, three branches for box B: laddoo, barfi and peda. That makes $4 \times 3 = 12$ paths.
2. List the sample space.
  Show solution
  Using $I_{1}$ , $I_{2}$ , $I_{3}$ , $V$ and $L$ , $B$ , $P$ : $S = {I_{1} L, I_{1} B, I_{1} P, I_{2} L, I_{2} B, I_{2} P, I_{3} L, I_{3} B, I_{3} P, V L, V B, V P}$ , so $n (S) = 12$ .
3. What is the probability of getting an idli and a peda?
  Show answer
  Answer 0.25
  The favourable paths are $I_{1} P$ , $I_{2} P$ and $I_{3} P$ , so $P = \frac{3}{12} = \frac{1}{4} = 0.25$ .
2
DiagramHardNCERT Exercise Set 7.4, Q2
A jar has 2 blue, 3 yellow and 5 white marbles. You pick a marble without looking, note its colour and return it to the jar. Your friend then takes a turn in the same way.
1. Draw a tree diagram showing the possible pairs of colours.
  Show solution
  First stage: blue, yellow or white. From each, the same three branches for your friend, giving 9 colour pairs from BB to WW. Because the marble goes back, the jar is the same for both picks.
2. Find the probability that you both pick the same colour.
  Show answer
  Answer 0.38
  There are $10 \times 10 = 100$ equally likely ordered pairs of marbles. Same colour: $2 \times 2 + 3 \times 3 + 5 \times 5 = 4 + 9 + 25 = 38$ pairs, so $P = \frac{38}{100} = 0.38$ .

End-of-chapter exercises

Practice · 19 questions

Practice questions with answers (NCERT “End-of-chapter exercises”)

Questions 1 to 16 practise the same skills as the NCERT end-of-chapter exercises, with new numbers and settings. Questions 9 to 16 are more challenging. Questions 17 to 19 are quick extra checks.

1
Short answerEasyNCERT End-of-chapter Exercises, Q1
Fill in the blanks.
1. An event that is sure to happen has probability .
  Show answer
  Answer 1
  Certain events sit at the top of the scale: probability 1.
2. Each single possible result of a random experiment is called an .
  Show answer
  Answer outcome
  A single possible result is an outcome; all of them together make the sample space.
3. The probability of rolling a 7 with one ordinary die is .
  Show answer
  Answer 0
  No face shows 7, so the event is impossible: probability 0.
4. A fair spinner has 5 equal sectors, one of them red. The probability that it stops on red is .
  Show answer
  Answer 1/5
  One favourable sector out of five: $\frac{1}{5} = 0.2$ .
2
Short answerEasyNCERT End-of-chapter Exercises, Q2
At a bus stop, 80 people were asked whether they drink chai every day, and 28 said yes.
1. The number 28 is the of “drinks chai daily”, and $\frac{28}{80}$ is its .
  Show answer
  Answer (1) frequency, (2) relative frequency
  A frequency is a count. A relative frequency divides the count by the total number of observations.
2. Write the relative frequency as a decimal.
  Show answer
  Answer 0.35
  $\frac{28}{80} = 0.35$ .
3
Quick answerMediumNCERT End-of-chapter Exercises, Q3
Which of these experiments have equally likely outcomes? Choose all that apply.
A drawing pin is dropped: it lands point up or point down.A spinner with 5 equal sectors is spun: it stops on 1, 2, 3, 4 or 5.A ball is picked from a bag of 6 red and 2 white balls: it is red or white.A card is drawn from a well-shuffled pack of 52: it is red or black.A student sits an exam: she passes or she fails.
Select all that apply
Show answer
Answer
(B) A spinner with 5 equal sectors is spun: it stops on 1, 2, 3, 4 or 5.
(D) A card is drawn from a well-shuffled pack of 52: it is red or black.
The spinner's sectors are equal, and a pack has 26 red and 26 black cards. A drawing pin is not symmetric, the bag has three times as many red balls as white ones, and passing depends on preparation, not on symmetry.
4
NumericalMediumNCERT End-of-chapter Exercises, Q4
Write the sample space, then find each probability as a decimal.
1. Two coins are tossed. Find the probability that both show the same face.
  Show answer
  Answer 0.5
  $S = {H H, H T, T H, T T}$ . Same face $= {H H, T T}$ , so $P = \frac{2}{4} = 0.5$ .
2. Twenty cards numbered 1 to 20 are shuffled and one is drawn. Find the probability of a multiple of 3.
  Show answer
  Answer 0.3
  $S = {1, 2, \dots, 20}$ . The multiples of 3 are 3, 6, 9, 12, 15 and 18, six of them, so $P = \frac{6}{20} = 0.3$ .
3. Roll a die once. Find the chance of getting a number smaller than 3.
  Show answer
  Answer 0.3333
  $S = {1, 2, 3, 4, 5, 6}$ and the event is ${1, 2}$ , so $P = \frac{2}{6} = \frac{1}{3} \approx 0.333$ .
4. A bag has 4 yellow, 3 green and 5 purple balls, and one is picked at random. Find the probability that it is not green.
  Show answer
  Answer 0.75
  There are 12 balls, and 9 of them are not green: $\frac{9}{12} = 0.75$ .
5. Three coins are tossed together. Find the probability of exactly one head.
  Show answer
  Answer 0.375
  $S$ has 8 outcomes, and exactly one head $= {H T T, T H T, T T H}$ , so $P = \frac{3}{8} = 0.375$ .
Quick answerEasyNCERT End-of-chapter Exercises, Q5
A pencil box holds four pens that feel the same: one red, one blue, one green and one black. One pen is taken out without looking. What is the probability that it is green? Give a decimal.
Show answer
Answer 0.25
One favourable pen out of four equally likely ones: $\frac{1}{4} = 0.25$ .
6
Short answerEasyNCERT End-of-chapter Exercises, Q6
Aarav has 3 kurtas (white, yellow and green) and 2 pyjamas (white and black).
1. Show every outfit of one kurta and one pyjama in a table. How many outfits are there?
  Show solution
  Table 7.5All six outfits
  Kurta with white pyjama with black pyjama
  white white–white white–black
  yellow yellow–white yellow–black
  green green–white green–black
2. He picks a kurta and a pyjama at random. What is the probability that both are white?
  Show answer
  Answer 0.1667
  Only one of the 6 equally likely outfits is white–white: $\frac{1}{6} \approx 0.167$ .
7
NumericalMediumNCERT End-of-chapter Exercises, Q7
A company tested 800 LED bulbs and recorded how many hours each one lasted.
Table 7.6Test results for 800 bulbs
Hours lasted under 5000 5000 to 9999 10 000 to 14 999 15 000 or more
Number of bulbs 36 164 280 320
1. Estimate the probability that a bulb lasts under 5000 hours.
  Show answer
  Answer 0.045
  $\frac{36}{800} = 0.045$ .
2. Estimate the probability that a bulb lasts from 5000 to 14 999 hours.
  Show answer
  Answer 0.555
  $\frac{164 + 280}{800} = \frac{444}{800} = 0.555$ .
3. Estimate the probability that a bulb lasts 15 000 hours or more.
  Show answer
  Answer 0.4
  $\frac{320}{800} = 0.4$ .
8
NumericalEasyNCERT End-of-chapter Exercises, Q8
The letters of NAGALAND are written on 8 identical cards, and one card is drawn without looking.
1. Find the probability that it shows N or G.
  Show answer
  Answer 0.375
  N appears twice and G once: 3 cards out of 8, so $\frac{3}{8} = 0.375$ .
2. Find the probability that it is not an A.
  Show answer
  Answer 0.625
  A appears 3 times, so 5 cards are not A: $\frac{5}{8} = 0.625$ .
9
NumericalMediumNCERT End-of-chapter Exercises, Q9
A spinner has 10 equal sectors numbered 1 to 10, all equally likely. Find the probability, as a decimal, that the pointer stops on:
1. the number 7
  Show answer
  Answer 0.1
  One sector out of ten: $0.1$ .
2. a prime number
  Show answer
  Answer 0.4
  2, 3, 5 and 7 are prime (1 is not): four sectors, so $\frac{4}{10} = 0.4$ .
3. a number greater than 7
  Show answer
  Answer 0.3
  8, 9 and 10: three sectors, so $0.3$ .
4. a number less than 11
  Show answer
  Answer 1
  Every sector is numbered below 11, so this is certain: probability 1.
5. a multiple of 4
  Show answer
  Answer 0.2
  4 and 8: two sectors, so $\frac{2}{10} = 0.2$ .
10
DiagramHardNCERT End-of-chapter Exercises, Q10
A pencil box holds 3 blue pens and 5 black pens. You take one pen out without looking and keep it aside, then take a second pen.
1. Draw a tree diagram with the probability on each branch.
  Show solution
  First pick: blue $\frac{3}{8}$ , black $\frac{5}{8}$ . After a blue pen: blue $\frac{2}{7}$ , black $\frac{5}{7}$ . After a black pen: blue $\frac{3}{7}$ , black $\frac{4}{7}$ .
2. Find the probability of a blue pen first and then a black pen.
  Show answer
  Answer 0.2679
  Multiply along the path: $\frac{3}{8} \times \frac{5}{7} = \frac{15}{56} \approx 0.268$ .
3. Find the probability that both pens are black.
  Show answer
  Answer 0.3571
  $\frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14} \approx 0.357$ .
Short answerMediumNCERT End-of-chapter Exercises, Q11
Two dice are rolled and the product of the two numbers is noted. Write down one event with probability 0 and one event with probability 1.
Show solution
AnswerFor example, “the product is 7” (probability 0) and “the product is at most 36” (probability 1).
Probability 0: “the product is 7”. Since 7 is prime and larger than 6, it cannot be made by multiplying two numbers from 1 to 6. Probability 1: “the product is at most 36”, because the largest product is $6 \times 6 = 36$ .
12
NumericalHardNCERT End-of-chapter Exercises, Q12
Write the sample space, then find each probability. Give decimals.
1. Two dice are rolled. Find the probability that the sum is a perfect square.
  Show answer
  Answer 0.1944
  The square sums are 4 and 9. Sum 4: (1, 3), (2, 2), (3, 1). Sum 9: (3, 6), (4, 5), (5, 4), (6, 3). That is 7 of 36 outcomes: $\frac{7}{36} \approx 0.194$ .
2. A bag holds 3 red, 2 green and 1 blue ball. Two balls are drawn one after the other without replacement. Find the probability that both are the same colour.
  Show answer
  Answer 0.2667
  There are $6 \times 5 = 30$ ordered pairs. Same colour: red–red $3 \times 2 = 6$ , green–green $2 \times 1 = 2$ , blue–blue 0. So $P = \frac{8}{30} = \frac{4}{15} \approx 0.267$ .
3. Three coins are tossed. Find the probability that the third coin shows tails and there is exactly one head in all.
  Show answer
  Answer 0.25
  The outcomes are $H T T$ and $T H T$ : 2 of 8, so $\frac{2}{8} = 0.25$ .
4. A three-digit number is made from the digits 2, 5 and 7, using each digit once. Find the probability that it is divisible by 5.
  Show answer
  Answer 0.3333
  There are $3 \times 2 \times 1 = 6$ numbers: 257, 275, 527, 572, 725 and 752. Divisible by 5 means ending in 5: 275 and 725. So $P = \frac{2}{6} = \frac{1}{3} \approx 0.333$ .
5. A quiz has 2 questions, each with 3 options and exactly one correct option. A student guesses both. Find the probability of exactly one correct answer.
  Show answer
  Answer 0.4444
  Each question can be answered right (1 way) or wrong (2 ways), so there are $3 \times 3 = 9$ equally likely answer pairs. Exactly one right: right-then-wrong $1 \times 2 = 2$ and wrong-then-right $2 \times 1 = 2$ . So $P = \frac{4}{9} \approx 0.444$ .
13
DiagramMediumNCERT End-of-chapter Exercises, Q13
A bag has three chits marked A, B and C. Use a tree diagram to write the sample space in each case.
1. A chit is drawn, noted and put back. Then a second chit is drawn.
  Show solution
  Each stage has 3 branches: $S = {A A, A B, A C, B A, B B, B C, C A, C B, C C}$ .
2. A chit is drawn and kept out. Then a second chit is drawn.
  Show solution
  The second stage has only the 2 chits that are left: $S = {A B, A C, B A, B C, C A, C B}$ .
3. With replacement, $n (S) =$ . Without replacement, $n (S) =$ .
  Show answer
  Answer (1) 9, (2) 6
  $3 \times 3 = 9$ with replacement and $3 \times 2 = 6$ without.
Short answerEasyNCERT End-of-chapter Exercises, Q14
A die is rolled and a chit is drawn from a box holding two chits marked X and Y. List the elements of the sample space.
Show solution
Answer12 elements, from 1X to 6Y.
$S = {1 X, 1 Y, 2 X, 2 Y, 3 X, 3 Y, 4 X, 4 Y, 5 X, 5 Y, 6 X, 6 Y}$ , so $n (S) = 6 \times 2 = 12$ .
Short answerMediumNCERT End-of-chapter Exercises, Q15
Two dice are rolled and we count how many of them show a six. Pick the correct sample space from the lists below, and explain what is wrong with each of the others.
${1, 2}$ ${0, 1, 2}$ ${0, 1, 2, 3}$ ${0, 1}$
Show solution
$S = {0, 1, 2}$ . The list ${1, 2}$ misses the case of no sixes, and ${0, 1}$ misses a double six. The list ${0, 1, 2, 3}$ includes 3, which is impossible with only two dice.
NumericalHardNCERT End-of-chapter Exercises, Q16
A square board of side 40 cm has a circle of radius 10 cm painted in the middle. A raindrop lands at a random point on the board. What is the probability that it lands inside the circle? Give a decimal.
Show solution
$P = \frac{π \times 10^{2}}{40 \times 40} = \frac{100 π}{1600} = \frac{π}{16} \approx 0.196$
Every point of the board is equally likely, so the probability is the fraction of the board's area that the circle covers.
Quick answerEasy
Which of these cannot be the probability of an event?
0 $\frac{7}{8}$ 1.250.001
Show answer
Answer (C) 1.25
Probabilities lie between 0 and 1, so 1.25 is too big. Even 0.001, though tiny, is allowed.
18
Quick answerMedium
For one roll of a fair die, arrange these events from least likely to most likely.
1. an even number
2. a number less than 10
3. a number greater than 1
4. a 6
5. a 7
Show answer
Answer
a 7
a 6
an even number
a number greater than 1
a number less than 10
Their probabilities are 0, $\frac{1}{6}$ , $\frac{1}{2}$ , $\frac{5}{6}$ and 1.
19
Quick answerEasy
Match each experiment with the size of its sample space.
Tap an item on the left, then its match on the right.
Show answer
Answer
one die : 6
two coins : 4
three coins : 8
two dice : 36
Multiply the number of outcomes at each stage: $6$ , $2 \times 2$ , $2 \times 2 \times 2$ and $6 \times 6$ .

Kurta	with white pyjama	with black pyjama
white	white–white	white–black
yellow	yellow–white	yellow–black
green	green–white	green–black

Hours lasted	under 5000	5000 to 9999	10 000 to 14 999	15 000 or more
Number of bulbs	36	164	280	320

Chapter summary

Recap

The Mathematics of Maybe: Introduction to Probability: key points and summary

Probability puts a number on how likely an event is: 0 for impossible, 1 for certain, and always $0 \leq P (E) \leq 1$ (Section 7.1.2).
In a random experiment we know every possible outcome, but we cannot predict the result of any single trial (Section 7.1.1).
Experimental probability, or relative frequency, is the number of trials in which the event happened divided by the total number of trials. It changes from one run to the next (Section 7.2.1).
Theoretical probability is the number of favourable outcomes divided by the number of possible outcomes. It works only when the outcomes are equally likely (Section 7.2.2).
A random sample lets us estimate probabilities, and counts, for a whole population. Larger, unbiased samples give better estimates (Section 7.2.3).
Over a very large number of trials, experimental probability settles near theoretical probability: the law of large numbers.
Tosses and rolls are independent: a streak never makes an outcome “due” (the gambler's fallacy).
A sample space $S$ lists every outcome exactly once, and $n (S)$ is its size (Section 7.3.1).
An event is a subset of $S$ . With equally likely outcomes, $P (E) = \frac{n (E)}{n (S)}$ (Section 7.3.2).
Tree diagrams list the outcomes of multi-step experiments; each path from start to tip is one outcome (Section 7.4).
When items are not put back, the branch probabilities change at the next stage; multiply along each path (the toffee example).

Fig. 7.16 ties the ideas of the chapter together, from a random experiment to a single number between 0 and 1.

5/5Over many trials the two routes agree on one number.

Fig. 7.16From a random experiment to a single number: the whole chapter in one picture.

A concept map built up in five steps. First, a box for a random experiment points down to a box for its sample space S. Second, a box for tree diagrams appears, with an arrow showing that a tree lists the sample space. Third, an arrow leads from the sample space to a box for an event, which is part of S. Fourth, the event splits into two routes: experimental probability, found from data, and theoretical probability, found by counting. Last, a dashed two-way arrow labelled many trials joins the two routes, and both point to a final box: the probability of the event, a number from 0 to 1.

Flash cards · 8

Words to remember

1 / 8

Chapter check

1 of 4

A spinner is spun 200 times and stops on red 46 times. Which statement is correct?

The theoretical probability of red is 0.23.The experimental probability of red is 0.23.Red will come up exactly 23 times in the next 100 spins.The spinner must be unfair.

Show answer

Answer (B) The experimental probability of red is 0.23.

\frac{46}{200} = 0.23

comes from data, so it is an experimental probability. It predicts roughly, not exactly, what happens next, and on its own it says nothing about fairness unless we know how many sectors are red.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise Set 7.1, Q1	Practice 7.1 · Q1
NCERT Exercise Set 7.2, Q1	Practice 7.2 · Q1
NCERT Exercise Set 7.2, Q2	Practice 7.2 · Q2
NCERT Exercise Set 7.2, Q3	Practice 7.2 · Q3
NCERT Exercise Set 7.2, Q4	Practice 7.2 · Q4
NCERT Exercise Set 7.2, Q5	Practice 7.2 · Q5
NCERT Exercise Set 7.2, Q6	Practice 7.2 · Q6
NCERT Exercise Set 7.3, Q1	Practice 7.3 · Q1
NCERT Exercise Set 7.3, Q2	Practice 7.3 · Q2
NCERT Exercise Set 7.3, Q3	Practice 7.3 · Q3
NCERT Exercise Set 7.4, Q1	Practice 7.4 · Q1
NCERT Exercise Set 7.4, Q2	Practice 7.4 · Q2
NCERT End-of-chapter Exercises, Q1	End-of-chapter exercises · Q1
NCERT End-of-chapter Exercises, Q2	End-of-chapter exercises · Q2
NCERT End-of-chapter Exercises, Q3	End-of-chapter exercises · Q3
NCERT End-of-chapter Exercises, Q4	End-of-chapter exercises · Q4
NCERT End-of-chapter Exercises, Q5	End-of-chapter exercises · Q5
NCERT End-of-chapter Exercises, Q6	End-of-chapter exercises · Q6
NCERT End-of-chapter Exercises, Q7	End-of-chapter exercises · Q7
NCERT End-of-chapter Exercises, Q8	End-of-chapter exercises · Q8
NCERT End-of-chapter Exercises, Q9	End-of-chapter exercises · Q9
NCERT End-of-chapter Exercises, Q10	End-of-chapter exercises · Q10
NCERT End-of-chapter Exercises, Q11	End-of-chapter exercises · Q11
NCERT End-of-chapter Exercises, Q12	End-of-chapter exercises · Q12
NCERT End-of-chapter Exercises, Q13	End-of-chapter exercises · Q13
NCERT End-of-chapter Exercises, Q14	End-of-chapter exercises · Q14
NCERT End-of-chapter Exercises, Q15	End-of-chapter exercises · Q15
NCERT End-of-chapter Exercises, Q16	End-of-chapter exercises · Q16

Quick check: MCQs on The Mathematics of Maybe: Introduction to Probability

7.1.1 What is Randomness?3 questions
7.1.2 The Probability Scale4 questions
7.2.1 Experimental Probability: Performing Observations or Experiments3 questions
7.2.2 Theoretical Probability3 questions
The gambler's fallacy4 questions
7.3.1 Sample Space3 questions
7.3.2 Events4 questions
7.4 Tree diagrams4 questions
Chapter check4 questions

Glossary · 15

Key terms (English–Hindi)

equally likelyसमप्रायिकsamprayik: Having the same chance of happening, because nothing favours one outcome over another.First seen in the chapter →
eventघटनाghatna: A collection of outcomes, that is, a subset of the sample space.First seen in the chapter →
experimental probabilityप्रायोगिक प्रायिकताprayogik prayikta: The fraction of trials in which an event actually happened.First seen in the chapter →

gambler's fallacy: The mistaken belief that a result becomes more likely because it has not happened for a while.First seen in the chapter →

law of large numbers: With a very large number of trials, the experimental probability of an event settles near its theoretical probability.First seen in the chapter →

outcomeपरिणामparinam: One possible result of a random experiment.First seen in the chapter →

probabilityप्रायिकताprayikta: A number from 0 to 1 that measures how likely an event is to happen.First seen in the chapter →

random experimentयादृच्छिक प्रयोगyadrichchhik prayog: A repeatable action whose possible results are known but whose next result is not, such as rolling a die.
relative frequencyसापेक्ष बारंबारताsapeksh barambarata: How often something happened divided by the total number of observations; the same as experimental probability.First seen in the chapter →

sampleप्रतिदर्शpratidarsh: A smaller group chosen from a population in order to study it. A good sample is large and chosen at random.First seen in the chapter →
sample spaceप्रतिदर्श समष्टिpratidarsh samashti: The set $S$ containing every outcome that a random experiment can produce.First seen in the chapter →

theoretical probabilityसैद्धांतिक प्रायिकताsaiddhantik prayikta: Favourable outcomes divided by all possible outcomes, when every outcome is equally likely.First seen in the chapter →
tree diagramवृक्ष आरेखvriksh aarekh: A branching diagram that lists the outcomes of a multi-step experiment; each path is one outcome.First seen in the chapter →
trialअभिप्रयोगabhiprayog: One performance of a random experiment, such as a single roll of a die.First seen in the chapter →

unbiasedअनभिनतanabhinat: Of a coin, die or spinner: made so that every outcome has the same chance.

The Mathematics of Maybe: Introduction to Probability: Class 9 Ganita Manjari · Chapter 7

What you will learn in The Mathematics of Maybe: Introduction to Probability

7.1 What is Probability?

7.1.1 What is Randomness?

7.1.2 The Probability Scale

Practice 7.1: questions with answers

7.2 Measuring Probability Objectively

7.2.1 Experimental Probability: Performing Observations or Experiments

7.2.2 Theoretical Probability

7.2.3 Analysing Statistical Data Using Probability

Experiment and theory in the long run

The gambler's fallacy

Practice 7.2: questions with answers

7.3 Elements of Probability: Sample Spaces and Events

7.3.1 Sample Space

7.3.2 Events

Practice 7.3: questions with answers

7.4 Tree diagrams

Practice 7.4: questions with answers

End-of-chapter exercises

Practice questions with answers (NCERT “End-of-chapter exercises”)

Chapter summary

The Mathematics of Maybe: Introduction to Probability: key points and summary

Find your NCERT exercise question

Quick check: MCQs on The Mathematics of Maybe: Introduction to Probability

Key terms (English–Hindi)

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