Introduction to Probability Class 9: Concepts, Tree Diagrams and Solved Questions

What Introduction to Probability is about

Weightage: This chapter is part of Unit VI, Statistics and Probability, which carries 10 of the 80 marks in the school annual exam, shared with Statistics.Source: CBSE Mathematics syllabus, Class IX 2026-27, course structure, Unit VI Statistics and Probability

Key concepts

Randomness and random experiments

NCERT section 7.1 What is Probability?

An action is random when you know every possible result but cannot predict which one will happen on a particular try. Tossing a coin, rolling a die and drawing a name from a box are random experiments (also called trials): repeat them and the result may change.

Randomness is useful: a coin toss to decide which cricket team bats first is fair because neither captain can control or predict it. Weather is also random in this sense: it depends on so many sensitive factors (temperature, humidity, wind, pressure) that it cannot be predicted with certainty, only estimated from patterns in data.

Guessing from feelings ("it is hot, so it may rain") is subjective probability. This chapter is about objective ways of measuring likelihood.

Worked example

Which of these are random experiments? Give a reason for each. (i) Drawing a card from a well-shuffled pack. (ii) Checking whether the sun rises in the east tomorrow. (iii) Spinning a fair spinner with 4 colours. (iv) Reading the number of days in March.

Try it first, then show the solutionSolution

1. (i) Random: all cards are possible but you cannot know which one you will draw.
2. (ii) Not random: the outcome is certain.
3. (iii) Random: four possible colours, none can be predicted.
4. (iv) Not random: March always has 31 days.

Answer: (i) and (iii) are random; (ii) and (iv) are not.

The probability scale from 0 to 1

NCERT section 7.1 What is Probability?

Probability is a number from 0 to 1 (or 0% to 100%) that measures how likely an event is. For any event E, $0\le P(E)\le 1$.

Probability	Word
0	impossible
between 0 and 0.5	less likely
0.5	even chance (equally likely)
between 0.5 and 1	more likely
1	certain

So 0.75 means a 75% chance, more likely than not. A number below 0 or above 1 can never be a probability.

Worked example

Place each event on the scale with a reason: (i) rolling a number greater than 6 on a die; (ii) getting heads on a fair coin; (iii) drawing a red sweet from a bag of only red sweets; (iv) rolling a 3 on a die.

Try it first, then show the solutionSolution

1. (i) A die shows only 1 to 6: impossible, 0.
2. (ii) Heads and tails are equally likely: even chance, 0.5.
3. (iii) Every sweet is red: certain, 1.
4. (iv) One face out of six: less likely, $\frac{1}{6}$.

Answer: (i) 0 (ii) 0.5 (iii) 1 (iv) $\frac{1}{6}$, less likely

Experimental probability (relative frequency)

NCERT section 7.2 Measuring Probability Objectively

Do the experiment many times and count:
$\text{Experimental probability}=\frac{\text{number of times the event occurred}}{\text{total number of trials}}$
This fraction is also called the relative frequency of the event. It is based on actual results, not on assumptions, so it is the only option when outcomes are not equally likely (for example, how a paper cup lands: on its bottom, top or side).

The outcomes of an experiment are listed in the sample space, written in braces: tossing a coin gives {H, T}; rolling a die gives {1, 2, 3, 4, 5, 6}.

Worked example

A spinner is spun 80 times and lands on red 28 times. Find the experimental probability of red, as a fraction, a decimal and a percentage.

Try it first, then show the solutionSolution

1. Experimental probability $=\frac{28}{80}$.
2. $=\frac{7}{20}=0.35=35\%$.

Answer: $\frac{7}{20}$, 0.35 or 35%

Theoretical probability

NCERT section 7.2 Measuring Probability Objectively

When all outcomes are equally likely (a fair coin, a fair die, a well-mixed bag), you can find the probability by reasoning, without any experiment:
$P(E)=\frac{\text{number of favourable outcomes}}{\text{number of possible outcomes}}$

'Fair' or 'unbiased' means there is no reason for one outcome to come up more often than another. Check this before using the formula: a marble drawn from 3 red and 7 blue marbles is not equally likely to be red or blue; but each marble is equally likely, so $P(\text{red})=\frac{3}{10}$.

Count repeated items separately: in 'PROBABILITY' there are 11 letters and 2 Bs, so $P(B)=\frac{2}{11}$.

Worked example

A letter is chosen at random from the word MATHEMATICS. Find the probability that it is (i) M (ii) a vowel (iii) not T.

Try it first, then show the solutionSolution

1. The word has 11 letters: M, A, T, H, E, M, A, T, I, C, S.
2. (i) Two Ms: $P(M)=\frac{2}{11}$.
3. (ii) Vowels A, E, A, I: $P(\text{vowel})=\frac{4}{11}$.
4. (iii) Two Ts, so 9 letters are not T: $P(\text{not T})=\frac{9}{11}$.

Answer: (i) $\frac{2}{11}$ (ii) $\frac{4}{11}$ (iii) $\frac{9}{11}$

Using survey data and samples

NCERT section 7.2 Measuring Probability Objectively

Businesses, scientists and planners estimate probabilities from statistical data. Usually you cannot ask everyone (the population), so you collect data from a sample and use its relative frequency as the probability.

To predict for the whole population: estimated number $=$ probability from the sample $\times$ population size.

The estimate is more trustworthy when the sample is larger and representative (for example, students from every class, not just one). Choosing such a sample is called sampling.

Worked example

In a sample of 40 students, 14 walk to school, 10 come by bus, 12 cycle and 4 come by car. The school has 1200 students. Estimate how many come by bus, and the probability that a randomly chosen student in the sample does not walk.

Try it first, then show the solutionSolution

1. $P(\text{bus})=\frac{10}{40}=0.25$.
2. Estimated bus users $=0.25\times 1200=300$.
3. Students in the sample who do not walk $=40-14=26$, so $P(\text{not walk})=\frac{26}{40}=0.65$.

Answer: About 300 students by bus; $P(\text{not walk})=0.65$.

Law of large numbers and the gambler's fallacy

NCERT section 7.2 Measuring Probability Objectively

Experimental and theoretical probabilities often differ, especially with few trials: 3 threes in 12 rolls gives $\frac{3}{12}=0.25$, not $\frac{1}{6}$. As the number of trials grows, the experimental probability tends to get closer to the theoretical one. This is the law of large numbers.

Probability describes the long run, not the next try. A fair coin has no memory: after six heads in a row, the chance of tails on the next toss is still $\frac{1}{2}$. Believing that tails is 'due' is the gambler's fallacy. Each roll of the dice in Snakes and Ladders (which grew out of India's Jñān-Chaupaḍ) is independent of the rolls before it.

Worked example

Riya rolls a fair die and gets a 6 three times in a row. She says, 'I can't possibly get a 6 next.' Is she right? What is the probability of a 6 on her next roll, and what happens to the proportion of 6s if she rolls 6000 times?

Try it first, then show the solutionSolution

1. Each roll is independent; the die does not remember earlier rolls.
2. So $P(6)=\frac{1}{6}$ on the next roll, as on every roll. Riya's reasoning is the gambler's fallacy.
3. Over 6000 rolls, the proportion of 6s should settle close to $\frac{1}{6}$, that is about 1000 sixes.

Answer: She is wrong; $P(6)=\frac{1}{6}$; over 6000 rolls about 1000 sixes are expected.

Sample space and events

NCERT section 7.3 Elements of Probability: Sample Spaces and Events

The sample space S is the list of all possible outcomes, each written once. The number of outcomes is the sample size $n(S)$.

• One coin: S = {H, T}, $n(S)=2$.
• One die: S = {1, 2, 3, 4, 5, 6}, $n(S)=6$.
• Two coins: S = {HH, HT, TH, TT}, $n(S)=4$ (HT and TH are different outcomes).

An event is any group of outcomes, a part (subset) of S. 'At least one head' with two coins is E = {HH, HT, TH}.

Since an outcome is either in E or not in E: $P(\text{not }E)=1-P(E)$. The sample space should be as detailed as the question needs: {No rain, Drizzle, Light rain, Heavy rain} instead of {Rain, No rain} if amounts matter.

Worked example

A coin is tossed and a die is rolled together. Write the sample space and find the probability of (i) a head and an even number (ii) not getting 'a head and an even number'.

Try it first, then show the solutionSolution

1. S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}, so $n(S)=12$.
2. (i) E = {H2, H4, H6}, so $P(E)=\frac{3}{12}=\frac{1}{4}$.
3. (ii) $P(\text{not }E)=1-\frac{1}{4}=\frac{3}{4}$.

Answer: (i) $\frac{1}{4}$ (ii) $\frac{3}{4}$

Tree diagrams for multi-step experiments

NCERT section 7.4 Tree diagrams

A tree diagram lists the outcomes of an experiment done in steps (toss a coin twice, pick a fruit from each of two baskets).
1. From a starting point, draw one branch for each outcome of step 1.
2. From the end of each branch, draw one branch for each outcome of step 2.
3. Each path from start to finish is one outcome; read them off to get the sample space.

You can write each branch's probability on it. When all paths are equally likely, $P(\text{event})=\frac{\text{favourable paths}}{\text{total paths}}$. For two coins there are $2\times 2=4$ paths; for three coins, $2\times 2\times 2=8$.

Worked example

Three fair coins are tossed. Use a tree diagram to list the outcomes and find the probability of (i) exactly two heads (ii) at least one tail.

Try it first, then show the solutionSolution

1. Each toss splits into H and T, giving $2\times 2\times 2=8$ paths: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
2. (i) Exactly two heads: HHT, HTH, THH, so $\frac{3}{8}$.
3. (ii) Only HHH has no tail, so $P(\text{at least one tail})=1-\frac{1}{8}=\frac{7}{8}$.

Answer: (i) $\frac{3}{8}$ (ii) $\frac{7}{8}$

Maths in real life: Introduction to Probability

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Experimental and Theoretical Probability
• Sample Space, Events and Tree Diagrams

Introduction to Probability formula sheet

Range of probability

$0\le P(E)\le 1$

0 impossible, 1 certain, 0.5 even chance.

Experimental probability

$P(E)\approx \frac{\text{number of times E occurred}}{\text{total number of trials}}$

Also called relative frequency.

Theoretical probability

$P(E)=\frac{\text{number of favourable outcomes}}{\text{number of possible outcomes}}$

Only when outcomes are equally likely.

Complement

$P(\text{not }E)=1-P(E)$

Estimate for a population

$\text{expected number}=P(E)\times \text{population size}$

P(E) from a representative sample.

Sample spaces to remember

$n(S):\text{one coin }2,\text{one die }6,\text{two coins }4,\text{three coins }8,\text{coin and die }12$

Outcomes of a two-step experiment

$n(S)=(\text{outcomes of step 1})\times (\text{outcomes of step 2})$

Count the paths of a tree diagram.

Sum over all outcomes

$P(o_1)+P(o_2)+\cdots +P(o_n)=1$

The probabilities of all outcomes in S add up to 1.

How the school exam asks Introduction to Probability

The question patterns to expect, and the method that works for each.

Tricks to remember Introduction to Probability

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Fair share over all shares

  Theoretical probability = favourable outcomes (your share) ÷ all outcomes (all shares).

  Use it when: Setting up any counting question.

• Zero to one, nothing outside

  Probability lives on a ruler marked 0 to 1. Anything off the ruler (negative or more than 1) is a mistake.

  Use it when: Checking every answer.

• The coin has no memory

  Past tosses do not change the next toss.

  Use it when: Spotting the gambler's fallacy.

• At least one? Subtract none

  P(at least one) = 1 − P(none).

  Use it when: Quick answers with two or three coins.

• Branch, then branch again

  Each step multiplies the number of paths: 2 coins give 4, 3 coins give 8.

  Use it when: Checking that a tree diagram or sample space is complete.

Introduction to Probability: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Introduction to Probability

The slips examiners see most often, and what to write instead.

• ✗Mistake: Treating HT and TH as the same outcome

  The order of two coins seems not to matter.

  Wrong working

  Two coins: S = {HH, HT, TT}, so P(one head) $=\frac{1}{3}$

  Right working

  S = {HH, HT, TH, TT}; P(exactly one head) $=\frac{2}{4}=\frac{1}{2}$

  ✓Do this: Imagine the coins in different colours: first coin, second coin.

  Usually costs: The whole answer.

• ✗Mistake: Counting a repeated letter only once

  Students count distinct letters instead of all letters.

  Wrong working

  BANANA: letters B, A, N, so P(A) $=\frac{1}{3}$

  Right working

  6 letters with three As: P(A) $=\frac{3}{6}=\frac{1}{2}$

  ✓Do this: Write out every letter with its position before counting.

  Usually costs: 1 mark.

• ✗Mistake: Assuming outcomes are equally likely when they are not

  Two possible results feel like a 50:50 chance.

  Wrong working

  A bag has 3 red and 7 blue marbles, so P(red) $=\frac{1}{2}$.

  Right working

  Each of the 10 marbles is equally likely: P(red) $=\frac{3}{10}$.

  ✓Do this: Count the individual objects, not the categories.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Expecting experimental results to match theory exactly

  The theoretical probability is taken as a guaranteed count.

  Wrong working

  10 tosses must give exactly 5 heads.

  Right working

  The number of heads in 10 tosses varies; the proportion gets close to $\frac{1}{2}$ only over many tosses.

  ✓Do this: Say 'about' for expected numbers and mention the law of large numbers.

  Usually costs: 1 mark in reasoning questions.

• ✗Mistake: Dividing by the wrong total in survey questions

  The population size is used instead of the sample size.

  Wrong working

  14 out of a sample of 40 walk; P(walk) $=\frac{14}{1200}$

  Right working

  P(walk) $=\frac{14}{40}=0.35$; estimated walkers $=0.35\times 1200=420$

  ✓Do this: Probability comes from the sample; the population is used only to scale up.

  Usually costs: Both parts of the question.

NCERT exercise guide: Introduction to Probability

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Introduction to Probability

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The probability of getting a number greater than 6 on a single roll of a die is
   (a) 0
   (b) $\frac{1}{6}$
   (c) 1
   (d) $\frac{5}{6}$

   Hint

   Which faces are greater than 6?

   Answer

   (a) 0

2. Q2 · 1 mark · MCQExam level

   When three coins are tossed together, the number of outcomes in the sample space is
   (a) 3
   (b) 6
   (c) 8
   (d) 9

   Hint

   $2\times 2\times 2$.

   Answer

   (c) 8

3. Q3 · 1 mark · Assertion-reasonExam level

   Assertion (A): If a coin is tossed 10 times, the experimental probability of heads must be $\frac{1}{2}$.
   Reason (R): The theoretical probability of heads for a fair coin is $\frac{1}{2}$.
   (a) Both A and R are true and R is the correct explanation of A.
   (b) Both A and R are true but R is not the correct explanation of A.
   (c) A is true but R is false.
   (d) A is false but R is true.

   Hint

   Do experiments always match theory?

   Answer

   (d). R is true, but in only 10 tosses the number of heads can easily differ from 5.

4. Q4 · 2 marks · Very short answerExam level

   A card is drawn at random from cards numbered 1 to 20. Find the probability that the number is a multiple of 3.

   Hint

   List 3, 6, 9, ...

   Answer

   Multiples of 3: 3, 6, 9, 12, 15, 18, so $\frac{6}{20}=\frac{3}{10}$.

5. Q5 · 2 marks · Very short answerHarder exam level

   A basketball player scores in 34 of her last 50 shots. Find the experimental probability that she scores, and estimate how many baskets she will make in her next 75 shots.

   Hint

   Relative frequency × number of shots.

   Answer

   $\frac{34}{50}=0.68$; about $0.68\times 75=51$ baskets.

6. Q6 · 3 marks · Short answerHarder exam level

   A fair spinner has 10 equal sectors numbered 1 to 10. Find the probability that it stops on (i) a prime number (ii) a number greater than 7 (iii) a multiple of 4.

   Hint

   Primes up to 10 are 2, 3, 5, 7.

   Answer

   (i) $\frac{4}{10}=\frac{2}{5}$ (ii) {8, 9, 10}: $\frac{3}{10}$ (iii) {4, 8}: $\frac{2}{10}=\frac{1}{5}$

7. Q7 · 2 marks · Very short answerHarder exam level

   A coin is tossed and a die is rolled. Find the probability of a tail and an odd number.

   Hint

   12 outcomes.

   Answer

   {T1, T3, T5}: $\frac{3}{12}=\frac{1}{4}$.

8. Q8 · 3 marks · Short answerHarder exam level

   Three coins are tossed. Find the probability of (i) exactly two heads (ii) at least two heads (iii) no heads.

   Hint

   List all 8 outcomes.

   Answer

   (i) {HHT, HTH, THH}: $\frac{3}{8}$ (ii) add HHH: $\frac{4}{8}=\frac{1}{2}$ (iii) {TTT}: $\frac{1}{8}$

9. Q9 · 3 marks · Short answerHarder exam level

   In a sample of 60 families, 21 own a two-wheeler only, 15 own a car only, 9 own both and 15 own neither. A town has 3000 families. Estimate the number of families that own (i) both (ii) at least one vehicle.

   Hint

   First find the relative frequencies.

   Answer

   (i) $\frac{9}{60}\times 3000=450$ (ii) $\frac{60-15}{60}\times 3000=2250$

10. Q10 · 5 marks · Long answerHarder exam level

    Box A has 2 red cards and 1 green card; box B has 1 red card and 1 yellow card. One card is drawn from each. Draw a tree diagram, list the sample space and find the probability that (i) both cards are red (ii) at least one card is red (iii) the cards are of different colours.

    Hint

    Label the two red cards in box A separately.

    Answer

    Paths: R1R, R1Y, R2R, R2Y, GR, GY (6). (i) $\frac{2}{6}=\frac{1}{3}$ (ii) $1-\frac{1}{6}=\frac{5}{6}$ (iii) R1Y, R2Y, GR, GY: $\frac{4}{6}=\frac{2}{3}$

11. Q11 · 3 marks · Short answerHOTS / competency

    A weather report gives a 35% chance of rain tomorrow. (i) What is the probability that it will not rain? (ii) Aarav says, 'Rain or no rain, it is 50:50.' Explain his mistake.

    Hint

    Are the two outcomes equally likely?

    Answer

    (i) $1-0.35=0.65$. (ii) Having two outcomes does not make them equally likely; the forecast, based on data, says no rain is much more likely (0.65 against 0.35).

12. Q12 · 3 marks · Short answerHOTS / competency

    Meena tosses a coin 10 times and gets 7 heads. She says the coin must be biased. Is she right? Design a better test and say what result would make you suspect a bias.

    Hint

    Think about the law of large numbers.

    Answer

    Not necessarily: 10 tosses is too few, and 7 heads can happen with a fair coin. Toss it many times (say 1000) and find the experimental probability. If the proportion of heads stays well away from 0.5 (for example, about 0.7) over many tosses, the coin is probably biased.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What is the probability of an impossible event?Show answer

   0

2. 2.What is the probability of a certain event?Show answer

   1

3. 3.What is a sample space?Show answer

   The list of all possible outcomes of a random experiment, each written once.

4. 4.Define experimental probability.Show answer

   Number of times the event occurred ÷ total number of trials (the relative frequency).

5. 5.When can you use favourable ÷ possible outcomes?Show answer

   When all the outcomes are equally likely.

6. 6.What does the law of large numbers say?Show answer

   With more and more trials, the experimental value settles nearer and nearer to the theoretical value.

7. 7.What is the gambler's fallacy?Show answer

   Wrongly believing that past results change the chance of the next independent result.

8. 8.How many outcomes are there when three coins are tossed?Show answer

   8

9. 9.If $P(E)=0.3$, what is $P(\text{not }E)$?Show answer

   0.7

10. 10.What is an event?Show answer

    A group of outcomes from the sample space (a subset of S).

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you name the five words on the probability scale and their positions?
• ☐Can you explain what makes an experiment random?
• ☐Can you find the experimental probability from a table of results?
• ☐Can you find the theoretical probability of a vowel from a given word?
• ☐Can you estimate a number for a whole population from a sample?
• ☐Can you explain why 10 tosses may not give 5 heads?
• ☐Can you spot the gambler's fallacy in a statement?
• ☐Can you write the sample space for a coin and a die together?
• ☐Can you draw a tree diagram for three coins?
• ☐Can you use $P(\text{not }E)=1-P(E)$?

Introduction to Probability: frequently asked questions