MathsCBSE Class 9 · Introduction to Probability

Sample Space, Events and Tree Diagrams: Class 9

Q: Is HT the same as TH?

As ordered outcomes of two tosses they are different, and counting them separately keeps all outcomes equally likely.

Q: What is a tree diagram for?

It lists every path of a multi-step experiment so that no outcome is missed or counted twice.

Q: Why is the sum 7 most likely with two dice?

Six of the 36 equally likely pairs give 7, more than for any other sum.

Updated 17 October 2026 · Fact-checked

The sample space is the set of all possible outcomes of an experiment, and an event is any subset of it. For equally likely outcomes, $P(E) = \frac{n(E)}{n(S)}$. For experiments in several steps, a tree diagram lists every path so that you can count outcomes without missing any: for example 4 outcomes for two coins, 8 for three coins and 36 for two dice. Draws without replacement have fewer branches at later steps.

Each path of the tree is one outcome, so S has 4 outcomes; the event 'exactly one head' picks HT and TH, giving 2/4 = 1/2.

Illustration drawn for QuestionBuddy.

Understand Sample Space, Events and Tree Diagrams

The sample space $S$ of an experiment is the set of all its possible outcomes. An event $E$ is a collection of outcomes from $S$ that we are interested in. If all outcomes in $S$ are equally likely, $P (E) = \frac{n ( E )}{n ( S )}$ , where $n$ counts outcomes.

Single-step spaces. A coin: $S = {H, T}$ . A die: $S = {1, 2, 3, 4, 5, 6}$ . A card from a pack: 52 outcomes.

Multi-step experiments and tree diagrams. In a tree diagram each stage is a set of branches; every path from the start to the end is one outcome. Tossing two coins gives paths $H H, H T, T H, T T$ : $4$ outcomes. For three coins there are $2 \times 2 \times 2 = 8$ outcomes. The number of outcomes of a multi-step experiment is the product of the number of options at each step. If one coin is tossed and one die rolled there are $2 \times 6 = 12$ outcomes. Two dice give $36$ ordered pairs $(a, b)$ , where $(2, 5)$ and $(5, 2)$ are different outcomes.

Reading events from the space. For two coins, "exactly one head" is ${H T, T H}$ , so its probability is $\frac{2}{4} = \frac{1}{2}$ , and "at least one head" is ${H H, H T, T H}$ with probability $\frac{3}{4}$ . For two dice, the sum $8$ arises from $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$ so $P = \frac{5}{36}$ ; the sum $7$ arises in $6$ ways, the most likely sum, with $P = \frac{1}{6}$ . In a table of the $36$ pairs, the sum $s$ has $6 - ∣ s - 7∣$ outcomes.

With or without replacement. If a card or ball is put back before the next draw (with replacement) the sample space stays the same at each step: drawing twice from 5 balls gives $5 \times 5 = 25$ ordered outcomes. If it is not put back (without replacement) the number of options drops: $5 \times 4 = 20$ ordered outcomes. From a bag of 3 red and 2 green balls, drawing twice with replacement gives $13$ same-colour outcomes out of $25$ (RR: $3 \times 3 = 9$ , GG: $2 \times 2 = 4$ ), a probability of $\frac{13}{25}$ ; without replacement it gives RR: $3 \times 2 = 6$ and GG: $2 \times 1 = 2$ out of $20$ , a probability of $\frac{8}{20} = \frac{2}{5}$ .

Complement and "at least". For three coins, "at least one head" is easiest as $1 - P (no head) = 1 - \frac{1}{8} = \frac{7}{8}$ . Always define the events precisely: "exactly 2 heads" in three tosses is ${H H T, H T H, T H H}$ , giving $\frac{3}{8}$ .

Method. Decide the stages, write the sample space or draw the tree, check that the outcomes are equally likely (otherwise weight them), identify the event, count, and write the fraction in lowest terms.

Sample Space, Events and Tree Diagrams: formulas to remember

Probability from a sample space: $P (E) = \frac{n ( E )}{n ( S )}$; Equally likely outcomes.
Outcomes of a multi-step experiment: $n (S) = n_{1} \times n_{2} \times \dots$; Product of options at each stage.
Two dice: $n (S) = 36, # (sum = s) = 6 - ∣ s - 7∣$; Ordered pairs.
Without replacement: $n (S) = n (n - 1) for two draws from n items$; With replacement it is $n^{2}$ .
Complement: $P (at least one) = 1 - P (none)$; Fast for several trials.
Event as a set: $E \subseteq S$; List the outcomes that satisfy the condition.

How to solve Sample Space, Events and Tree Diagrams questions

Build the sample space first; then count the event.

1Identify the stages and the options at each stage.
2Draw a tree or list the outcomes in order, such as $H T$ or $(3, 5)$ .
3Check the total against the product of the options (and reduce it for "without replacement").
4Write the event as a list of outcomes that satisfy the condition.
5Compute $P (E) = \frac{n ( E )}{n ( S )}$ and simplify.
6For "at least one" or "not", use the complement.

Sample Space, Events and Tree Diagrams in real life

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.

Counting outcomes with a tree

Suppose a canteen offers a meal combo of one soup (tomato, sweet corn or lemon coriander) and one dessert (fruit or ice cream), and a combo is picked at random.

How many combinations are there, and what is the probability of tomato soup with fruit?

Try it first, then show the solutionSolution

Options: $3 \times 2 = 6$ combinations.
Favourable: tomato with fruit, only $1$ .
Probability $= \frac{1}{6}$ .

Answer: There are 6 combinations and the probability of tomato soup with fruit is $\frac{1}{6}$ .

Memory hook: Multiply the choices at each step.

Tricks to remember Sample Space, Events and Tree Diagrams

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

Multiply the stages
The total number of outcomes is the product of options at each stage, reduced for non-replacement.
Use it when: Sizing a sample space quickly.
Order matters in a pair
$(2, 5)$ and $(5, 2)$ are different outcomes for two dice or two draws.
Use it when: Avoiding undercounting.
At least means one minus none
Subtract the "none" case from 1.
Use it when: Several coins or draws.

Sample Space, Events and Tree Diagrams: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Two-digit numbers from cards · 3 marks · Short answerBoard level

Three cards numbered 1, 2 and 3 are drawn one after another without replacement, and the first two cards form a two-digit number (the first card is the tens digit). Write the sample space and find the probability that the number is (i) even, (ii) greater than 20, (iii) divisible by 3.

Try it first, then show the solutionSolution

The first card has 3 options and the second 2, so $S = {12, 13, 21, 23, 31, 32}$ with $n (S) = 6$ . (1)
(i) Even: $12, 32$ , so $\frac{2}{6} = \frac{1}{3}$ . (½)
(ii) Greater than 20: $21, 23, 31, 32$ , so $\frac{4}{6} = \frac{2}{3}$ . (½)
(iii) Divisible by 3: $12, 21$ (digit sum 3), so $\frac{2}{6} = \frac{1}{3}$ . (1)

Answer: (i) $\frac{1}{3}$ ; (ii) $\frac{2}{3}$ ; (iii) $\frac{1}{3}$ .

Example 2: Sum on two dice · 3 marks · Short answerBoard level

Two dice are rolled. Find the probability that (i) the sum is 8, (ii) the sum is 7, (iii) both dice show the same number.

Try it first, then show the solutionSolution

The sample space has $6 \times 6 = 36$ ordered pairs. (½)
(i) Sum 8: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$ , so $\frac{5}{36}$ . (1)
(ii) Sum 7: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$ , so $\frac{6}{36} = \frac{1}{6}$ . (1)
(iii) Doubles: $(1, 1)$ to $(6, 6)$ , so $\frac{6}{36} = \frac{1}{6}$ . (½)

Answer: (i) $\frac{5}{36}$ ; (ii) $\frac{1}{6}$ ; (iii) $\frac{1}{6}$ .

Example 3: Three coins · 5 marks · Long answerHarder board level

Three coins are tossed. Write the sample space and find the probability of (i) exactly two heads, (ii) at least two heads, (iii) all three the same, (iv) a head on the first coin and exactly two heads altogether.

Try it first, then show the solutionSolution

$S = {H H H, H H T, H T H, H T T, T H H, T H T, T T H, T T T}$ , $n (S) = 8$ . (1)
(i) Exactly two heads: ${H H T, H T H, T H H}$ , so $\frac{3}{8}$ . (1)
(ii) At least two heads: add $H H H$ , giving $4$ outcomes: $\frac{4}{8} = \frac{1}{2}$ . (1)
(iii) All the same: ${H H H, T T T}$ , so $\frac{2}{8} = \frac{1}{4}$ . (1)
(iv) First coin a head and exactly two heads: ${H H T, H T H}$ , so $\frac{2}{8} = \frac{1}{4}$ . (1)

Answer: (i) $\frac{3}{8}$ ; (ii) $\frac{1}{2}$ ; (iii) $\frac{1}{4}$ ; (iv) $\frac{1}{4}$ .

Example 4: Draws without replacement · 3 marks · Short answerHarder board level

A bag has 2 red and 3 blue balls. Two balls are drawn one after another without replacement. Using a tree diagram, find the probability that (i) both are red, (ii) the balls are of different colours.

Try it first, then show the solutionSolution

Label the balls $R_{1}, R_{2}, B_{1}, B_{2}, B_{3}$ . The first draw has 5 options and the second has 4, so $n (S) = 5 \times 4 = 20$ . (1)
(i) Both red: $R_{1} R_{2}$ and $R_{2} R_{1}$ , so $\frac{2}{20} = \frac{1}{10}$ . (1)
(ii) Different colours: first red then blue $2 \times 3 = 6$ , first blue then red $3 \times 2 = 6$ , total $12$ : $\frac{12}{20} = \frac{3}{5}$ . (1)

Answer: (i) $\frac{1}{10}$ ; (ii) $\frac{3}{5}$ .

Example 5: A coin-and-die game · 4 marks · Case-basedHOTS / competency

Case: a game is played by tossing a coin and then rolling a die. You win if the coin shows a head and the die shows a prime number, or the coin shows a tail and the die shows 6. (i) How many outcomes are there? (ii) Find the probability of a head with a prime. (iii) Find the probability of winning. (iv) Is the game fair, meaning a win probability of $\frac{1}{2}$ ?

Try it first, then show the solutionSolution

(i) $2 \times 6 = 12$ outcomes. (½)
(ii) Head with prime ( $2, 3, 5$ ): $H 2, H 3, H 5$ , so $\frac{3}{12} = \frac{1}{4}$ . (1)
(iii) Winning outcomes: $H 2, H 3, H 5, T 6$ , so $\frac{4}{12} = \frac{1}{3}$ . (1½)
(iv) $\frac{1}{3} < \frac{1}{2}$ , so the game is not fair; it favours the house. (1)

Answer: (i) 12; (ii) $\frac{1}{4}$ ; (iii) $\frac{1}{3}$ ; (iv) not fair.

Where students lose marks in Sample Space, Events and Tree Diagrams

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

Mistake: Treating (a, b) and (b, a) as one outcome
Undercounting two-dice outcomes.
Wrong working
Sum 7 has 3 outcomes: (1,6), (2,5), (3,4).
Right working
Sum 7 has 6 ordered outcomes.
Do this: Use ordered pairs for two dice or two draws.
Usually costs: The answer.
Mistake: Using 5 x 5 without replacement
Forgetting that the first item is gone.
Wrong working
$n (S) = 25$ for two draws without replacement.
Right working
$n (S) = 5 \times 4 = 20$ .
Do this: Read the word "without replacement".
Usually costs: 2 marks.
Mistake: Counting a sample space of unequal outcomes
Using "H or T" outcomes that are not equal.
Wrong working
Number of heads can be 0, 1 or 2, so each has probability $\frac{1}{3}$ .
Right working
Outcomes $H H, H T, T H, T T$ are equally likely, so $P (1) = \frac{1}{2}$ .
Do this: List the equally likely outcomes, not the counts.
Usually costs: The question.

Exam tips

Write the sample space explicitly for at most 8 to 12 outcomes; otherwise count by products.
Use ordered pairs for two dice, two cards and two draws.
Read "with" or "without" replacement before counting.
Cross-check by computing a complement.

Practice questions on Sample Space, Events and Tree Diagrams

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

Q1 · 1 mark · MCQBoard level
Two dice are thrown together. The probability that the sum is 12 is
(a) $\frac{1}{36}$
(b) $\frac{1}{12}$
(c) $\frac{1}{6}$
(d) $\frac{1}{18}$
Hint
Only one pair gives 12.
Answer
(a) $\frac{1}{36}$ . The only pair is $(6, 6)$ among 36 outcomes.
Q2 · 3 marks · Short answerBoard level
A die is thrown twice. Find the probability that (i) the first throw is even and the second is odd, (ii) the sum is less than 5, (iii) both throws give the same number.
Hint
Count ordered pairs out of 36.
Answer
(i) $3 \times 3 = 9$ pairs, $\frac{9}{36} = \frac{1}{4}$ . (ii) Sums 2, 3, 4 occur in $1 + 2 + 3 = 6$ ways, $\frac{6}{36} = \frac{1}{6}$ . (iii) $6$ pairs, $\frac{6}{36} = \frac{1}{6}$ .
Q3 · 3 marks · Short answerHarder board level
Two dice are rolled together. Find the probability that the product of the two numbers is a perfect square.
Hint
List pairs whose product is 1, 4, 9, 16, 25 or 36.
Answer
Product 1: $(1, 1)$ ; product 4: $(1, 4), (4, 1), (2, 2)$ ; product 9: $(3, 3)$ ; product 16: $(4, 4)$ ; product 25: $(5, 5)$ ; product 36: $(6, 6)$ . That is $1 + 3 + 1 + 1 + 1 + 1 = 8$ pairs, so the probability is $\frac{8}{36} = \frac{2}{9}$ .
Q4 · 5 marks · Long answerHarder board level
Two cards are drawn one after another, without replacement, from five cards numbered 1 to 5. List how many ordered outcomes there are, and find the probability that (i) the sum is even, (ii) the first card is greater than the second, (iii) the sum is 6, (iv) both numbers are odd.
Hint
There are $5 \times 4 = 20$ ordered pairs, none with equal numbers.
Answer
$n (S) = 20$ . (i) Both odd: $3 \times 2 = 6$ ; both even: $2 \times 1 = 2$ ; total $8$ , so $\frac{8}{20} = \frac{2}{5}$ . (ii) By symmetry half the pairs have the first card larger: $\frac{10}{20} = \frac{1}{2}$ . (iii) $(1, 5), (5, 1), (2, 4), (4, 2)$ : $\frac{4}{20} = \frac{1}{5}$ . (iv) $\frac{6}{20} = \frac{3}{10}$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case: a bag holds 3 red and 2 green balls. Two balls are drawn one after another. (i) With replacement, find the probability that both balls are the same colour. (ii) With replacement, find the probability of at least one green ball. (iii) Without replacement, find the probability that both are the same colour. (iv) Why does removing a ball change the answer?
Hint
With replacement there are $5 \times 5 = 25$ ordered outcomes; without, $5 \times 4 = 20$ .
Answer
(i) RR: $3 \times 3 = 9$ ; GG: $2 \times 2 = 4$ ; total $13$ , so $\frac{13}{25}$ . (ii) No green: $3 \times 3 = 9$ outcomes, so at least one green is $1 - \frac{9}{25} = \frac{16}{25}$ . (iii) RR: $3 \times 2 = 6$ ; GG: $2 \times 1 = 2$ ; total $8$ out of $20$ : $\frac{2}{5}$ . (iv) Without replacement the bag changes after the first draw (fewer balls, fewer of that colour), so the second draw is not from the same sample space.

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.Outcomes for three coins?Show answer
8.
2.Outcomes for two dice?Show answer
36.
3.Most likely sum of two dice?Show answer
7, with probability $\frac{1}{6}$ .
4.What is an event?Show answer
A subset of the sample space.
5.Number of ordered draws of 2 from 5 without replacement?Show answer
20.

Sample Space, Events and Tree Diagrams: frequently asked questions

Is HT the same as TH?

As ordered outcomes of two tosses they are different, and counting them separately keeps all outcomes equally likely.

What is a tree diagram for?

It lists every path of a multi-step experiment so that no outcome is missed or counted twice.

Why is the sum 7 most likely with two dice?

Six of the 36 equally likely pairs give 7, more than for any other sum.