MathsICSE Class 10 · Probability

Probability of an Event: Sample Space, Favourable Outcomes and the Formula (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

For equally likely outcomes, the probability of an event $E$ is $P(E) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$. It always lies between 0 (impossible) and 1 (certain), and $P(\text{not } E) = 1 - P(E)$. List the sample space first, then count the favourable outcomes.

Every probability sits between 0 and 1; count favourable outcomes over all outcomes, like 3 even faces out of 6 giving 1/2.

Illustration drawn for QuestionBuddy.

Understand Probability of an Event

The CISCE 2027 syllabus covers random experiments, sample space, events, the definition of probability and simple problems on single events.

Key words.

A random experiment has known possible outcomes, but you cannot say in advance which one will occur (drawing a card, picking a ball).
The sample space $S$ is the set of all possible outcomes; $n (S)$ is how many there are.
An event $E$ is a set of outcomes we are interested in; these are the favourable outcomes, $n (E)$ .
Outcomes are equally likely when none is more likely than another ('at random', 'well shuffled', 'fair').

Definition. $P (E) = \frac{n ( E )}{n ( S )}$ . Give the answer as a fraction in lowest terms (or as a decimal or percentage if asked).

Range. $0 \leq P (E) \leq 1$ .

$P = 0$ : impossible event (a 7 on an ordinary die).
$P = 1$ : sure event (a number less than 7 on a die).
Any decimal or percentage between 0 and 1 is possible: 67% $= 0.67$ is a valid probability; $\frac{5}{4}$ and $- 0.2$ are not (2025 Q1(iii)).

Complementary event. 'Not $E$ ' contains every outcome not in $E$ , so $P (not E) = 1 - P (E)$ , and $P (E) + P (not E) = 1$ .

Counting favourable outcomes carefully.

1 is a perfect square ( $1 = 1^{2}$ ) but not prime (2025 Q9(ii)).
'Multiple of both 4 and 6' means multiple of 12 (their LCM).
For letters of a word, every card counts, including repeats: in ENGINEERING there are 11 cards, three of them N.
'Common to two words' means letters in both words, not all letters of either (2024 Q8(ii)).

Unknown quantities. If a probability is given, write $\frac{favourable}{total}$ with the unknown and solve. When items are added or removed, recount the total before using the formula again.

Probability of an Event: formulas to remember

Probability of an event: $P (E) = \frac{n ( E )}{n ( S )}$; Equally likely outcomes.
Range: $0 \leq P (E) \leq 1$; 0 impossible, 1 sure.
Complement: $P (not E) = 1 - P (E)$; Quicker for 'not' or 'at least' events.
Multiples of two numbers: $multiple of a and b = multiple of lcm (a, b)$; 4 and 6: multiples of 12.

How to solve Probability of an Event questions

Write $S$ , list the favourable outcomes, divide, simplify.

1Write the sample space or its size $n (S)$ ; for numbered cards from $a$ to $b$ , $n (S) = b - a + 1$ .
2List the favourable outcomes explicitly (do not just count in your head).
3Check the traps: 1 is a square but not prime; repeated letters count; 'and' needs both conditions.
4Divide: $P (E) = \frac{n ( E )}{n ( S )}$ .
5Simplify the fraction; for 'not', use $1 - P (E)$ .
6For an unknown, set up the fraction with the unknown and solve; recount totals after any change.

Probability of an Event 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.

A random quiz question

Suppose a quiz app picks one question at random from 120: 30 history, 50 science and 40 geography.

What is the probability of a science question, and of a question that is not history?

Try it first, then show the solutionSolution

$P (science) = \frac{50}{120} = \frac{5}{12}$ .
$P (not history) = 1 - \frac{30}{120} = \frac{3}{4}$ .

Answer: $\frac{5}{12}$ and $\frac{3}{4}$

Memory hook: 'Not' is 1 minus.

Probability as a percentage

Imagine a pencil box with 20 pencils, 3 of which have broken leads, and you take one without looking.

What is the chance of a broken pencil, as a fraction, decimal and percentage?

Try it first, then show the solutionSolution

$n (S) = 20$ and 3 outcomes are favourable: $P = \frac{3}{20}$ .
$\frac{3}{20} = 0.15 = 15%$ , which lies between 0 and 1 (0% and 100%) as it must.

Answer: $\frac{3}{20}$ = 0.15 = 15%

Memory hook: Same probability, three outfits.

Tricks to remember Probability of an Event

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

Write S first
No total, no probability: the denominator is always the whole sample space.
Use it when: Avoiding the 2024 sample-space error.
One is square, not prime
$1 = 1^{2}$ belongs with the squares; primes start at 2.
Use it when: Numbered-card questions.
Between nought and one
Any probability, written any way, must fit between 0 and 1 (0% to 100%).
Use it when: 'Which cannot be a probability?' MCQs.

Probability of an Event: solved examples

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

Example 1: Which is not a probability? · 1 mark · MCQBoard level

Which of the following cannot be the probability of an event?
(a) 0
(b) $\frac{3}{7}$
(c) $- 0.2$
(d) 85%

Try it first, then show the solutionSolution

A probability lies between 0 and 1; 85% $= 0.85$ is allowed; $- 0.2 < 0$ is not.

Answer: (c)

Example 2: Numbered cards 11 to 50 · 3 marks · Short answerBoard level

Cards numbered 11 to 50 are shuffled and one is drawn at random. Find the probability that the number on it is (a) a multiple of 7, (b) a perfect square, (c) a prime number.

Try it first, then show the solutionSolution

$n (S) = 50 - 11 + 1 = 40$ .
(a) 14, 21, 28, 35, 42, 49: $\frac{6}{40} = \frac{3}{20}$ .
(b) 16, 25, 36, 49: $\frac{4}{40} = \frac{1}{10}$ .
(c) 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47: $\frac{11}{40}$ .

Answer: (a) $\frac{3}{20}$ (b) $\frac{1}{10}$ (c) $\frac{11}{40}$

Example 3: Letters of a word · 3 marks · Short answerHarder board level

Each letter of the word ENGINEERING is written on a separate card, and the 11 cards are shuffled. One card is drawn at random. Find the probability that it shows (a) a vowel, (b) the letter N, (c) a letter of the word RING.

Try it first, then show the solutionSolution

$n (S) = 11$ (E, N, G, I, N, E, E, R, I, N, G): E 3, N 3, G 2, I 2, R 1.
(a) Vowels: E, E, E, I, I: $\frac{5}{11}$ .
(b) N appears three times: $\frac{3}{11}$ .
(c) Cards showing R, I, N or G: $1 + 2 + 3 + 2 = 8$ , so $\frac{8}{11}$ .

Answer: (a) $\frac{5}{11}$ (b) $\frac{3}{11}$ (c) $\frac{8}{11}$

Example 4: Unknown number of balls · 3 marks · Short answerHarder board level

A bag has 4 red balls, 6 blue balls and some green balls. The probability of drawing a green ball at random is $\frac{1}{3}$ . Find the number of green balls, and the probability that the ball drawn is not blue.

Try it first, then show the solutionSolution

Let there be $g$ green balls: $\frac{g}{10 + g} = \frac{1}{3}$ , so $3 g = 10 + g$ and $g = 5$ .
Total $= 15$ ; $P (not blue) = 1 - \frac{6}{15} = \frac{9}{15} = \frac{3}{5}$ .

Answer: 5 green; $\frac{3}{5}$

Example 5: Changing the box · 3 marks · Short answerHOTS / competency

A box contains 60 bulbs, of which 9 are defective. One bulb is drawn at random.
(a) Find the probability that it is not defective.
(b) The bulb drawn is found to be good and is kept aside. Another bulb is now drawn from the box. Find the probability that it is defective.

Try it first, then show the solutionSolution

(a) $P (good) = 1 - \frac{9}{60} = \frac{51}{60} = \frac{17}{20}$ .
(b) The box now has 59 bulbs, still 9 defective: $P = \frac{9}{59}$ .

Answer: (a) $\frac{17}{20}$ (b) $\frac{9}{59}$

Example 6: Case study: a lucky dip of 60 discs · 4 marks · Case-basedHOTS / competency

A lucky dip has 60 discs numbered 1 to 60. One disc is drawn at random. Find the probability that the number is:
(a) a multiple of both 4 and 6;
(b) a perfect cube;
(c) a number whose digits add up to 6;
(d) not a multiple of 5.

Try it first, then show the solutionSolution

(a) Multiples of $lcm (4, 6) = 12$ : 12, 24, 36, 48, 60: $\frac{5}{60} = \frac{1}{12}$ .
(b) 1, 8, 27 (64 is too big): $\frac{3}{60} = \frac{1}{20}$ .
(c) 6, 15, 24, 33, 42, 51, 60: $\frac{7}{60}$ .
(d) 12 multiples of 5, so $1 - \frac{12}{60} = \frac{4}{5}$ .

Answer: (a) $\frac{1}{12}$ (b) $\frac{1}{20}$ (c) $\frac{7}{60}$ (d) $\frac{4}{5}$

Where students lose marks in Probability of an Event

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

Mistake: Using part of the sample space as the total
The denominator is every possible outcome, not just the favourable ones.
Wrong working
$P (even on a die) = \frac{2}{3}$ , using ${2, 4, 6}$ as the total
Right working
$S = {1, 2, 3, 4, 5, 6}$ , so $P = \frac{3}{6} = \frac{1}{2}$
Do this: Write $S$ in full before counting.
Usually costs: Reported in the CISCE examiners' comments on 2024 Q1(xiii).
Mistake: 1 counted as prime or left out of the squares
1 has only one factor, so it is not prime; $1 = 1^{2}$ is a perfect square.
Wrong working
Squares from 1 to 30: 4, 9, 16, 25
Right working
1, 4, 9, 16, 25
Do this: Start every list from the first card.
Usually costs: Reported in the CISCE examiners' comments on 2025 Q9(ii).
Mistake: Unsimplified answers
The final answer should be in lowest terms.
Wrong working
$\frac{6}{8}$
Right working
$\frac{3}{4}$
Do this: Cancel common factors at the last step.
Usually costs: Reported in the CISCE examiners' comments on 2024 Q8(ii) and 2025 Q9(ii).
Mistake: Rejecting a percentage
67% is 0.67, which lies between 0 and 1.
Wrong working
'67% cannot be a probability'
Right working
$\frac{5}{4} > 1$ cannot be a probability
Do this: Convert every option to a decimal before comparing.
Usually costs: Reported in the CISCE examiners' comments on 2025 Q1(iii).

Exam tips

Asked in 2025 Q9(ii) (3 marks): cards 01 to 30 with prizes for a perfect square, an even multiple of 3 and a prime; the expected answers were $\frac{1}{6}$ , $\frac{1}{6}$ and $\frac{1}{3}$ .
Asked in 2024 Q8(ii) (3 marks): letters A, D, M, N, O, S, U, Y on cards; letters of MONDAY, not in MONDAY, and common to SUNDAY and MONDAY.
The CISCE examiners suggest writing the sample space and the favourable outcomes in every answer.
Write each set of favourable outcomes in braces, then the count, then the fraction.

Practice questions on Probability of an Event

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
The probability of an impossible event is:
(a) $- 1$
(b) 0
(c) $\frac{1}{2}$
(d) 1
Hint
No favourable outcomes.
Answer
(b)
Q2 · 2 marks · Short answerBoard level
The letters of the word STATISTICS are written on 10 cards and one is drawn at random. Find the probability that it is (a) the letter S, (b) a vowel.
Hint
Count repeated letters separately.
Answer
(a) S appears 3 times: $\frac{3}{10}$ (b) A, I, I: $\frac{3}{10}$
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If $P (E) = 0.38$ , then $P (not E) = 0.62$ .
Reason (R): $P (E) + P (not E) = 1$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
$1 - 0.38$ .
Answer
(c)
Q4 · 3 marks · Short answerHarder board level
Tickets numbered 1 to 75 are put in a box and one is drawn at random. Find the probability that the number is (a) a multiple of 9, (b) a number ending in 5, (c) a perfect square.
Hint
List each set; do not forget 1 in (c).
Answer
(a) 9, 18, …, 72: $\frac{8}{75}$ (b) 5, 15, …, 75: $\frac{8}{75}$ (c) 1, 4, 9, …, 64: $\frac{8}{75}$
Q5 · 2 marks · Short answerHarder board level
A bag contains 3 white balls and some black balls. The probability of drawing a black ball is three times that of drawing a white ball. Find the number of black balls.
Hint
Same total, so compare counts.
Answer
Black $= 3 \times 3 = 9$ (then $P (black) = \frac{9}{12} = \frac{3}{4}$ and $P (white) = \frac{1}{4}$ ).
Q6 · 2 marks · Short answerHarder board level
A class has 24 boys and some girls. A student is chosen at random, and the probability that it is a girl is $\frac{3}{7}$ . How many girls are there?
Hint
$\frac{g}{24 + g} = \frac{3}{7}$ .
Answer
$7 g = 72 + 3 g$ , so $g = 18$ .
Q7 · 3 marks · Short answerHOTS / competency
A bag contains only red and blue balls, and the probability of drawing a red ball is $\frac{1}{4}$ . When 6 more red balls are added, the probability becomes $\frac{1}{2}$ . Find the original numbers of red and blue balls.
Hint
Let there be $r$ red and $n$ balls in all; then $n = 4 r$ .
Answer
$\frac{r + 6}{4 r + 6} = \frac{1}{2}$ gives $2 r + 12 = 4 r + 6$ , so $r = 3$ and $n = 12$ : 3 red and 9 blue.

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.Formula for $P (E)$ ?Show answer
$\frac{n ( E )}{n ( S )}$
2.Probability of a sure event?Show answer
1
3. $P (not E)$ ?Show answer
$1 - P (E)$
4.Is 1 prime? Is it a perfect square?Show answer
Not prime; yes, a perfect square.
5.Number of cards from 11 to 50?Show answer
40

Probability of an Event: frequently asked questions

What is the formula for the probability of an event?

For equally likely outcomes, P(E) = number of favourable outcomes ÷ total number of outcomes. The answer always lies between 0 and 1.

What is a sample space?

The sample space is the set of all possible outcomes of a random experiment. For one die it is {1, 2, 3, 4, 5, 6}, with 6 outcomes.

How do I find the probability of an event not happening?

Subtract the probability of the event from 1: P(not E) = 1 − P(E).