MathsCBSE Class 10 · Probability

Probability Basic Terms and Classical Definition for Class 10

Updated 14 October 2026 · Fact-checked

Classical probability measures chance as a fraction: favourable outcomes divided by all equally likely outcomes. You must know random experiment, outcome, sample space, event, elementary event and equally likely outcomes. The board asks MCQs on valid probability values, assertion-reason on equally likely outcomes, and 2 to 5 mark counting problems with numbers, letters, coins and unknowns.

Colours are not always equally likely; split into equal parts first, then count favourable over total.

Illustration drawn for QuestionBuddy.

Understand Probability: Basic Terms and Classical Definition

Probability puts a number on chance. You start with a random experiment: an action whose result you cannot predict in advance, such as spinning a fair spinner or picking a token from a box. Each possible result is an outcome, and the full list of outcomes is the sample space. An event is any collection of outcomes you care about, for example 'the token shows an even number'. An event with just one outcome, such as getting a 4 on a die, is an elementary event.

The classical definition says that when all outcomes are equally likely, $P (E) = \frac{favourable outcomes}{total outcomes}$ . It works because equally likely outcomes share the chance equally. If there are 10 outcomes, each carries $\frac{1}{10}$ of the chance, so an event made of 3 of them gets $\frac{3}{10}$ .

The condition equally likely is the heart of the definition. Words like 'fair', 'unbiased', 'well-shuffled', 'at random' and 'identical balls' tell you the condition holds. If it fails, the formula gives wrong answers. A spinner with one sector of $18 0^{\circ}$ and two of $9 0^{\circ}$ has three colours, but the colours are not equally likely, so the chance of the big sector is $\frac{1}{2}$ , not $\frac{1}{3}$ . Fix such cases by splitting into equal parts: think of four equal $9 0^{\circ}$ pieces, two of which form the big sector.

Because favourable outcomes are between $0$ and the total, every probability lies from $0$ to $1$ . A sure event has probability $1$ and an impossible event has $0$ . The other topic pages in this chapter build on this: complements, coins and dice, cards, and finding unknowns in bags all use this one formula with careful counting.

So your real job in every question is to count correctly: write a total of equally likely outcomes, list the favourable ones without missing or repeating any, divide and simplify. When counting letters of a word, count each letter position separately, so PEPPER has 6 equally likely letters, not 3.

Probability: Basic Terms and Classical Definition: formulas to remember

Classical probability: $P (E) = \frac{number of outcomes favourable to E}{total number of equally likely outcomes}$; Use only when every outcome is equally likely.
Range of probability: $0 \leq P (E) \leq 1$; Reject any negative value or value above 1 (or 100%).
Sure and impossible events: $P (sure) = 1, P (impossible) = 0$; Sure: all outcomes favourable. Impossible: none favourable.
Elementary events add to 1: $P (E_{1}) + P (E_{2}) + \dots + P (E_{n}) = 1$; Each elementary event has exactly one outcome.
Count of whole numbers from a to b: $b - a + 1$; Use when tickets start at a number other than 1.

How to solve Probability: Basic Terms and Classical Definition questions

Every question on this topic follows the same counting routine. Write each step so the examiner can award method marks.

1Identify the random experiment and check that outcomes are equally likely (look for 'fair', 'at random', 'well-shuffled').
2Write the total number of outcomes, listing the sample space if it is small; split unequal outcomes into equal parts if needed.
3Read the event carefully: 'or', 'and', 'not', 'at least', 'at most' change what counts.
4List the favourable outcomes explicitly, counting overlaps only once.
5Divide favourable by total and simplify to lowest terms.
6If an unknown is involved, set the fraction equal to the given probability and solve.
7Check: the answer lies between 0 and 1, and any unknown is a sensible whole number.

Probability: Basic Terms and Classical Definition 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.

Classical definition of probability

Suppose your school canteen runs a lucky-coupon scheme during sports week. Coupons numbered 101 to 200 are mixed in a box and each student picks one without looking. A coupon wins a free samosa if its number is a multiple of 8.

Find the probability that a coupon picked at random wins a samosa. Also find the probability that the coupon number has the same first and last digit.

Try it first, then show the solutionSolution

Total coupons $= 200 - 101 + 1 = 100$ .
The multiples of 8 in this range run from 104 to 200, so their number is $\frac{200 - 104}{8} + 1 = 13$ .
So 13 of the 100 equally likely coupons win.
$P (samosa) = \frac{13}{100}$ .
Same first and last digit: the numbers 101, 111, 121, ..., 191 qualify, which is 10 numbers; 200 does not (its first digit is 2 and its last digit is 0).
$P = \frac{10}{100} = \frac{1}{10}$ .

Answer: P(samosa) = 13/100; P(same first and last digit) = 1/10

Memory hook: Coupon box rule: count the range with b − a + 1 before anything else.

Equally likely outcomes

Imagine your class makes a spinner for a festival stall. It is a circle divided into sectors: red takes $18 0^{\circ}$ , green takes $12 0^{\circ}$ and yellow takes $6 0^{\circ}$ . A player wins a prize if the arrow stops on green or yellow.

Your friend says there are 3 colours, so the chance of winning is $\frac{2}{3}$ . Is this correct? Find the correct probability of winning. The stall owner then redraws the circle, keeping yellow at $6 0^{\circ}$ , so that the probability of winning really is $\frac{2}{3}$ . Find the new green and red angles.

Try it first, then show the solutionSolution

The three colours are not equally likely because the sectors are unequal.
Split the circle into equal $6 0^{\circ}$ parts: $36 0^{\circ} \div 6 0^{\circ} = 6$ equally likely parts.
Red $= 3$ parts, green $= 2$ parts, yellow $= 1$ part.
Winning parts $= 2 + 1 = 3$ , so $P (win) = \frac{3}{6} = \frac{1}{2}$ , not $\frac{2}{3}$ .
For $P (win) = \frac{2}{3}$ , green and yellow together must take $\frac{2}{3} \times 36 0^{\circ} = 24 0^{\circ}$ .
Yellow stays $6 0^{\circ}$ , so green $= 24 0^{\circ} - 6 0^{\circ} = 18 0^{\circ}$ and red $= 36 0^{\circ} - 24 0^{\circ} = 12 0^{\circ}$ .

Answer: No; the correct probability of winning is 1/2. For a winning chance of 2/3, green must be $18 0^{\circ}$ and red $12 0^{\circ}$ (yellow $6 0^{\circ}$ ).

Memory hook: Unequal slices of a spinner: cut them into equal slices before you count.

Tricks to remember Probability: Basic Terms and Classical Definition

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

F over T
Probability is 'Favourite over Total': what you want on top, everything possible below.
Use it when: Write the fraction the right way up every time.
Zero to One fence
Probability lives inside a fence from 0 to 1; anything outside is wrong.
Use it when: Eliminate MCQ options and check your final answers.
FAIR check
No 'fair', 'random' or 'well-shuffled'? Ask: are outcomes really equally likely?
Use it when: Assertion-reason questions on match results, spinners and coin lists.

Probability: Basic Terms and Classical Definition: solved examples

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

Example 1: Sample space, event and elementary events on spinners · 3 marks · Short answerBoard level

(i) A fair spinner has 8 equal sectors numbered 1 to 8. Write the sample space and the event E: 'the number is prime'. Find $P (E)$ .

(ii) A second spinner has only the numbers 1, 2, 3 and 4 on unequal sectors. $P (1) = \frac{1}{3}$ , $P (2) = \frac{1}{4}$ and $P (3) = \frac{1}{6}$ . Find $P (4)$ .

(iii) Find the angle of the sector numbered 4 on the second spinner.

Try it first, then show the solutionSolution

Sample space $S = {1, 2, 3, 4, 5, 6, 7, 8}$ , so total outcomes $= 8$ . (½ mark)
$E = {2, 3, 5, 7}$ , so favourable outcomes $= 4$ . (½ mark)
$P (E) = \frac{4}{8} = \frac{1}{2}$ . (½ mark)
The elementary events 1, 2, 3, 4 cover every outcome, so $P (1) + P (2) + P (3) + P (4) = 1$ . (½ mark)
$P (4) = 1 - (\frac{4}{12} + \frac{3}{12} + \frac{2}{12}) = 1 - \frac{9}{12} = \frac{1}{4}$ . (½ mark)
Angle of sector 4 $= \frac{1}{4} \times 36 0^{\circ} = 9 0^{\circ}$ . (½ mark)

Answer: (i) S = {1, 2, ..., 8}, E = {2, 3, 5, 7}, P(E) = 1/2 (ii) P(4) = 1/4 (iii) 90°

Example 2: Impossible and sure events · 2 marks · Very short answerBoard level

A box has 12 identical cards numbered 1 to 12. One card is drawn at random. Find the probability that the number on the card is (i) 13, (ii) less than 13, (iii) a prime number or 1.

Try it first, then show the solutionSolution

Total equally likely outcomes $= 12$ . (½ mark)
No card shows 13, so this is an impossible event: $P = \frac{0}{12} = 0$ . (½ mark)
Every card shows a number less than 13, so this is a sure event: $P = \frac{12}{12} = 1$ . (½ mark)
Prime or 1: 1, 2, 3, 5, 7, 11, that is 6 cards; $P = \frac{6}{12} = \frac{1}{2}$ . (½ mark)

Answer: (i) 0 (ii) 1 (iii) 1/2

Example 3: Letters of a word · 1 mark · MCQBoard level

A letter is chosen at random from the letters of the word ASSESSMENT (all 10 letter positions equally likely). The probability that the letter is S is

(a) $\frac{1}{10}$

(b) $\frac{3}{10}$

(d) $\frac{1}{2}$

Try it first, then show the solutionSolution

Total letters: A, S, S, E, S, S, M, E, N, T $= 10$ .
Number of S $= 4$ .
$P (S) = \frac{4}{10} = \frac{2}{5}$ .

Answer: (c) 2/5

Example 4: Two-digit tickets with or · 3 marks · Short answerHarder board level

Tickets numbered 10 to 69 are mixed in a box. One ticket is drawn at random. Find the probability that the number is (i) a multiple of 5 or a perfect square, (ii) a number whose digits add to 9.

Try it first, then show the solutionSolution

Total $= 69 - 10 + 1 = 60$ . (½ mark)
Multiples of 5: 10, 15, ..., 65, that is 12 numbers. (½ mark)
Perfect squares: 16, 25, 36, 49, 64, that is 5; 25 is also a multiple of 5, so new ones $= 4$ . (½ mark)
Favourable $= 12 + 4 = 16$ , $P = \frac{16}{60} = \frac{4}{15}$ . (½ mark)
Digit sum 9: 18, 27, 36, 45, 54, 63, that is 6 numbers. (½ mark)
$P = \frac{6}{60} = \frac{1}{10}$ . (½ mark)

Answer: (i) 4/15 (ii) 1/10

Example 5: Assertion-reason on equally likely outcomes · 1 mark · Assertion-reasonBoard level

A spinner has four sectors with angles $12 0^{\circ}$ (red), $6 0^{\circ}$ (green), $6 0^{\circ}$ (blue) and $12 0^{\circ}$ (red).

Assertion (A): The probability that the arrow stops on green is $\frac{1}{3}$ , because the spinner shows three colours.

Reason (R): The classical formula can be used only when all outcomes are equally likely.

(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.

(d) A is false but R is true.

Try it first, then show the solutionSolution

The three colours are not equally likely, as the sectors are unequal. All angles are multiples of $6 0^{\circ}$ , so split into equal $6 0^{\circ}$ parts: $36 0^{\circ} \div 6 0^{\circ} = 6$ parts.
Red $= 2 + 2 = 4$ parts, green $= 1$ part, blue $= 1$ part.
$P (green) = \frac{1}{6}$ , not $\frac{1}{3}$ , so A is false.
R is a true statement of the condition for the classical definition.

Answer: (d) A is false but R is true.

Example 6: Finding an unknown from a probability · 5 marks · Long answerHarder board level

A box has 24 identical pens: some blue, some black, and 6 red. The probability of drawing a blue pen at random is $\frac{1}{3}$ . (i) Find the number of black pens. (ii) If $x$ black pens are removed, the probability of a red pen becomes $\frac{2}{7}$ . Find $x$ . (iii) After this removal, find the probability of drawing a blue pen.

Try it first, then show the solutionSolution

Blue pens $= \frac{1}{3} \times 24 = 8$ . (1 mark)
Black pens $= 24 - 8 - 6 = 10$ . (1 mark)
After removal total $= 24 - x$ , so $\frac{6}{24 - x} = \frac{2}{7}$ . (1 mark)
$42 = 48 - 2 x \Rightarrow x = 3$ ; check $\frac{6}{21} = \frac{2}{7}$ . (1 mark)
$P (blue) = \frac{8}{21}$ . (1 mark)

Answer: (i) 10 black pens (ii) x = 3 (iii) 8/21

Example 7: Case: school quiz draw · 4 marks · Case-basedHOTS / competency

Suppose for a school quiz, 40 students write their roll numbers 1 to 40 on identical slips put in a box. The quiz master draws one slip at random to choose who answers first.

(i) Find the probability that the roll number is a prime number. (1 mark)

(ii) Find the probability that the roll number is a two-digit number. (1 mark)

(iii) Slips of roll numbers 1 to 10 are taken out because those students have already answered. From the rest, find the probability that the roll number drawn is a multiple of 3 or 4. (2 marks)

Try it first, then show the solutionSolution

Primes up to 40: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, that is 12; $P = \frac{12}{40} = \frac{3}{10}$ . (1 mark)
Two-digit numbers: 10 to 40, that is 31; $P = \frac{31}{40}$ . (1 mark)
Remaining slips: 11 to 40, total $30$ . (½ mark)
Multiples of 3: 12, 15, ..., 39, that is 10. Multiples of 4: 12, 16, ..., 40, that is 8. (½ mark)
Common multiples of 12: 12, 24, 36, that is 3; favourable $= 10 + 8 - 3 = 15$ . (½ mark)
$P = \frac{15}{30} = \frac{1}{2}$ . (½ mark)

Answer: (i) 3/10 (ii) 31/40 (iii) 1/2

Where students lose marks in Probability: Basic Terms and Classical Definition

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

Mistake: Counting only distinct letters in a word
Repeated letters are separate equally likely positions.
Wrong working
BALLOON has letters B, A, L, O, N, so P(L) = 1/5.
Right working
7 letter positions, L appears twice, so P(L) = 2/7.
Do this: Count letters by position, not by type.
Usually costs: Full 1 mark in an MCQ.
Mistake: Wrong total for a range of numbers
Students subtract without adding 1.
Wrong working
Tickets 20 to 80: total = 60.
Right working
Total = 80 − 20 + 1 = 61.
Do this: Always use b − a + 1 and write it as a step.
Usually costs: Every part that uses the total, often 2 to 3 marks.
Mistake: Double counting with 'or'
Numbers in both groups get counted twice.
Wrong working
1 to 20, multiple of 2 or 5: 10 + 4 = 14, P = 14/20.
Right working
10 and 20 are in both: 10 + 4 − 2 = 12, P = 12/20 = 3/5.
Do this: List the favourable numbers or subtract the common ones.
Usually costs: 1 to 1½ marks.
Mistake: Using the formula on outcomes that are not equally likely
Students count categories instead of equally likely outcomes.
Wrong working
Spinner with sectors 180°, 90°, 90°: P(big sector) = 1/3.
Right working
Split into four 90° parts: P = 2/4 = 1/2.
Do this: Ask whether outcomes are equally likely before dividing.
Usually costs: Assertion-reason mark and method marks.

Exam tips

Write the total number of outcomes as its own line; it usually carries a step mark.
List favourable outcomes in your answer, so the examiner can see the method even if a count slips.
Always reduce the fraction to lowest terms, for example 12/40 = 3/10, so that it matches an MCQ option.
In assertion-reason items, test A by actual counting, then judge separately whether R explains it.
For an unknown, substitute your value back into the given probability as a final check.

Practice questions on Probability: Basic Terms and Classical Definition

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
Which of the following cannot be the probability of an event?
(a) $0.07$
(b) $\frac{13}{12}$
(c) $45%$
(d) $\frac{2}{3}$
Hint
Compare each value with 1.
Answer
(b) 13/12, because it is greater than 1.
Q2 · 2 marks · Very short answerBoard level
A number is chosen at random from 1 to 60. Find the probability that it is a multiple of 4 but not a multiple of 6.
Hint
Count multiples of 4, then remove multiples of 12.
Answer
Multiples of 4: 15; multiples of 12: 5; favourable = 10; P = 10/60 = 1/6.
Q3 · 3 marks · Short answerHarder board level
A letter is chosen at random from the letters of the word MATHEMATICS. Find the probability that it is (i) a vowel, (ii) a letter that appears more than once in the word, (iii) a letter that comes before H in the English alphabet.
Hint
There are 11 letters; count repeated letters by position.
Answer
Total 11. (i) Vowels A, E, A, I = 4, P = 4/11. (ii) M, M, A, A, T, T = 6, P = 6/11. (iii) A, E, A, C = 4, P = 4/11.
Q4 · 3 marks · Short answerHarder board level
A bag holds 5 red balls and some green balls. The probability of drawing a green ball is three times the probability of drawing a red ball. Find (i) the number of green balls, (ii) the number of green balls that must be removed so that the probability of a red ball becomes 1/2.
Hint
Let green = g; the ratio of probabilities equals the ratio of counts.
Answer
(i) g/(5+g) = 3 × 5/(5+g) so g = 15. (ii) 5/(20 − y) = 1/2 gives y = 10 green balls removed.
Q5 · 2 marks · Short answerHOTS / competency
Meera says: 'A cricket match can be won, lost or drawn, so the probability our team wins is 1/3.' Arjun says: 'A card drawn from cards numbered 1 to 9 is either odd or even, so P(odd) = 1/2.' Is each student's reasoning correct? Explain, and give the correct P(odd) where possible.
Hint
Check whether the listed outcomes are equally likely.
Answer
Both are wrong. Match results are not equally likely, so 1/3 cannot be claimed. Odd and even are not equally likely from 1 to 9: there are 5 odd numbers out of 9, so P(odd) = 5/9.
Q6 · 5 marks · Long answerHarder board level
A bag contains 30 identical marbles that are red, blue or green. The probability of drawing a red marble is $\frac{1}{5}$ , and the probability of drawing a blue marble is twice the probability of drawing a green marble. (i) Find the number of marbles of each colour. (ii) 5 blue marbles are removed and $y$ green marbles are added, after which the probability of drawing a green marble is $\frac{1}{2}$ . Find $y$ . (iii) Find the new probability of drawing a red marble.
Hint
Find red first; then blue + green = 24 with blue = 2 × green. After the change, the total is 30 − 5 + y.
Answer
(i) Red = (1/5) × 30 = 6; blue + green = 24 and blue = 2 × green, so green = 8, blue = 16. (ii) New total = 25 + y, green = 8 + y; (8 + y)/(25 + y) = 1/2 gives 16 + 2y = 25 + y, so y = 9 (check: 17/34 = 1/2). (iii) P(red) = 6/34 = 3/17.
Q7 · 4 marks · Case-basedHOTS / competency
Suppose a housing society holds a lucky draw at its Diwali fair. Identical tokens numbered 1 to 60 are put in a box and one token is drawn at random.
(i) Find the probability that the digits of the number on the token add up to 6. (1 mark)
(ii) Find the probability that the number is a prime number. (1 mark)
(iii) A family wins a gift hamper if the number is a multiple of 6 or a multiple of 9. Find the probability of winning the hamper. (2 marks)
Hint
Single-digit numbers count too; for (iii) subtract the multiples of 18 counted twice.
Answer
Total = 60. (i) 6, 15, 24, 33, 42, 51, 60: 7 numbers, P = 7/60. (ii) Primes 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59: 17 numbers, P = 17/60. (iii) Multiples of 6: 10; multiples of 9: 9, 18, 27, 36, 45, 54, that is 6; common (multiples of 18): 18, 36, 54, that is 3; favourable = 10 + 6 − 3 = 13, P = 13/60.
Q8 · 3 marks · Short answerHarder board level
A bag contains $(x^{2} - 4)$ red balls and $(2 x + 4)$ green balls, where $x$ is a whole number. A ball is drawn at random and the probability that it is red is $\frac{2}{3}$ . Find (i) the value of $x$ , (ii) the number of balls of each colour, (iii) the probability that the ball drawn is green.
Hint
Total = x² + 2x. Cross-multiply to get a quadratic and reject any value that gives an impossible number of balls.
Answer
(i) Total = x² + 2x. 3(x² − 4) = 2(x² + 2x) gives x² − 4x − 12 = 0, so (x − 6)(x + 2) = 0; x = −2 makes the total 0, so x = 6. (ii) Red = 32, green = 16 (total 48; check 32/48 = 2/3). (iii) P(green) = 16/48 = 1/3.

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.What is a sample space?Show answer
The list of all possible outcomes of a random experiment.
2.What is an event?Show answer
A collection of outcomes of an experiment that you are interested in.
3.State the classical definition of probability.Show answer
P(E) = favourable outcomes ÷ total equally likely outcomes.
4.Range of any probability?Show answer
From 0 to 1, both included.
5.Probability of an impossible event?Show answer
0
6.How many numbers from 15 to 45 inclusive?Show answer
45 − 15 + 1 = 31

Probability: Basic Terms and Classical Definition: frequently asked questions

What does equally likely mean in probability?

It means no outcome has a better chance of happening than any other. A fair die or a well-shuffled deck gives equally likely outcomes. The classical formula can be used only in such cases.

Can probability be written as a percentage or decimal?

Yes. A probability of one fourth can be written as 0.25 or 25 percent. It must still lie from 0 to 1, or 0 to 100 percent.

Is experimental probability in the Class 10 syllabus?

The 2026-27 syllabus covers the classical definition of probability and simple problems on it. Focus your preparation on theoretical probability based on equally likely outcomes.

What is the difference between an outcome and an event?

An outcome is a single result, such as getting a 4 on a die. An event can contain one or many outcomes, such as getting an even number, which includes 2, 4 and 6.