Probability Class 10 Notes PDF: Formulas, Examples and Questions

These Probability Class 10 Notes PDF explain every board-level concept using formulas, sample spaces, solved examples and practice questions. Probability is Chapter 14 in the current NCERT Class 10 Mathematics textbook, although some older books and websites still call it Chapter 15.

This revision guide covers probability formulas, complementary events, coins, dice, playing cards, marbles, number-selection questions, common mistakes, MCQs and quick revision plans.

Probability Class 10 Notes Overview

Class 10 Probability measures how likely an event is by comparing favourable outcomes with all equally likely possible outcomes.

Main Probability Formula

P(E) = Number of outcomes favourable to E / Total number of equally likely outcomes

Here, E represents the event whose probability must be found.

Complementary Probability Formula

The complement of event E means that E does not happen.

P(not E) = 1 - P(E)

Therefore:

P(E) + P(not E) = 1

Probability Class 10 Notes PDF Free Download

The downloadable Probability Class 10 notes PDF should provide the same core teaching content as this page in a compact, printable format.

Essential Probability Rules

The current NCERT exercise directly tests complementary probability, impossible and certain events, the range from 0 to 1 and the sum of elementary-event probabilities.

One-Minute Example

A fair die has the following outcomes:

S = {1, 2, 3, 4, 5, 6}

Let E be the event “an even number appears.”

Favourable outcomes:

E = {2, 4, 6}

Therefore:

P(E) = 3 / 6 = 1 / 2

Is Probability Chapter 14 or Chapter 15 in Class 10?

Probability is Chapter 14 in the current rationalised NCERT Class 10 Mathematics textbook. Chapter 15 is the numbering found in some older editions and legacy resources.

The official NCERT chapter PDF is titled Probability 14 and is marked as a 2026–27 reprint.

Use these labels:

• Current NCERT: Chapter 14 – Probability
• Older NCERT resources: Chapter 15 – Probability
• In an examination: Follow the chapter numbering used in your prescribed textbook

Both phrases may appear in online searches:

• Probability Class 10 Chapter 14 notes
• Probability Class 10 Chapter 15 notes

The mathematical concepts remain the same, but current educational content should identify it primarily as Chapter 14.

Probability Class 10 Syllabus for 2026–27

The 2026–27 CBSE Class 10 Probability syllabus covers the classical definition of probability and simple problems involving the probability of an event.

CBSE also states that students should apply probability to everyday likelihood and simple real-life situations.

Topics Included

• Classical or theoretical probability
• Equally likely outcomes
• Probability of a simple event
• Favourable and total outcomes
• Complementary events
• Sure and impossible events
• Coins, dice and cards
• Random selection from groups of objects
• Simple real-life probability problems

Topics Not Required as Core Class 10 Content

The following belong to later or more advanced study and should not be treated as required Class 10 Probability topics:

• Conditional probability
• Bayes’ theorem
• Probability distributions
• Permutations and combinations as a full topic
• Advanced independent-event formulas
• Advanced dependent-event formulas

Unit Weightage

Statistics and Probability form a combined CBSE unit. Do not claim that Probability alone always carries a fixed number of marks because individual paper patterns may vary.

Also Check: Latest CBSE Class 10 Maths syllabus

Basic Maths vs Standard Maths

Both Basic and Standard Maths students study the same core Probability concepts, but Standard Maths places more emphasis on applying and analysing ideas.

The 2026–27 Standard Maths paper design assigns approximately 24% to application and 22% to analysing, evaluating and creating across the complete paper.

Important Terms in Class 10 Probability

The key Probability terms describe an uncertain activity, its possible results and the group of outcomes being studied.

Equally Likely Outcomes Matter

The formula:

P(E) = Favourable outcomes / Total outcomes

works directly only when the outcomes being counted are equally likely.

For example, a car either starts or does not start, but those two outcomes are not automatically equally likely. The condition of the car affects their likelihood. NCERT includes questions asking students to decide whether listed outcomes are genuinely equally likely.

Experimental and Theoretical Probability

Theoretical probability is calculated from possible equally likely outcomes, while experimental probability is estimated from results observed in actual trials.

Experimental Probability Example

A coin is tossed 50 times and lands on heads 28 times.

Experimental probability of heads:

P(H) = 28 / 50 = 14 / 25

This does not mean the theoretical probability of a fair coin has changed.

The theoretical probability remains:

P(H) = 1 / 2

Examination Focus

The current CBSE syllabus explicitly lists the classical definition of probability and simple event-probability questions. Experimental probability is useful background, but revision should prioritize classical problems unless a teacher or question paper asks otherwise.

How to Solve Probability Questions Step by Step

A Probability question can usually be solved by identifying the experiment, listing the outcomes and applying the favourable-outcomes formula.

1. Identify the experiment.

Determine what is being tossed, drawn, selected or observed.

2. Write the sample space.

List all possible outcomes when the set is small.

3. Check equal likelihood.

Confirm that each counted outcome has the same chance.

4. Identify favourable outcomes.

Select the outcomes that satisfy the event.

5. Apply the formula.

P(E) = Number of favourable outcomes / Total number of equally likely outcomes

6. Simplify the answer.

Reduce the fraction where possible.

7. Check the range.

The final answer must lie between 0 and 1.

Model Solution

Question:

A bag contains 4 blue balls and 6 yellow balls. Find the probability of selecting a blue ball at random.

Total outcomes:

4 + 6 = 10

Favourable outcomes:

4

Probability:

P(blue) = 4 / 10 = 2 / 5

Five-Second Answer Check

• Are all counted outcomes equally likely?
• Is the denominator the total number of outcomes?
• Is the numerator the number of favourable outcomes?
• Is the fraction simplified?
• Is the answer between 0 and 1?

“At Least,” “At Most” and Other Probability Wording

Probability wording must be translated into exact mathematical cases before any calculation begins.

At Least One

When two coins are tossed:

S = {HH, HT, TH, TT}

“At least one head” includes:

{HH, HT, TH}

Therefore:

P(at least one head) = 3 / 4

NCERT uses this exact type of reasoning and treats HT and TH as separate ordered outcomes.

At Most One

For two coin tosses, “at most one head” includes zero heads or exactly one head:

{TT, HT, TH}

Therefore:

P(at most one head) = 3 / 4

Using a Complement for “At Least One”

It is often faster to calculate:

P(at least one success) = 1 - P(no successes)

For two coin tosses:

P(at least one head) = 1 - P(TT)

P(at least one head) = 1 - 1 / 4

P(at least one head) = 3 / 4

Coin Toss Probability Questions

Coin-toss questions require ordered sample spaces, which means HT and TH are treated as different outcomes.

One Coin

S = {H, T}

P(H) = 1 / 2

P(T) = 1 / 2

Two Coins

S = {HH, HT, TH, TT}

Three Coins

Three coin tosses have:

2 x 2 x 2 = 8 possible outcomes

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Example:

Find the probability of exactly two heads.

Favourable outcomes:

{HHT, HTH, THH}

Therefore:

P(exactly two heads) = 3 / 8

Common Coin Mistakes

• Writing only HH, HT and TT and omitting TH
• Treating “exactly one” as “at least one”
• Assuming two tosses have only two outcomes
• Forgetting that order matters
• Counting the number of heads instead of complete outcomes

Dice Probability Questions

A fair die has six equally likely outcomes, while two distinguishable dice produce 36 ordered outcomes.

One Die

S = {1, 2, 3, 4, 5, 6}

NCERT includes one-die questions involving prime numbers, odd numbers and numbers lying within a stated range.

Two Dice

For a blue die and a red die, each outcome is an ordered pair:

(Blue die result, Red die result)

There are:

6 x 6 = 36 equally likely ordered outcomes

Probability of a Sum of 7

Favourable outcomes:

(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

Therefore:

P(sum of 7) = 6 / 36 = 1 / 6

Probability of a Sum of 8

Favourable outcomes:

(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)

Therefore:

P(sum of 8) = 5 / 36

Why the Sums Are Not Equally Likely

The possible sums are 2 through 12, but they do not each occur in the same number of ways.

Therefore, using 1 / 11 as the probability of every sum is incorrect. NCERT specifically asks students to examine and reject that argument.

Common Dice Mistakes

• Using 11 as the denominator for two-dice sums
• Treating (2, 5) and (5, 2) as one outcome
• Forgetting that two dice produce 36 outcomes
• Including an endpoint excluded by “greater than” or “less than”
• Treating repeated labels as a single outcome when they appear on different faces

Playing Cards Probability Questions

A standard deck has 52 cards divided into four suits, and knowing the deck structure prevents most card-probability errors.

Standard Deck Facts

Face cards are jacks, queens and kings. An ace is not normally counted as a face card in Class 10 questions.

NCERT’s exercise includes kings, face cards, red face cards, individual named cards and suits.

Example 1: Red Card

P(red card) = 26 / 52 = 1 / 2

Example 2: Face Card

There are 12 face cards.

P(face card) = 12 / 52 = 3 / 13

Example 3: Red King

There are two red kings:

• King of hearts
• King of diamonds

P(red king) = 2 / 52 = 1 / 26

Example 4: Jack of Hearts

There is exactly one jack of hearts.

P(jack of hearts) = 1 / 52

Without Replacement

When a card is drawn and not replaced, the total number of cards decreases.

Suppose one queen is removed from a set containing the ten, jack, queen, king and ace of diamonds.

Four cards remain, including one ace.

P(ace on second draw) = 1 / 4

The removed queen cannot be drawn again.

P(queen on second draw) = 0

NCERT includes this form of without-replacement question in Exercise 14.1.

Common Card Mistakes

• Counting an ace as a face card
• Confusing 13 hearts with 26 red cards
• Saying there are four red kings instead of two
• Forgetting that each suit has 13 cards
• Keeping the denominator at 52 after a card is removed

Probability with Marbles, Balls and Objects in a Bag

Bag-selection probability is found by dividing the number of required objects by the total number of objects, provided every object is equally likely to be drawn.

Example

A box contains:

• 3 blue marbles
• 2 white marbles
• 4 red marbles

Total marbles:

3 + 2 + 4 = 9

Therefore:

P(blue) = 3 / 9 = 1 / 3

P(white) = 2 / 9

P(red) = 4 / 9

The three probabilities add to 1.

1 / 3 + 2 / 9 + 4 / 9 = 1

NCERT uses this exact structure to explain random selection and the sum of mutually exhaustive outcomes.

Probability of “Not Red”

Using a complement:

P(not red) = 1 - P(red)

If a bag contains 3 red and 5 black balls:

P(red) = 3 / 8

P(not red) = 1 - 3 / 8

P(not red) = 5 / 8

Does Object Size Matter?

The phrase “drawn at random” normally assumes that every object is equally likely to be selected.

If objects have noticeably different sizes, shapes or positions and the question does not establish equal likelihood, the simple counting formula may not be justified.

Number-Selection Probability Questions

Number-selection questions require accurate counting of primes, multiples, perfect squares and inclusive ranges.

Prime Numbers

A prime number has exactly two positive factors:

• 1
• The number itself

Important facts:

• 1 is not prime.
• 2 is the only even prime number.

Example:

A number is chosen from 1 to 10.

Prime numbers:

2, 3, 5, 7

Therefore:

P(prime) = 4 / 10 = 2 / 5

Perfect Squares

From 1 to 30, the perfect squares are:

1, 4, 9, 16, 25

Therefore:

P(perfect square from 1 to 30) = 5 / 30 = 1 / 6

Multiples

From 1 to 20, the multiples of 5 are:

5, 10, 15, 20

Therefore:

P(multiple of 5) = 4 / 20 = 1 / 5

Counting Inclusive Ranges

The number of integers from a to b, including both endpoints, is:

b - a + 1

From 11 to 30:

30 - 11 + 1 = 20

Common Number-Selection Mistakes

• Treating 1 as prime
• Forgetting that 1 is a perfect square
• Omitting one endpoint in an inclusive range
• Including the endpoint in “less than”
• Miscounting multiples
• Assuming zero is included when the range begins at 1

NCERT includes a numbered-disc problem involving two-digit numbers, perfect squares and divisibility by 5.

Spinner and Real-Life Probability Questions

Spinner and real-life problems use the same probability formula only when the represented outcomes are equally likely.

Equal Spinner Sections

A spinner is divided into eight equal sections numbered 1 to 8.

Probability of landing on an odd number:

Favourable outcomes:

1, 3, 5, 7

Therefore:

P(odd) = 4 / 8 = 1 / 2

NCERT includes an eight-section spinner question involving a specific number, odd numbers and numerical inequalities.

Unequal Spinner Sections

When sectors have different sizes, they are not equally likely merely because each section has a different label.

A larger sector has a greater chance of being selected.

Defective Product Example

A batch contains:

• 12 defective pens
• 132 good pens

Total pens:

12 + 132 = 144

Probability of choosing a good pen:

P(good) = 132 / 144 = 11 / 12

NCERT includes defective pens, bulbs and products as practical applications of random selection.

Original Case Study

A school checks 200 calculators before distributing them.

1. Probability of selecting a fully working calculator

P(working) = 176 / 200 = 22 / 25

2. Probability of selecting a calculator that is not working

P(not working) = 8 / 200 = 1 / 25

3. Probability that a calculator has no complete failure

This includes fully working calculators and calculators with a minor display issue.

176 + 16 = 192

P(no complete failure) = 192 / 200 = 24 / 25

Complementary Events in Probability

Complementary events cover every possible outcome between them, so their probabilities always add to 1.

If event E means “a die shows 6,” then not E means “the die does not show 6.”

P(E) = 1 / 6

P(not E) = 1 - 1 / 6

P(not E) = 5 / 6

When to Use the Complement

Use 1 - P(E) when the opposite event is easier to count.

Common examples:

• At least one head
• Not selecting a red object
• A product is not defective
• A player does not win
• A particular number does not appear
• At least one success in repeated trials

Winning and Losing

If a match cannot end in a draw and:

P(Player A wins) = 0.62

Then:

P(Player B wins) = 1 - 0.62

P(Player B wins) = 0.38

NCERT uses a tennis-match example to demonstrate this complementary relationship.

Common Complement Mistakes

• Subtracting from 100 instead of 1 when probabilities are written as decimals
• Choosing an event that is not the exact opposite
• Forgetting that the two events must cover all possibilities
• Using a complement when direct counting is simpler
• Writing a negative result because the original probability was invalid

Solved Examples of Probability for Class 10

Solved examples should show the sample space, favourable outcomes, formula and final check rather than presenting only an answer.

Example 1: Sure Event

Question:

A die is thrown. Find the probability of obtaining a number less than 7.

All outcomes 1, 2, 3, 4, 5 and 6 satisfy the event.

P(number less than 7) = 6 / 6 = 1

This is a sure event.

Example 2: Impossible Event

Question:

A card is drawn from a standard deck. Find the probability that it is both a heart and a club.

No card belongs to two suits.

P(heart and club) = 0 / 52 = 0

This is an impossible event.

Example 3: Exactly One Head

Two coins are tossed.

S = {HH, HT, TH, TT}

Favourable outcomes:

{HT, TH}

P(exactly one head) = 2 / 4 = 1 / 2

Example 4: At Least One 5 in Two Die Throws

It is faster to find the complement:

No 5 appears in either throw.

Probability of not obtaining 5 in one throw:

5 / 6

For two ordered throws:

P(no 5) = 5 / 6 x 5 / 6

P(no 5) = 25 / 36

Therefore:

P(at least one 5) = 1 - 25 / 36

P(at least one 5) = 11 / 36

NCERT treats throwing one die twice as equivalent to throwing two distinguishable dice for this type of problem.

Example 5: Card Without Replacement

Five cards are shuffled:

• 10 of diamonds
• Jack of diamonds
• Queen of diamonds
• King of diamonds
• Ace of diamonds

The queen is drawn and put aside.

Four cards remain.

P(ace next) = 1 / 4

No queen remains.

P(queen next) = 0

Example 6: Incorrect Reasoning

Claim:

Two dice have 11 possible sums, so the probability of each sum is 1 / 11.

Why it is wrong:

The sums are not equally likely.

A sum of 2 has one ordered outcome.

A sum of 7 has six ordered outcomes.

Correct probabilities:

P(sum of 2) = 1 / 36

P(sum of 7) = 6 / 36 = 1 / 6

Probability Class 10 Important Questions

The most useful Class 10 Probability practice set mixes direct calculation, interpretation, complementary events and reasoning.

Foundation Questions

1. A die is thrown once. Find the probability of obtaining a multiple of 2.
2. A card is drawn from a standard deck. Find the probability of obtaining a black card.
3. A bag contains 5 green and 7 yellow balls. Find the probability of drawing a green ball.
4. Two coins are tossed. Find the probability of exactly two tails.
5. A number is chosen from 1 to 20. Find the probability that it is prime.

Board-Ready Questions

6. Two dice are thrown. Find the probability that their sum is 9.
7. A card is drawn from a deck. Find the probability that it is neither a king nor a queen.
8. Three coins are tossed. Find the probability of exactly two heads.
9. A number is selected from 1 to 50. Find the probability that it is a multiple of 6.
10. A batch contains 15 defective and 135 good bulbs. Find the probability of selecting a bulb that is not defective.

Reasoning Questions

11. Explain why “getting a sum of 2” and “getting a sum of 7” are not equally likely when two dice are thrown.
12. A student says the probability of rain tomorrow is 1.3. Explain the error.
13. A spinner has four sections of unequal area. Can each labelled outcome be assigned probability 1 / 4? Justify.
14. Explain why HT and TH are separate outcomes when two coins are tossed.
15. A card is removed from a deck and not replaced. Explain why the denominator changes for the next draw.

Also Check: Probability Class 10 important questions with solutions

Probability Class 10 MCQs

These MCQs test formulas, terminology, counting and common Probability misconceptions.

1. The Probability of a Sure Event Is:

A. 0

B. 1 / 2

C. 1

D. Greater than 1

Answer: C

A sure event must occur, so its probability is 1.

2. Which Value Cannot Represent a Probability?

A. 0.4

B. 3 / 5

C. 110%

D. 1

Answer: C

A probability cannot exceed 1, and 110% equals 1.1.

3. A Fair Die Is Thrown Once. The Probability of Obtaining a Prime Number Is:

A. 1 / 6

B. 1 / 3

C. 1 / 2

D. 2 / 3

Answer: C

The prime outcomes are 2, 3 and 5.

P(prime) = 3 / 6 = 1 / 2

4. Two Coins Are Tossed. The Probability of At Least One Head Is:

A. 1 / 4

B. 1 / 2

C. 3 / 4

D. 1

Answer: C

The favourable outcomes are HH, HT and TH.

5. A Standard Deck Contains How Many Face Cards?

A. 4

B. 8

C. 12

D. 16

Answer: C

Each suit has a jack, queen and king.

3 x 4 = 12

6. If P(E) = 0.35, Then P(not E) Equals:

A. 0.35

B. 0.65

C. 1.35

D. -0.35

Answer: B

P(not E) = 1 - 0.35

P(not E) = 0.65

7. Two Dice Are Thrown. The Total Number of Ordered Outcomes Is:

A. 6

B. 11

C. 12

D. 36

Answer: D

Each result on the first die can be paired with six results on the second die.

8. Which Card Is Not a Face Card?

A. Jack

B. Queen

C. King

D. Ace

Answer: D

Standard Class 10 problems count jacks, queens and kings as face cards.

9. “At Most One Head” in Two Coin Tosses Includes:

A. HH only

B. HT and TH only

C. TT, HT and TH

D. Every outcome

Answer: C

At most one means zero or one.

10. The Probability of an Impossible Event Is:

A. 0

B. 1 / 2

C. 1

D. Undefined

Answer: A

Common Mistakes in Class 10 Probability

Most Probability errors come from incorrect counting or misreading the event rather than from the formula itself.

Mistake-Proof Checklist

Before submitting an answer, confirm:

1. The event has been translated correctly.
2. Every possible outcome has been counted.
3. Favourable outcomes satisfy the complete condition.
4. Outcomes are equally likely.
5. The denominator reflects any removal.
6. The answer is between 0 and 1.

How to Write Probability Answers for Full Marks

A complete Probability solution should identify the outcomes, show the formula and state a simplified final probability.

Recommended Answer Format

1. Write the total number of possible outcomes.
2. Write the number of favourable outcomes.
3. State the formula.
4. Substitute the values.
5. Simplify the fraction.
6. Write a final sentence.

Model Answer

Question:

A die is thrown once. Find the probability of obtaining an odd number.

Possible outcomes:

S = {1, 2, 3, 4, 5, 6}

Favourable outcomes:

{1, 3, 5}

Probability formula:

P(odd) = Number of favourable outcomes / Total number of outcomes

P(odd) = 3 / 6

P(odd) = 1 / 2

Therefore, the probability of obtaining an odd number is 1 / 2.

When to Write the Sample Space

Write the full sample space when:

• The number of outcomes is small
• Order matters
• The question asks for justification
• Two coins or two dice are involved
• Listing outcomes prevents a counting error

A full list may be unnecessary when selecting one object from a clearly counted group, but the total and favourable counts should still be shown.

Is NCERT Enough for Class 10 Probability?

NCERT should be completed first, but Standard Maths students generally benefit from additional reasoning questions, PYQs and sample-paper practice.

The NCERT exercise covers rules, equal likelihood, complements, bags, spinners, dice, cards, defective objects, number selection, two-dice sums and repeated trials.

Basic Maths Preparation

Complete:

1. Every NCERT example
2. Exercise 14.1
3. Direct formula questions
4. Basic complement questions
5. Selected MCQs
6. Recent official sample-paper questions

Standard Maths Preparation

Complete:

1. NCERT examples and exercise
2. NCERT Exemplar or equivalent reasoning problems
3. Questions involving misleading wording
4. Two-dice reasoning
5. Case-study questions
6. Previous-year and sample-paper questions
7. Timed mixed practice

Probability Class 10 Quick Revision Plan

A focused revision plan should combine formulas, sample spaces, common traps and a small set of mixed questions.

15-Minute Revision

1. Review the main formula.
2. Review P(not E) = 1 - P(E).
3. Memorize the standard card-deck facts.
4. List the two-coin sample space.
5. Review why two dice have 36 outcomes.
6. Read the at-least and at-most table.
7. Solve two quick questions.

45-Minute Revision

1. Spend 10 minutes on definitions and formulas.
2. Spend 10 minutes on coins and dice.
3. Spend 10 minutes on cards and object selection.
4. Spend 10 minutes solving five mixed questions.
5. Spend 5 minutes correcting mistakes.

Full Chapter Revision

1. Read the complete Class 10 Probability notes.
2. Solve all NCERT examples and Exercise 14.1.
3. Complete a 15-question mixed worksheet.
4. Attempt 10 MCQs.
5. Solve recent official-paper or sample-paper questions.
6. Review every incorrect answer by mistake type.
7. Retake the questions without notes.

Related Class 10 Maths Resources

These related resources support chapter revision, examination practice and cross-topic preparation.

• Class 10 Statistics Notes PDF
• Class 10 Maths Formula Sheet
• Class 10 Maths Chapter-Wise Notes
• Class 10 Maths MCQs
• CBSE Class 10 Maths Important Questions
• Class 10 Maths PYQs
• Class 10 Maths Sample Papers
• NCERT Solutions for Class 10 Maths