NCERT Class 10 Probability Notes – Concepts, Formulas & Quick Revision

Chapter 14 · CBSE Class X

🎲

Probability

Probability Random Experiments Events Outcomes CBSE Class X NCERT

📘 Definition

Probability is a branch of mathematics that quantifies uncertainty. It provides a systematic way to measure how likely an event is to occur in a well-defined experiment. The value of probability always lies in the interval:

\[ 0 \leq P(E) \leq 1 \]

where P(E) = 0 indicates an impossible event and P(E) = 1 indicates a certain event.

Classical (Theoretical) Definition

In Class X, probability is based on the assumption of equally likely outcomes. If an experiment has a finite number of outcomes, then:

\[ P(E) = \dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \]

This definition applies only when all outcomes have equal chance of occurring, such as tossing a fair coin or rolling a fair die.

🔎 Key Fact

Key Terms We Must Know

🎨 SVG Diagram

Visual Understanding

Illustration of probability as a ratio of favourable outcomes to total outcomes:

🔢 Formula

✏️ Example

A die is thrown once. Find the probability of getting an even number.

Identify sample space and favourable outcomes.

Write sample space
Identify even numbers
Apply formula

Sample Space: \( S = \{1,2,3,4,5,6\} \)

Favourable Outcomes: \( E = \{2,4,6\} \)

\[ P(E) = \frac{3}{6} = \frac{1}{2} \]

📐 Derivation

Derivation Insight

The formula \( P(E) = \frac{n(E)}{n(S)} \) arises from the principle of equal likelihood. If each outcome has equal chance, then probability becomes a ratio of counts rather than weights. This simplifies analysis in finite experiments.

🌟 Importance

⚡ Exam Tip

⚠️ Warning

Common Mistake

📋 Case Study

A card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is: (i) a red card, (ii) a king.

Concept: Understanding classification of outcomes.

Solution:

Total outcomes = 52

(i) Red cards = 26

\[ P(\text{Red}) = \frac{26}{52} = \frac{1}{2} \]

(ii) Kings = 4

\[ P(\text{King}) = \frac{4}{52} = \frac{1}{13} \]

🎲

Important Aspects of Probability

Probability Random Experiments Events Outcomes CBSE Class X NCERT

📌 Note

Random Experiment

A random experiment is a process whose outcome cannot be predicted with certainty, even when performed under identical conditions.

Key Property: Uncertainty with a well-defined set of outcomes.

Examples:

Tossing a coin → Outcomes: \(H, T\)
Rolling a die → Outcomes: \(1,2,3,4,5,6\)

📌 Note

Sample Space

The sample space is the complete set of all possible outcomes of a random experiment.

Formula Representation:

\[ S = \{\text{all possible outcomes}\} \]

Example:

For a die: \( S = \{1,2,3,4,5,6\} \)

Critical Note: Incorrect sample space leads to wrong probability.

📌 Note

Event (E)

An event is a subset of the sample space representing outcomes of interest.

Types:

Simple Event: Single outcome (e.g., getting 3)
Compound Event: Multiple outcomes (e.g., even numbers)

Example:

Even numbers on a die: \( E = \{2,4,6\} \)

📌 Note

Equally Likely Outcomes

Outcomes are equally likely when each outcome has the same probability of occurring.

Importance:

Essential condition for classical probability
Ensures fairness and uniform distribution

Warning: Do not assume equal likelihood without justification.

📖 Theory

Classical (Theoretical) Probability

🔢 Formula

\[ P(E) = \frac{n(E)}{n(S)} \]

Board Focus: Most CBSE questions are based on this model.

📌 Note

Range of Probability Values

📌 Note

Probability of an Impossible Event

📌 Note

Probability of a Sure Event

📌 Note

Complementary Events

🎨 SVG Diagram

📌 Note

Logical Counting & Systematic Listing

🌟 Importance

Probability connects theoretical mathematics with real-world decision-making.

Used in statistics, economics, and science
Helps in risk analysis and predictions
Foundation for higher classes (Class XI & XII)

Board Relevance: Frequently appears in case-study based questions.

📋 Case Study

A bag contains 3 red, 4 blue, and 5 green balls. One ball is drawn at random. Find the probability that the ball drawn is not green.

Concept: Use complement probability.

Solution:

Total balls = 12

Green balls = 5

\[ P(\text{Not Green}) = 1 - P(\text{Green}) = 1 - \frac{5}{12} = \frac{7}{12} \]

🎲

Example-1

Probability Random Experiments Events Outcomes CBSE Class X NCERT

❓ Question

Find the probability of getting a head when a coin is tossed once. Also, find the probability of getting a tail.

💡 Concept

🗺️ Roadmap

Identify the sample space
Count total outcomes
Identify favourable outcomes for each event
Apply probability formula

🎨 SVG Diagram

🧩 Solution

Sample Space
\[ S = \{H, T\} \]
Total Outcomes
\[ n(S) = 2 \]
Probability of Head
Favourable outcomes for Head:
\[ \{H\} \]
, so
\[ n(E) = 1 \]
\[P(\text{Head}) = \frac{1}{2}\]
Probability of Tail
Favourable outcomes for Tail:
\[ \{T\} \]
, so
\[ n(F) = 1 \]
\[P(\text{Tail}) = \frac{1}{2}\]

🔢 Formula

Formula Used

\[ P(E) = \frac{n(E)}{n(S)} \]

⚡ Exam Tip

⚠️ Warning

Common Mistakes

📋 Case Study

A coin is tossed twice. What is the probability of getting at least one head?

Solution:

Sample Space: \( S = \{HH, HT, TH, TT\} \)

Favourable outcomes: \( \{HH, HT, TH\} \)

\[ P(\text{at least one head}) = \frac{3}{4} \]

🎲

Example-2

Probability Random Experiments Events Outcomes CBSE Class X NCERT

❓ Question

A bag contains a red ball, a blue ball and a yellow ball, all of the same size. Kritika draws one ball without looking. Find the probability that she draws:
(i) Yellow ball
(ii) Red ball
(iii) Blue ball

💡 Concept

🗺️ Roadmap

Define sample space
Count total outcomes
Identify favourable outcome for each case
Apply probability formula

🎨 SVG Diagram

🧩 Solution

Sample Space
\[ S = \{\text{Red}, \text{Blue}, \text{Yellow}\} \]
Total Outcomes
\[ n(S) = 3 \]
Probability of Yellow Ball
Favourable outcome: \( \{\text{Yellow}\} \), so \( n(E) = 1 \)

\[ P(\text{Yellow}) = \frac{1}{3} \]
Probability of Red Ball
Favourable outcome: \( \{\text{Red}\} \), so \( n(E) = 1 \)

\[ P(\text{Red}) = \frac{1}{3} \]
Probability of Blue Ball
(iii) Probability of Blue Ball

Favourable outcome: \( \{\text{Blue}\} \), so \( n(E) = 1 \)

\[ P(\text{Blue}) = \frac{1}{3} \]

🔢 Formula

Formula Used

\[ P(E) = \frac{n(E)}{n(S)} \]

⚡ Exam Tip

⚠️ Warning

Common Mistakes

📋 Case Study

If one more red ball is added to the bag, what is the probability of drawing a red ball?

Solution:

Total balls = 4 (2 Red, 1 Blue, 1 Yellow)

Favourable outcomes (Red) = 2

\[ P(\text{Red}) = \frac{2}{4} = \frac{1}{2} \]

🎲

Example-3

Probability Random Experiments Events Outcomes CBSE Class X NCERT

❓ Question

Suppose a die is thrown once.
(i) What is the probability of getting a number greater than 4?
(ii) What is the probability of getting a number less than or equal to 4?

💡 Concept

🗺️ Roadmap

Write sample space of the die
Count total outcomes
Identify favourable outcomes for each case
Apply probability formula

🎨 SVG Diagram

🧩 Solution

Sample Space
\[ S = \{1,2,3,4,5,6\} \]
Total Outcomes:
\[ n(S) = 6 \]
(i) Probability of getting a number greater than 4

Favourable outcomes: \( \{5,6\} \), so \( n(E) = 2 \)
\[P(E) = \frac{2}{6} = \frac{1}{3}\]
(ii) Probability of getting a number ≤ 4

Favourable outcomes: \( \{1,2,3,4\} \), so \( n(F) = 4 \)
\[P(F) = \frac{4}{6} = \frac{2}{3}\]
Shortcut Using Complement
Since events are complementary:
\[\begin{aligned}P(\text{≤ 4}) &= 1 - P(\text{> 4}) \\&= 1 - \frac{1}{3} \\&= \frac{2}{3}\end{aligned}\]

🔢 Formula

Formula Used

\[ P(E) = \frac{n(E)}{n(S)} \]

⚡ Exam Tip

⚠️ Warning

Common Mistakes

📋 Case Study

Find the probability of getting a number that is neither divisible by 2 nor by 3.

Solution:

Numbers divisible by 2: \( \{2,4,6\} \)

Numbers divisible by 3: \( \{3,6\} \)

Numbers satisfying neither: \( \{1,5\} \)

\[ P = \frac{2}{6} = \frac{1}{3} \]

🎲

Example-4

Probability Random Experiments Events Outcomes CBSE Class X NCERT

❓ Question

One card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card:
(i) is an ace
(ii) is not an ace

💡 Concept

🗺️ Roadmap

Identify total number of cards
Count number of favourable outcomes
Apply probability formula
Use complement for second part

🎨 SVG Diagram

🧩 Solution

Step 1: Total Outcomes

Total number of cards =
\[ n(S) = 52 \]
Probability of getting an Ace

Number of aces = 4
\[\begin{aligned} P(E) &= \frac{4}{52} \\&= \frac{1}{13} \end{aligned}\]
Probability of not getting an Ace

Using complement:
\[\begin{aligned} P(\overline{E}) &= 1 - P(E) \\&= 1 - \frac{1}{13} \\&= \frac{12}{13} \end{aligned}\]

📐 Derivation

Derivation Insight

Since total outcomes = 52 and aces = 4, non-aces = 48. Direct computation also gives:

\[ P(\text{not ace}) = \frac{48}{52} = \frac{12}{13} \]

This verifies the complement rule.

🔢 Formula

Formula Used

\[ \begin{aligned} P(E) &= \frac{n(E)}{n(S)}, \\ P(\overline{E}) &= 1 - P(E) \end{aligned} \]

⚡ Exam Tip

⚠️ Warning

Common Mistakes

📋 Case Study

Find the probability that a card drawn is neither an ace nor a king.

Solution:

Total aces = 4, kings = 4 → total excluded = 8

Remaining cards = 44

\[ P = \frac{44}{52} = \frac{11}{13} \]

🎲

Example-5

Probability Random Experiments Events Outcomes CBSE Class X NCERT

❓ Question

Two players, Sangeeta and Reshma, play a tennis match. The probability of Sangeeta winning is 0.62. Find the probability that Reshma wins the match.

💡 Concept

🗺️ Roadmap

Define events clearly
Recognize complementary relationship
Apply complement formula

🎨 SVG Diagram

🧩 Solution

Let:

\( S \): Event that Sangeeta wins

\( R \): Event that Reshma wins
Given:

\( P(S) = 0.62 \)
Since both events are complementary:
\[P(S) + P(R) = 1\]
\[0.62 + P(R) = 1\]
\[\begin{aligned]P(R) &= 1 - 0.62 \\&= 0.38\end{aligned}\]

🔢 Formula

Formula Used

⚡ Exam Tip

⚠️ Warning

Common Mistakes

📋 Case Study

If the probability of a team losing a match is 0.25, find the probability that the team does not lose.

Solution:

\[ P(\text{not lose}) = 1 - 0.25 = 0.75 \]

🎲

Example-6

Probability Random Experiments Events Outcomes CBSE Class X NCERT

❓ Question

There are 40 students in a class, 25 girls and 15 boys. One student is selected at random. Find the probability that the selected student is:
(i) a girl
(ii) a boy

💡 Concept

🗺️ Roadmap

Identify total number of students
Determine favourable outcomes for each case
Apply probability formula
Verify using complement if needed

🎨 SVG Diagram

🧩 Solution

Total Outcomes:

\[ n(S) = 40 \]
(i) Probability of selecting a girl

Favourable outcomes = 25
\[ \begin{aligned} P(G) &= \frac{25}{40} \\&= \frac{5}{8} \end{aligned} \]
Probability of selecting a boy

Favourable outcomes = 15
\[ \begin{aligned} P(B) &= \frac{15}{40} \\&= \frac{3}{8} \end{aligned} \]
Verification Using Complement
Since selecting a girl and selecting a boy are complementary events:
\[\begin{aligned}P(G) + P(B) &= \frac{5}{8} + \frac{3}{8} \\&= 1\end{aligned}\]

🔢 Formula

Formula Used

\[ P(E) = \frac{n(E)}{n(S)} \]

⚡ Exam Tip

⚠️ Warning

Common Mistakes

📋 Case Study

If 5 more girls join the class, what is the probability of selecting a girl now?

Solution:

New total students = 45

Girls = 30

\[ P(G) = \frac{30}{45} = \frac{2}{3} \]

🎲

Example-7

Probability Random Experiments Events Outcomes CBSE Class X NCERT

❓ Question

A box contains 3 blue, 2 white, and 4 red marbles. One marble is drawn at random. Find the probability that the marble drawn is:
(i) white
(ii) blue
(iii) red

💡 Concept

🗺️ Roadmap

Calculate total number of marbles
Identify favourable outcomes for each color
Apply probability formula

🎨 SVG Diagram

🧩 Solution

Total Outcomes:

\[ n(S) = 3 + 2 + 4 = 9 \]
Probability of drawing a white marble

Favourable outcomes = 2
\[P(W) = \frac{2}{9}\]
Probability of drawing a blue marble

Favourable outcomes = 3
\[P(B) = \frac{3}{9} = \frac{1}{3}\]
Probability of drawing a red marble

Favourable outcomes = 4
\[P(R) = \frac{4}{9}\]
Verification Check
Since all outcomes are covered:
\[P(W) + P(B) + P(R) = \frac{2}{9} + \frac{3}{9} + \frac{4}{9} = 1\]

🔢 Formula

Formula Used

\[ P(E) = \frac{n(E)}{n(S)} \]

⚡ Exam Tip

⚠️ Warning

Common Mistakes

📋 Case Study

What is the probability that the marble drawn is not blue?

Solution:

\[ \begin{aligned} P(\text{not blue}) &= 1 - P(B) \\&= 1 - \frac{1}{3} \\&= \frac{2}{3} \end{aligned} \]

🎲

Example-8

Probability Random Experiments Events Outcomes CBSE Class X NCERT

❓ Question

Harpreet tosses two different coins simultaneously (₹1 and ₹2 coin). Find the probability that she gets at least one head.

💡 Concept

🗺️ Roadmap

List all possible outcomes
Count total outcomes
Identify favourable outcomes
Apply probability formula
Optional: verify using complement

🎨 SVG Diagram

🧩 Solution

Step 1: Sample Space

\[ S = \{(H,H), (H,T), (T,H), (T,T)\} \]
Total Outcomes:
\[ n(S) = 4 \]
Favourable Outcomes (at least one head)
\[ E = \{(H,H), (H,T), (T,H)\} \]
\[n(E) = 3 \]
\[P(E) = \frac{3}{4}\]
Shortcut Using Complement
Instead of counting directly, find probability of no head (both tails):
\[ \begin{aligned} P(\text{at least one head}) &= 1 - P(\text{no head}) \\&= 1 - \frac{1}{4} \\&= \frac{3}{4} \end{aligned} \]

🔢 Formula

Formula Used

\[ \begin{aligned}P(E) &= \frac{n(E)}{n(S)}, \\ P(\overline{E}) &= 1 - P(E)\end{aligned} \]

⚡ Exam Tip

⚠️ Warning

Common Mistakes

📋 Case Study

Find the probability that exactly one head appears.

Solution:

Favourable outcomes: \( \{(H,T), (T,H)\} \)

\[ P = \frac{2}{4} = \frac{1}{2} \]

🎲

Example-9

Probability Random Experiments Events Outcomes CBSE Class X NCERT

❓ Question

A carton has 100 shirts: 88 good, 8 with minor defects, and 4 with major defects. One shirt is drawn at random. Find the probability that:
(i) it is acceptable to Jimmy (only good shirts)
(ii) it is acceptable to Sujatha (rejects only major defects)

💡 Concept

🗺️ Roadmap

Identify total outcomes
Define acceptance criteria for each trader
Count favourable outcomes
Apply probability formula

🎨 SVG Diagram

🧩 Solution

Total Outcomes:

\[ n(S) = 100 \]
(i) Probability that shirt is acceptable to Jimmy

Jimmy accepts only good shirts
Favourable outcomes = 88
\[P(J) = \frac{88}{100} = 0.88\]
(ii) Probability that shirt is acceptable to Sujatha

Sujatha rejects only major defective shirts
\[P(S) = \frac{96}{100} = 0.96\]
Alternative (Complement Method)
Sujatha rejects only major defective shirts:
\[P(\text{reject}) = \frac{4}{100}\]
\[P(\text{accept}) = 1 - \frac{4}{100} = 0.96\]

🔢 Formula

Formula Used

\[ P(E) = \frac{n(E)}{n(S)} \]

⚡ Exam Tip

⚠️ Warning

Common Mistakes

📋 Case Study

Find the probability that a randomly selected shirt has any defect.

Solution:

Total defective = \( 8 + 4 = 12 \)

\[ P(\text{defective}) = \frac{12}{100} = 0.12 \]

🎲

Example-10

Probability Random Experiments Events Outcomes CBSE Class X NCERT

❓ Question

Two dice (one blue and one grey) are thrown simultaneously. Find the probability that the sum of numbers obtained is:
(i) 8
(ii) 13
(iii) less than or equal to 12

💡 Concept

🗺️ Roadmap

Construct sample space of ordered pairs
Count total outcomes
Identify favourable pairs based on sum
Apply probability formula

🎨 SVG Diagram

🧩 Solution

Sample Space

\[ \begin{aligned} S = \{(i,j) \mid i,j &= 1,2,3,4,5,6\}, \\\\ n(S) &= 36 \end{aligned} \]
Probability of sum = 8

Favourable outcomes:
\[(2,6), (3,5), (4,4), (5,3), (6,2)\]
\[P(E) = \frac{5}{36}\]
Probability of sum = 13

Maximum possible sum = 12, so no such outcome exists
P(F) = 0
Probability of sum ≤ 12

All possible outcomes satisfy this condition
\[P(G) = \frac{36}{36} = 1\]

📐 Derivation

Conceptual Insight

The maximum sum when two dice are thrown is \(6 + 6 = 12\), hence any sum ≤ 12 always occurs. Therefore, the event is a sure event with probability 1.

🔢 Formula

Formula Used

\[ P(E) = \frac{n(E)}{n(S)} \]

⚡ Exam Tip

🗒️ Warning

Common Mistakes

Ignoring ordered pairs (missing cases like (2,6) vs (6,2)).
Assuming sum 13 is possible.
Not recognizing sure events.

📋 Case Study

Find the probability that the sum is a prime number.

Solution:

Prime sums: 2, 3, 5, 7, 11

Total favourable outcomes = 15

\[ P = \frac{15}{36} = \frac{5}{12} \]

✦ NCERT Mathematics · Class X · Chapter 14 ✦

Probability Engine

A complete learning system — concepts, solver, practice & interactive modules

🔭 What is Probability?

Probability is a numerical measure of the likelihood of an event occurring. It bridges the gap between certainty and impossibility, giving us a precise language for uncertainty.

In Class X, we study Classical (Theoretical) Probability, where all outcomes are assumed to be equally likely — the most fundamental and clean form.

Core Definition

P(E) = Number of Favourable Outcomes / Total Number of Outcomes

Assumes all outcomes are equally likely

📐 Key Terms

EXPERIMENT

An action that produces a well-defined set of outcomes (e.g., tossing a coin, rolling a die).

SAMPLE SPACE (S)

The set of ALL possible outcomes. For a coin: S = {H, T}. For a die: S = {1,2,3,4,5,6}.

EVENT (E)

Any subset of the sample space. "Getting a head" is an event: E = {H}.

COMPLEMENTARY EVENT

All outcomes NOT in E. If E = {H}, then Ē = {T}. Always: P(E) + P(Ē) = 1.

🎯 Range & Certainty

Impossible Event P = 0

0

0.5

1

Sure Event P = 1

For any event E: 0 ≤ P(E) ≤ 1

Sum of probabilities of all elementary events = 1

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

🃏 Standard Sample Spaces to Memorise

🪙

ONE COIN

S = {H, T}
n(S) = 2

🪙🪙

TWO COINS

S = {HH,HT,TH,TT}
n(S) = 4

🎲

ONE DIE

S = {1,2,3,4,5,6}
n(S) = 6

🎲🎲

TWO DICE

All (a,b) pairs
n(S) = 36

🃏

DECK OF CARDS

52 cards total
n(S) = 52

🎴

PLAYING CARD SUITS

♠ ♥ ♦ ♣ · 13 each
Face: J, Q, K = 12 total

∑ Complete Formula Sheet

All formulae required for NCERT Class X, Chapter 14 — Probability.

Core Formulae

Formula 1 — Classical Probability

P(E) = n(E) / n(S)

n(E) = number of favourable outcomes · n(S) = total outcomes in sample space

Formula 2 — Complementary Events

P(not E) = 1 − P(E) or P(Ē) = 1 − P(E)

P(E) + P(Ē) = 1 for every event E

Formula 3 — Probability Bounds

0 ≤ P(E) ≤ 1

P(E) = 0 → Impossible Event · P(E) = 1 → Sure/Certain Event

Formula 4 — Sum of All Probabilities

P(E₁) + P(E₂) + · · · + P(Eₙ) = 1

When E₁, E₂, …, Eₙ are all elementary events of the experiment

Formula 5 — Empirical (Experimental) Probability

P(E) = Number of times E occurs / Total number of trials

As trials → ∞, empirical probability → theoretical probability

🃏 Playing Cards — Quick Reference

Suit Breakdown (52 total)

♠ Spades (Black) 13

♣ Clubs (Black) 13

♥ Hearts (Red) 13

♦ Diamonds (Red) 13

Category Counts

Face cards (J, Q, K) 12

Aces 4

Number cards (2–10) 36

Black cards 26

Red cards 26

🎲 Two Dice — Sum Reference Table

Total outcomes = 36. The table below shows the number of ways to get each sum:

Sum	Ways	P(sum)	Combinations

💡 Tips & Tricks

🔄

Use Complement First

When asked for "at least one" or "not…", compute P(complement) first. It's almost always easier: P(at least one H in 2 coins) = 1 − P(no H) = 1 − 1/4 = 3/4.

📋

List Systematically

For two dice, always write all 36 outcomes as ordered pairs (a, b). Never guess — listing prevents missing outcomes like (3,4) and (4,3) being distinct.

🃏

Memorise Card Numbers

52 total, 4 suits of 13. Face cards = 12 (3 per suit). Aces = 4. These come up in every card problem — know them cold to save time.

✂️

Simplify Fractions Always

P = 13/52 must be written as 1/4. Boards and exams deduct marks for unsimplified probability fractions. Always reduce to lowest terms.

🎯

Verify with Sum = 1

After finding all event probabilities, check they sum to 1. This instantly catches arithmetic errors without re-solving the problem.

🧩

Identify the Experiment

Always state the sample space before computing. "A card is drawn from a well-shuffled deck" → n(S) = 52. Never skip this step under exam pressure.

⚠️ Common Mistakes to Avoid

❌ Confusing (3,4) and (4,3)

In two-dice problems, (3,4) and (4,3) are DIFFERENT outcomes. Many students count this as one, halving their sample space — always use ordered pairs.

❌ Writing P > 1

Probability can never exceed 1. If your answer is greater than 1, you've made an error — either in the numerator or denominator. Recheck immediately.

❌ Including the Ace as a Face Card

The Ace is NOT a face card. Face cards are only Jack, Queen, King (12 total). Ace is a separate category entirely — a very common exam slip.

❌ Forgetting Red Jacks vs Black Jacks

When asked for P(red face card), remember 2 red Jacks + 2 red Queens + 2 red Kings = 6. Not all 12 face cards are red — only 6 are.

❌ Not Checking Exhaustive Events

When mutually exclusive events are given probabilities, verify they sum to 1 (if exhaustive). Many problems give extra info to trap students who don't check.

❌ Using Experimental as Theoretical

Experimental probability is based on actual trials, not calculation. Never substitute one for the other unless the question explicitly states large-trial convergence.

⚙ Step-by-Step Problem Solver

Select a problem type, fill in the values, and get a complete worked solution with every step explained.

🔧 Solver Configuration

📖 Concept-wise Practice Questions

Original concept-building questions with complete worked solutions — organised by topic. Click any question to reveal the full step-by-step solution.

🎯 Adaptive Quiz

Test your understanding with 20 carefully crafted MCQs. Each answer reveals an explanation.

Question

Score

Correct

Total

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🎲 Probability Simulator

Run virtual experiments to see how experimental probability converges to theoretical probability as the number of trials increases. This is the Law of Large Numbers in action.

EXPERIMENT TYPE

NUMBER OF TRIALS

🃏 Concept Flashcards

Flip each card to reveal the answer. Use these for last-minute revision — key definitions, formulae, and quick facts.

Card 1 of 18

QUESTION — tap to flip

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ANSWER

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🗺 Interactive Concept Map

Click any concept node to explore it in detail — definition, examples, and connections to other concepts.

🎲

Random Experiment

The foundation

🗂

Sample Space

All possible outcomes

🎯

Event

Subsets of outcomes

⚖️

Classical Probability

Equally likely outcomes

🔄

Complementary Events

P(E) + P(Ē) = 1

🔬

Experimental Probability

Based on actual trials

📚

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NCERT Class 10 Probability Notes – Concepts, Formulas & Quick Revision

NCERT Class 10 Probability Notes – Concepts, Formulas & Quick Revision — Complete Notes & Solutions · academia-aeternum.com

Probability marks a significant transition in mathematical learning by introducing students to the systematic study of uncertainty. Unlike earlier chapters that deal with fixed values and definite results, this chapter explores situations where outcomes are not predictable with certainty but can still be analysed logically. Through probability, students learn how mathematics models chance, fairness, and likelihood in a precise and reasoned manner. This chapter develops the ability to…

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Probability

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Why This Chapter Matters for Exams

Key Concept Highlights

Important Formula Capsules

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Classical (Theoretical) Definition

Key Terms We Must Know

Visual Understanding

Derivation Insight

Common Mistake

Random Experiment

Sample Space

Event (E)

Equally Likely Outcomes

Classical (Theoretical) Probability

Range of Probability Values

Probability of an Impossible Event

Probability of a Sure Event

Complementary Events

Logical Counting & Systematic Listing

Formula Used

Common Mistakes

Formula Used

Common Mistakes

Shortcut Using Complement

Formula Used

Common Mistakes

Derivation Insight

Formula Used

Common Mistakes

Formula Used

Common Mistakes

Verification Using Complement

Formula Used

Common Mistakes

Verification Check

Formula Used

Common Mistakes

Shortcut Using Complement

Formula Used

Common Mistakes

Alternative (Complement Method)

Formula Used

Common Mistakes

Sample Space

Conceptual Insight

Formula Used

Common Mistakes

Probability Engine

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