Illustration of probability as a ratio of favourable outcomes to total outcomes:
Chapter 14 · CBSE Class X
📘 Definition
🔎 Key Fact
Key Terms We Must Know
🎨 SVG Diagram
Visual Understanding
🔢 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
\[ P(\text{King}) = \frac{4}{52} = \frac{1}{13} \]
📌 Note
📌 Note
Random Experiment
📌 Note
Sample Space
📌 Note
Event (E)
📌 Note
Equally Likely Outcomes
📖 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
📋 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
\[ P(\text{Not Green}) = 1 - P(\text{Green}) = 1 - \frac{5}{12} = \frac{7}{12} \]
❓ 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: \( \{HH, HT, TH\} \)
\[ P(\text{at least one head}) = \frac{3}{4} \]
❓ 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
(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
\[ P(\text{Red}) = \frac{2}{4} = \frac{1}{2} \]
❓ 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?
(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: \( \{1,5\} \)
\[ P = \frac{2}{6} = \frac{1}{3} \]
❓ 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
(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
\[ P = \frac{44}{52} = \frac{11}{13} \]
❓ 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:
\[ P(\text{not lose}) = 1 - 0.25 = 0.75 \]
❓ 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
(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
\[ P(G) = \frac{30}{45} = \frac{2}{3} \]
❓ 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
(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:
\[ \begin{aligned} P(\text{not blue}) &= 1 - P(B) \\&= 1 - \frac{1}{3} \\&= \frac{2}{3} \end{aligned} \]
❓ 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: \( \{(H,T), (T,H)\} \)
\[ P = \frac{2}{4} = \frac{1}{2} \]
❓ 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)
(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 = \( 8 + 4 = 12 \)
\[ P(\text{defective}) = \frac{12}{100} = 0.12 \]
❓ 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
(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
n(S) = 2
🪙🪙
TWO COINS
S = {HH,HT,TH,TT}
n(S) = 4
n(S) = 4
🎲
ONE DIE
S = {1,2,3,4,5,6}
n(S) = 6
n(S) = 6
🎲🎲
TWO DICE
All (a,b) pairs
n(S) = 36
n(S) = 36
🃏
DECK OF CARDS
52 cards total
n(S) = 52
n(S) = 52
🎴
PLAYING CARD SUITS
♠ ♥ ♦ ♣ · 13 each
Face: J, Q, K = 12 total
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
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📖 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
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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.
🃏 Concept Flashcards
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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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तमसो मा ज्योतिर्गमय · Est. 2025
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Probability | Mathematics Class 10 | Academia Aeternum
Probability | Mathematics Class 10 | Academia Aeternum — 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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