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MathematicsCh-14 15 min comprehensive revision
NCERT Class 10 Mathematics — Chapter 14

Probability

Classical theoretical probability, sample spaces, complementary events (P(E)+P(Eˉ)=1P(E) + P(\bar{E}) = 1), 52-card deck distribution, simultaneous 2-dice throws, and geometry-based probability.

Quick Key Takeaways:
Classical Definition: For equally likely outcomes: P(E)=Number of outcomes favorable to ETotal number of possible outcomes in sample space S=n(E)n(S)P(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of possible outcomes in sample space } S} = \frac{n(E)}{n(S)}
Probability Bounds: 0≤P(E)≤10 \le P(E) \le 1. Impossible event: P(E)=0P(E) = 0; Sure / Certain event: P(E)=1P(E) = 1.
Complementary Events: P(E)+P(not E)=1  ⟹  P(Eˉ)=1−P(E)P(E) + P(\text{not } E) = 1 \implies P(\bar{E}) = 1 - P(E).
Playing Card Deck (52 Cards): 26 Red (13 Hearts, 13 Diamonds) and 26 Black (13 Spades, 13 Clubs). 12 Face Cards (4 Kings, 4 Queens, 4 Jacks). 4 Aces are honour cards (not face cards).
Two Dice Sample Space: 6×6=366 \times 6 = 36 outcomes. Sum ranges from 2 to 12.
Jump to Concept:1. 1. Sample Space Architecture & Fundamental Probability Laws2. 2. Geometric Probability & Real-World Decision Modeling3. 3. Important Solved Board Examination Questions (3-Mark & 5-Mark)4. 4. CBSE Case-Study Modeling: Quality Control Inspection in an LED Bulb Factory5. 5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric
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1. Sample Space Architecture & Fundamental Probability Laws

Axioms & Foundational Theory

Comprehensive mathematical foundations, definitions, formulas, and visual conceptual model for Probability.

• Complete 52-Card Deck Breakdown
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Total Cards: 52
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Two Colours: 26 Red cards and 26 Black cards.
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Four Suits (13 cards each):
- Hearts (♥ - Red), Diamonds (♦ - Red), Spades (♠ - Black), Clubs (♣ - Black).
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Card Denominations in each suit: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.
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Face Cards: 3 per suit (Jack, Queen, King) ×4=12 Face Cards\times 4 = \mathbf{12\text{ Face Cards}} (6 Red, 6 Black).
• Two-Dice Simultaneous Throw (n(S)=36n(S) = 36)
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(1,1),(1,2),…,(1,6)(1,1), (1,2), \dots, (1,6)
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(2,1),(2,2),…,(2,6)(2,1), (2,2), \dots, (2,6)
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(3,1),(3,2),…,(3,6)(3,1), (3,2), \dots, (3,6)
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(4,1),(4,2),…,(4,6)(4,1), (4,2), \dots, (4,6)
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(5,1),(5,2),…,(5,6)(5,1), (5,2), \dots, (5,6)
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(6,1),(6,2),…,(6,6)(6,1), (6,2), \dots, (6,6)
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Doublets (same number on both): (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)  ⟹  6/36=1/6(1,1), (2,2), (3,3), (4,4), (5,5), (6,6) \implies 6/36 = 1/6.
📊 Probability: Sample Space & Outcome ScaleVisual Model
0 (Impossible)P(E) = 00.5 (Even Chance)Fair Coin Toss1 (Certain Event)P(E) = 1PLAYING CARDS (52 PACK): 26 RED (♥, ♦) & 26 BLACK (♠, ♣)12 Face Cards (4 Kings, 4 Queens, 4 Jacks) | 4 Aces | P(E) + P(Ē) = 1

Visual schematic mapping theoretical probability bounds from impossible ($P=0$) to sure event ($P=1$), 52-card deck hierarchy, and 2-dice distribution.

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2. Geometric Probability & Real-World Decision Modeling

Theorem Proofs & Derivations

Step-by-step rigorous geometric and algebraic theorem proofs and formula derivations for Probability.

• Geometric Probability Formula
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When target is a 2D geometric region:
P(Landing in Target)=Area of Specified Target RegionTotal Area of Bounding RegionP(\text{Landing in Target}) = \frac{\text{Area of Specified Target Region}}{\text{Total Area of Bounding Region}}
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3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

Topper Step Solutions

Standard CBSE board exam numericals and riders with complete step-by-step mathematical working and justifications.

• 3-Mark Standard Board Question: One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting: (i) a king of red colour, (ii) a face card, (iii) a red face card, (iv) a spade.
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Total Outcomes: n(S)=52n(S) = 52.
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(i) A king of red colour: 2 red kings (King of Hearts, King of Diamonds).
P=252=126P = \frac{2}{52} = \mathbf{\frac{1}{26}}
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(ii) A face card: 12 face cards (4 Kings, 4 Queens, 4 Jacks).
P=1252=313P = \frac{12}{52} = \mathbf{\frac{3}{13}}
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(iii) A red face card: 6 red face cards (2 Kings, 2 Queens, 2 Jacks).
P=652=326P = \frac{6}{52} = \mathbf{\frac{3}{26}}
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(iv) A spade: 13 spades.
P=1352=14P = \frac{13}{52} = \mathbf{\frac{1}{4}}
• 5-Mark Heavyweight Board Problem / Rider: Two dice are thrown at the same time. What is the probability that the sum of the two numbers appearing on the top of the dice is: (i) 8, (ii) 13, (iii) less than or equal to 12?
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Sample Space: n(S)=6×6=36n(S) = 6 \times 6 = 36.
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(i) Sum is 8: Favourable outcomes are (2,6),(3,5),(4,4),(5,3),(6,2)  ⟹  n(E)=5(2,6), (3,5), (4,4), (5,3), (6,2) \implies n(E) = 5.
P(Sum 8)=536P(\text{Sum } 8) = \mathbf{\frac{5}{36}}
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(ii) Sum is 13: Maximum possible sum is 6+6=126 + 6 = 12. No outcomes give 13 (impossible event)   ⟹  n(E)=0\implies n(E) = 0.
P(Sum 13)=0P(\text{Sum } 13) = \mathbf{0}
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(iii) Sum is less than or equal to 12: All 36 outcomes have sum ≤12\le 12 (sure event)   ⟹  n(E)=36\implies n(E) = 36.
P(Sum ≤12)=3636=1P(\text{Sum } \le 12) = \frac{36}{36} = \mathbf{1}
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4. CBSE Case-Study Modeling: Quality Control Inspection in an LED Bulb Factory

Case Study Mastery (4 Marks)

Real-world mathematical modeling scenario with structured multi-part questions and full step solutions.

• Practical Application Context: Quality Control Inspection in an LED Bulb Factory
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her.
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Q1: What is the total number of good pens in the lot? →\rightarrow Good pens =144−20=124 pens= 144 - 20 = \mathbf{124\text{ pens}}.
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Q2: What is the probability that she will buy it? →\rightarrow She buys if the pen is good. P(Buys)=124144=3136P(\text{Buys}) = \frac{124}{144} = \mathbf{\frac{31}{36}}.
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Q3: What is the probability that she will not buy it? →\rightarrow P(Not buys)=1−P(Buys)=1−3136=536P(\text{Not buys}) = 1 - P(\text{Buys}) = 1 - \frac{31}{36} = \mathbf{\frac{5}{36}} (or 20144=536\frac{20}{144} = \frac{5}{36}).
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5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Important Solved Board Questions

Examiner step-marking allocations, common algebraic/arithmetic deduction traps, and presentation guidelines.

• Step-by-Step Marking Rubric & Presentation Guidelines
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1 Mark: Stating total sample space n(S)n(S) clearly.
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1 Mark: Listing favourable outcomes n(E)n(E) explicitly.
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1 Mark: Quoting formula P(E)=n(E)/n(S)P(E) = n(E)/n(S).
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1 Mark: Reduced fraction in simplest form with verification 0≤P≤10 \le P \le 1.
• Common Calculation Traps & Verification Checklist
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Trap 1: Counting Aces as face cards (Aces have letters, but no facial portrait; face cards are only Jack, Queen, King).
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Trap 2: In 2-dice problems, treating (2,6)(2, 6) and (6,2)(6, 2) as a single outcome instead of two distinct outcomes.
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Trap 3: Writing probability greater than 1 or negative.
Authentic Board Question (3 Marks)Topic: Probability Mathematical Proof & Computations
State the governing mathematical theorem/formula, show complete geometric or computational steps, and find the exact result for Probability.

Official CBSE Step-by-Step Marking Breakdown:

Step 1: Given / To Prove / Formula Setup: State the governing theorem (BPT, Tangents, Pythagoras), mensuration formula, or trigonometric ratios with given values.
1 Mark
Step 2: Step-by-Step Derivation / Arithmetic Working: Show complete deductive steps or numerical calculations with π=22/7\pi = 22/7 or trigonometric standard values.
1 Mark
Step 3: Final Boxed Answer with Units / Q.E.D.: State the final numerical result with proper square/cubic units (cm2,m3\text{cm}^2, \text{m}^3) or formal proof conclusion.
1 Mark
Model Student Answer (Target: Full 3/3 Marks):
To score full 3 marks on Probability in CBSE Mathematics:

1. Theorem / Formula: Write the standard governing formula or state the geometric theorem clearly.
2. Calculation / Proof: Substitute dimensions systematically or provide deductive step justifications.
3. Final Result: Box the final answer with required units (cm,cm2,m3\text{cm}, \text{cm}^2, \text{m}^3) or conclude with "Hence Proved".
Examiner Mark Deduction Traps:
•Always write intermediate calculation lines to earn partial credit under CBSE step-marking.
•State reasons in parentheses (e.g., "[Tangents from an external point are equal]") during geometric proofs.

High-Frequency Conceptual Doubts & FAQs

Curated answers to the most common questions asked by Class 10 students.
P(E)=Number of outcomes favorable to ETotal number of possible outcomes=n(E)n(S)P(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of possible outcomes}} = \frac{n(E)}{n(S)}. For all events, 0≤P(E)≤10 \le P(E) \le 1. An impossible event has P(E)=0P(E) = 0, and a sure/certain event has P(E)=1P(E) = 1.

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