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Chapter 25: Probability
ICSE Class 10 Mathematics • Complete Chapter Notes • Theory, Diagrams & Step-by-Step Practice Masterclass
Official ICSE Syllabus & Scope of Examination
Prescribed Syllabus: Classical definition of probability; simple problems on single events; sample space of coin toss ($1, 2, 3$ coins), rolling a die / two dice, playing cards deck ($52$ cards), numbered cards / discs, and colored balls in a bag.
Dice: Throwing $1$ die ($n=6$), $2$ dice ($n=36$ grid), sum and product analysis.
Playing Cards: $52$-card architecture (Suits, Colors, Face cards, Honours, Number cards).
Numbers & Divisibility: Primes, composites, perfect squares, multiples of $a$ or $b$.
Bags & Unknowns: Algebra-based probability equations ($x$ unknown balls).
PART (A): Fundamental Terminology & Theoretical Foundations
1. Basic Terms & Concepts
Term
Formal Definition
Standard Example
Experiment
An operation or process which produces well-defined results/outcomes.
Tossing a coin, throwing a die, drawing a card.
Random Experiment
An experiment where all possible outcomes are known in advance, but the exact outcome of a particular trial cannot be predicted beforehand.
Rolling a balanced 6-sided die: we know outcomes are $\{1, 2, 3, 4, 5, 6\}$, but cannot guarantee which one appears.
Sample Space ($S$)
The set of all possible outcomes of a random experiment. The total number of outcomes is denoted as $n(S)$.
Tossing two coins: $S = \{HH, HT, TH, TT\}$, where $n(S) = 4$.
Event ($E$)
Any subset of the sample space $S$. In simple words, an event is the collection of one or more favorable outcomes.
Event of getting an odd number on a die: $E = \{1, 3, 5\}$.
Equally Likely Outcomes
Outcomes are said to be equally likely when none of them is preferred or has a greater chance of occurring over any other.
In an unbiased coin, getting a Head ($H$) or a Tail ($T$) are equally likely.
Elementary Event
An event having only one favorable outcome from the sample space.
Getting a $6$ in a single throw of a die: $E = \{6\}$.
Compound Event
An event containing more than one favorable outcome.
Getting a prime number on a die: $E = \{2, 3, 5\}$ (3 outcomes).
2. Classical (Theoretical) Probability Formula
CORE PROBABILITY FORMULA
If a random experiment has $n(S)$ total exhaustive, mutually exclusive, and equally likely outcomes, out of which $n(E)$ outcomes are favorable to an event $E$, then:
$$ P(E) = \frac{n(E)}{n(S)} = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of all possible outcomes}} $$
Sample Space $S$ containing Event $E$ and its Complementary Event $\bar{E}$ ($E'$).
3. Fundamental Laws and Axioms of Probability
Inviolable Probability Axioms
Range of Probability: The probability of any event $E$ is always a real number between $0$ and $1$ inclusive:
$$ 0 \le P(E) \le 1 $$
A probability can NEVER be negative ($P < 0$) and can NEVER be greater than $1$ ($P > 1$ or $> 100\%$).
Sure (or Certain) Event: An event that is 100% guaranteed to happen:
$$ P(\text{Sure Event}) = 1 $$
Example: Getting a number less than $7$ on a standard die: $P = 6/6 = 1$.
Impossible Event: An event that can never occur under any circumstances:
$$ P(\text{Impossible Event}) = 0 $$
Example: Getting a number $8$ on a standard 6-sided die: $P = 0/6 = 0$.
Sum of Elementary Events: The sum of probabilities of all individual elementary events of an experiment is always equal to $1$:
$$ P(E_1) + P(E_2) + P(E_3) + \dots + P(E_n) = 1 $$
Whenever a question asks for the probability of "at least one" occurrence, computing all direct cases can be lengthy. Use the complementary shortcut:
$$ P(\text{At least one}) = 1 - P(\text{None}) $$
4. Classroom Practice Problems (Fundamental Concepts & Axioms)
✍ PRACTICE PROBLEM A1: RANGE & VALIDITY OF PROBABILITY VALUES
Q. Which of the following numbers cannot represent the probability of an event? Give reasons for each:
(a) $\frac{3}{5}$, (b) $-1.5$, (c) $15\%$, (d) $2.4$, (e) $0$, (f) $\frac{7}{4}$.
Core Rule: For any valid probability $E$, $0 \le P(E) \le 1$.
(d) $2.4 > 1 \implies$ Cannot be a probability (probability cannot exceed 1).
(e) $0 \implies$ Valid (represents an Impossible Event).
(f) $\frac{7}{4} = 1.75 > 1 \implies$ Cannot be a probability (value $> 1$).
✍ PRACTICE PROBLEM A2: COMPLEMENTARY EVENTS RULE
Q. In a game of badminton between Aarav and Kabir, the probability of Aarav winning the match is $0.62$.
(i) Find the probability of Aarav losing the match.
(ii) Find the probability of Kabir winning the match (assuming no ties).
(i) Probability of Aarav losing:
$$ P(\text{Aarav not winning}) = 1 - P(\text{Aarav winning}) = 1 - 0.62 = \mathbf{0.38} $$
(ii) Probability of Kabir winning:
Kabir wins when Aarav loses:
$$ P(\text{Kabir winning}) = P(\text{Aarav losing}) = \mathbf{0.38} $$
A fair coin has 2 outcomes: Head ($H$) and Tail ($T$). For $n$ coins tossed simultaneously (or 1 coin tossed $n$ times), total outcomes $= \mathbf{2^n}$.
"Exactly 2 Heads" in 3 Coins: Must have precisely 2 heads $\to \{HHT, HTH, THH\} \implies 3$ outcomes.
"At least 2 Heads" in 3 Coins: Means 2 or more heads ($2$ or $3$ heads) $\to \{HHT, HTH, THH, HHH\} \implies 4$ outcomes.
"At most 2 Heads" in 3 Coins: Means 2 or fewer heads ($0, 1,$ or $2$ heads; i.e. everything except $HHH$) $\implies 7$ outcomes.
2. Classroom Practice Problems (Coin Toss Models)
✍ PRACTICE PROBLEM B1: TWO COINS SIMULTANEOUS TOSS
Q. Two unbiased coins are tossed together. Find the probability of getting:
(i) Exactly one head, (ii) At least one head, (iii) At most one tail, (iv) No head.
(ii) At least one head ($1$ or $2$ heads): Favorable outcomes $= \{HT, TH, HH\} \implies n(E_2) = 3$.
$$ P(E_2) = \mathbf{\frac{3}{4}} $$
(iii) At most one tail ($0$ or $1$ tail): Favorable outcomes $= \{HH, HT, TH\} \implies n(E_3) = 3$.
$$ P(E_3) = \mathbf{\frac{3}{4}} $$
(iv) No head ($0$ heads $\to TT$): Favorable outcomes $= \{TT\} \implies n(E_4) = 1$.
$$ P(E_4) = \mathbf{\frac{1}{4}} $$
✍ PRACTICE PROBLEM B2: THREE COINS SIMULTANEOUS TOSS
Q. Three coins are tossed once. Find the probability of obtaining:
(i) All heads, (ii) Exactly two heads, (iii) At least two heads, (iv) At most two tails, (v) All tails.
PART (C): Dice Experiments (Single Die & Two Dice $6\times 6$ Grid)
1. Single Die Throw ($n=6$)
Sample space for a single throw of a six-faced die: $S = \{1, 2, 3, 4, 5, 6\} \implies n(S) = 6$.
Classification of Single Die Numbers
Prime Numbers on a Die: $\{2, 3, 5\} \implies 3$ outcomes $\implies P(\text{Prime}) = \frac{3}{6} = \frac{1}{2}$. (Note: 1 is NEITHER prime nor composite).
Composite Numbers on a Die: $\{4, 6\} \implies 2$ outcomes $\implies P(\text{Composite}) = \frac{2}{6} = \frac{1}{3}$.
When two dice are thrown simultaneously (or one die is thrown twice), the sample space consists of $6^2 = 36$ ordered pairs $(x, y)$:
Complete $6 \times 6 = 36$ outcomes grid when throwing two dice simultaneously.
3. Two Dice Sum Distribution Table
Sum ($S$)
Favourable Pairs
Count ($n(E)$)
Probability $P(S)$
$\mathbf{2}$
$(1,1)$
$1$
$1/36$
$\mathbf{3}$
$(1,2), (2,1)$
$2$
$2/36 = 1/18$
$\mathbf{4}$
$(1,3), (2,2), (3,1)$
$3$
$3/36 = 1/12$
$\mathbf{5}$
$(1,4), (2,3), (3,2), (4,1)$
$4$
$4/36 = 1/9$
$\mathbf{6}$
$(1,5), (2,4), (3,3), (4,2), (5,1)$
$5$
$5/36$
$\mathbf{7}$ (Max freq)
$(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$
$\mathbf{6}$
$6/36 = \mathbf{1/6}$
$\mathbf{8}$
$(2,6), (3,5), (4,4), (5,3), (6,2)$
$5$
$5/36$
$\mathbf{9}$
$(3,6), (4,5), (5,4), (6,3)$
$4$
$4/36 = 1/9$
$\mathbf{10}$
$(4,6), (5,5), (6,4)$
$3$
$3/36 = 1/12$
$\mathbf{11}$
$(5,6), (6,5)$
$2$
$2/36 = 1/18$
$\mathbf{12}$
$(6,6)$
$1$
$1/36$
4. Classroom Practice Problems (Dice Models)
✍ PRACTICE PROBLEM C1: TWO DICE SUMS & DOUBLETS
Q. Two dice are rolled simultaneously. Find the probability of getting:
(i) A doublet (same number on both dice),
(ii) A sum of at least $10$,
(iii) A doublet of prime numbers,
(iv) The product of numbers on top faces is $12$.
Q. Two dice are thrown simultaneously. What is the probability that:
(i) $5$ will not come up on either die?
(ii) $5$ will come up on at least one die?
(i) Favorable pairs where $5$ comes up on at least one die:
$$\{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)\} \implies 11 \text{ pairs}$$
(i) $5$ will NOT come up on either die:
$$ n(\text{No } 5) = 36 - 11 = 25 \implies P(\text{No } 5) = \mathbf{\frac{25}{36}} $$
(ii) $5$ will come up on at least one die:
$$ P(\text{At least one } 5) = 1 - P(\text{No } 5) = 1 - \frac{25}{36} = \mathbf{\frac{11}{36}} $$
PART (D): 52 Playing Cards Hierarchy & Suit Architecture
1. Complete 52-Card Deck Breakdown
A standard deck consists of 52 cards divided into 4 suits of 13 cards each:
Complete 52-card hierarchy breakdown by color, suit, face cards, and ranks.
Category
Specific Cards Included
Count in 52 Deck
Standard Probability $P(E)$
Face Cards (Picture Cards)
Kings ($4$), Queens ($4$), Jacks ($4$)
$\mathbf{12}$
$12/52 = \mathbf{3/13}$
Red Face Cards
$2$ Kings, $2$ Queens, $2$ Jacks of Hearts & Diamonds
$\mathbf{6}$
$6/52 = \mathbf{3/26}$
Black Face Cards
$2$ Kings, $2$ Queens, $2$ Jacks of Spades & Clubs
$\mathbf{6}$
$6/52 = \mathbf{3/26}$
Aces (Honour Cards)
$1$ Ace per suit (A♠, A♣, A♥, A♦) — Not face cards!
$\mathbf{4}$
$4/52 = \mathbf{1/13}$
Number (Digit) Cards
Cards from $2$ to $10$ in all 4 suits ($9 \times 4$)
$\mathbf{36}$
$36/52 = \mathbf{9/13}$
Any Specific Rank
e.g. All Kings, All Queens, All 7s
$\mathbf{4}$
$4/52 = \mathbf{1/13}$
Any Specific Suit
e.g. All Spades or All Hearts
$\mathbf{13}$
$13/52 = \mathbf{1/4}$
2. Classroom Practice Problems (Playing Card Models)
✍ PRACTICE PROBLEM D1: SINGLE CARD DRAW FROM FULL DECK
Q. One card is drawn at random from a well-shuffled deck of 52 cards. Find the probability that the card drawn is:
(i) A king of red color,
(ii) A face card,
(iii) A red face card,
(iv) The jack of hearts,
(v) A spade,
(vi) The queen of diamonds,
(vii) Neither a spade nor a king.
Total Outcomes: $n(S) = 52$.
(i) King of red color: $2$ cards (King of Hearts, King of Diamonds) $\implies P = \frac{2}{52} = \mathbf{\frac{1}{26}}$.
(ii) A face card: $12$ cards $\implies P = \frac{12}{52} = \mathbf{\frac{3}{13}}$.
(iii) A red face card: $6$ cards $\implies P = \frac{6}{52} = \mathbf{\frac{3}{26}}$.
(iv) Jack of hearts: Only $1$ card $\implies P = \mathbf{\frac{1}{52}}$.
(v) A spade: $13$ cards $\implies P = \frac{13}{52} = \mathbf{\frac{1}{4}}$.
(vi) Queen of diamonds: Only $1$ card $\implies P = \mathbf{\frac{1}{52}}$.
(vii) Neither a spade nor a king:
Spades $= 13$, other Kings $= 3 \implies 13 + 3 = 16$ cards.
$$ \text{Favorable cards} = 52 - 16 = 36 \implies P = \frac{36}{52} = \mathbf{\frac{9}{13}} $$
✍ PRACTICE PROBLEM D2: CARDS REMOVED WITHOUT REPLACEMENT
Q. All the four face cards of spades (Jack, Queen, King) and the Ace of spades are removed from a pack of 52 cards. The remaining cards are well-shuffled, and then one card is drawn at random. Find the probability that the drawn card is:
(i) A black card, (ii) A face card, (iii) A spade, (iv) A red card.
(i) Black card remaining: Original $26 - 4 \text{ spades removed} = 22 \text{ black cards}$.
$$ P(\text{Black}) = \frac{22}{48} = \mathbf{\frac{11}{24}} $$
(ii) Face card remaining: Original $12 - 3 \text{ spade face cards} = 9 \text{ face cards}$.
$$ P(\text{Face Card}) = \frac{9}{48} = \mathbf{\frac{3}{16}} $$
(iii) A spade remaining: Original $13 - 4 = 9 \text{ spades}$.
$$ P(\text{Spade}) = \frac{9}{48} = \mathbf{\frac{3}{16}} $$
(iv) A red card remaining: All $26$ red cards are intact.
$$ P(\text{Red Card}) = \frac{26}{48} = \mathbf{\frac{13}{24}} $$
PART (E): Numbered Cards, Set Theory & Number Properties
1. Important Mathematical Number Categories
Classification Rules for Numbered Slips / Cards ($1$ to $N$)
Prime Numbers (from $1$ to $100$): Numbers greater than 1 having exactly two factors (1 and itself).
Primes up to 30: $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\} \implies 10$ primes.
Composite Numbers: Numbers having more than two factors ($4, 6, 8, 9, 10, 12, 14, 15, \dots$).
Crucial note: 1 is NEITHER prime nor composite.
Divisible by '$a$ OR $b$': Apply Principle of Inclusion-Exclusion:
$$ n(A \text{ or } B) = n(A) + n(B) - n(A \text{ and } B) $$
where $n(A \text{ and } B)$ is count of numbers divisible by $\text{LCM}(a, b)$.
2. Classroom Practice Problems (Numbered Cards & Sets)
✍ PRACTICE PROBLEM E1: CARDS NUMBERED 1 TO 100 (INCLUSION-EXCLUSION)
Q. One hundred cards are numbered from $1$ to $100$ and placed in a box. A card is drawn at random. Find the probability that the number on the card is:
(i) A perfect square,
(ii) A prime number less than $30$,
(iii) Divisible by $4$ or $6$,
(iv) Divisible by both $4$ and $6$,
(v) Neither divisible by $4$ nor by $6$.
(ii) Prime number less than $30$: $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\} \implies 10$ numbers.
$$ P(\text{Prime } < 30) = \frac{10}{100} = \mathbf{\frac{1}{10}} $$
(iii) Divisible by $4$ or $6$:
Divisible by $4$: $\lfloor 100/4 \rfloor = 25$ numbers.
Divisible by $6$: $\lfloor 100/6 \rfloor = 16$ numbers.
Divisible by both ($\text{LCM}(4,6) = 12$): $\lfloor 100/12 \rfloor = 8$ numbers.
$\therefore n(4 \text{ or } 6) = 25 + 16 - 8 = 33 \implies P = \mathbf{\frac{33}{100}}$.
(iv) Divisible by both $4$ and $6$ (multiples of $12$): $8$ numbers $\implies P = \frac{8}{100} = \mathbf{\frac{2}{25}}$.
(v) Neither divisible by $4$ nor by $6$:
$$ P(\text{Neither } 4 \text{ nor } 6) = 1 - P(4 \text{ or } 6) = 1 - \frac{33}{100} = \mathbf{\frac{67}{100}} $$
PART (F): Bags, Unknowns ($x$ Variables), Quality & Geometric Models
1. Problems Involving Unknown Number of Balls ($x$ Variables)
In ICSE board exams, several high-scoring questions involve setting up an algebraic equation from a given probability relationship.
✍ PRACTICE PROBLEM F1: ALGEBRAIC EQUATION ON UNKNOWN BALLS
Q. A box contains some black balls and $30$ white balls. If the probability of drawing a black ball is $\frac{2}{5}$ of the probability of drawing a white ball, find the number of black balls in the box.
Step 1: Define Variables & Total Outcomes:
Let the number of black balls $= x$.
Number of white balls $= 30$.
Total balls in box $n(S) = x + 30$.
Step 3: Set up & Solve Algebraic Equation:
Given: $P(\text{Black}) = \frac{2}{5} \times P(\text{White})$
$$ \frac{x}{x + 30} = \frac{2}{5} \times \frac{30}{x + 30} $$
Since $x + 30 \ne 0$, multiply both sides by $(x + 30)$:
$$ x = \frac{2 \times 30}{5} = 2 \times 6 = \mathbf{12} $$
Conclusion: The box contains $12$ black balls.
✍ PRACTICE PROBLEM F2: GEOMETRIC PROBABILITY MODEL (TARGET IN RECTANGLE)
Q. A circle with diameter $20\text{ cm}$ is drawn inside a rectangular paper of dimensions $40\text{ cm} \times 30\text{ cm}$. A small dart/die is dropped randomly on the paper without looking. Find the probability that it lands:
(i) Inside the circle, (ii) Outside the circle. (Take $\pi = 3.14$)
Step 1: Calculate Total & Favourable Areas:
Total Area of Rectangle $= \text{Length} \times \text{Breadth} = 40 \times 30 = 1200\text{ cm}^2$.
Radius of Circle $r = \frac{20}{2} = 10\text{ cm}$.
Area of Circle $= \pi r^2 = 3.14 \times 10^2 = 314\text{ cm}^2$.
(i) Probability that it lands inside the circle:
$$ P(\text{Inside}) = \frac{\text{Area of Circle}}{\text{Total Area of Rectangle}} = \frac{314}{1200} = \frac{157}{600} \approx \mathbf{0.2617} $$
(ii) Probability that it lands outside the circle:
$$ P(\text{Outside}) = 1 - P(\text{Inside}) = 1 - \frac{157}{600} = \mathbf{\frac{443}{600}} $$
PART (G): ICSE Board Examination Masterclass (Past 10 Years PYQs)
✍ BOARD MASTERCLASS 1: BIRTHDAYS IN LEAP VS NON-LEAP YEAR
Q. Two friends Ananya and Tanya were born in the year 2024. Find the probability that:
(i) They share the same birthday.
(ii) They have different birthdays.
Identify Year Type: Year $2024$ is divisible by $4 \implies$ it is a Leap Year ($366$ days).
(i) Probability of sharing the SAME birthday:
Ananya can have her birthday on any of the 366 days. For Tanya to have her birthday on the exact same day, there is only $1$ favorable day:
$$ P(\text{Same Birthday}) = \mathbf{\frac{1}{366}} $$
(ii) Probability of having DIFFERENT birthdays:
$$ P(\text{Different Birthdays}) = 1 - P(\text{Same Birthday}) = 1 - \frac{1}{366} = \mathbf{\frac{365}{366}} $$
✍ BOARD MASTERCLASS 2: DEFECTIVE ITEMS DRAWN WITHOUT REPLACEMENT
Q. A box contains $20$ light bulbs, out of which $4$ are defective. A bulb is drawn at random from the box.
(i) What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this second bulb is defective?
(i) First Draw:
Total bulbs $= 20$, Defective $= 4 \implies P(\text{Defective}) = \frac{4}{20} = \mathbf{\frac{1}{5}}$.
(ii) Second Draw (Without Replacement):
Since the first bulb was non-defective and not replaced:
Remaining total bulbs $= 20 - 1 = 19$.
Remaining defective bulbs $= 4$ (all defective bulbs are still in the box).
$$ P(\text{Second is Defective}) = \mathbf{\frac{4}{19}} $$
Q. A game consists of spinning an arrow which comes to rest pointing at one of the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$. If all outcomes are equally likely, find the probability that the arrow will point at:
(i) $9$, (ii) An even number, (iii) A prime number, (iv) A number greater than $8$, (v) A number which is a multiple of $3$.
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