Probability — Class 10 IMO

Theoretical Probability

What is the classical definition of probability? Probability is the measure of how likely an event is to occur. The classical definition (also called theoretical probability) states:

P(E) = Number of favorable outcomes / Total number of possible outcomes

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

where:

P(E) = probability of event E
n(E) = number of outcomes favorable to event E
n(S) = total number of all possible outcomes (sample space)

Key properties:

Probability is always between 0 and 1 (inclusive)
0 means impossible event (will never happen)
1 means certain event (will always happen)
Sum of probabilities of all possible outcomes = 1
P(not E) = 1 − P(E) (complementary event)

Real-life examples:

Tossing a coin: P(getting heads) = 1/2 = 0.5
Rolling a die: P(getting 4) = 1/6 ≈ 0.1667
Drawing an ace from a deck of 52 cards: P(ace) = 4/52 = 1/13

Important assumptions:

All outcomes are equally likely
The experiment is random (no bias)
Only classical probability is in Grade 10 syllabus (not experimental/statistical)

┌─────────────────────────────────────────────────────────────┐
│         CLASSICAL PROBABILITY - VISUAL REPRESENTATION        │
└─────────────────────────────────────────────────────────────┘

PROBABILITY SCALE:

    0                    0.5                    1
    │───────────────────────│───────────────────────│
    Impossible          Even Chance           Certain
    
    Examples:
    0 = P(sun rising in west)
    0.5 = P(getting heads on coin toss)
    1 = P(sun rising in east)


SAMPLE SPACE VISUALIZATION:

COIN TOSS:               DIE ROLL:

      ┌───┐                  ┌───┐
      │ H │                  │ 1 │
      └───┘                  └───┘
      ┌───┐                  ┌───┐
      │ T │                  │ 2 │
      └───┘                  └───┘
    n(S) = 2                ┌───┐
                            │ 3 │
                            └───┘
                            ┌───┐
                            │ 4 │
                            └───┘
                            ┌───┐
                            │ 5 │
                            └───┘
                            ┌───┐
                            │ 6 │
                            └───┘
                          n(S) = 6


PROBABILITY AS A FRACTION:

    P(E) = Favorable Outcomes / Total Outcomes
    
    Example: Drawing a red card from deck of 52 cards
    
    Total outcomes = 52 cards
    Favorable outcomes = 26 red cards (hearts + diamonds)
    
    P(red) = 26/52 = 1/2 = 0.5


COMPLEMENTARY EVENTS:

         P(E) + P(not E) = 1
    
    ┌─────────────────────────────────────────┐
    │              SAMPLE SPACE                │
    │  ┌─────────────┐  ┌─────────────────────┐│
    │  │             │  │                     ││
    │  │    Event    │  │   Complement        ││
    │  │      E      │  │      E'             ││
    │  │  P(E)       │  │  P(not E) = 1-P(E)  ││
    │  │             │  │                     ││
    │  └─────────────┘  └─────────────────────┘│
    └─────────────────────────────────────────┘

Example 1: A fair coin is tossed once. Find the probability of getting a head.

Step 1: Total possible outcomes = {H, T} → n(S) = 2
Step 2: Favorable outcomes (head) = {H} → n(E) = 1
Step 3: P(E) = n(E)/n(S) = 1/2

Answer: P(H) = 1/2 = 0.5

Example 2: A die is rolled once. Find the probability of getting an even number.

Step 1: Total outcomes = {1,2,3,4,5,6} → n(S) = 6
Step 2: Even numbers = {2,4,6} → n(E) = 3
Step 3: P(even) = 3/6 = 1/2

Answer: P(even) = 1/2

Example 3: One card is drawn from a well-shuffled deck of 52 playing cards. Find the probability that it is either a king or a queen.

Step 1: Total outcomes = 52
Step 2: Number of kings = 4, number of queens = 4
Step 3: Favorable outcomes = 4 + 4 = 8 (no overlap — a card cannot be both king and queen)
Step 4: P(king or queen) = 8/52 = 2/13

Answer: P(king or queen) = 2/13

Quick recap

Classical probability: P(E) = Number of favorable outcomes / Total outcomes
Probability is always between 0 and 1 (inclusive)
Sum of probabilities of all outcomes = 1
P(not E) = 1 − P(E) (complement rule)
All outcomes must be equally likely for classical definition to apply
Probability can be expressed as fraction, decimal, or percentage

✓ Quick check

A card is drawn from a pack of 52. The probability of getting a face card is:

12 face cards out of 52 give 12/52 = 3/13.

If P(E) = 0.3, then P(not E) equals:

P(not E) = 1 − P(E) = 1 − 0.3 = 0.7.

Coins, Dice and Cards

What are simple problems on single events? A single event involves one experiment (like tossing a coin once, rolling a die once, drawing one card). These problems use the basic probability formula with:

n(S) = total number of equally likely outcomes
n(E) = number of outcomes favorable to the event

Types of single event problems:

Type	Example	Sample Space Size
Coin toss (1 coin)	Getting head	2
Coin toss (2 coins)	Getting at least one head	4
Die roll (1 die)	Getting prime number	6
Deck of cards	Getting a spade	52
Drawing a ball from bag	Getting red ball	Total balls
Birth month	Born in December	12
Two-digit number	Multiple of 5	90 (10 to 99)

Important note for overlapping events: If two events cannot happen together (mutually exclusive), add their probabilities: P(A or B) = P(A) + P(B)

┌─────────────────────────────────────────────────────────────┐
│          SIMPLE PROBABILITY PROBLEMS - VISUAL GUIDE          │
└─────────────────────────────────────────────────────────────┘

PROBLEM TYPE 1: TOSSING TWO COINS

    Coin 1    Coin 2    Outcome
    ┌────┐    ┌────┐
    │ H  │    │ H  │  →  HH
    ├────┤    ├────┤
    │ H  │    │ T  │  →  HT
    ├────┤    ├────┤
    │ T  │    │ H  │  →  TH
    ├────┤    ├────┤
    │ T  │    │ T  │  →  TT
    └────┘    └────┘
    
    Sample Space S = {HH, HT, TH, TT} → n(S) = 4


PROBLEM TYPE 2: ROLLING TWO DICE

    Die 1 →  1   2   3   4   5   6
    Die 2 ↓
      1    (1,1)(2,1)(3,1)(4,1)(5,1)(6,1)
      2    (1,2)(2,2)(3,2)(4,2)(5,2)(6,2)
      3    (1,3)(2,3)(3,3)(4,3)(5,3)(6,3)
      4    (1,4)(2,4)(3,4)(4,4)(5,4)(6,4)
      5    (1,5)(2,5)(3,5)(4,5)(5,5)(6,5)
      6    (1,6)(2,6)(3,6)(4,6)(5,6)(6,6)
    
    n(S) = 36


PROBLEM TYPE 3: DRAWING FROM A BAG

    Bag contains: 3 Red, 4 Blue, 5 Green balls
    
    Total = 3 + 4 + 5 = 12 balls
    
    P(Red) = 3/12 = 1/4
    P(Blue) = 4/12 = 1/3
    P(Green) = 5/12
    P(not Red) = 1 - 1/4 = 3/4


PROBLEM TYPE 4: DECK OF CARDS

    Standard deck: 52 cards
    ┌─────────────────────────────────────────┐
    │  4 suits: ♠ Spades, ♥ Hearts, ♦ Diamonds, ♣ Clubs │
    │  Each suit: 13 cards (A,2-10,J,Q,K)    │
    └─────────────────────────────────────────┘
    
    P(Heart) = 13/52 = 1/4
    P(Face card) = 12/52 = 3/13 (J,Q,K in 4 suits)
    P(Ace) = 4/52 = 1/13

Example 1: Two coins are tossed simultaneously. Find the probability of getting exactly one head.

Step 1: Sample space S = {HH, HT, TH, TT} → n(S) = 4
Step 2: Favorable outcomes (exactly one head) = {HT, TH} → n(E) = 2
Step 3: P(exactly one head) = 2/4 = 1/2

Answer: 1/2

Example 2: A bag contains 5 red, 4 blue, and 3 green balls. One ball is drawn at random. Find the probability that it is: (i) red (ii) not blue

Step 1: Total balls = 5 + 4 + 3 = 12 → n(S) = 12
Step 2: (i) Red balls = 5 → P(red) = 5/12
Step 3: (ii) Blue balls = 4 → P(blue) = 4/12 = 1/3
Step 4: P(not blue) = 1 − P(blue) = 1 − 1/3 = 2/3

Answer: (i) 5/12; (ii) 2/3

Example 3: A number is chosen at random from the numbers 10 to 99 (two-digit numbers). Find the probability that the number is a multiple of 5.

Step 1: Two-digit numbers from 10 to 99: total count = 99 − 10 + 1 = 90 → n(S) = 90
Step 2: Multiples of 5 end with 0 or 5: 10,15,20,...,95
Step 3: Number of multiples = (95 − 10)/5 + 1 = 85/5 + 1 = 17 + 1 = 18
Step 4: P(multiple of 5) = 18/90 = 1/5 = 0.2

Answer: 1/5

Quick recap

For single events: P(E) = n(E)/n(S)
List sample space carefully — count all possible outcomes
For "at least one" type problems, consider using complement: 1 − P(none)
For mutually exclusive events: P(A or B) = P(A) + P(B)
Always simplify fractions to lowest terms
Probability of complementary event: P(not E) = 1 − P(E)

✓ Quick check

One card is drawn from 52 cards. The probability that it is a king is:

4 kings out of 52 give 4/52 = 1/13.

A card is drawn from a pack of 52. The probability that it is a spade is:

13 spades out of 52 give 1/4.

Complementary Events

The event "not E" is the complement of E, and P(E) + P(not E) = 1. So P(not E) = 1 − P(E) — often the quickest route to an answer.

Example 1: If P(E) = 0.3, P(not E)?

1 − 0.3 = 0.7.

Example 2: P(not getting a king) from a deck?

1 − 1/13 = 12/13.

Quick recap

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

✓ Quick check

A child has cards numbered 1 to 20. The probability of drawing a multiple of 5 is:

Multiples of 5 are 5, 10, 15, 20 — four of 20 give 1/5.

A spinner has 8 equal sectors numbered 1 to 8. The probability of landing on a number greater than 5 is:

Numbers 6, 7, 8 are three of 8, so 3/8.