🎲 Probability — Class 10

Theoretical probability, equally likely outcomes, cards, dice, and word problems

1. Introduction

In Class 9, we studied experimental (empirical) probability based on actual experiments. In Class 10, we study theoretical (classical) probability based on equally likely outcomes.

📖 Theoretical Probability

P(E) = Number of outcomes favourable to E / Total number of equally likely outcomes

This formula is valid only when all outcomes are equally likely.

🌟 Real-Life Connection

When you flip a fair coin, both heads and tails are equally likely. You don't need to flip it 1000 times — you can mathematically say: P(Heads) = 1/2. That's theoretical probability! It's based on logic, not actual experiments.

2. Key Terms

📖 Important Definitions

Random Experiment: An experiment where outcome cannot be predicted with certainty. Example: tossing a coin, rolling a die.

Sample Space (S): The set of all possible outcomes of a random experiment.

Event (E): A subset of the sample space (one or more outcomes).

Elementary Event: An event with only one outcome.

Compound Event: An event with more than one outcome.

⚡ Range of Probability

0 ≤ P(E) ≤ 1

P(E) = 0 → Impossible event (e.g., getting 7 on a die)

P(E) = 1 → Sure/Certain event (e.g., getting a number <7 on a die)

P(E) + P(Ē) = 1 where Ē is the complementary event (E not occurring)

3. Problems Based on Coins

📖 Sample Space for Coins

1 coin: S = {H, T} → Total = 2 outcomes

2 coins: S = {HH, HT, TH, TT} → Total = 4 outcomes

3 coins: S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} → Total = 8 outcomes

n coins: Total outcomes = 2ⁿ

💡 Example: Two Coins

Q: Two unbiased coins are tossed. Find P(exactly one head).

S = {HH, HT, TH, TT} → n(S) = 4

Favourable outcomes (exactly 1 head): HT, TH → n(E) = 2

P(E) = 2/4 = 1/2

4. Problems Based on Dice

📖 Sample Space for Dice

1 die: S = {1, 2, 3, 4, 5, 6} → Total = 6 outcomes

2 dice: Total = 6 × 6 = 36 outcomes

For 2 dice, sample space is written as (d₁, d₂) ordered pairs.

💡 Example: Single Die

Q: A die is thrown once. Find probability of getting: (a) a prime number, (b) a number divisible by 3.

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

(a) Prime numbers: 2, 3, 5 → P = 3/6 = 1/2

(b) Divisible by 3: 3, 6 → P = 2/6 = 1/3

💡 Example: Two Dice

Q: Two dice are thrown. Find P(sum = 7).

Favourable outcomes with sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes

P(sum=7) = 6/36 = 1/6

5. Problems Based on Playing Cards

📖 Deck of 52 Cards

4 suits: Hearts ♥ (red), Diamonds ♦ (red), Clubs ♣ (black), Spades ♠ (black)

Each suit has 13 cards: A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K

Face cards: J (Jack), Q (Queen), K (King) → 3 per suit → 12 total

Aces: 4 (one per suit)

Total: 52 cards

💡 Example: Cards

Q: One card is drawn from a well-shuffled pack. Find:

(a) P(red card) = 26/52 = 1/2

(b) P(face card) = 12/52 = 3/13

(d) P(neither a king nor a queen) = (52–4–4)/52 = 44/52 = 11/13

6. Problems Based on Bags/Balls

💡 Example: Balls in a Bag

Q: A bag contains 3 red, 5 black, and 7 white balls. A ball is drawn at random. Find probability that it is: (a) black, (b) not white.

Total balls = 3 + 5 + 7 = 15

(a) P(black) = 5/15 = 1/3

(b) P(not white) = (15–7)/15 = 8/15

🔑 Important Notes

Probability is always between 0 and 1 (inclusive).
Sum of probabilities of all elementary events = 1.
P(Ē) = 1 – P(E) (complementary rule).
All outcomes must be equally likely for classical probability.
For 2 dice: always list ordered pairs to avoid missing outcomes.

📋 Probability Formula Sheet

Basic Formula

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

Favourable / Total

Range

0 ≤ P(E) ≤ 1

P(impossible) = 0

P(certain) = 1

Complement

P(Ē) = 1 – P(E)

Sample Spaces

1 coin: 2 outcomes

2 coins: 4 outcomes

1 die: 6 outcomes

2 dice: 36 outcomes

Cards: 52 outcomes