1. Chapter at a Glance
"It will probably rain today." "The chance of getting a six is one in six." "That team is likely to win." We use words like maybe, likely, certain all the time. Probability is the mathematics that turns these words into numbers between 0 and 1. In this chapter we learn to measure chance in two ways — by doing experiments (empirical probability) and by reasoning about equally likely outcomes (theoretical probability) — and we organise outcomes using lists, tables and tree diagrams.
- Random experiments, outcomes, sample space and events.
- The probability scale from 0 (impossible) to 1 (certain).
- Empirical (experimental) probability from repeated trials.
- Theoretical probability for equally likely outcomes.
- Complementary events: P(not E) = 1 − P(E).
- Two-step experiments with tables and tree diagrams.
- Why experimental results get closer to theory as trials increase.
Note: experiment results in this chapter are practice data for learning.
2. Basic Terms
| Term | Meaning | Example (rolling a die) |
|---|---|---|
| Random experiment | Action whose result cannot be predicted with certainty | Rolling a die |
| Outcome | One possible result | Getting 4 |
| Sample space S | Set of all possible outcomes | {1, 2, 3, 4, 5, 6} |
| Event E | A collection of outcomes | "Even number" = {2, 4, 6} |
| Equally likely | No outcome is favoured | Each face on a fair die |
3. The Probability Scale
- P(impossible event) = 0, e.g. getting 7 on a normal die.
- P(sure event) = 1, e.g. getting a number less than 7.
- For any event E: 0 ≤ P(E) ≤ 1.
4. Empirical (Experimental) Probability
P(E) = (number of trials in which E happened) ÷ (total number of trials)
Practice data — a coin tossed in batches; running proportion of heads:
| Total tosses | 10 | 50 | 100 | 500 | 1000 |
|---|---|---|---|---|---|
| Heads | 7 | 29 | 46 | 258 | 497 |
| Proportion | 0.70 | 0.58 | 0.46 | 0.516 | 0.497 |
With few trials the proportion jumps around; with many trials it settles near 0.5. This tendency is called the law of large numbers.
Practice data: of 400 bulbs tested, 12 were defective. Estimate the probability that a bulb chosen at random is defective, and the number of defective bulbs expected in a batch of 2,500.
Solution: P = 12/400 = 0.03. Expected = 0.03 × 2500 = 75.
5. Theoretical Probability
P(E) = (number of outcomes favourable to E) ÷ (total number of outcomes)
A fair die is rolled. Find P(prime), P(multiple of 3), P(number > 4).
Solution: Primes {2, 3, 5}: 3/6 = 1/2. Multiples of 3 {3, 6}: 1/3. Greater than 4 {5, 6}: 1/3.
One card is drawn from a well-shuffled deck of 52. Find P(king) and P(red face card).
Solution: 4 kings → 1/13. Red face cards: J, Q, K of hearts and diamonds = 6 → 3/26.
A bag has 5 red, 3 blue and 2 green marbles. One is drawn at random. Find P(blue) and P(not green).
Solution: P(blue) = 3/10. P(green) = 2/10 so P(not green) = 4/5.
6. Complementary Events
The event "not E" (written Ē or E′) happens exactly when E does not. P(E) + P(not E) = 1.
The probabilities of all the outcomes of an experiment add up to 1.
7. Two-step Experiments: Tables and Trees
Two coins: the outcomes are HH, HT, TH, TT — four equally likely outcomes. (HT and TH are different!) So P(exactly one head) = 2/4 = 1/2.
Two dice: a 6 × 6 table gives 36 equally likely outcomes.
| Sum | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Ways | 1 | 2 | 3 | 4 | 5 | 6 | 5 | 4 | 3 | 2 | 1 |
7 is the most likely sum (6/36 = 1/6); 2 and 12 are least likely (1/36 each).
Two dice are rolled. Find P(sum = 9) and P(doublet).
Solution: Sum 9: (3,6), (4,5), (5,4), (6,3) → 4/36 = 1/9. Doublets: 6 → 1/6.
8. Solved Problems (Board Pattern)
A spinner has 8 equal sectors numbered 1–8. Find P(number is a factor of 8).
Solution: Factors {1, 2, 4, 8} → 4/8 = 1/2.
Three coins are tossed. Find P(at least two heads).
Solution: 8 outcomes. At least two heads: HHH, HHT, HTH, THH → 4/8 = 1/2.
A box has 24 balls, some red and some white. P(red) = 3/8. How many white balls are there?
Solution: Red = 3/8 × 24 = 9. White = 15.
9. Master Formula Table
| Idea | Formula |
|---|---|
| Empirical probability | P(E) = trials with E ÷ total trials |
| Theoretical probability | P(E) = favourable ÷ total (equally likely) |
| Range | 0 ≤ P(E) ≤ 1 |
| Complement | P(not E) = 1 − P(E) |
| Expected count | P(E) × number of trials |
| Outcomes of n coins | 2ⁿ |
| Outcomes of two dice | 36 |
10. Master Summary Table
| Topic | Key idea | Exam focus |
|---|---|---|
| Sample space | List all outcomes | Coins, dice, cards |
| Empirical P | From data | Defective items, survey data |
| Theoretical P | Equally likely outcomes | Dice, bags, spinners |
| Complement | 1 − P(E) | "Not" and "at least" questions |
| Two-step | Table or tree | Two coins/dice sums |
| Law of large numbers | More trials → closer to theory | Interpret experiment tables |
11. Mind Map
- Outcome, sample space
- 0 to 1 scale
- From trials
- Large numbers
- Favourable ÷ total
- Complement
- Tree diagram
- 6 × 6 table
12. Tips and Tricks
- Always write the full sample space first, then circle the favourable outcomes.
- For "at least one", use 1 − P(none).
- Know a deck: 52 cards, 4 suits × 13; 26 red, 26 black; 12 face cards (J, Q, K).
- Treating HT and TH as one outcome.
- Giving a probability greater than 1 or negative.
- Using the theoretical formula when outcomes are not equally likely (e.g. a weighted die).
13. Important One-liners
- Probability is always between 0 and 1.
- P(E) + P(not E) = 1.
- The probabilities of all outcomes of an experiment sum to 1.
- Empirical probability approaches theoretical probability as trials increase.
- With two dice, 7 is the most likely sum.
- Tossing n coins gives 2ⁿ outcomes.
14. Exam-pattern Questions with Detailed Answers
Which cannot be the probability of an event?
- 0
- 0.7
- 2/3
- 1.2
Show answer
Probability cannot exceed 1.
Answer: (D)
A die is rolled. Find P(getting an odd number).
Show answer
{1, 3, 5}: 3/6 = 1/2.
Answer: 1/2
If P(E) = 0.35, find P(not E).
Show answer
1 − 0.35 = 0.65.
Answer: 0.65
Practice data: in 60 days, it rained on 18 days. Estimate the probability that it will not rain on a given day.
Show answer
No rain on 42 days: 42/60 = 0.7.
Answer: 0.7
A card is drawn from a deck of 52. Find P(neither a heart nor a king).
Show answer
Hearts or kings: 13 + 3 = 16. Neither: 36/52 = 9/13.
Answer: 9/13
Two coins are tossed. Find P(at least one tail).
Show answer
1 − P(HH) = 1 − 1/4 = 3/4.
Answer: 3/4
Two dice are rolled. Find P(sum is less than 5).
Show answer
Sums 2, 3, 4: 1 + 2 + 3 = 6 ways. 6/36 = 1/6.
Answer: 1/6
A number is chosen at random from 1 to 30. Find P(it is a multiple of 3 or 5).
Show answer
Multiples of 3: 10; of 5: 6; of 15: 2. 10 + 6 − 2 = 14. P = 14/30 = 7/15.
Answer: 7/15
A bag has x red and 12 blue balls. If P(red) is half of P(blue), find x.
Show answer
x/(x + 12) = ½ × 12/(x + 12) → x = 6.
Answer: x = 6
Two dice are rolled. Find P(the product of the numbers is even).
Show answer
Product is odd only if both are odd: 3 × 3 = 9 ways. Even: 36 − 9 = 27. P = 3/4.
Answer: 3/4
Three coins are tossed. Find P(more heads than tails) and P(all show the same face).
Show answer
More heads: HHH, HHT, HTH, THH → 4/8 = 1/2. All same: HHH, TTT → 2/8 = 1/4.
Answer: 1/2 and 1/4
Riya says: "When two coins are tossed, there are three outcomes — two heads, two tails, one of each — so P(two heads) = 1/3." Is she right? Explain.
Show answer
No. Those three outcomes are not equally likely. "One of each" can happen in two ways (HT, TH). The equally likely outcomes are HH, HT, TH, TT, so P(two heads) = 1/4.
Answer: No — the correct probability is 1/4
15. Quick Revision Summary
- Probability measures chance on a 0–1 scale.
- Empirical: from trials; theoretical: favourable ÷ total for equally likely outcomes.
- P(not E) = 1 − P(E); all outcome probabilities sum to 1.
- Use trees and tables for two-step experiments (4 outcomes for 2 coins, 36 for 2 dice).
- More trials → experimental results approach theory.