Understanding Chance
This guide explains the why behind every probability rule — not just what to memorise, but how to reason about randomness. Work through each section in order, try every checkpoint question, then reveal the answer to check yourself.
Before you can calculate any probability, you need precise language. "I might roll a high number" is vague — mathematics requires you to say exactly which outcomes you mean. The vocabulary below makes that precision possible.
Words like "outcome," "event," and "sample space" might sound like jargon, but each one refers to a distinct concept. Mixing them up leads to miscounts. An event can contain multiple outcomes — confusing the two is the number-one source of probability errors. Getting the vocabulary right first makes every calculation straightforward.
Key terms
| Term | Meaning | Example (rolling a die) |
|---|---|---|
| Experiment | A process that produces a random result | Rolling a 6-sided die once |
| Outcome | A single possible result of the experiment | Rolling a 4 |
| Sample space (S) | The set of all possible outcomes | S = {1, 2, 3, 4, 5, 6} |
| Event (A) | A set of one or more outcomes you are interested in | A = rolling an even number = {2, 4, 6} |
| Favourable outcomes | The outcomes within the event A | 2, 4, 6 — there are 3 of them |
P(A) = 0 → impossible event (will never happen).
P(A) = 1 → certain event (will always happen).
0 < P(A) < 1 → possible but not guaranteed. The closer to 1, the more likely.
Each flip produces either Heads (H) or Tails (T). There are two flips, so we need to combine every possibility from flip 1 with every possibility from flip 2.
Flip 1 = T: (T, H) (T, T)
There are 4 equally likely outcomes. Note that (H,T) and (T,H) are different outcomes even though both have one head and one tail — the order matters here.
b) Event A = "the spinner lands on an odd number." List the favourable outcomes of A.
c) Can the probability of any event ever equal 1.5? Explain why or why not.
a) S = {1, 2, 3, 4} — 4 outcomes.
b) Favourable outcomes for A: {1, 3}.
c) No. Probability is always between 0 and 1 inclusive. 1.5 would mean something is 150% certain, which is impossible. A probability greater than 1 always signals a calculation error.
Theoretical probability is what we predict should happen based on logic and counting — before running any experiment. It relies on one critical assumption: all outcomes in the sample space are equally likely.
The formula P(A) = favourable / total only works when every outcome has the same chance of occurring. For a fair die, each face has the same 1-in-6 chance, so the formula is valid. For a loaded die that lands on 6 twice as often, the formula breaks down — you can't just count outcomes. Always check the "equally likely" assumption before applying the formula.
Resist the urge to jump straight to a fraction. Writing out S (or drawing a diagram) first prevents missing outcomes or double-counting. Once S is listed, counting favourable outcomes is easy.
P(blue) = 5/10 = 1/2 = 0.5 = 50%
Outcomes that are NOT red = blue + green = 5 + 2 = 7.
There are no yellow marbles in the bag, so this event has 0 favourable outcomes.
b) A bag has 4 orange, 3 purple, and 1 white ball. Find P(purple).
c) Using the same bag as (b), what is the probability of drawing either orange or white?
a) S contains all 52 playing cards; |S| = 52. P(ace) = 4/52 = 1/13 ≈ 7.7%.
b) Total = 4 + 3 + 1 = 8. P(purple) = 3/8 = 0.375 = 37.5%.
c) Orange or white = 4 + 1 = 5 favourable outcomes. P(orange or white) = 5/8 = 0.625 = 62.5%.
Theoretical probability tells us what should happen in a perfect world. Experimental probability records what actually happens when we run an experiment many times. The two rarely match exactly — and that's perfectly normal.
Theoretical probability assumes all outcomes are equally likely and that the experiment is "fair." In the real world, a coin might be slightly unbalanced, a spinner might have unequal sections, or an event might be too complex to count outcomes for (like the chance of rain). In these cases, we measure probability by actually conducting many trials and recording results.
Theoretical vs Experimental
| Theoretical | Experimental | |
|---|---|---|
| Based on | Logic and equally likely outcomes (counting) | Actual data collected from trials |
| Formula | favourable outcomes / total outcomes in S | times A occurred / total trials |
| Exact? | Yes — gives one precise value for a fair experiment | Varies — changes each time you repeat the experiment |
| Improves with | More careful counting of the sample space | More trials (Law of Large Numbers) |
Difference = 56% − 50% = 6 percentage points
The experimental result is slightly higher than theoretical, but this is normal variation with only 50 trials. With many more flips, the experimental probability would converge toward 50%.
Each section is equal (1 of 4), so theoretical P = 1/4 = 0.25 = 25% for all four.
P(B) = 8/40 = 0.200 = 20% (expected: 25%)
P(C) = 11/40 = 0.275 = 27.5% (expected: 25%)
P(D) = 7/40 = 0.175 = 17.5% (expected: 25%)
None of the sections match theory exactly. A scored highest (35% vs 25%), but with only 40 spins, this variation is expected. With 4,000 spins, each would be much closer to 25%.
b) Could you find the theoretical probability of the thumbtack landing point-up the same way as a fair coin? Explain.
c) A die is rolled 120 times. The number 3 appears 25 times. What is the experimental probability of rolling a 3? Is this higher or lower than the theoretical probability?
a) P(point-up) = 52/80 = 0.65 = 65%.
b) No. Unlike a coin, a thumbtack does not have equally likely outcomes — the point-up and point-down positions are not symmetric. We cannot count outcomes to find theoretical probability; we can only measure it experimentally.
c) P(3) = 25/120 = 5/24 ≈ 20.8%. Theoretical P(3) = 1/6 ≈ 16.7%. The experimental result is slightly higher than theoretical.
The complement of an event A (written A' or À) is the event that A does not happen. Together, A and A' cover every possible outcome — so their probabilities must add up to 1.
Imagine asking: "What is the probability of rolling at least one 6 in two rolls of a die?" Counting every favourable case (one 6 on roll 1, one 6 on roll 2, a 6 on both) is tricky. But the complement — "no 6 on either roll" — is easy to calculate. Then P(at least one 6) = 1 − P(no 6 at all). Always ask: is the complement simpler to count?
Whenever you see "at least one," "at least two," or "one or more," the complement approach almost always saves effort. The complement of "at least one" is "none at all," which is usually a single simple calculation.
Event A = "at least one 6 in two rolls."
Complement A' = "no 6 on either roll" (both rolls show 1, 2, 3, 4, or 5).
P(A') = 5/6 × 5/6 = 25/36 ≈ 0.694
b) P(A) = 3/8. Find P(A').
c) A spinner has probability 0.2 of landing on red and 0.5 of landing on blue. What is the probability of landing on neither red nor blue?
a) P(fail) = 1 − 0.78 = 0.22 = 22%.
b) P(A') = 1 − 3/8 = 5/8 = 0.625.
c) P(red or blue) = 0.2 + 0.5 = 0.7. P(neither) = 1 − 0.7 = 0.3 = 30%.
When an experiment has two or more steps, you need a systematic method to list every possible combination. Two powerful tools: tree diagrams (great for sequential experiments) and outcome tables (great for two simultaneous experiments).
Human intuition for counting is unreliable with multiple steps. People routinely undercount — for example, thinking two dice have 12 outcomes instead of 36. Being systematic (drawing every branch, filling every cell of a table) guarantees you won't miss any outcome. A missed outcome means a wrong denominator, which means a wrong probability.
How to build a tree diagram
- Draw one branch for each outcome of the first event.
- From each branch, draw one sub-branch for each outcome of the second event.
- Each complete path from root to leaf is one outcome of the combined experiment.
- Multiply probabilities along a path to get the probability of that combined outcome.
- Add probabilities across paths to get the probability of an event spanning multiple paths.
How to build an outcome table
- Put all outcomes of event 1 as row headers, all outcomes of event 2 as column headers.
- Fill each cell with the combined outcome (e.g., the sum, the pair, etc.).
- Total cells = total outcomes. Count the cells matching your event for the numerator.
Coin: Tails → Die: 1, 2, 3, 4, 5, 6 → 6 paths
Total paths = 2 × 6 = 12 outcomes
Only one path satisfies both conditions: Heads on the coin AND 3 on the die.
Method 2: P(H) × P(3) = 1/2 × 1/6 = 1/12
Rows = die 1 (1 through 6), columns = die 2 (1 through 6). Each cell shows the sum. The complete table has 6 × 6 = 36 cells.
D1=1: 2 3 4 5 6 7
D1=2: 3 4 5 6 7 8
D1=3: 4 5 6 7 8 9
D1=4: 5 6 7 8 9 10
D1=5: 6 7 8 9 10 11
D1=6: 7 8 9 10 11 12
→ 6 favourable outcomes
Each die has 6 faces. Because every outcome on die 1 can pair with every outcome on die 2, the total is 6 × 6 = 36 outcomes — not 6 + 6 = 12. Using 12 as the denominator gives completely wrong probabilities. Always multiply the number of outcomes for each step to get the total.
b) Using the two-dice outcome table from the example above, find P(sum = 2) and P(sum = 12).
c) A bag has 2 red and 3 blue marbles. A marble is drawn, its colour noted, and then a coin is flipped. How many outcomes are in the combined sample space? Draw a quick tree diagram structure and find P(red marble and tails).
a) Each flip has 2 outcomes. Three flips: 2 × 2 × 2 = 8 outcomes.
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
b) Sum = 2: only (1,1) → P = 1/36. Sum = 12: only (6,6) → P = 1/36 each. Both are the least likely sums.
c) 5 marble outcomes × 2 coin outcomes = 10 outcomes total.
Tree: {R1, R2, B1, B2, B3} each branch to {H, T}.
P(red marble) = 2/5. P(tails) = 1/2.
P(red and tails) = 2/5 × 1/2 = 2/10 = 1/5 = 0.2 = 20%.