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R6.1 Stage 6 Probability

Quantifying probability

Quantify probability on the 0–1 scale; “a chances in b”; Laplace’s rule / equiprobability; list outcomes with tables, trees, and Venn diagrams / sets; the frequentist approach; sample space; probabilities sum to 1.

Probability turns “will it happen?” into a number. Instead of saying “unlikely” or “pretty sure”, we measure chance on a scale from 0 to 1.

What it means

Every probability is a number from 0 to 1. A probability of 0 means the event is impossible; 1 means it is certain; ½ means it is exactly as likely to happen as not. Nothing can ever be more likely than certain, so no probability is below 0 or above 1.

To find a probability we first list the sample space — every possible outcome, with none missing and none counted twice. Rolling one ordinary die has sample space {1, 2, 3, 4, 5, 6}: six outcomes. When those outcomes are equally likely (a fair die has no reason to favour one face), we use Laplace’s rule:

probability = (number of favourable outcomes) ÷ (total number of outcomes)

So the probability of rolling a 4 is 1 ÷ 6. The probability of rolling an even number is 3 ÷ 6 = ½, because three of the six outcomes (2, 4, 6) are favourable. People often say this as “3 chances in 6”.

Because the sample space covers everything that can happen, the probabilities of all outcomes must add up to exactly 1. On a die, 1/6 × 6 = 1. This gives a shortcut: the chance an event does not happen is 1 minus the chance it does. If rain has probability 0.3, then no rain has probability 0.7.

Laplace’s rule only works when outcomes are equally likely. When they aren’t — a bent coin, a real drawing pin — we use the frequentist approach instead: do the experiment many times and take probability ≈ (times it happened) ÷ (times you tried). The more trials, the closer this estimate settles toward the true value. To organise outcomes we use tables (for two dice), tree diagrams (for step-by-step choices), and Venn diagrams (to see how events overlap).

Worked examples

A bag of marbles. A bag holds 3 red, 2 blue and 5 green marbles — 10 in total, each equally likely to be drawn.

P(red)   = 3/10 = 0.3
P(blue)  = 2/10 = 0.2
P(green) = 5/10 = 0.5
total    = 1.0   (they sum to 1)
P(not green) = 1 − 0.5 = 0.5

Two coins, using a table. Flip a 5p and a 10p. The table lists all four equally likely outcomes:

10p H10p T
5p HHHHT
5p TTHTT

Exactly one head appears in HT and TH, so P(exactly one head) = 2/4 = ½.

Frequentist estimate. A drawing pin is tossed 50 times and lands point-up 30 times. We can’t reason about “fairness” here, so we estimate P(point-up) ≈ 30/50 = 0.6.

The generative-art connection

Probability is where art stops being fixed and starts to breathe. In mathartcademy, a program that draws dots can be told: “each dot is green with probability 0.5, red with 0.3, blue with 0.2.” Run it once and you get one scatter; run it again and you get a different one — yet the proportions stay near 3:2:5. The picture is the sample space made visible, and watching many dots appear is the frequentist law happening in front of you: the colours settle toward their probabilities as the canvas fills. Seeing Theory’s animations show exactly this convergence.

The internal dot-multiplier tool is a natural stage for this: instead of every arm painting the same dot, let each dot pick its colour by a weighted coin-flip, and the starburst becomes a portrait of a probability distribution — symmetric in structure, random in detail. Change the weights and the whole mood of the artwork shifts, because you are literally editing the numbers that sum to 1.

Common misconceptions

  • “It hasn’t come up, so it’s due.” A fair die has no memory. After five throws with no 6, the chance of a 6 next throw is still 1/6. Past results don’t change equally-likely outcomes.
  • Confusing “how many favourable” with “the probability”. Two red marbles is a count, not a probability. You must divide by the total to land on the 0–1 scale.
  • Using Laplace’s rule when outcomes aren’t equal. “It rains or it doesn’t, so P = ½” is wrong — those two outcomes aren’t equally likely. When you can’t justify equal likelihood, count real trials instead.

What a learner should be able to solve

The benchmark for mastering this concept — the problems a child at this stage is expected to be able to work through.

  1. Two fair dice — one red, one blue — are rolled and the two numbers are added. Imagine the full `6×6` table of the `36` equally likely pairs. Find `P(sum = 10)` and `P(sum is at least 10)` as fractions in simplest form. Then use the complement to find `P(sum is 9 or less)` and check it fits with your answer.

    Answer

    There are `36` equally likely (red, blue) pairs. Sum `= 10`: `(4,6),(5,5),(6,4)` → 3 pairs, so `P(sum = 10) = 3/36 = 1/12`. Sum at least 10 means 10, 11 or 12: add `(5,6),(6,5)` for 11 and `(6,6)` for 12 → `3 + 2 + 1 = 6` pairs, so `P(≥10) = 6/36 = 1/6`. By the complement, `P(9 or less) = 1 − 1/6 = 5/6`, and indeed `1/6 + 5/6 = 1` because every roll is either at-least-10 or 9-or-less.

    Art hook Draw the 6×6 grid of dice-pairs as a coloured heat-grid where each cell's hue is set by its sum (cool for small sums, hot for large). Let the learner toggle a threshold line so all cells with 'sum ≥ 10' light up, and a counter shows favourable/36 as a fraction — the anti-diagonal band of big sums glows as a corner of colour.

  2. In a class of `30` pupils, `18` play football, `12` play chess, and `7` play both. From these counts, work out how many play football only, chess only, and neither. Then, if one pupil is chosen at random, give `P(plays football only)` and `P(plays neither)` as fractions in simplest form, and check that the probabilities of the four Venn regions add to `1`.

    Answer

    Football only `= 18 − 7 = 11`; chess only `= 12 − 7 = 5`; both `= 7`; neither `= 30 − (11 + 5 + 7) = 30 − 23 = 7`. So `P(football only) = 11/30` and `P(neither) = 7/30`. The four regions have probabilities `11/30, 5/30, 7/30, 7/30`, and `11 + 5 + 7 + 7 = 30`, so they sum to `30/30 = 1` — the Venn diagram partitions the whole class exactly once.

    Art hook Draw two overlapping circles as translucent colour fields (orange 'football', teal 'chess'); the overlap blends to a third colour. Drop one labelled dot per pupil into its correct region — football-only, chess-only, both, or the surrounding 'neither' box — so the four counts and their fractions become a living Venn you can literally count.

  3. A bottle-cap is flipped `50` times and lands 'top-up' `32` times. Explain why you should NOT assume `P(top-up) = 1/2` here, and give the frequentist estimates of `P(top-up)` and `P(top-down)` as decimals. Then say what you could do to make the estimate more trustworthy.

    Answer

    A bottle-cap is not symmetric, so there is no reason the two outcomes are equally likely — Laplace's rule needs equally likely outcomes, so we can't just say `1/2` each. Use the data instead: `P(top-up) ≈ 32/50 = 0.64` and `P(top-down) ≈ 18/50 = 0.36` (which is `1 − 0.64`, as it must be). Flipping the cap many more times would reduce the random wobble and let the estimate settle closer to its true long-run value.

    Art hook Animate flips one at a time as caps dropping into an 'up' or 'down' bin; plot the running fraction top-up/total as a line that jitters wildly in the first few flips then steadies toward its true value as the trial count climbs past 50, 100, 200 — the frequentist law drawn live.

Training exercises

Practice problems, easier first, that build toward the bar above. Each doubles as a seed for an interactive artwork.

  1. Match each event to the best number on the `0` to `1` scale: 'the sun will rise tomorrow', 'a tossed fair coin shows heads', 'a dog will read this sentence aloud'. Choose from `0`, `1/2`, `1`.

    Answer

    Sun rises = `1` (certain). Fair coin heads = `1/2` (as likely as not). Dog reads aloud = `0` (impossible). Every probability sits somewhere on this 0-to-1 line.

    Art hook Draw a horizontal number line from 0 to 1 as a colour gradient (red at 0, green at 1). Let the learner drag event labels to a spot; the label snaps to a dot and shows its value.

  2. A fair die is rolled. List the sample space, then give `P(rolling a 5)` as a fraction.

    Answer

    Sample space `{1, 2, 3, 4, 5, 6}` — 6 equally likely outcomes. `P(5) = 1/6`, since exactly one outcome is favourable.

    Art hook Place 6 dots evenly around a circle, one per die face. On each 'roll', light up one dot at random and keep a tally arc around each; the arcs grow toward equal length as rolls accumulate.

  3. A bag holds `3` red and `7` green sweets, all equally likely to be picked. Write `P(red)` and `P(green)` as fractions, then as decimals.

    Answer

    10 sweets total. `P(red) = 3/10 = 0.3`. `P(green) = 7/10 = 0.7`. (Together they make `1`, since every sweet is red or green.)

    Art hook Scatter 10 dots and colour 3 red, 7 green; then re-draw with the same probabilities as a random 100-dot cloud so the 3:7 ratio becomes a visible field of colour.

  4. The probability it rains on Saturday is `0.3`. What is the probability it does NOT rain? Explain the rule you used.

    Answer

    `P(no rain) = 1 − 0.3 = 0.7`. Rain and no-rain are the only two outcomes and together they are certain, so their probabilities must sum to `1`; subtract to find the complement.

    Art hook Draw a single bar of length 1 split into a wet-blue segment (0.3) and a dry-yellow segment (0.7). Dragging the split point live updates both numbers so they always add to 1.

  5. A fair spinner has `5` equal sections coloured red, red, blue, blue, yellow. Find `P(red)`, `P(blue)` and `P(yellow)`, and check they sum to `1`.

    Answer

    `P(red) = 2/5`, `P(blue) = 2/5`, `P(yellow) = 1/5`. Sum: `2/5 + 2/5 + 1/5 = 5/5 = 1`, which it must, as the spinner always lands on some section.

    Art hook Render a 5-sector wheel with the given colours; a spin animation rotates a needle and, over many spins, a stacked bar of results grows toward the 2:2:1 proportions.

  6. Spot the mistake. Sam says: 'A football match is either won or lost, so the probability my team wins is `1/2`.' What is wrong with Sam's reasoning?

    Answer

    Sam used Laplace's rule, but that only works when outcomes are equally likely. Winning and losing are not equally likely for most teams, and a draw is possible too. You'd need past results (a frequentist estimate) to judge the real chance.

    Art hook Show two mislabelled equal bars ('win' and 'lose') then let the user re-weight them by dragging, plus add a 'draw' slice — reshaping the wrong 50/50 into a realistic, unequal split.

  7. Two fair dice (red and blue) are rolled and their numbers added. Using a `6×6` table of the sums, find `P(sum = 7)` as a fraction in simplest form.

    Answer

    The table has `36` equally likely (red, blue) pairs. A sum of 7 comes from `(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)` — 6 favourable. `P(sum = 7) = 6/36 = 1/6`.

    Art hook Draw the 6x6 grid of dice-pairs as a coloured heat-grid where each cell's hue = its sum; the diagonal band of 7s glows brightest, making 'most common sum' visible as a stripe of colour.

  8. A coin is flipped `20` times and lands heads `13` times. Estimate `P(heads)` from this data as a decimal. Then say why this estimate might differ from `0.5`, and what would make it more trustworthy.

    Answer

    Frequentist estimate: `P(heads) ≈ 13/20 = 0.65`. It differs from `0.5` because with only 20 trials there is a lot of random wobble. Flipping many more times would let the estimate settle closer to the true `0.5` for a fair coin (the long-run / frequentist law).

    Art hook Animate a growing scatter of coin-flip dots (gold=heads, grey=tails) while a line chart tracks the heads-fraction; the line jitters near the start and flattens toward 0.5 as the dot count climbs into the hundreds.

  9. In a class of `20` pupils, `12` like art, `8` like music, and `5` like both. Picture two overlapping circles. How many like art only? If one pupil is picked at random, find `P(likes music)` as a fraction in simplest form.

    Answer

    Art only = `12 − 5 = 7`. `P(likes music) = 8/20 = 2/5`. (Filling the Venn: 7 art-only, 5 both, 3 music-only, and `20 − 15 = 5` who like neither.)

    Art hook Draw two overlapping circles as translucent colour fields (magenta 'art', cyan 'music'); the overlap blends to a third colour. Drop a labelled dot per pupil into the correct region so the counts become a living Venn.

  10. A fair spinner has `8` equal sections numbered `1` to `8`. Find `P(a number greater than 5)` and `P(an odd number)` as fractions in simplest form, then check whether `P(greater than 5) + P(5 or less)` equals `1`.

    Answer

    Greater than 5: `6, 7, 8` → `P = 3/8`. Odd: `1, 3, 5, 7` → `P = 4/8 = 1/2`. Complement check: `P(5 or less) = 5/8`, and `3/8 + 5/8 = 8/8 = 1`. Correct — an event and its complement sum to 1.

    Art hook An 8-sector wheel where sectors matching the chosen event flip to a highlight colour on demand; a side counter shows favourable/total updating as you toggle between 'greater than 5' and 'odd'.

  11. Design a fair spinner (equal sections) so that `P(blue) = 1/3` and `P(red) = 1/2`. What is the smallest number of equal sections you can use, how many are blue and how many red, and what is `P(the remaining colour)`?

    Answer

    Use `6` equal sections (the smallest common multiple of 3 and 2). Blue = `1/3` of 6 = `2` sections; red = `1/2` of 6 = `3` sections; that leaves `1` section for a third colour, say green. `P(green) = 1/6`, and `2/6 + 3/6 + 1/6 = 1`.

    Art hook An interactive wheel-builder: choose number of sections and paint each; a live readout shows each colour's probability as a fraction and flags when they don't sum to 1, turning design into a probability puzzle.

  12. A tree diagram is drawn for flipping a fair coin twice. There are two branches (H, T) at the first flip and two at each of those. How many end-paths are there, what is the probability of the single path HH, and what do all four path-probabilities add up to?

    Answer

    There are `2 × 2 = 4` end-paths: HH, HT, TH, TT. Each is equally likely, so `P(HH) = 1/4`. All four paths sum to `1/4 × 4 = 1`, because the tree lists every possible outcome exactly once.

    Art hook Draw the branching tree left-to-right with animated dots travelling down each branch; each of the 4 leaf nodes collects dots, and over many runs the four piles stay near-equal, showing 1/4 each visually.

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