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Learn: Probability

A quick, calm intro — then you choose when to climb.

The idea

Probability measures chance — how likely a result is, from 0 (never) to 1 (always). It runs on one fraction and two moves.

The two moves:

  1. 1

    P = good ÷ all — every chance starts as a fraction — count the outcomes you want, divide by every possible outcome

  2. 2

    AND → multiply — this AND that both happen — independent events multiply their chances

  3. 3

    OR → add — either this or that happens — add the chances when the events do not overlap (they cannot both happen)

Quick memory tips

  • A percent is just the fraction × 100 — 1/4 = 25%
  • “Both” is rarer than either alone: a product of fractions shrinks
  • S = sample space (all outcomes) · A = event (the ones you want) · P(A) = n(A) ÷ n(S)

Set notation

The three symbols that carry every probability problem.

SSS

sample space — every possible outcome (the 12 tickets)

AAA

event — the outcomes we want (the 3 winners)

P(A)=n(A)n(S)P(A) = \frac{n(A)}{n(S)}P(A)=n(S)n(A)​

good ÷ all — lucky dip: 3 ÷ 12 = 1/4

✓✓✓

n(S) = 12 · n(A) = 3 → P(A)=312=14P(A) = \frac{3}{12} = \frac{1}{4}P(A)=123​=41​

Worked example

Every step explained — why we do it, and what happens.

  1. 1
    P=312=14P = \frac{3}{12} = \frac{1}{4}P=123​=41​

    Why: Single event: 3 winning tickets out of 12 — the good outcomes over every outcome: n(A) over n(S).

    What happens: P = 1/4 — the fraction of chance.

  2. 2
    =25%= 25\%=25%

    Why: A percent is the fraction × 100: 1/4 = 0.25 → 25%.

    What happens: 25% — the same chance written “out of 100”.

  3. 3
    12×12=14=25%\frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 25\%21​×21​=41​=25%

    Why: Independent AND: two coins both landing heads — two independent events, and “both” is AND, so multiply the separate chances.

    What happens: 25% — rarer than either coin alone, because both must cooperate.

  4. 4
    Trap: 12+12=1=100% ?\text{Trap: } \frac{1}{2} + \frac{1}{2} = 1 = 100\%\,?Trap: 21​+21​=1=100%?

    Why: The trap: adding the chances. Adding is for OR — this OR that, when the events do not overlap. “Both” is AND, and AND multiplies.

    What happens: 100% would be a guarantee — two coin flips can never promise double heads.

  5. 5
    AND→multiplyOR→add\text{AND} \to \text{multiply} \qquad \text{OR} \to \text{add}AND→multiplyOR→add

    Rule summary: AND → multiply (independent events) · OR → add (events that do not overlap).

Counting (Optional Toolbox)

Fast ways to count without listing every outcome. Not needed for the basics — come back when the questions get big.

optional

Factorial

n!=n×(n−1)×⋯×1n! = n \times (n-1) \times \cdots \times 1n!=n×(n−1)×⋯×1

Multiply the countdown: 5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 1205!=5×4×3×2×1=120 — five books on a shelf, spot by spot.

Permutations (order matters)

P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n - r)!}P(n,r)=(n−r)!n!​

Arrangements: who is first changes the answer. Five books on a shelf → 5!=1205! = 1205!=120 orderings.

Combinations (order does not)

C(n,r)=n!r!(n−r)!C(n, r) = \frac{n!}{r!(n - r)!}C(n,r)=r!(n−r)!n!​

Selections: the team is the team whoever you pick first. A team of 3 from 5 → C(5,3)=10C(5, 3) = 10C(5,3)=10.

Shelf = order matters → permutation · team = order does not → combination. The factorial is the building block of both.

Real-world example

The school fair has a lucky dip: a jar holds 12 tickets and 3 of them win prizes. A second jar holds 5 tickets with 1 winner. What is the chance of grabbing a winner from the first jar — and of winning BOTH jars?

jar 1jar 2win · 3/12lose · 9/12win · 1/5lose · 4/5win · 1/5lose · 4/5

both jars: 312×15=120=5%\text{both jars: } \frac{3}{12} \times \frac{1}{5} = \frac{1}{20} = 5\%both jars: 123​×51​=201​=5%

Follow the win → win path: both jars is one path, and one path multiplies.

  1. 1
    P1=312P_1 = \frac{3}{12}P1​=123​

    Why: 3 winning tickets out of 12 in the jar — n(A) over n(S): good outcomes over all outcomes.

    What happens: P₁ = 3/12.

  2. 2
    =14=25%= \frac{1}{4} = 25\%=41​=25%

    Why: Simplify 3/12 to 1/4; × 100 gives the percentage.

    What happens: 25% on the first jar.

  3. 3
    Trap: 14+15=920=45% ?\text{Trap: } \frac{1}{4} + \frac{1}{5} = \frac{9}{20} = 45\%\,?Trap: 41​+51​=209​=45%?

    Why: The trap: adding the chances. The tree shows both jars as ONE path — win, then win again — and one path multiplies.

    What happens: 45% would be “first OR second” — not “both”.

  4. 4
    14×15=120=5%\frac{1}{4} \times \frac{1}{5} = \frac{1}{20} = 5\%41​×51​=201​=5%

    Why: Both jars: multiply. The second jar has 1 winner in 5, so 1/4 × 1/5.

    What happens: 5% — one chance in twenty for the double win.

  5. 5
    120<14 and 15\frac{1}{20} < \frac{1}{4} \text{ and } \frac{1}{5}201​<41​ and 51​

    Why: The check: two wins must be rarer than either win alone — a product of fractions shrinks, a sum would grow.

    What happens: 5% < 25% and 5% < 20% ✓ — multiplication was right.

Common glitch

Two classic glitches:

1.AND vs OR — adding independent chances

Wrong:P(both) = P(A) + P(B) — two chances, so add them

Right:“Both” is AND → multiply. Addition is for OR — either one — and only when the events do not overlap.

2.Turning a fraction into a percent

Wrong:3/12 = 3% — just swap the fraction for a percent

Right:Percent = fraction × 100: 3/12 = 1/4 = 25%. “Per cent” means “out of 100”.

Try a related calculator

  • Probability Calculator — Count outcomes and calculate simple probabilities.
  • Percentage Calculator — Express a probability as a percentage.

Next up: Statistics →

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