Visual walkthrough — Probability basics — sample space, events, P(E) = favourable - total

This page rebuilds the parent formula from absolute zero, using pictures at every step. By the end you will see why counting favourable outcomes and dividing by total outcomes gives a probability — not because a teacher said so, but because it is forced on us by three tiny, obvious rules.

We will never use a symbol before it is drawn.

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Step 1 — What "an outcome" and "the sample space" look like

WHAT: We lay every possible outcome out as a separate box.

WHY: Before we can talk about "chance", we must be certain we have listed everything. If even one outcome is missing, every number we compute afterwards is wrong. Picturing them as boxes stops us from forgetting any.

PICTURE: Below, a die's six outcomes are six equal boxes. The count of boxes is a number we will call $n(S)$ — read it as "the size of $S$", i.e. how many boxes there are.

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Step 2 — What "an event" looks like

WHAT: We circle the boxes $\{5,6\}$ and name that circle $E$ = "the roll is greater than 4".

WHY: A question like "what's the chance of rolling more than 4?" is really the question "how big is this circle compared to all the boxes?" Turning a sentence into a circle is the whole trick — it converts language into counting.

PICTURE: The circled boxes are the favourable outcomes. Their count is $n(E)$ — the size of the circle.

$P(E)=\frac{\stackrel{\text{boxes inside the circle}}{\overbrace{n(E)}}}{\underset{\text{all boxes}}{\underbrace{n(S)}}}$

We have not proven this yet — we've only guessed the shape of the answer. Steps 3–6 earn it.

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Step 3 — The three rules every probability must obey

To derive anything we need a starting point. These three rules (the axioms) are the seed. Each is obvious once pictured.

WHAT: We paint a "certainty bar" of total length 1 and cut it into pieces.

WHY these three? Axiom 2 says the bar has length exactly 1 (total certainty). Axiom 3 says non-overlapping pieces add up — you can lay them end to end. Axiom 1 says no piece is negative or longer than the whole bar. Nothing else is needed.

PICTURE: The full bar = all of $S$, length 1. Later we slice it.

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Step 4 — Equally likely means "equal slices"

WHAT: We cut the length-1 bar into $n(S)$ pieces of equal width and call each width $p$.

WHY equal? If we believed one face were more likely, we'd give it a fatter slice — but a fair die gives no face a reason to be fatter. Symmetry forces equal slices. This is the one assumption the whole formula rests on.

PICTURE: Six equal slices, each labelled $p$, filling the whole bar.

By Axiom 3 (glue all slices) and Axiom 2 (total = 1):

$\underset{n(S)\text{slices}}{\underbrace{p+p+\cdots +p}}=1⟹n(S)\cdot p=1⟹p=\frac{1}{n(S)}$

Reading it term by term:

• $p$ — the width of one slice = the chance of one outcome.
• $n(S)\cdot p$ — all slices glued back into the full bar.
• $=1$ — because the full bar is total certainty (Axiom 2).
• $p=\frac{1}{n(S)}$ — solving: each single outcome has chance one over the number of boxes.

For a die: $p=\frac{1}{6}$. For a coin: $p=\frac{1}{2}$. It just falls out.

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Step 5 — Add up the slices inside the circle

WHAT: Our event $E$ circles $n(E)$ boxes. Each box owns a slice of width $\frac{1}{n(S)}$. We glue only the circled slices.

WHY Axiom 3 applies: two different boxes never share an outcome — they are mutually exclusive — so their slices lay end to end with no overlap. We are allowed to simply add.

PICTURE: The certainty bar again, but the slices belonging to circled boxes are highlighted and pushed together into one shaded block.

$P(E)=\underset{n(E)\text{times}}{\underbrace{\frac{1}{n(S)}+\frac{1}{n(S)}+\cdots +\frac{1}{n(S)}}}=\frac{n(E)}{n(S)}$

Term by term:

• Each $\frac{1}{n(S)}$ — one circled box's slice.
• "$n(E)$ times" — because the circle contains exactly $n(E)$ boxes.
• $\frac{n(E)}{n(S)}$ — $n(E)$ copies of the same width is just $n(E)$ over $n(S)$.

That is the formula — and now we know exactly why it is true. It is glued slices of a bar of certainty. See the full Sample Space and Events and Set Theory machinery this rests on.

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Step 6 — The two edge cases fall out for free

WHAT: Check the smallest and largest possible circles.

WHY: A derivation you trust must survive its extremes. If the formula breaks at the boundary, it's wrong in the middle too.

PICTURE: Left — an empty circle (the impossible event, no boxes). Right — a circle swallowing every box (the sure event).

• Impossible event $E=\varnothing$: it circles 0 boxes, so $n(E)=0$ and $P(\varnothing )=\frac{0}{n(S)}=0.$ Zero slices glued = length 0. Correct.
• Sure event $E=S$: it circles all boxes, so $n(E)=n(S)$ and $P(S)=\frac{n(S)}{n(S)}=1.$ Every slice glued back = the full bar = 1. This is exactly Axiom 2 reappearing — a good sign the logic closed on itself.
• In between: since $0\le n(E)\le n(S)$, dividing gives $0\le P(E)\le 1$, recovering Axiom 1. Nothing can ever land outside the certainty bar.

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Worked check with the picture in mind

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The one-picture summary

The whole story in one image: boxes → equal slices → glue the circled ones → divide.

Recall Feynman retelling — say it like a story

Imagine a chocolate bar exactly one unit long; that "one" means total certainty — something is guaranteed to happen. Now count how many outcomes your experiment has and snap the bar into that many equal pieces, one per outcome, because a fair setup has no favourite. Each piece is one over the count: that's the chance of a single outcome. To find the chance of an event, you just circle the outcomes you care about, gather up their pieces, and lay them side by side — no overlaps, so you're allowed to add. The length you gathered, out of the whole bar, is your probability: circled boxes over total boxes, $n(E)/n(S)$. Circle nothing and you gathered length zero — impossible. Circle everything and you're back to the whole bar — certain. That is the entire theory of classical probability: organised counting on a bar of certainty.

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Where to go next: Conditional Probability (shrinking the sample space), Law of Large Numbers (why real dice obey these slices), Random Variables (attaching numbers to boxes).