Probability — classical, empirical, axiomatic (Kolmogorov axioms): Visual walkthrough

This page is the picture companion to the parent note. It leans on the language of Set Theory — Union, Intersection, Complement.

Step 1 — The sample space is a board of equal tiles

WHAT. Imagine a random experiment. Every possible outcome is one small square tile. Lay all of them side by side into one big rectangle. That whole rectangle is the sample space, written $S$ — "everything that could happen."

WHY tiles. We want to see probability as area. If every tile is the same size, then counting tiles and measuring area are the same thing — and area is something eyes are good at. This is the classical idea ("equally likely outcomes") turned into a picture.

PICTURE. Look at figure below: 36 tiles (the $6 \times 6$ outcomes of rolling two dice — the parent's running example). The whole peach rectangle is $S$ .

Figure — Probability — classical, empirical, axiomatic (Kolmogorov axioms)

Here the symbol $P (\cdot)$ just means "fraction of the board's area that this region covers." That is all $P$ will ever mean on this page.

Step 2 — An event is a coloured-in region

WHAT. Pick a question with a yes/no answer, like "is the dice-sum even?" The tiles where the answer is yes form a region. Colour them. That coloured region is an event, and we name it with a letter — say $A$ .

WHY. Turning a question into a shape means we can now do geometry instead of logic. "How likely is $A$ ?" becomes "what fraction of the board is coloured $A$ ?"

PICTURE. The magenta region $A$ below.

If $A$ covers no tiles, $P (A) = 0$ ; if it covers all of them, $P (A) = 1$ . The number can never dip below $0$ (you can't have negative area) — that is Kolmogorov's non-negativity axiom, seen.

Step 3 — Two events: overlap is the whole story

WHAT. Now colour a second event $B$ (violet) on the same board. Because $A$ and $B$ are both regions on one board, they can share tiles.

WHY. Almost every real question involves two events — "King or Heart", "rain or wind". To reason about " $A$ or $B$ " we must first look at how their regions sit together. There are exactly two possibilities: they touch (overlap) or they don't. Step 4 handles the easy no-overlap case; Steps 5–6 handle overlap.

PICTURE. Magenta $A$ and violet $B$ with a shared purple sliver where they cross.

Step 4 — The easy case: no overlap, just add

WHAT. Suppose the two regions do not touch (disjoint). Then to find the area of " $A$ or $B$ ", you simply add the two areas — there's no double-booked tile.

WHY. When nothing is shared, no tile gets counted twice, so areas add cleanly. This is Kolmogorov's additivity axiom shown as a picture: separate blobs, add their sizes.

PICTURE. Two blobs that don't touch; their areas snap together with a "+".

This is the whole rule only when they don't touch. The moment they do touch, this line over-counts — which is the trap in the parent note's first [!mistake]. Next we fix it.

Step 5 — The overlap gets counted twice

WHAT. Let the regions overlap again (Step 3's picture). Try the naive move: add $P (A) + P (B)$ . Watch the purple sliver.

WHY. $P (A)$ already includes the purple sliver (it's part of $A$ ). $P (B)$ also includes that same sliver (it's part of $B$ ). So when you write $P (A) + P (B)$ , the sliver is measured twice — you paid for it once inside $A$ and once inside $B$ .

PICTURE. The sliver is shaded with double stripes, flagged "counted 2×".

Reading that equation, the fix jumps out: the sum is too big by exactly one sliver. Subtract it.

Step 6 — Subtract the overlap once: the rule appears

WHAT. Rearrange the Step-5 equation to isolate what we want.

WHY. We remove the one extra copy of the shared sliver so every tile is counted exactly once. That single subtraction is the correction.

PICTURE. The sliver's second copy is lifted away with a "− $P (A \cap B)$ " arrow, leaving the clean union.

Consistency check with Step 4. If the events are disjoint, $A \cap B = \emptyset$ , so $P (A \cap B) = 0$ , and the box collapses to $P (A \cup B) = P (A) + P (B)$ — exactly the easy case. One formula rules both situations.

Step 7 — Degenerate & edge cases (so no scenario surprises you)

WHAT & WHY. A formula you trust must survive its extreme inputs. We test each corner on the tile picture.

Not one corner breaks the boxed rule — because subtracting the sliver once is the right thing in every configuration.

The one-picture summary

Two overlapping blobs; the annotation reads area of A ∪ B = A + B − (the sliver). That single caption is the whole derivation.

Recall Feynman: tell it like a story

Draw the world as a rug made of equal squares — that rug is everything that can happen, and its total size is 1. Colour the squares where "event A" is true, and separately colour the squares where "event B" is true. Want to know how much rug is coloured by A or B? Naively you'd add A's amount and B's amount. But look — the squares that are true for both got coloured twice, so you paid for that little shared patch two times. So subtract that shared patch once. That's it: A or B = A plus B minus the shared bit. If A and B never touch, the shared bit is nothing, and you just add. If B is hidden inside A, the shared bit is all of B, so adding B changes nothing. The one little subtraction fixes every case.

Connections

Set Theory — Union, Intersection, Complement — the $\cup, \cap, \emptyset$ algebra these tiles obey.
Conditional Probability & Bayes' Theorem — zoom the board into a single event and re-measure.
Independence of Events — when the sliver's area equals $P (A) P (B)$ .
Permutations & Combinations — how we count the equal tiles in the first place.