Sample spaces, events, and axioms of probability: Visual walkthrough

Before Step 1, let us fix the pictures and words we will lean on the whole way.

The whole derivation is a fight against one enemy: double counting. Keep that word in mind.

Step 1 — The picture that starts everything: two overlapping blobs

WHAT. We draw the sample space as a rectangle, and place two events $A$ and $B$ inside so that they overlap. The overlap is $A \cap B$ .

WHY. Overlap is the only interesting case. If the blobs don't touch, the answer is trivially "add the areas." We deliberately start with the hard case so the fix is visible.

PICTURE. Below, blob $A$ (blue) and blob $B$ (yellow) share a green lens in the middle — that lens is $A \cap B$ .

Figure — Sample spaces, events, and axioms of probability

Notice the trap: if you paint $A$ 's area, then paint $B$ 's area on top, the green lens gets painted twice. That is the double counting we must correct.

Step 2 — Slice $A \cup B$ into pieces that never touch

WHAT. We cut the union into two disjoint (non-overlapping) pieces: $A \cup B = A \cup (B \cap A^{c})$

WHY. Our only tool for adding probabilities is Axiom 3, and Axiom 3 is only allowed when the pieces are mutually exclusive (never happen together). So before we can add, we must rewrite the union as non-touching pieces. Here:

$A$ = the whole blue blob, and
$B \cap A^{c}$ = "the part of $B$ that is outside $A$ " — the yellow crescent minus the green lens.

PICTURE. The blue blob and the yellow crescent tile the union perfectly, with no shared area.

Applying Axiom 3 (add disjoint pieces): $P (A \cup B) = P (A) + P (B \cap A^{c}) (⋆)$

We are not done: $P (B \cap A^{c})$ is not something we were given. Step 3 fixes that.

Step 3 — Slice $B$ the same way to name the crescent

WHAT. Cut blob $B$ itself into its inside- $A$ part and its outside- $A$ part: $B = (B \cap A) \cup (B \cap A^{c})$

WHY. We want a formula for the mystery crescent $B \cap A^{c}$ . The cleanest way is to notice that $B$ is made of exactly two disjoint slabs: the part sitting inside $A$ (the green lens, $B \cap A$ ) and the part sitting outside $A$ (the crescent, $B \cap A^{c}$ ). Since these two slabs are disjoint, Axiom 3 applies to them too.

PICTURE. Blob $B$ split along the border of $A$ : green lens on one side, yellow crescent on the other.

Now solve this for the crescent (just algebra — subtract $P (A \cap B)$ from both sides): $P (B \cap A^{c}) = P (B) - P (A \cap B) (⋆ ⋆)$

Step 4 — Substitute and watch the double count cancel

WHAT. Put the crescent formula $(⋆ ⋆)$ back into the union formula $(⋆)$ .

WHY. $(⋆)$ still had the mystery crescent in it; $(⋆ ⋆)$ tells us what that crescent's probability equals in terms of things we do know ( $P (B)$ and $P (A \cap B)$ ). Substituting eliminates the mystery.

PICTURE. The green lens, once painted twice, gets subtracted exactly once — leaving each region painted exactly once.

That $- P (A \cap B)$ is not a mysterious correction — it is the precise undoing of the double paint we spotted in Step 1. Inclusion–exclusion is derived. ∎

Step 5 — Edge case: disjoint events (blobs don't touch)

WHAT. Suppose $A$ and $B$ never happen together, i.e. $A \cap B = \emptyset$ .

WHY. We must check our formula survives the degenerate case. The parent note's Axiom 3 was stated only for disjoint events — our formula should reduce to it, or something is wrong.

PICTURE. No overlap: no green lens to subtract.

(That $P (\emptyset) = 0$ itself follows from the axioms: $Ω \cup \emptyset = Ω$ with $Ω, \emptyset$ disjoint gives $P (Ω) = P (Ω) + P (\emptyset)$ , so $P (\emptyset) = 0$ .)

Step 6 — Edge case: nested events (one blob inside the other)

WHAT. Suppose $A \subseteq B$ — blob $A$ sits entirely inside blob $B$ .

WHY. Another degenerate shape. Here the overlap is all of $A$ : $A \cap B = A$ . We check the formula still behaves and, as a bonus, it hands us monotonicity (bigger event, bigger-or-equal probability).

PICTURE. Blue blob fully swallowed by yellow blob; the union is just the yellow blob.

Step 7 — Worked numbers: the two-dice check

WHAT. Use the parent's example. Two fair dice, $∣Ω∣ = 36$ . Let $A =$ "first die is 6", $B =$ "sum is 10".

WHY. A concrete count proves the formula and shows how the naive add-only answer is wrong.

PICTURE. The $6 \times 6$ grid of outcomes; blue column = $A$ , yellow diagonal = $B$ , green cell = their single shared outcome $(6, 4)$ .

The one-picture summary

One rectangle, two blobs, one green lens — and the whole story is: paint $A$ , paint $B$ , then un-paint the lens once so nothing is double-counted. Every axiom used is labelled where it entered.

Recall Feynman retelling (say it like you're explaining to a friend)

Imagine a whiteboard that is your whole world of outcomes — its total "amount of certainty" is exactly $1$ . Draw two clouds on it, $A$ and $B$ , that overlap a bit. You want to know how much of the board is covered by at least one cloud, which we call $A \cup B$ . Naively you'd measure cloud $A$ , measure cloud $B$ , and add. But the little sliver where they overlap — call it $A \cap B$ — got measured in both clouds, so you counted it twice. The cure is to subtract that overlap once: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ . To prove it honestly we only used three rules: probabilities are never negative, the whole board is $1$ , and non-overlapping pieces just add. We chopped the union into "all of $A$ " plus "the part of $B$ outside $A$ " (those don't overlap, so we can add), then figured out that outside-part equals $P (B)$ minus the overlap. Substitute, and out pops the formula. Two sanity checks seal it: if the clouds don't touch, the overlap is zero and we're back to plain adding; if one cloud sits fully inside the other, the union is just the bigger cloud and — bonus — the smaller cloud can never have more probability than the bigger one.

Recall Quick self-test

Why is the correction term subtracted, not added? ::: The overlap $A \cap B$ is included inside both $P (A)$ and $P (B)$ , so it is counted twice; subtracting it once leaves it counted exactly once. Which axiom lets us split $A \cup B$ into disjoint pieces and add them? ::: Axiom 3 (countable additivity) — but only after we make the pieces mutually exclusive. What does the formula reduce to when $A \cap B = \emptyset$ ? ::: $P (A \cup B) = P (A) + P (B)$ , i.e. Axiom 3 itself. For $A \subseteq B$ , what inequality drops out? ::: Monotonicity: $P (A) \leq P (B)$ .

Where this goes next: the overlap term $P (A \cap B)$ is the seed of Conditional Probability (which normalizes by $P (B)$ ) and hence Bayes' Theorem. Counting outcomes on the dice grid is the finite-space version of Random Variables and Probability Distributions.