Conditional probability — P(A - B) = P(A∩B) - P(A): Visual walkthrough

This is a picture-first companion to the parent note. If a word here feels new, it will have a picture next to it.

Step 1 — Draw the whole world of outcomes

WHAT. An experiment (roll a die, draw a card) can end in several ways. Each possible ending is an outcome. Collect all of them into one box — that box is the sample space, written $S$ . See Sample space and events for the full idea; here we only need " $S$ = the box of all dots."

WHY. Before we can talk about "the chance of something", we need a fixed universe to measure inside. You cannot say "half" without knowing "half of what". $S$ is the "of what".

PICTURE. Each little chalk dot is one outcome. We will pretend for now that every dot is equally likely — same chance of being the winner. We write $N$ for the total number of dots.

Figure — Conditional probability — P(A - B) = P(A∩B) - P(A)

Step 2 — Mark the two events $A$ and $B$

WHAT. An event is just a chosen loop drawn around some of the dots — a subset of $S$ . We draw two loops: $A$ and $B$ . They usually overlap.

WHY. Conditional probability is a relationship between two events. We need both on stage before we can ask how one affects the other.

PICTURE. Blue loop $A$ , pink loop $B$ . The lens-shaped overlap — dots inside both loops — is written $A \cap B$ and read " $A$ and $B$ ". The symbol $\cap$ is a little cup that catches only what the two share.

$A \cap B$ ::: the overlap region; dots that satisfy both events at once.
$n (A \cap B)$ ::: how many dots sit in that overlap.

Step 3 — Someone tells you " $B$ happened" → the world shrinks

WHAT. Now the key move. A friend peeks at the result and says: "I promise the outcome is inside $B$ ." Every dot outside $B$ is now impossible — rub it off the board.

WHY. This single fact — " $B$ happened" — is information. Information kills possibilities. The dots outside $B$ can no longer win, so they should no longer count. The pink loop $B$ becomes your new, smaller box.

PICTURE. Watch the greyed-out dots fade. What survives is exactly $B$ : a shrunken sample space with $n (B)$ dots in it.

Step 4 — Ask the question inside the new box

WHAT. We are inside $B$ . Now ask: of the dots that survived, which also lie in $A$ ? Those are exactly the overlap dots $A \cap B$ — they are the only $A$ -dots that were not rubbed away.

WHY. Any $A$ -dot that lived outside $B$ died in Step 3. So inside our shrunken world, " $A$ happens" can only mean "an $A \cap B$ dot". This is why the overlap, not all of $A$ , is the star of the formula.

PICTURE. Yellow highlights the overlap dots — the winners inside the survivors.

Applying the count rule of Step 1, but inside the box $B$ (so the total is now $n (B)$ , not $N$ ): $P (A ∣ B) = \frac{n ( A \cap B )}{n ( B )}$

numerator $n (A \cap B)$ ::: yellow dots — survivors that are also in $A$ .
denominator $n (B)$ ::: all surviving dots — the new "total".
the fraction ::: the slice of the shrunken box taken up by $A$ .

Notice: this is our first real formula, and it fell straight out of the picture.

Step 5 — Trade dot-counts for probabilities

WHAT. Counting dots only works when all dots are equally likely. To free ourselves from that assumption, divide the top and the bottom of the fraction by $N$ (the total dots in the whole board).

WHY. Dividing top and bottom of a fraction by the same number never changes its value — but it converts every count $n (\cdot)$ into a probability $P (\cdot) = n (\cdot) / N$ from Step 1. That swap is what makes the formula survive into worlds where outcomes are not equally likely.

PICTURE. Same fraction, relabelled: counts on the left morph into probabilities on the right.

$P (A ∣ B) = \frac{n ( A \cap B )}{n ( B )} = \frac{n ( A \cap B ) / N}{n ( B ) / N} = \frac{P ( A \cap B )}{P ( B )}$

$\frac{n ( A \cap B )}{N} = P (A \cap B)$ ::: overlap slice of the whole board.
$\frac{n ( B )}{N} = P (B)$ ::: pink slice of the whole board.
their ratio ::: rescales "overlap out of everything" into "overlap out of $B$ ."

Step 6 — Why divide by $P (B)$ and not by $P (A)$ ?

WHAT. Test the formula on the one case whose answer we already know for sure: what is $P (B ∣ B)$ ? If you know $B$ happened, the chance $B$ happened must be $1$ — dead certain.

WHY. A correct rescaling must make the new box add up to full certainty. Divide by $P (B)$ and it does; divide by $P (A)$ and it does not — that alone forces the choice.

PICTURE. The pink box, fully certain, meter pinned at $1$ .

$P (B ∣ B) = \frac{P ( B \cap B )}{P ( B )} = \frac{P ( B )}{P ( B )} = 1 ✓$

$B \cap B = B$ ::: a set overlapped with itself is itself.
result $= 1$ ::: exactly the certainty we demanded. Dividing by $P (A)$ would have given $P (B) / P (A)$ — no reason to equal $1$ , so it must be wrong.

Step 7 — The forbidden case: $P (B) = 0$

WHAT. What if $B$ is impossible — an empty loop, $n (B) = 0$ , so $P (B) = 0$ ?

WHY. Then the denominator is $0$ , and $\frac{anything}{0}$ is undefined. But there is a deeper reason than "no dividing by zero": you cannot be told that an impossible event happened. There is no smaller box to stand inside — the box is empty. The question itself is meaningless.

PICTURE. An empty pink loop with a chalk cross through it — no dots to condition on.

Step 8 — Rearrange into the multiplication rule (bonus picture)

WHAT. Multiply both sides of the boxed formula by $P (B)$ .

WHY. Sometimes you do not want $P (A ∣ B)$ ; you want $P (A \cap B)$ — the chance both happen. This is exactly what a tree diagram chains, and what the multiplication rule states.

PICTURE. Two-branch tree: first branch weighted $P (B)$ , second (conditional) branch weighted $P (A ∣ B)$ ; multiply along the path.

$P (A \cap B) = P (B) P (A ∣ B) = P (A) P (B ∣ A)$

$P (B)$ ::: probability of the first thing (root branch).
$P (A ∣ B)$ ::: probability of the second given the first (child branch).
product ::: probability of walking the whole path, i.e. both.

And independence appears instantly: if learning $B$ changes nothing, $P (A ∣ B) = P (A)$ , so $P (A \cap B) = P (A) P (B)$ . (For flipping a condition correctly, that is Bayes' Theorem; for summing conditionals over a partition, Law of total probability.)

The one-picture summary

Everything compressed: whole board $\to$ shrink to $B$ $\to$ count overlap out of $B$ $\to$ rescale to probabilities.

Recall Feynman: tell the whole walkthrough to a 12-year-old

Picture a chalkboard covered in dots — every dot is a way the game could end (Step 1). Draw a blue loop and a pink loop around some dots; the little football-shaped bit where they cross is "both" (Step 2). Now a friend peeks and swears the real dot is inside the pink loop. Instantly, every dot outside pink is impossible — wipe them off. Your board just got smaller; the pink loop is your whole world now (Step 3). Ask: of the dots still alive, which are also in blue? Only the ones in that football-shaped crossing. So the chance is (football dots) ÷ (all pink dots) — that's $P (A ∣ B) = n (A \cap B) / n (B)$ (Step 4). To make it work even when dots aren't equally likely, divide top and bottom by the total dot count; the counts turn into probabilities and you get $P (A \cap B) / P (B)$ (Step 5). Why divide by pink and not blue? Because if the friend says "it's in pink" then "chance it's in pink" must be a rock-solid $1$ — and only dividing by $P (B)$ gives that (Step 6). If the pink loop is empty, there's nothing to peek at, so the question makes no sense — that's why we need $P (B) > 0$ (Step 7). Flip the formula around and you get the "both" rule that trees use (Step 8). Done — you drew the definition.

Recall Quick self-check

Why is the denominator $P (B)$ and not $P (A)$ ? ::: Because $B$ is the new box we stand inside; only dividing by $P (B)$ makes $P (B ∣ B) = 1$ . Which dots survive after " $B$ happened"? ::: Only those inside $B$ ; the new total is $n (B)$ . Why does only $A \cap B$ count, not all of $A$ ? ::: The $A$ -dots outside $B$ were wiped away, so inside $B$ "in $A$ " means "in the overlap". What breaks when $P (B) = 0$ ? ::: Division by zero, and there is no non-empty box to condition on — the question is meaningless.

Connections

Sample space and events — the board of dots we started from.
Multiplication rule of probability — Step 8's rearranged form.
Independence of events — when $P (A ∣ B) = P (A)$ .
Tree diagrams — the picture of Step 8.
Bayes' Theorem — flipping the condition safely.
Law of total probability — summing conditionals over a partition.