Visual walkthrough — Complementary events — P(A') = 1 − P(A)

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Step 1 — What even is an "outcome"? Draw the box of possibilities

WHAT. Roll a fair six-sided die. The things that can happen are $1,2,3,4,5,6$. Put each one as a dot inside a box.

WHY. Before we can talk about the chance of anything, we need the full list of what "anything" even means. No box → nothing to measure. The box is our stage.

PICTURE. Six chalk dots sit inside one rectangle. That rectangle is $S$. Nothing lives outside it.

• The rectangle $=S$ — the whole stage.
• Each dot $=$ one outcome. Six dots because a die has six faces.

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Step 2 — What is an "event"? Circle the outcomes you care about

WHAT. Say we care about "rolling a $6$". Draw a loop around just the dot labelled $6$. That loop is the event $A$.

WHY. Real questions ("did I win?", "is it a six?") never ask about one raw outcome in isolation — they ask about a chosen group. Circling the dots is exactly how we turn a question into maths.

PICTURE. One blue loop hugs the dot $6$. The other five dots ($1$–$5$) sit outside that loop but still inside the box.

• $A$ $=$ the blue loop $=$ "a $6$ shows up".
• Inside $A$: one dot. Outside $A$ but inside $S$: five dots.

Recall Why "inside/outside" is the whole trick

Question ::: Every dot is either inside the loop or outside it — there is no third place to be. That single fact is the seed of the complement rule.

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Step 3 — Meet the complement: everything the loop left behind

WHAT. Shade every dot that is not inside the blue loop. Here that is $\{1,2,3,4,5\}$. That pink region is $A^{\prime }$.

WHY. We want a name for "the other stuff", because very often the other stuff is easier to count than the stuff we want. The complement is that name.

PICTURE. The box splits into two coloured zones with no overlap and no gap: blue $A$ on one side, pink $A^{\prime }$ on the other, together filling the box exactly.

• Blue zone $=A$ (the $6$).
• Pink zone $=A^{\prime }$ (everything else: $1$–$5$).
• $A^{\prime }=S-A$: read the arrow — take the whole box, remove the loop, keep the rest.

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Step 4 — The two magic properties, shown by the picture

WHAT. Look again at Step 3's picture and check two things: (1) is there any dot painted both blue and pink? No. (2) Is there any dot painted neither? No.

WHY. These two facts are what let us add and what let us reach "$1$" in the next steps. Without "no overlap" we would double-count; without "no gap" we would miss outcomes.

PICTURE. A zoom on the border between blue and pink: the line is clean — every dot is on exactly one side.

• $A\cap A^{\prime }=\varnothing$ — no dot is in both zones (no overlap).
• $A\cup A^{\prime }=S$ — the two zones together are the whole box (no gap).

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Step 5 — Add the probabilities: why we can just sum them

WHAT. Give each dot a weight — a slice of chance. The die is fair, so each of the six dots carries weight $\frac{1}{6}$. The chance of a zone is just the total weight of the dots inside it.

WHY this tool? We use the addition rule and not something fancier because Step 4 proved there is no overlap. When zones don't overlap, "either" can't double-count anything, so plain $+$ is exactly right. If they did overlap we would have to subtract the shared part — but they don't.

PICTURE. Blue weight (one slice $=\frac{1}{6}$) sits beside pink weight (five slices $=\frac{5}{6}$); the two bars butt together with no overlap, and their lengths just add.

• $P(A)=\frac{1}{6}$ — total blue weight (one dot).
• $P(A^{\prime })=\frac{5}{6}$ — total pink weight (five dots).
• $P(A\cup A^{\prime })$ — the two bars laid end to end, i.e. the sum.

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Step 6 — The whole box weighs exactly $1$

WHAT. Add up the weight of every dot in the box: $\frac{1}{6}\times 6=1$. The full box always weighs $1$.

WHY. "$1$" is our chalk mark for certain. Since the box $S$ contains every possible outcome, the die must land on one of them — so the box's total chance is full, i.e. $1$. This is not a calculation, it is the rule we agree to start from (an axiom).

PICTURE. The whole rectangle is filled edge to edge with a "$=1$" tag stamped across it — one full, complete bar.

• The full rectangle $=S$.
• Its total shaded length $=P(S)=1$ (completely full, no empty strip).

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Step 7 — Snap the three facts together to read off the rule

WHAT. We now own three true sentences from the pictures:

$$\underset{\text{Step 4: no gap}}{\underbrace{A\cup A^{\prime }=S}}\Rightarrow \underset{\text{Step 6: box weighs }1}{\underbrace{P(A\cup A^{\prime })=P(S)=1}}$$ $$\underset{\text{Step 5: no overlap, so add}}{\underbrace{P(A\cup A^{\prime })=P(A)+P(A^{\prime })}}$$

Set the two expressions for $P(A\cup A^{\prime })$ equal:

$$P(A)+P(A^{\prime })=1$$

WHY. Both lines describe the same thing — the total chance of "either $A$ or not-$A$", which is the whole box. Two names for one quantity must be equal. Now just slide $P(A)$ across:

$$\boxed{P(A^{\prime })=1-P(A)}$$

• $P(A^{\prime })$ — the pink-zone weight we want.
• $1$ — the full box (Step 6).
• $-P(A)$ — carve away the blue loop's weight (Step 5).

PICTURE. A full bar of length $1$ with the blue $P(A)$ chunk sliced off the left; what physically remains on the right is $P(A^{\prime })$.

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Step 8 — Edge cases: the certain event and the impossible event

WHAT. Push the loop to its two extremes.

• Loop swallows the whole box: $A=S$, so $P(A)=1$. Then the pink complement is nothing left: $A^{\prime }=\varnothing$ and $P(A^{\prime })=1-1=0$. ✔ A certain event has an impossible complement.
• Loop shrinks to nothing: $A=\varnothing$, so $P(A)=0$. Then the complement is the whole box: $A^{\prime }=S$ and $P(A^{\prime })=1-0=1$. ✔ An impossible event has a certain complement.

WHY. A good formula must survive its extremes. At both ends the rule gives an answer that lands neatly in the legal range $0\le P\le 1$ — it never spits out something below $0$ or above $1$.

PICTURE. Left panel: the loop fills the box (all blue, no pink). Right panel: the loop is a dot-sized nothing (all pink). The bar reading $1-P(A)$ moves smoothly between them.

• Left: $P(A)=1\Rightarrow P(A^{\prime })=0$ (nothing left over).
• Right: $P(A)=0\Rightarrow P(A^{\prime })=1$ (everything left over).
• In between: as blue grows, pink shrinks by exactly the same amount — they trade, always summing to $1$.

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The one-picture summary

One bar of total length $1$. Cut it once. The left piece is $P(A)$, the right piece is $P(A^{\prime })$, and the cut can slide anywhere from all-blue to all-pink — but the bar is always full. That single sliding cut is the complement rule.

Recall Feynman retelling — the whole walk in plain words

Everything that can happen lives in one box (Step 1). We circle the results we care about and call that circle $A$ (Step 2). Everything the circle missed we call $A^{\prime }$ — the leftovers (Step 3). Two facts are obvious from the drawing: nothing is in both the circle and the leftovers, and nothing is in neither — the circle and the leftovers perfectly tile the box (Step 4). Because they never overlap, their chances just add (Step 5). Because the box holds all possibilities, its total chance is a full $1$ (Step 6). Put those together: circle-chance plus leftover-chance equals the full box, which is $1$; so leftover-chance is just $1$ minus circle-chance — $P(A^{\prime })=1-P(A)$ (Step 7). Test the extremes: a sure thing leaves $0$ over, an impossible thing leaves the whole $1$ over (Step 8). And the picture that holds it all is one bar of length $1$ with a single sliding cut (summary). That's why "at least one" problems (1.3.12-At-least-one-problems) become easy — you measure the small leftover "none" piece and subtract.

Connections

• Parent: Complementary events — P(A') = 1 − P(A)
• 1.3.01-Sample-space-and-events — the box $S$ and the dots we circled here
• 1.3.03-Addition-rule-for-probabilities — the "no overlap → just add" step
• 1.3.09-Independent-events — complements combine with independence
• 1.3.12-At-least-one-problems — where this bar-cut trick pays off
• 1.4.05-Binomial-probability — $P(X\ge 1)=1-P(X=0)$ is this same cut

#flashcards/maths

What does the sample space $S$ represent in the bar picture? :: The full bar of total length 1 — every possible outcome, weighing 1 altogether.

Why can we write $P(A\cup A^{\prime })=P(A)+P(A^{\prime })$? :: Because $A$ and $A^{\prime }$ never overlap ($A\cap A^{\prime }=\varnothing$), so the addition rule applies with no double-counting.