Mutually exclusive events — addition rule: Visual walkthrough

We build everything on one honest idea: probability is fair counting of dots.

Step 1 — Draw the whole world as dots

WHAT. Imagine every possible outcome of an experiment as a single dot inside a big rectangle. The rectangle is the sample space, written $S$ . If we roll one fair die, there are six dots: the faces $1, 2, 3, 4, 5, 6$ .

WHY. Before we can talk about "the chance of this or that", we need a place where everything that could happen lives. Counting is only fair when we can see the full pool we're counting from — that pool is $S$ . See Probability — basic definitions & sample space.

PICTURE. Six dots, each equally likely, floating in the box $S$ .

Figure — Mutually exclusive events — addition rule

Step 2 — An event is a fenced-off group of dots

WHAT. An event is just some of those dots, circled off. Let $A =$ "roll an even number" $= {2, 4, 6}$ . Draw a blue loop around exactly those three dots.

WHY. We rarely care about a single outcome; we care about a question like "is it even?". That question is answered by a subset of dots — the ones that make the answer "yes". Circling them turns a vague question into countable dots.

PICTURE. The blue loop $A$ captures $2, 4, 6$ ; the dots $1, 3, 5$ stay outside.

Step 3 — Two events, and the overlap you can literally see

WHAT. Add a second event $B =$ "roll a number $> 3$ " $= {4, 5, 6}$ , drawn as an orange loop. Now look where blue and orange cross: the dots $4$ and $6$ sit inside both loops. That shared region is the intersection $A \cap B = {4, 6}$ .

WHY. We're heading toward "chance of $A$ or $B$ ". The whole difficulty of that question lives in this crossing region — the dots that belong to two clubs at once. Naming it now ( $A \cap B$ ) lets us handle it later. This is a Venn diagram.

PICTURE. Blue loop and orange loop overlapping; the lens-shaped middle (green shading) holds $4$ and $6$ .

Step 4 — Why naive adding over-counts (the double-count made visible)

WHAT. Try to count the union by simply adding: $n (A) + n (B) = 3 + 3 = 6$ . But the true union ${2, 4, 5, 6}$ has only 4 dots. We got 6. The extra 2 are exactly $4$ and $6$ .

WHY. When you count "dots in the blue loop" you already counted $4$ and $6$ . When you then count "dots in the orange loop" you count $4$ and $6$ again. Each overlap dot got tallied twice. That is where the error is born — and you can point at the two guilty dots.

PICTURE. The two shared dots lit red, each shown being tallied once for blue and once for orange — a "counted twice" tag on each.

Step 5 — Subtract the overlap once: inclusion–exclusion

WHAT. We over-added the shared dots by exactly one copy each. So remove one copy of the overlap: $n (A \cup B) = n (A) + n (B) - n (A \cap B)$ Check: $3 + 3 - 2 = 4$ . ✓ That matches the four dots we can see.

WHY. We included every dot when we added the loops, but included the overlap twice; so we exclude one copy of it. Add everything, take back what you double-added — inclusion–exclusion. See General addition rule & inclusion–exclusion.

PICTURE. Same Venn, with the green lens being "lifted out once" — an arrow pulling one copy of ${4, 6}$ away, leaving a clean count of 4.

Step 6 — The special case: pull the loops apart

WHAT. Now change the events so they share no dot. Take $A = {2}$ (blue) and $B = {5}$ (orange) on the die. The loops no longer touch: $A \cap B = \emptyset$ (the empty set, no dots).

WHY. If there is nothing in the crossing region, there is nothing that got counted twice — so there is nothing to subtract. The correction term isn't ignored; it is genuinely $0$ because $n (A \cap B) = 0$ .

PICTURE. Two separated loops with a gap between them; the crossing region is empty, marked $\emptyset$ .

Step 7 — Degenerate & edge cases (never hit a scene we didn't show)

WHAT & WHY & PICTURE — four boundary situations, each with its Venn look:

One loop swallows the other ( $A \subseteq B$ , e.g. $A = {4}$ , $B = {4, 5, 6}$ ). Overlap $= A$ itself, so $P (A \cup B) = P (B)$ . Picture: small blue loop entirely inside orange.
Loops are identical ( $A = B$ ). Then $P (A \cup B) = P (A)$ — adding twice and subtracting once leaves one copy.
Empty event ( $A = \emptyset$ , $P (A) = 0$ ). Adding an event with no dots changes nothing: $P (\emptyset \cup B) = P (B)$ .
Loops fill the box — complement case: $A$ and $A^{'}$ share nothing and together cover all of $S$ . Then $P (A) + P (A^{'}) = \frac{n ( S )}{n ( S )} = 1$ , giving $P (A^{'}) = 1 - P (A)$ .

Recall Why "mutually exclusive" is the

opposite of "independent" Question ::: If $A$ and $B$ are mutually exclusive with $P (A), P (B) > 0$ , are they independent? Answer ::: No. Exclusive means "if $A$ happens, $B$ cannot" — knowing $A$ tells you everything about $B$ (it's impossible). That's strong dependence. Independence (Independent events — multiplication rule) means $A$ tells you nothing about $B$ . Exclusive lives in the world of adding unions; independent lives in multiplying intersections.

Step 8 — Sanity check: probabilities can't exceed 1

WHAT. Suppose $P (A) = 0.6$ , $P (B) = 0.7$ . Could they be mutually exclusive?

WHY. If they were, $P (A \cup B) = 0.6 + 0.7 = 1.3$ — bigger than the whole box, which is impossible ( $P \leq 1$ ). The picture forces the loops to overlap: their shared share is at least $0.6 + 0.7 - 1 = 0.3$ .

PICTURE. A box that is only "size 1" with two loops so large they must intersect, the forced overlap $\geq 0.3$ shaded red.

The one-picture summary

Everything above, compressed: two loops, the overlap subtracted once (general rule), and the same picture with loops pulled apart so the overlap vanishes (mutually exclusive rule).

Recall Feynman retelling — say it to a 12-year-old

Picture a table covered in dots — every possible thing that could happen. Loop a blue lasso around the "yes" dots for question A, and an orange lasso around the "yes" dots for question B. Want the chance of "A or B"? Count blue dots, count orange dots, add them. But any dot caught by both lassos just got counted twice — so cross it out once. That single "cross out the overlap once" is the whole general rule: $P (A) + P (B) - P (A \cap B)$ . Now slide the two lassos apart until they don't touch — there's no dot in both, nothing to cross out, so you just add. Those pulled-apart lassos are what "mutually exclusive" means. Same rule, one with a zero. And if the lassos ever grow so big their shares add past 1, they cannot be pulled apart — they're forced to overlap.

Active-recall

Union in dot-language

dots in either loop;

n (A \cup B)

Where does the double-count come from?

overlap dots (

A \cap B

) get tallied once for each loop.

Why subtract

P (A \cap B)

exactly once?

we added it twice, so we remove one extra copy.

Mutually exclusive in the picture

loops don't touch;

A \cap B = \emptyset

, so the subtracted term is

0

Can

P (A) = 0.6, P (B) = 0.7

be exclusive?

No —

1.3 > 1

; they must overlap by

\geq 0.3

Complement from the picture

A

and

A^{'}

share nothing and fill

S

, so

P (A^{'}) = 1 - P (A)

Connections

Mutually exclusive events — addition rule
Probability — basic definitions & sample space
General addition rule & inclusion–exclusion
Venn diagrams in probability
Independent events — multiplication rule
Complementary events