Independent events — multiplication rule: Worked examples

The scenario matrix

Every problem in this topic lands in exactly one of these cells. The right move differs cell to cell.

Cell	What makes it that cell	Right move	Worked in
A. Genuinely independent	Physically separate trials; one cannot touch the other	Multiply $P (A) P (B)$	Ex 1
B. With replacement	The setup is reset between draws	Multiply — independence restored	Ex 2
C. Without replacement	First event changes the setup	Use $P (A) P (B ∣ A)$ , not a product	Ex 3
D. "At least one" (complement)	Many trials, want $\geq 1$ success/failure	$1 - \prod (1 - p_{i})$	Ex 4
E. Degenerate: a certain / impossible event	Some $P = 1$ or $P = 0$	Multiply — but watch what it forces	Ex 5
F. Mutually exclusive trap	Events can't co-occur ( $P (A \cap B) = 0$ )	Never multiply — they're dependent	Ex 6
G. Real-world word problem	Independence hidden in the story	Test independence, then multiply	Ex 7
H. Exam twist: solve backwards	Given the joint answer, find a missing $P$	Rearrange the rule	Ex 8
I. Limiting behaviour	"How many trials to be almost sure?"	Push $1 - (1 - p)^{n} \to 1$	Ex 9

The picture below is the whole page on one board — a decision tree that routes any "AND" problem into its cell. Every example that follows is just one leaf of this tree, so refer back here whenever you feel lost.

Figure — Independent events — multiplication rule

Cell A — Genuinely independent

The grid below is that verification: 12 equal cells, and the multiply just picks the single pale-yellow one. Count the highlighted cell — that ratio $1/12$ is the answer, seen not computed.

Cell B — With replacement (independence restored)

Cell C — Without replacement (the trap)

The two side-by-side trees below make the contrast unmissable: the blue (replaced) tree keeps $4/10$ on its second branch, the pink (not replaced) tree drops to $3/9$ . Same first branch, different second branch — that single changed number is the entire difference between $0.16$ and $0.133$ .

Cell C⁺ — More than two events: the chain rule

Cell D — "At least one" via the complement

Cell E — Degenerate: a certain or impossible event

Worked example Ex 5 · Multiplying by

$1$ and by $0$ A drawer light turns on with probability $1$ (it's certain) whenever you open the drawer, and independently a bulb inside is already dead with probability $0$ (brand-new, never dead). Let $A = light on$ ( $P (A) = 1$ ), $B = bulb dead$ ( $P (B) = 0$ ). Find $P (A \cap B)$ .

Forecast: if one event is impossible, what must "both happen" be?

$P (A) = 1, P (B) = 0$ . Why this step? Given — these are the degenerate endpoints of probability.
Here we cannot lean on $P (A ∣ B) = P (A)$ , because $P (B) = 0$ makes the conditional $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ a $\frac{0}{0}$ — literally undefined. Why this step? The bar-formula needs a positive denominator; with an impossible condition there is no "world where $B$ happened" to shrink into.
So argue directly instead: since $A \cap B \subseteq B$ , we have $0 \leq P (A \cap B) \leq P (B) = 0$ , forcing $P (A \cap B) = 0$ . And the product $P (A) P (B) = 1 \times 0 = 0$ agrees — so the multiplication rule still holds, even though the conditional route is unavailable. Why this step? We prove the answer with monotonicity, not with an undefined quantity.

Verify: "both happen" cannot exceed the chance of the impossible part, so $P (A \cap B) = 0$ . ✓ Sanity check at the other endpoint: a certain event $A$ ( $P (A) = 1$ ) satisfies $P (A \cap C) = 1 \cdot P (C) = P (C)$ for any independent $C$ — here $P (C) > 0$ keeps every conditional well-defined, and certainty adds no constraint, exactly right.

Cell F — Mutually exclusive is NOT independent

Worked example Ex 6 · The classic confusion, quantified

One roll of a fair die. Let $A = roll a 2$ and $B = roll a 5$ . Is $P (A \cap B) = P (A) P (B)$ ?

Forecast: the words "separate outcomes" tempt you toward independence — but check it.

$P (A) = \frac{1}{6}, P (B) = \frac{1}{6}$ . Why this step? One face each.
$A$ and $B$ cannot both happen on one roll → $P (A \cap B) = 0$ . Why this step? A single die shows one face; $2$ and $5$ are Mutually Exclusive Events.
Test the rule: $P (A) P (B) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36} \neq = 0 = P (A \cap B) .$ Why this step? The product rule fails → these events are dependent, not independent.
Intuition: knowing you rolled a $2$ makes a $5$ impossible — that's information transfer, the opposite of independence.

Verify: $\frac{1}{36} \neq = 0$ , so the equality that defines independence is violated. ✓ Exclusive (both possible) $\Rightarrow$ dependent — always.

Cell G — Real-world word problem (independence hidden)

Worked example Ex 7 · Two commuters, one storm

On any morning, Asha catches her bus with probability $0.8$ and, on a separate route, Ben catches his with probability $0.75$ . Find $P (both catch their bus)$ — then re-read for the trap.

Forecast: does the multiply apply, or is there a hidden link?

Test independence. Separate routes, separate buses → normally one doesn't affect the other → treat as independent. Why this step? In word problems you must hunt for a shared cause (a storm, a shared bus) before multiplying.
$P (both) = 0.8 \times 0.75 = 0.6.$ Why this step? Independent AND → product.
The twist: if a single storm could delay both buses, the misses would be correlated and the product would be wrong. The problem states separate routes with no common cause → the assumption stands.

Verify: $0.6$ lies between $0$ and the "surely both" cap $min (0.8, 0.75) = 0.75$ , and equals $0.8 \cdot 0.75$ . Units are pure probabilities (dimensionless), $0 \leq 0.6 \leq 1$ . ✓

Cell H — Exam twist: solve the rule backwards

Cell I — Limiting behaviour: "how many until almost sure?"

Worked example Ex 9 · Keep rolling for a six

Each roll of a fair die shows a $6$ with probability $p = \frac{1}{6}$ , rolls independent. How many rolls $n$ make $P (at least one 6) \geq 0.9$ ? What happens as $n \to \infty$ ?

Forecast: guess the number of rolls before computing.

$P (no 6 in n rolls) = (\frac{5}{6})^{n}$ . Why this step? Independent failures multiply — the Binomial Distribution "all-fail" corner.
$P (at least one 6) = 1 - (\frac{5}{6})^{n}$ . Why this step? Complement Rule again — "at least one" is "not none."
Need $1 - (\frac{5}{6})^{n} \geq 0.9 ⟺ (\frac{5}{6})^{n} \leq 0.1$ . Take logs: $n \geq \frac{ln 0.1}{ln ( 5/6 )} \approx \frac{- 2.3026}{- 0.1823} \approx 12.63$ , so $n = 13$ . Why this step? The log turns the repeated product into a solvable line.
Limit: as $n \to \infty$ , $(\frac{5}{6})^{n} \to 0$ , so $P (at least one 6) \to 1$ . Why this step? A base below $1$ raised to a growing power vanishes — you become almost certain eventually, but never exactly $1$ in finite rolls.

Verify: $1 - (5/6)^{12} \approx 0.888 < 0.9$ but $1 - (5/6)^{13} \approx 0.907 \geq 0.9$ , so $13$ is the first roll count that works. ✓

The curve below shows this climb: read where the blue staircase first pokes above the pink $0.9$ line (that's $n = 13$ ), and watch it flatten toward the yellow ceiling at $1$ but never touch it.

Recall Which cell am I in? (quick triage)

Did the first event change the setup for the second? ::: If yes → Cell C (use $P (B ∣ A)$ ); if no → multiply. The problem says "at least one" — what's the move? ::: Complement: $1 - \prod (1 - p_{i})$ (Cells D, I). Two outcomes of one single trial can't co-occur — independent? ::: No — mutually exclusive with positive prob ⇒ dependent (Cell F). Given $P (A \cap B)$ and $P (A)$ , both independent, find $P (B)$ ? ::: Divide: $P (B) = P (A \cap B) / P (A)$ (Cell H). Three dependent stages — what rule? ::: Chain rule $P (A) P (B ∣ A) P (C ∣ A \cap B)$ (Cell C⁺).

Connections

General Multiplication Rule — the Cell C / chain-rule engine: $P (A \cap B) = P (A) P (B ∣ A)$ .
Conditional Probability — supplies $P (B ∣ A)$ and the meaning of the " $∣$ " bar used all over this page.
Complement Rule — powers every "at least one" cell (D, I).
Mutually Exclusive Events — the Cell F contrast to independence.
Binomial Distribution — repeated independent trials, the natural home of Cell I.
Bayes' Theorem — the deeper reversal engine behind Cell H.