LESSON PLAN

Probability — Two Independent Events

Apothem Team

Grade 8 · Data & Probability

LESSON AT A GLANCE

Warm-up

5 min

Explore

15 min

Formalize

10 min

Practice

12 min

Exit ticket

3 min

Warm-up

The two-event wager: "Flip a coin AND roll a die. I'll pay 3:1 if you get heads-and-six." Take votes on fairness, then count: $\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$ — the fair payout is 11:1. The house was robbing them.

Grade 8 probability is the multiplication rule for independent events, wielded with tree diagrams and tables — and used to audit games, apps, and claims.

Explore

Tree-diagram atelier: (1) coin-then-die drawn in full — 12 leaves, each probability a product along its branch; (2) the two-child family (GG, GB, BG, BB — and why "at least one girl" is $\frac{3}{4}$ ); (3) the three-shot free-throw sequence for a 60% shooter: leaves carry $0.6^3$ , $0.6^2 \times 0.4$ , etc., and the class discovers the leaves' probabilities SUM TO 1 — the tree's built-in audit.

Then dependence sabotage: draw two names from a hat WITHOUT replacement and watch the second branch's probabilities shift ( $\frac{1}{23}$ became $\frac{1}{22}$ or 0). The multiplication rule survives, but the second factor must be the UPDATED probability — independence was a special case all along.

Formalize

Formalize the general and special multiplication rules:

$P(A \text{ then } B) = P(A) \times P(B \mid A) \qquad \text{independent} \Rightarrow P(B \mid A) = P(B)$

Independence is a claim about the WORLD, not a default: coins and dice don't communicate (independent); cards dealt from one deck do (dependent). The question "does the first event change the second's setup?" decides which factor enters the product.

Practice

Practice: two tree diagrams (one independent, one without-replacement); three multiplication-rule computations with independence justified in a sentence; one "at least one" via the complement ( $1 - P(\text{none})$ ); one game-fairness audit with payout recommendation.

Exit ticket: bag has 3 red, 2 blue. Draw two without replacement: $P(\text{both red})$ ? ( $\frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$ .)

Exit ticket

Exit ticket: bag has 3 red, 2 blue. Draw two without replacement: $P(\text{both red})$ ? ( $\frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$ .)

TIP The complement trick for "at least one" (

1 - P(\text{none})

) feels like cheating and is actually the professional route — one clean product instead of a forest of branches. Teach it as the lazy person's rigour.

WORKED EXAMPLES

Example 1 — The password audit: independence as security

The setup: a 4-digit PIN, digits 0–9, chosen randomly. A thief gets ONE guess.

Step 1: Each digit guess: $\frac{1}{10}$ , independent (the pad doesn't react).

Step 2: Whole PIN: $\left(\frac{1}{10}\right)^4 = \frac{1}{10{,}000}$ .

Step 3: The "three tries before lockout" upgrade: $P(\text{in within 3}) \approx \frac{3}{10{,}000}$ (the tries are on different guesses — addition of disjoint tiny chances is honest here).

Step 4: Now the human factor: if the thief knows the PIN is a birth year (19XX/20XX), the space collapses to ~120 candidates — $\frac{3}{120} = \frac{1}{40}$ . Randomness was doing all the work; predictability gave it away. The multiplication rule measures exactly how much security each independent random digit buys — and how much a pattern refunds to the attacker.

Example 2 — At least one rainy day: the complement shortcut

The forecast: each of the three camping days has an independent 30% rain chance. $P(\text{at least one wet day})$ ?

Step 1: The direct route's price: wet-day patterns number 7 of the 8 branches — a bookkeeping swamp.

Step 2: The complement: "at least one wet" fails only if ALL THREE are dry: $P(\text{dry day}) = 0.7$ , so $P(\text{all dry}) = 0.7^3 = 0.343$ .

Step 3: Flip: $P(\text{at least one wet}) = 1 - 0.343 = 0.657$ — about 66%.

Step 4: Calibrate the surprise: each day was only 30%, yet the trip is two-to-one to catch rain. Small per-trial risks COMPOUND across trials — the same mathematics as defect rates, side effects, and why backups exist. "Unlikely each time" and "unlikely ever" are different claims, separated by an exponent.

Example 3 — Without replacement, felt in the fingers: the sock drawer

The drawer: 6 black socks, 4 white, grabbed blind, two pulls, no returns. $P(\text{matching pair})$ ?

Step 1: Split by first pull. Black first: $\frac{6}{10}$ ; then black again from the depleted drawer: $\frac{5}{9}$ → $\frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$ .

Step 2: White pair: $\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$ .

Step 3: Matching either way (disjoint cases — now addition is legal): $\frac{1}{3} + \frac{2}{15} = \frac{7}{15} \approx 0.47$ .

Step 4: The audit: mismatch probability should complete it: $1 - \frac{7}{15} = \frac{8}{15}$ ; direct check: $\frac{6}{10}\times\frac{4}{9} + \frac{4}{10}\times\frac{6}{9} = \frac{48}{90} = \frac{8}{15}$ ✓ the tree balances.

Step 5: Note where each rule fired: multiplication along branches (with UPDATED second factors), addition across disjoint branches, complement as the audit. One drawer, the whole toolkit.

MATERIALS

•Coins, dice, hats with names

•Tree diagram templates

•Card decks

•Game audit sheets

•Practice set (PDF)

WATCH FOR

!Probabilities added where the story says AND. Addition is for either-or (mutually exclusive); multiplication chains sequential/joint events.

!Without-replacement treated as independent — the hat demo makes the shift visible.

!Tree leaf-probabilities not summing to 1 and nobody noticing. The audit habit: total every tree.