Week 4 · Unit 4

The Case Files on Chance

Detective Bayes needs a sharp new partner. Every mystery in this unit is solved with one tool: probability. No tricky math brain required — just careful thinking, one clue at a time.

⚠️Field Warning: this unit has the most calculations of the semester. Go slow, re-read each case twice, and trust the process — every formula here is just common sense written in symbols.

🎲 🃏 🔍 📊

Basic Probability Rules

addition · complement

Conditional Probability

& independence

Random Variables

E(X) & Var(X)

Binomial Distribution

B(n, p)

Geometric Distribution

first success

★

Case Summary

full cheat sheet

CASE №1

The Loaded Dice Rumor

A rumor says the dice at the school festival booth are unfair. Before accusing anyone, Detective Bayes needs the basics: what probability actually means, and two simple rules that solve almost every "what are the odds" question.

🔎 The Core Idea

Probability of an event A is just a fraction: how many outcomes make A happen, divided by all possible outcomes (when every outcome is equally likely).

P(A) = n(A) / n(S) , 0 ≤ P(A) ≤ 1

S is the sample space — every possible result. n(S) counts all of them.

📏 Two Rules That Solve Everything

Rule 1Complement Rule — "everything except A" always equals 1 minus A.

P(A′) = 1 − P(A)

Rule 2Addition Rule — for "A or B", don't double-count the overlap.

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

If A and B can never happen together (mutually exclusive), then P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B).
"At least one" questions are almost always faster solved using the complement rule.

Example 1 — The Fair Die Check try it, then flip

A fair six-sided die is rolled once. Find P(the result is even or greater than 4).

SOLVED

Let A = {2,4,6} (even), B = {5,6} (greater than 4).

P(A) = 3/6, P(B) = 2/6.

Overlap A ∩ B = {6}, so P(A ∩ B) = 1/6.

P(A ∪ B) = 3/6 + 2/6 − 1/6 = 4/6 = 2/3.

Answer: 2/3

Example 2 — At Least One Prize try it, then flip

A booth has 10 tickets: 3 are winning tickets. You draw 2 tickets without replacement. Find P(at least one winning ticket).

SOLVED

"At least one" is the complement of "none win."

P(no winners) = (7/10) × (6/9) = 42/90 = 7/15.

P(at least one) = 1 − 7/15 = 8/15.

Answer: 8/15

CASE №2

The Locked Locker Clue

A clue was found — but only if a certain condition is true. Detective Bayes explains that probability can change once you already know something happened. That's conditional probability.

🔎 Conditional Probability

P(A|B) means "the probability of A, given that B has already happened." You shrink the sample space down to just B, then ask how much of that is also A.

P(A|B) = P(A ∩ B) / P(B) (P(B) ≠ 0)

Rearranging gives the multiplication rule, useful for "and" questions in sequence:

P(A ∩ B) = P(A) × P(B|A)

🤝 Independence

Two events are independent when knowing one happened tells you nothing new about the other — the condition doesn't change anything.

A, B independent ⟺ P(A ∩ B) = P(A) × P(B)

Equivalently: P(A|B) = P(A), and P(B|A) = P(B).
Drawing with replacement → independent. Drawing without replacement → dependent (the sample space shrinks).

Example 1 — Two Dice, One Clue try it, then flip

Two fair dice are rolled. Given that the sum is 8, find the probability that at least one die shows a 6.

SOLVED

B = "sum is 8": outcomes (2,6),(3,5),(4,4),(5,3),(6,2) → n(B) = 5.

A ∩ B = "sum is 8 AND at least one 6": (2,6),(6,2) → n(A ∩ B) = 2.

P(A|B) = n(A∩B)/n(B) = 2/5.

Answer: 2/5

Example 2 — Cards With and Without Replacement try it, then flip

A bag has 4 red and 6 blue balls. Two balls are drawn. Find P(both red) (a) with replacement, (b) without replacement.

SOLVED

(a) With replacement (independent): P = (4/10) × (4/10) = 16/100 = 4/25.

(b) Without replacement (dependent): P = (4/10) × (3/9) = 12/90 = 2/15.

Notice (b) is smaller — after removing one red ball, red balls become relatively scarcer.

Answers: (a) 4/25 (b) 2/15

CASE №3

The Fortune Teller's Table

A carnival "fortune wheel" claims to be generous. Detective Bayes builds a table of every possible outcome and its probability — a probability distribution — to reveal the wheel's true average payout.

🔎 Discrete Random Variable

A random variable X assigns a number to each outcome (e.g. "number of heads," "points won"). Its probability distribution lists every value of X with its probability, and all probabilities must add to 1.

X	x₁	x₂	⋯	xₙ
P(X)	p₁	p₂	⋯	pₙ

p₁ + p₂ + ⋯ + pₙ = 1

📐 Expected Value & Variance

Expected value E(X) is the long-run average — a weighted average of every value by its probability.

E(X) = Σ x·P(x)

Variance Var(X) measures spread — how far values wander from the mean.

Var(X) = E(X²) − [E(X)]²

where E(X²) = Σ x²·P(x). The standard deviation is just σ(X) = √Var(X).

Example 1 — Two Coin Flips try it, then flip

Let X = number of heads in 2 fair coin flips. Build the distribution and find E(X) and Var(X).

SOLVED

X	0	1	2
P(X)	1/4	2/4	1/4

E(X) = 0(1/4) + 1(2/4) + 2(1/4) = 4/4 = 1.

E(X²) = 0²(1/4) + 1²(2/4) + 2²(1/4) = 0 + 2/4 + 4/4 = 6/4 = 1.5.

Var(X) = E(X²) − [E(X)]² = 1.5 − 1 = 0.5.

Answers: E(X)=1, Var(X)=0.5

Example 2 — The Fortune Wheel Payout try it, then flip

A game costs 1,000 won to play. You win 5,000 won with probability 0.1, 1,000 won (a refund) with probability 0.3, and nothing otherwise. Find the expected net profit.

SOLVED

Net profit X (winnings − 1,000 cost): 4,000 with p=0.1; 0 with p=0.3; −1,000 with p=0.6.

E(X) = 4000(0.1) + 0(0.3) + (−1000)(0.6) = 400 + 0 − 600 = −200.

On average, a player loses 200 won per play — the booth has the edge.

Answer: E(X) = −200 won

CASE №4

The Repeated Suspects Pattern

The same kind of "trial" repeats over and over — a coin flipped 10 times, a shot attempted 5 times — always with just two outcomes: success or failure. This repeating pattern has a name: the binomial distribution.

🔎 Bernoulli Trial → Binomial

A Bernoulli trial has exactly two outcomes: success (probability p) or failure (probability q = 1 − p). Repeat it n independent times and count the number of successes X — that's a binomial random variable, written X ~ B(n, p).

P(X = k) = C(n, k) · p^k · q^n−k

C(n, k) = n! / [k!(n−k)!] counts how many ways to arrange k successes among n trials.

📐 Shortcut Formulas

No need to sum every term to find the average or spread — binomial has ready-made shortcuts:

E(X) = np · Var(X) = npq

Example 1 — Guessing on a Quiz try it, then flip

A quiz has 5 true/false questions. A student guesses randomly on every question. Find P(exactly 3 correct).

SOLVED

X ~ B(n=5, p=0.5). Want P(X=3).

C(5,3) = 10.

P(X=3) = 10 × (0.5)³ × (0.5)² = 10 × (1/32) = 10/32 = 5/16.

Answer: 5/16

Example 2 — Free Throw Average try it, then flip

A player makes free throws with probability 0.8. She attempts 10 shots. Find E(X) and Var(X) for the number made.

SOLVED

X ~ B(n=10, p=0.8), so q = 0.2.

E(X) = np = 10 × 0.8 = 8.

Var(X) = npq = 10 × 0.8 × 0.2 = 1.6.

Answers: E(X)=8, Var(X)=1.6

CASE №5

The First Break in the Case

Sometimes what matters isn't "how many successes," but "how long until the very first one." That waiting game is the geometric distribution.

🔎 Waiting for the First Success

Repeat independent trials (success probability p) until the first success appears. Let X = the trial number on which that first success happens.

P(X = k) = q^k−1 · p (k = 1, 2, 3, ...)

This means: (k−1) failures in a row, then finally one success.

📐 Expected Waiting Time

The average number of trials needed for the first success is simply the reciprocal of p — the rarer the success, the longer the wait.

E(X) = 1 / p

Example 1 — Waiting for a Six try it, then flip

A fair die is rolled repeatedly. Find the probability that the first 6 appears exactly on the 4th roll.

SOLVED

p = 1/6 (rolling a six), q = 5/6, k = 4.

P(X=4) = q³ · p = (5/6)³ × (1/6) = (125/216) × (1/6) = 125/1296.

Answer: 125/1296 ≈ 0.096

Example 2 — Average Attempts to Pass try it, then flip

A driving test has a pass probability of 0.25 per attempt (independent attempts). On average, how many attempts are needed to pass for the first time?

SOLVED

p = 0.25 = 1/4.

E(X) = 1/p = 1 ÷ (1/4) = 4.

On average, expect it to take 4 attempts to pass for the first time.

Answer: 4 attempts

FINAL REPORT

Detective Bayes' Case Summary

Every formula from every case, filed in one place. Read this the night before a test — it's your whole unit in one page.

№1 · Basic Probability Rules

P(A) = n(A)/n(S) P(A′) = 1 − P(A) P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

№2 · Conditional Probability & Independence

P(A|B) = P(A ∩ B) / P(B) P(A ∩ B) = P(A) × P(B|A) Independent ⟺ P(A ∩ B) = P(A)P(B)

№3 · Random Variables

E(X) = Σ x·P(x) Var(X) = E(X²) − [E(X)]² σ(X) = √Var(X)

№4 · Binomial Distribution B(n, p)

P(X=k) = C(n,k)·p^k·q^n−k E(X) = np · Var(X) = npq

№5 · Geometric Distribution

P(X=k) = q^k−1·p E(X) = 1/p

🧭 Detective's Golden Rule

See "or" → think addition rule. See "and," "given," "already know" → think conditional/multiplication rule. See "exactly k out of n" → think binomial. See "until the first" → think geometric. Always check what's independent vs. dependent first — it decides which formula fits.