IN TRAINING
The Case Files of
Probability
Every event leaves a clue. Every clue has a number attached to it. Your job, Rookie, is to learn how to read them. Eight case files stand between you and your detective's badge.
Sample Space & Events
Before a detective can solve anything, she must list every possible outcome. That list is called the sample space.
Sample Space (S) is the set of ALL possible outcomes of a random process. Think of it as the complete suspect list — nothing left out, nothing repeated.
Event is any subset of the sample space — one suspect, or a group of suspects who share something in common.
Event A ⊆ S
That's it. No calculation yet — just careful, complete listing. Most probability mistakes start with an incomplete sample space, so this is the detective's first and most important habit.
A single six-sided die is rolled. Write the sample space S, and the event A = "an even number is rolled."
The die can land on any of six faces, so:
A = {2, 4, 6}
A rookie mistake is forgetting an outcome (like leaving out 1 or 6). List them in order, every time.
Two coins are flipped. Write the sample space, and event B = "at least one head."
Each coin has 2 outcomes, and the coins are flipped together, so there are 2 × 2 = 4 total outcomes:
B = {HH, HT, TH}
Notice TT is the ONLY outcome missing from B — "at least one head" excludes just the all-tails case.
A card is drawn and only its suit is recorded. Write S, and event C = "the suit is red."
C = {♥, ♦}
Only hearts and diamonds are red suits, so C contains exactly those two outcomes.
Probability Axioms & the Complement Rule
Every good agency has rules. These three are the ones probability can never break.
Axiom 1: Every probability is between 0 and 1.
Axiom 2: The probability of the whole sample space is 1 — something in the suspect list must happen.
Complement Rule: "Not A" (written A') is everything in S that isn't in A. Since A and A' together make up the whole sample space:
The complement is the detective's favorite shortcut: sometimes it's far easier to find the probability of "it does NOT happen" and subtract from 1.
A fair die is rolled. P(rolling a 4) = 1/6. Find P(not rolling a 4).
Five of the six faces are "not a 4," which matches perfectly.
In a bag of marbles, P(red) = 0.35. What is the probability the marble drawn is NOT red?
A weather report says P(rain tomorrow) = 0.20. What's the probability it does NOT rain?
An 80% chance of a dry day — good news for any stakeout.
The Addition Rule
"Or" cases: how often does A happen, OR B happen, OR both?
Mutually Exclusive Events can never happen at the same time (no overlap). For these, just add:
Non-Mutually Exclusive Events CAN overlap. If you just add, you count the overlap twice — so subtract it back out once:
Picture two witness lists with some names appearing on both. Add both lists, then remove the duplicate names once.
Roll a die. A = "roll a 1 or 2," B = "roll a 5 or 6." Find P(A or B).
A and B share no outcomes, so they're mutually exclusive:
P(A ∪ B) = 2/6 + 2/6 = 4/6 = 2/3
A standard deck of 52 cards. A = "draw a heart" (13 cards), B = "draw a face card" (12 cards: J, Q, K in every suit). 3 cards are both a heart AND a face card. Find P(heart or face card).
= 22/52 = 11/26 ≈ 0.423
Without subtracting the overlap (the heart face cards: J♥, Q♥, K♥), we'd have double-counted those 3 cards.
Out of 30 students: 12 play soccer, 10 play basketball, and 5 play both. Find P(soccer or basketball).
= 17/30 ≈ 0.567
The Multiplication Rule & Independence
"And" cases: how often do A and B BOTH happen?
Independent Events: one outcome has zero effect on the other (a coin doesn't remember the last flip). For independent events, simply multiply:
General Multiplication Rule (works even when events are NOT independent):
Checking independence: A and B are independent only if P(A ∩ B) really does equal P(A) · P(B). If it doesn't match, the events are dependent.
A coin is flipped and a die is rolled. Find P(heads AND rolling a 6).
The coin and die don't influence each other — independent events:
Two cards are drawn from a 52-card deck WITHOUT replacement. Find P(both are aces).
After the first ace is removed, only 3 aces remain among 51 cards — the second draw depends on the first:
= 12/2652 = 1/221 ≈ 0.0045
Roll one die. A = "even number," B = "greater than 4." Is A independent of B?
A = {2,4,6} so P(A) = 1/2. B = {5,6} so P(B) = 1/3. A ∩ B = {6}, so P(A ∩ B) = 1/6.
This equals P(A ∩ B) = 1/6 ✓ Independent!
Conditional Probability
New evidence changes the odds. Conditional probability is how a detective updates her thinking once she knows something is already true.
P(A | B) reads as "the probability of A, GIVEN that B has already happened." Because B has already occurred, B becomes the new, smaller sample space:
Read the bar "|" as "given." Everything to the right of it has already happened — it's a fact, not a question.
A card is drawn from a 52-card deck. Given that it's a face card (12 cards), what's the probability it's a King (4 of the 12 face cards)?
Once we know it's a face card, we only care about those 12 cards — and 4 of them are Kings.
Two dice are rolled. Given that the first die shows a 3, find the probability the sum of both dice is 8.
Since the first die is fixed at 3, only the second die is still random (6 equally likely outcomes). Only a second roll of 5 gives a sum of 8:
In a class, P(likes math) = 0.4 and P(likes math AND likes coding) = 0.28. Find P(likes coding | likes math).
Among students who like math, 70% also like coding.
Two-Way Tables
A detective's favorite kind of evidence: a grid that organizes every suspect by two traits at once.
A two-way table sorts a group by two categories at once. Once it's built, every probability question becomes a fraction: (the cell or row/column you want) ÷ (the total that matches your condition).
Marginal probability uses a row/column total ÷ grand total. Joint probability uses one inside cell ÷ grand total. Conditional probability uses one inside cell ÷ its row or column total.
| Prefers Math | Prefers English | Total | |
|---|---|---|---|
| Boys | 40 | 60 | 100 |
| Girls | 50 | 50 | 100 |
| Total | 90 | 110 | 200 |
Q1 (Marginal): P(Boy)?
Q2 (Joint): P(Boy AND prefers Math)?
Q3 (Conditional): P(prefers Math | Boy)?
Given the student is a boy, we only look within the "Boys" row (100 students total):
Compare Q2 and Q3 closely — same numerator (40), but a different, smaller denominator, because "given Boy" shrinks our world down to just the Boys row.
Tree Diagrams
When a case unfolds in stages, a tree diagram maps every path a suspect could take.
Draw one branch per possible outcome at each stage. Label every branch with its probability. To find the probability of an entire path, multiply along the branches. To find the probability of several paths, add the path totals. All path totals in the whole tree must add up to 1 — that's how a detective double-checks her work.
A bag holds 5 red and 3 blue marbles. Two are drawn without replacement. Map out the tree and find P(exactly one red).
First draw: P(Red) = 5/8, P(Blue) = 3/8. Since there's no replacement, the second draw's pool changes:
If 1st = Blue → 5 red, 2 blue left (7 total)
P(RB) = 5/8 × 3/7 = 15/56
P(BR) = 3/8 × 5/7 = 15/56
P(BB) = 3/8 × 2/7 = 6/56
Check: 20/56 + 15/56 + 15/56 + 6/56 = 56/56 = 1 ✓
= 15/56 + 15/56 = 30/56 = 15/28 ≈ 0.536
Introduction to Bayes' Theorem
The Agency's most powerful tool: working BACKWARD from the evidence to the true cause.
So far, we've mostly asked "given the cause, what's the probability of this evidence?" Bayes' Theorem flips the question around: "given the evidence, what's the probability of each possible cause?"
The tricky part is always finding P(B) — the total probability of the evidence across every possible cause. A tree diagram from Case 1.7 is the perfect tool to compute it: add up every branch that ends in the evidence you observed.
A rare disease affects 1% of a population: P(Disease) = 0.01. A test correctly detects the disease 95% of the time (P(+|Disease) = 0.95), but also gives a false positive 5% of the time on healthy people (P(+|No Disease) = 0.05). If a random person tests positive, what's the probability they actually have the disease?
First, find P(+) — the total chance of testing positive, from BOTH groups (sick and healthy):
= (0.95)(0.01) + (0.05)(0.99)
= 0.0095 + 0.0495 = 0.059
Now apply Bayes' Theorem:
The twist every detective must learn: even with a "95% accurate" test, a positive result only means about a 16% chance of truly having the disease — because the disease is so rare that false positives from the huge healthy group pile up. Rarity matters as much as accuracy.
Factory A makes 60% of a company's widgets, with a 2% defect rate. Factory B makes the other 40%, with a 5% defect rate. A widget is picked at random and found defective. What's the probability it came from Factory A?
= 0.012 + 0.02 = 0.032
Even though Factory A makes most of the widgets, its low defect rate means it's actually the less likely source of a defective one — only 37.5%, compared to 62.5% from Factory B.
The Detective's Handbook
Every formula from Unit 1, filed and cross-referenced for your final review.
| Case | Concept | Formula |
|---|---|---|
| 1.1 | Sample Space & Events | S = all outcomes; Event A ⊆ S |
| 1.2 | Axioms & Complement | 0 ≤ P(A) ≤ 1; P(S)=1; P(A') = 1 − P(A) |
| 1.3 | Addition (mutually exclusive) | P(A ∪ B) = P(A) + P(B) |
| 1.3 | Addition (non-mutually exclusive) | P(A ∪ B) = P(A) + P(B) − P(A ∩ B) |
| 1.4 | Multiplication (independent) | P(A ∩ B) = P(A) · P(B) |
| 1.4 | Multiplication (general) | P(A ∩ B) = P(A) · P(B|A) |
| 1.5 | Conditional Probability | P(A|B) = P(A ∩ B) / P(B) |
| 1.6 | Two-Way Tables | Marginal = row or column total ÷ grand total; Joint = cell ÷ grand total; Conditional = cell ÷ row/column total |
| 1.7 | Tree Diagrams | Multiply along a branch; add across matching branches |
| 1.8 | Bayes' Theorem | P(A|B) = [P(B|A)·P(A)] / P(B) |
The Detective's Golden Rules:
1. List the full sample space before doing anything else.
2. "Or" usually means add (watch for double-counted overlap). "And" usually means multiply.
3. The word "given" means the sample space has shrunk — you're now only looking inside one row, column, or branch.
4. A tree diagram's paths must always sum to 1. If they don't, a branch is missing or a probability is wrong.
5. Bayes' Theorem is just conditional probability solved backward — from the effect back to the cause.
🏆 Case Closed, Detective.
You've reviewed all eight case files of Unit 1. Foundations of Probability are now part of your permanent record. Time to put the badge on and move to Unit 2.