Lesson 5: Probability Rules and Venn DiagramsAP

Probability is the mathematics of "what proportion of the time, in the long run?" A random phenomenon has outcomes we cannot predict in any single trial, but which settle into a predictable pattern over many repetitions — the law of large numbers. CED Unit 4 builds probability from rules, not intuition, because intuition fails precisely where the exam probes: overlapping events, "at least one" problems, and invented independence.

The basic rules

Every probability satisfies 0 ≤ P(A) ≤ 1, and the probabilities of all outcomes in the sample space sum to 1. Two workhorse rules follow.

• Complement rule: P(not A) = 1 − P(A). Whenever a problem says "at least one", compute the probability of *none* and subtract from 1.
• Addition rule (general): P(A or B) = P(A) + P(B) − P(A and B). The subtraction matters whenever the events can happen together. If they cannot — mutually exclusive (disjoint) events — the overlap term is zero and the rule simplifies to P(A) + P(B).
• Multiplication rule for independent events: if A and B are independent (knowing one occurred does not change the probability of the other), then P(A and B) = P(A) × P(B).

Venn diagrams make overlaps visible: draw the overlap once, and every region is counted exactly once — the entire addition rule is the picture.

Worked example 1. From a standard 52-card deck, one card is drawn at random. What is P(heart or face card)? There are 13 hearts and 12 face cards, but 3 cards (jack, queen, king of hearts) are both. Adding blindly double-counts them: P = 13/52 + 12/52 − 3/52 = 22/52 ≈ 0.423. Skip the subtraction and you claim 25/52 — the exact error the answer choices will offer you.

Worked example 2 (Venn). Of 300 students, 180 play a sport, 90 play an instrument, and 60 do both. The Venn regions are: both = 60; sport only = 180 − 60 = 120; instrument only = 90 − 60 = 30; neither = 300 − (120 + 60 + 30) = 90. So P(sport or instrument) = 210/300 = 0.70, and P(neither activity) = 90/300 = 0.30. Notice the check: the four regions must sum to 300. If yours sum to 330, you have added the "both" group twice — the single most common Venn error.

Simulation: probability by experiment

Some probabilities cannot be computed directly, so we simulate. A valid simulation description has four parts: (1) state how the random process is modelled (for example, random digits 0–9, where 0–6 represent a made free throw for a 70% shooter and 7–9 a miss); (2) define a trial precisely (two digits = one trip to the line); (3) repeat many times (at least 20–30 trials for a rough estimate, hundreds for stability); (4) compute the relative frequency of the outcome of interest. The FRQ scoring for simulations is mechanical: model, trial, repetitions, answer — four components, four scoring opportunities.

Real-world application

Insurers price risk with exactly these rules. The probability that a driver has *at least one* claim in five years is computed via the complement of "no claims"; weather services chain daily rain probabilities the same way; quality engineers multiply independent stage reliabilities with the multiplication rule. When a headline says a screening programme "doubles your chance of detection", it is leaning on probabilities the public rarely checks against base rates — Lesson 6 will arm you for that.

Common mistakes

• Adding overlapping probabilities without subtracting the overlap. If both events can occur together, P(A or B) ≠ P(A) + P(B).
• Translating "or" as "and" (and vice versa). In probability, "or" means union (either, or both); "and" means intersection (both).
• Assuming independence because it would be convenient. Independence must come from the design (physically separate trials) or be checked numerically — never asserted.
• Answering the "at least one" question directly with a long addition instead of the one-line complement.

AP Exam Experience

Question 1. Events A and B satisfy P(A) = 0.5, P(B) = 0.4 and P(A and B) = 0.15. Then P(A or B) is:

• (A) 0.55
• (B) 0.65
• (C) 0.75
• (D) 0.90
• (E) 0.20

Answer: (C). P(A or B) = 0.5 + 0.4 − 0.15 = 0.75. Choice (D) is the answer for students who forget the events overlap; (A) subtracts the overlap twice.

Question 2. A machine produces a defective part with probability 0.02, independently each time. In a run of 5 parts, the probability of at least one defective part is closest to:

• (A) 0.02
• (B) 0.10
• (C) 0.90
• (D) 0.096
• (E) 0.104

Answer: (D). P(at least one) = 1 − P(none) = 1 − (0.98)^5 = 1 − 0.9039 ≈ 0.096. Choice (B) multiplies 5 × 0.02 — a tempting shortcut that overstates the risk and ignores overlap.

Question 3. A simulation will model a basketball player who makes 75% of free throws, taking 2 shots per trial. Which assignment of random digits is correct?

• (A) Digits 0–6 = make, 7–9 = miss
• (B) Pairs 00–74 = make, 75–99 = miss
• (C) Digits 0–7 = make, 8–9 = miss
• (D) Digits 0–4 = make, 5–9 = miss
• (E) One digit per trial, 0–6 = make

Answer: (B). 75% needs 75 of 100 equally likely outcomes, so two-digit numbers 00–74 represent a make. (A) gives 70%, (C) gives 80%, (D) gives 50%, and (E) uses the wrong number of digits per shot.

Free-response question. At a school of 250 surveyed students, 140 read fiction regularly, 110 read nonfiction regularly, and 55 read both. (a) What is the probability that a randomly chosen student reads fiction or nonfiction (or both)? (b) What is the probability that a randomly chosen student reads neither? (c) What is the probability that a randomly chosen student reads nonfiction, given that the student reads fiction? (d) Describe a simulation, using random digits, to estimate the probability that in a random group of 4 students, at least one reads both fiction and nonfiction. (You may use that 55/250 = 0.22 of students read both.)

Scoring guideline.

• Part (a): Essentially correct if (140 + 110 − 55)/250 = 195/250 = 0.78, with the overlap subtracted. Partially correct if the Venn regions are right but the final probability uses a wrong denominator. Incorrect if the answer is 250/250 or 0.99+ (overlap ignored entirely).
• Part (b): Essentially correct if it obtains 250 − 195 = 55 students, giving 55/250 = 0.22 (or 1 − 0.78). Partially correct if the count is right but not converted to a probability.
• Part (c): Essentially correct if the denominator is restricted to fiction readers: 55/140 ≈ 0.393. Incorrect if 55/250 is used — that is the joint probability, not the conditional one; readers watch for exactly this distinction.
• Part (d): Essentially correct if all four components appear: digits assigned to match 0.22 (for example, two-digit numbers 00–21 = "reads both", 22–99 = "does not"); a trial defined as 4 students (8 digits) with the outcome "at least one reads both" recorded; many repetitions stated (at least 20, ideally 100+); and the estimate computed as the relative frequency of successful trials. Partially correct if exactly one component is missing or the digit assignment gives the wrong probability.

Deeper dive: two-way tables as Venn diagrams in disguise

Every two-way table is a probability machine. Divide any cell by the grand total and you have a joint probability; divide by a row or column total and you have a conditional probability; the margins give the simple probabilities. Students who see this unity stop memorising separate formulas: the addition rule is "row total + column total − cell", and the multiplication check for independence is "does the cell equal the product of its margins divided by N?" Consider the travel table from Lesson 1: P(junior) = 90/190 ≈ 0.474, P(car) = 115/190 ≈ 0.605, and their product is ≈ 0.287 — but the actual P(junior and car) = 45/190 ≈ 0.237. The mismatch (0.237 ≠ 0.287) is a numerical independence test performed in one line, previewing Lesson 6 and, ultimately, the chi-square test of Lesson 17, which is nothing more than this comparison, aggregated over every cell and judged against chance.

One more discipline: say what the event is

Probability notation is a compression system, and sloppy decompression loses marks. Before computing anything, translate the words into events with names: let A = "the part is defective", B = "the part came from the night shift". Then "defective or night shift" is A ∪ B, "defective and night shift" is A ∩ B, "not defective" is the complement of A. The translation step feels pedantic until the problem's parts (b) and (c) quietly switch from "or" to "given" — the students who named their events notice the switch; the students who dove into arithmetic often do not. Readers cannot award method marks to unnamed numbers, so the naming is not decoration: it is how the method becomes visible on the page.

Second worked example: dice, exactly counted

Two fair dice are rolled; the sample space has 36 equally likely outcomes, so probabilities are just careful counting. P(sum = 7): the pairs (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) give 6/36 = 1/6 ≈ 0.167 — the most likely sum, because it has the most arrangements; sums are not equally likely even though outcomes are. P(sum ≥ 10) = (3 + 2 + 1)/36 = 6/36 = 1/6 as well (10 has three pairs, 11 has two, 12 has one). P(at least one six in four rolls of a single die) is a complement problem, not a counting one: 1 − (5/6)^4 = 1 − 0.4823 ≈ 0.518 — better than even, the fact that made Chevalier de Méré's fortune in the 1650s and helped provoke Pascal and Fermat's invention of probability theory. De Méré's companion bet — at least one double-six in 24 rolls of two dice — is 1 − (35/36)^24 ≈ 0.491, just *under* half, and the tiny gap between his two bets (0.518 vs 0.491) is a standing warning: probability arithmetic, done exactly, routinely defeats gamblers' intuition by a few decisive percentage points.