PR-1: Solutions — Basic Probability Concepts

Problem 1 — Classical Probability and Complement (Variants 0–4)

Each variant applies and the complement rule .

• Variant 0 — 52-card deck, A = “draw a heart”: ; hearts , so , .
• Variant 1 — 8 red, 5 blue, 7 green marbles, A = “red”: ; , .
• Variant 2 — 10-section spinner, A = “multiple of 3”: ; multiples of 3 in are , so , .
• Variant 3 — fair die, A = “prime”: ; primes in are (1 is not prime), so , .
• Variant 4 — 12 defective, 88 non-defective, A = “defective”: ; , .

Problem 2 — Applying the Complement Rule (Variants 0–4)

Each applies .

• Variant 0 — .
• Variant 1 — .
• Variant 2 — .
• Variant 3 — .
• Variant 4 — .

Problem 3 — General Addition Rule (Generator)

Problem 4 — Venn Diagram and Mutual Exclusivity (80 students)

Given: 80 students; 48 play a team sport; 35 play an instrument; 12 do both.

Breakdown: both = 12; sport only = ; instrument only = ; neither = .

(a) Play neither: 9 (since 36 + 12 + 23 = 71 accounted for).

(b) : . Without subtracting the intersection, — an impossible count that flags the double-count error.

(c) Mutually exclusive? No. — 12 students do both. Mutual exclusivity requires .

(d) Independence claim: false. Independence has a precise definition (PR-2): . “Completely different activities” is a semantic argument, not a test. Note too: these events are not mutually exclusive, and mutually exclusive events with positive probability are always dependent — knowing A occurred tells you B did not. “Mutually exclusive” and “independent” are not synonyms.

Problem 1 — Empirical Probability from Two-Way Tables (Variants 0–4)

For every variant: = row total / grand total; = intersection cell / grand total. Never divide the cell by a row or column total — that gives a conditional probability.

• Variant 0 — 150 students, L = “library”, T = “tutor”: (row total, not the cell 42); .
• Variant 1 — 200 people, P = “phone”, V = “video”: ; .
• Variant 2 — 160 adults, E = “exercises”, G = “good sleep”: ; .
• Variant 3 — 200 commuters, C = “car”, O = “on time”: ; .
• Variant 4 — 180 gym members, S = “student”, W = “weekly”: ; .

Problem 2 — Addition Rule from a Two-Way Table (Variants 0–4)

Read , , (all over the grand total), then apply the General Addition Rule.

• Variant 0 — C = “car”, F = “full-time”: , , → .
• Variant 1 — S = “streaming”, L = “live sports”: .
• Variant 2 — V = “vitamins”, G = “7+ h sleep”: .
• Variant 3 — N = “reads news”, V = “votes”: .
• Variant 4 — H = “cooks”, T = “tracks calories”: .

Problem 3 — Complement Strategy: At-Least-One (Generator)

Problem 4 — Find the Error (Variants 0–4)

• Variant 0 — assumed for “even or > 4” on a die: 6 is both, so , . Correct: .
• Variant 1 — claimed two coin flips are mutually exclusive “because different times”: HH shows both occur, .
• Variant 2 — wrote : copied instead of applying the rule. Correct: (and ✓).
• Variant 3 — assigned : sum violates . Correct: .
• Variant 4 — used , ignoring the given : the product form requires independence (must be established, not assumed). Use the given value: .

Problem 5 — Multi-Step Synthesis (120 students)

Given: 120 students; submitted all: 80 yes / 40 no; passed: 90 yes / 30 no; among submitters, 75 passed.

	Passed	Failed	Total
Submitted all	75	5	80
Did not submit	15	25	40
Total	90	30	120

Fill order: submitted & passed = 75 (given); submitted & failed = ; not-submitted & passed = ; not-submitted & failed = . Cross-check: ✓.

(a) Submitted all AND failed: 5.

(b) : .

(c) : bottom-right cell . Via complement: ✓.

(d) Mutually exclusive? No. .

(e) Independence claim: cannot conclude without a formal test (PR-2). Observed: the events are not mutually exclusive, and the high overlap (75 of 80 submitters passed vs. 15 of 40 non-submitters) suggests a strong positive association. “Different things ⟹ independent” is a semantic argument, not a mathematical one.

Problem 1 — Feynman Test: Why in General

When two events can both occur in a single trial, some outcomes belong to both A and B. Adding counts those shared outcomes twice. The General Addition Rule subtracts the double-counted region exactly once.

Example: in a class of 30, 12 play piano, 10 play guitar, 4 play both. Naively , but the 4 are double-counted; the true count is , so .

is valid only when A and B are mutually exclusive () — no double-counting to correct.

Problem 2 — Apply: Colored Blocks

Setup: 12 red, 8 blue, 5 yellow (25 total). R = “red”; W = “warm” (red or yellow) = 17 blocks.

Key observation: every red block is warm, so , meaning and . So — adding R to W contributes nothing new.

Problem 3 — Error Analysis: Simple Addition Rule on Overlapping Events

Student’s claim: ”, are unrelated, so .”

Error: applied the simplified rule without checking . It is valid only for mutually exclusive events.

Why 1.10 is impossible: all probabilities lie in ; a value above 1 always signals the intersection was not subtracted.

Correction: without the exact value is undetermined, but it is bounded: upper ; lower . “Unrelated” is not a criterion for mutual exclusivity.

Path A — The Pollster (500 residents)

	Healthy Eating	Not Healthy Eating	Total
Gym Member	110	90	200
Not a Member	130	170	300
Total	240	260	500

Cross-check: ✓.

Task 1 — Marginals: ; ; (intersection cell, not a total).

Task 2 — Addition rule: — 66% do at least one of the two behaviors.

Task 3 — Neither: . Direct check: “Not a Member AND Not Healthy” cell , and ✓.

Task 4 — Mutual exclusivity claim: false. The table shows 110 who are gym members and eat healthily, so . The semantic argument doesn’t substitute for reading the intersection cell.

Path B — The Referee (Sports Commentator)

Claim: “Home team wins 60% and scores first 55% — so winning OR scoring first is 115%!”

Task 1 — Assumption: the commentator used the simplified rule , which requires — almost certainly false and never established.

Task 2 — Why 115% is impossible: every probability is in ; 1.15 violates this. Many winning games also had the team scoring first — those are counted in both percentages, producing the overcount.

Task 3 — Correct formula: . With : — 75%, not 115%. Since , the union lies in , capped at 1 by the axioms → true range .

Task 4 — “Different things” ≠ mutually exclusive: one game can be both a win and a score-first — same trial, not different ones. Mutual exclusivity requires the events cannot co-occur in a single trial, which is not the case here.

Challenge 1 — Derive the General Addition Rule from Counting

(a) . Forming counts each outcome of twice; subtracting removes one copy, leaving each union outcome counted once.

(b) Dividing by : — derived directly from counting, with no assumptions beyond equally likely outcomes.

(c) If A and B are mutually exclusive, , so and . Dividing by gives the simplified rule — valid only when the intersection is empty.

Challenge 2 — At-Least-One via Complement

Setup: 3 items selected independently, each , so .

(a) , so . The naive overcounts.

(b) Mutually exclusive? Yes — “zero defectives” and “at least one defective” share no outcomes and together cover ; this is exactly the complement relationship: .

(c) by enumeration: exactly 2 defective in 3 items has arrangements, each , so . . Thus . The complement strategy in (a) was one subtraction; enumeration here needed several cases — complement scales far better.

Challenge 3 — Non-Uniform Probability Space

Setup: biased die with ; faces 1–5 equally share the rest.

(a) Remaining probability , split among 5 faces: . Verify: ✓.

(b) . The fair-die shortcut does not apply — each face must be weighted by its actual probability.

(c) (6 is not prime).

(d) Even , prime , intersection , so → not mutually exclusive.