PR-1 Solutions: Basic Probability Concepts

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

For every variant: \( P(A) \) = row total / grand total; \( P(A \cap B) \) = intersection cell / grand total. Never divide the cell count by a row or column total — that produces a conditional probability, not a marginal or joint probability.

Variant 0 — 150 students; L = "library"; T = "tutor help":

• \( P(L) = 70/150 \approx 0.467 \) — use the library row total (70), not the cell (42).
• \( P(L \cap T) = 42/150 = 0.28 \) — use the intersection cell divided by grand total 150.

Variant 1 — 200 people; P = "phone"; V = "prefers video":

• \( P(P) = 120/200 = 0.60 \)
• \( P(P \cap V) = 80/200 = 0.40 \)

Variant 2 — 160 adults; E = "exercises regularly"; G = "good sleep":

• \( P(E) = 80/160 = 0.50 \)
• \( P(E \cap G) = 52/160 = 0.325 \)

Variant 3 — 200 commuters; C = "car"; O = "on time":

• \( P(C) = 120/200 = 0.60 \)
• \( P(C \cap O) = 90/200 = 0.45 \)

Variant 4 — 180 gym members; S = "student"; W = "uses gym weekly":

• \( P(S) = 90/180 = 0.50 \)
• \( P(S \cap W) = 54/180 = 0.30 \)

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

General approach: read \( P(A) \), \( P(B) \), \( P(A \cap B) \) from the table (all divided by the grand total), then apply the General Addition Rule.

Variant 0 — 200 adults; C = "owns car"; F = "full-time":

• \( P(C) = 120/200 = 0.60 \); \( P(F) = 130/200 = 0.65 \); \( P(C \cap F) = 80/200 = 0.40 \)
• \( P(C \cup F) = 0.60 + 0.65 - 0.40 = \mathbf{0.85} \)

Variant 1 — 200 adults; S = "streaming subscription"; L = "watches live sports":

• \( P(S) = 120/200 = 0.60 \); \( P(L) = 110/200 = 0.55 \); \( P(S \cap L) = 70/200 = 0.35 \)
• \( P(S \cup L) = 0.60 + 0.55 - 0.35 = \mathbf{0.80} \)

Variant 2 — 200 adults; V = "takes vitamins"; G = "7+ hours sleep":

• \( P(V) = 110/200 = 0.55 \); \( P(G) = 100/200 = 0.50 \); \( P(V \cap G) = 60/200 = 0.30 \)
• \( P(V \cup G) = 0.55 + 0.50 - 0.30 = \mathbf{0.75} \)

Variant 3 — 200 adults; N = "reads news daily"; V = "votes regularly":

• \( P(N) = 120/200 = 0.60 \); \( P(V) = 110/200 = 0.55 \); \( P(N \cap V) = 75/200 = 0.375 \)
• \( P(N \cup V) = 0.60 + 0.55 - 0.375 = \mathbf{0.775} \)

Variant 4 — 200 adults; H = "cooks at home"; T = "tracks calories":

• \( P(H) = 120/200 = 0.60 \); \( P(T) = 80/200 = 0.40 \); \( P(H \cap T) = 55/200 = 0.275 \)
• \( P(H \cup T) = 0.60 + 0.40 - 0.275 = \mathbf{0.725} \)

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

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

Variant 0 — Student assumed \( P(A \cap B) = 0 \) for "even or greater than 4" on a fair die:

Error: Assumed mutual exclusivity without checking. The outcome 6 is both even and greater than 4, so \( A \cap B = \{6\} \) and \( P(A \cap B) = 1/6 \neq 0 \). Correct: \( P(A \cup B) = 3/6 + 2/6 - 1/6 = 4/6 = 2/3 \).

Variant 1 — Student claimed "flip 1 = H" and "flip 2 = H" are mutually exclusive because they happen at different times:

Error: "Different times" does not imply mutual exclusivity. Outcome HH shows both events occur simultaneously: \( P(A \cap B) = P(HH) = 1/4 \neq 0 \). The events are not mutually exclusive.

Variant 2 — Student wrote \( P(A') = P(A) = 0.35 \):

Error: Copied \( P(A) \) instead of applying the complement rule. Correct: \( P(A') = 1 - P(A) = 1 - 0.35 = 0.65 \). Verify: \( 0.35 + 0.65 = 1 \). ✓

Variant 3 — Student assigned \( P(\text{pass}) = 0.70 \) and \( P(\text{fail}) = 0.70 \):

Error: Sum \( 0.70 + 0.70 = 1.40 > 1 \) violates Axiom 2 (\( P(S) = 1 \)). For complementary exhaustive events: \( P(\text{fail}) = 1 - 0.70 = 0.30 \).

Variant 4 — Student used \( P(A \cap B) = P(A) \times P(B) = 0.12 \) ignoring the given \( P(A \cap B) = 0.10 \):

Error: Assumed independence to compute the intersection. The formula \( P(A \cap B) = P(A) \times P(B) \) requires independence — a condition that must be established, not assumed. The problem provides \( P(A \cap B) = 0.10 \) directly; always use the given value. Correct: \( P(A \cup B) = 0.30 + 0.40 - 0.10 = 0.60 \).

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Problem 5 — Multi-Step Synthesis (120 students, Parts a–e)

Given: 120 students total. Submitted all assignments: 80 (yes), 40 (no). Passed final: 90 (yes), 30 (no). Among those who submitted all assignments, 75 passed.

Completed 2×2 table — build this first:

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

How to fill the table: "Submitted and passed" = 75 (given). "Submitted and failed" = 80 − 75 = 5. "Did not submit and passed" = 90 − 75 = 15. "Did not submit and failed" = 30 − 5 = 25 (or 40 − 15 = 25). ✓ Cross-check: 75 + 5 + 15 + 25 = 120.

Part (a) — Students who submitted all AND failed:

Correct answer: 5. From the table: submitted row, failed column = 5.

Part (b) — \( P(\text{submitted OR passed}) \):

\[ P(\text{submitted} \cup \text{passed}) = \frac{80}{120} + \frac{90}{120} - \frac{75}{120} = \frac{95}{120} \approx 0.792 \]

Part (c) — \( P(\text{did not submit AND did not pass}) \):

Read the bottom-right cell directly: \( P = 25/120 \approx 0.208 \).

Alternatively via complement: \( P(\text{did not submit AND did not pass}) = 1 - P(\text{submitted OR passed}) = 1 - 95/120 = 25/120 \). ✓

Part (d) — Are "submitted all" and "passed" mutually exclusive?

No. \( P(\text{submitted AND passed}) = 75/120 \neq 0 \). The events clearly co-occur — 75 students did both.

Part (e) — Evaluate the independence claim:

We cannot conclude independence without a formal test (covered in PR-2). What we can observe: the events are not mutually exclusive, and the very high overlap (75 of the 80 submitters passed, compared to only 15 of the 40 non-submitters) suggests a strong positive association. The claim "different things therefore independent" is a semantic argument, not a mathematical one.

Question 1 — Feynman Test: Why \( P(A \cup B) \neq P(A) + P(B) \) in General

Model answer:

When two events can both occur in a single trial, some outcomes belong to both A and B simultaneously. Adding \( P(A) + P(B) \) counts those shared outcomes twice — once in the count for A and once in the count for B. The General Addition Rule \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) corrects for this by subtracting the double-counted region exactly once.

Concrete example: In a class of 30 students, 12 play piano, 10 play guitar, and 4 play both. Adding naively: \( 12 + 10 = 22 \), giving 22/30. But those 4 students were counted twice — once in "piano" and once in "guitar." The true count of students playing at least one instrument is \( 12 + 10 - 4 = 18 \), so \( P(\text{at least one}) = 18/30 = 0.60 \).

The only situation where \( P(A \cup B) = P(A) + P(B) \) is valid is when A and B are mutually exclusive — when \( P(A \cap B) = 0 \) and no outcome can belong to both events. In that case there is no double-counting to correct.

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Question 2 — Apply: Colored Blocks

Setup: Bag contains 12 red, 8 blue, 5 yellow blocks (25 total). R = "red"; W = "warm color" (red or yellow). Warm blocks = 12 + 5 = 17.

Key observation: All red blocks are warm-colored, so \( R \subseteq W \). This means \( R \cap W = R \), and \( R \cup W = W \).

\[ P(R \cup W) = P(R) + P(W) - P(R \cap W) = \frac{12}{25} + \frac{17}{25} - \frac{12}{25} = \frac{17}{25} \]

Correct answer: \( P(R \cup W) = P(W) = 17/25 \). Equivalently, because \( R \subseteq W \), the union is just W — every red block is already warm-colored, so adding R to W adds nothing new.

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Question 3 — Error Analysis: Applied Simple Addition Rule to Overlapping Events

Student's claim: "Since \( P(A) = 0.60 \) and \( P(B) = 0.50 \) are unrelated events, \( P(A \cup B) = 0.60 + 0.50 = 1.10 \)."

Error identified: Applied the simplified Addition Rule \( P(A \cup B) = P(A) + P(B) \) without verifying that \( P(A \cap B) = 0 \) (mutual exclusivity). The simplified rule is only valid for mutually exclusive events.

Why 1.10 is impossible: Axiom 1 requires all probabilities to lie in \( [0, 1] \). A result greater than 1 is always a signal that the simplified rule was applied to overlapping events — the intersection was not subtracted.

Correction: Without \( P(A \cap B) \), the exact value of \( P(A \cup B) \) cannot be determined. We can establish bounds:

• Upper bound: \( P(A \cup B) \leq \min(1,\; P(A) + P(B)) = 1.00 \)
• Lower bound: \( P(A \cup B) \geq \max(P(A), P(B)) = 0.60 \)

The true answer \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) requires knowing \( P(A \cap B) \) from the data. "Unrelated" is not a mathematical criterion for mutual exclusivity.

Path A — The Pollster (500 residents)

Survey of 500 city residents:

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

Cross-check: \( 110 + 90 + 130 + 170 = 500 \). ✓ All four cells sum to n = 500.

Task 1 — Marginal probabilities:

• \( P(\text{gym}) = 200/500 = 0.40 \) — use the "Gym Member" row total.
• \( P(\text{healthy}) = 240/500 = 0.48 \) — use the "Healthy Eating" column total.
• \( P(\text{gym AND healthy}) = 110/500 = 0.22 \) — use the intersection cell (top-left). Not a row or column total.

Task 2 — General Addition Rule:

\[ P(\text{gym OR healthy}) = P(\text{gym}) + P(\text{healthy}) - P(\text{gym AND healthy}) \]

\[ = 0.40 + 0.48 - 0.22 = 0.66 \]

Interpretation: 66% of the 500 residents are gym members, or eat healthily, or both (at least one of the two behaviors).

Task 3 — Neither (complement) and cell verification:

\[ P(\text{neither}) = 1 - P(\text{gym OR healthy}) = 1 - 0.66 = 0.34 \]

Verify directly: "Not a Member AND Not Healthy Eating" cell = 170. \( 170/500 = 0.34 \). ✓ The complement calculation and the direct cell read agree.

Task 4 — Evaluate the mutual exclusivity claim:

The colleague's claim is false. Mutual exclusivity requires \( P(\text{gym AND healthy}) = 0 \) — the formal definition: \( A \cap B = \emptyset \). But the table shows 110 residents who are gym members and eat healthily, so \( P(\text{gym AND healthy}) = 110/500 = 0.22 \neq 0 \). The semantic argument ("exercise focus vs. diet focus are separate things") does not substitute for the mathematical check: read the intersection cell and compare to zero.

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Path B — The Referee (Sports Commentator)

Commentator's claim: "The home team wins 60% of the time and scores first 55% of the time — so their chance of winning OR scoring first is 115%!"

Task 1 — Mathematical assumption:

The commentator applied the simplified Addition Rule: \( P(W \cup S) = P(W) + P(S) = 0.60 + 0.55 = 1.15 \). This rule requires \( P(W \cap S) = 0 \) — that winning and scoring first cannot both occur in the same game. This condition is almost certainly false and was never established.

Task 2 — Why 115% is impossible:

Axiom 1 requires \( 0 \leq P(\text{any event}) \leq 1 \). A probability of 1.15 violates this axiom directly. Conceptually, in many games where the home team wins, they also scored first — those games are counted in both the 60% and the 55%, producing the overcount. The overlap was not subtracted.

Task 3 — Correct formula and numerical answer:

The missing piece is \( P(W \cap S) \) — the probability that the team wins AND scores first in the same game. The correct formula:

\[ P(W \cup S) = P(W) + P(S) - P(W \cap S) = 0.60 + 0.55 - P(W \cap S) \]

With \( P(W \cap S) = 0.40 \):

\[ P(W \cup S) = 0.60 + 0.55 - 0.40 = \mathbf{0.75} \]

The correct probability is 75%, not 115%.

Bound derivation: Since \( 0 \leq P(W \cap S) \leq \min(P(W), P(S)) = \min(0.60, 0.55) = 0.55 \):

\[ P(W \cup S) = 0.60 + 0.55 - P(W \cap S) \in [0.60 + 0.55 - 0.55,\; 0.60 + 0.55 - 0] = [0.60,\; 1.15] \]

But \( P(W \cup S) \) is also bounded above by 1 (Axiom 1), so the true range is \( [0.60, 1.00] \). The commentator's 1.15 is the uncorrected ceiling — it is impossible as a probability.

Task 4 — "Measuring different things" does not imply mutual exclusivity:

A single game can result in both a home-team win and the home team scoring first — these happen in the same trial, not different ones. Mutual exclusivity requires that both events cannot occur in the same trial, which is clearly not the case here. "Measuring different statistics" is a semantic observation, not a mathematical proof of \( P(W \cap S) = 0 \).

Challenge 1 — Derive the General Addition Rule from Counting

Part (a) — Counting \( |A \cup B| \):

Correct answer: \( |A \cup B| = |A| + |B| - |A \cap B| \).

When you form \( |A| + |B| \), every outcome in \( A \cap B \) gets counted twice — once in \( |A| \) and once in \( |B| \). Subtracting \( |A \cap B| \) removes one copy of the double-counted outcomes, leaving each outcome in the union counted exactly once.

Part (b) — Dividing by \( |S| \):

Dividing both sides of \( |A \cup B| = |A| + |B| - |A \cap B| \) by \( |S| \) yields:

\[ \frac{|A \cup B|}{|S|} = \frac{|A|}{|S|} + \frac{|B|}{|S|} - \frac{|A \cap B|}{|S|} \]

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

This is the General Addition Rule — derived directly from the counting argument, with no additional assumptions beyond equally likely outcomes and the definition of classical probability.

Part (c) — Mutually exclusive case:

If A and B are mutually exclusive, then \( A \cap B = \emptyset \), so \( |A \cap B| = 0 \). The formula reduces to:

\[ |A \cup B| = |A| + |B| \]

Dividing by \( |S| \): \( P(A \cup B) = P(A) + P(B) \). This is the simplified addition rule — valid only when the intersection is empty.

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Challenge 2 — At-Least-One via Complement

Setup: 3 items selected independently. Each has \( P(\text{defective}) = 0.05 \), so \( P(\text{not defective}) = 0.95 \).

Part (a) — \( P(\text{at least one defective}) \):

Complement strategy: "at least one defective" and "zero defective" are complementary events.

\[ P(\text{none defective}) = (0.95)^3 = 0.857375 \]

\[ P(\text{at least one defective}) = 1 - (0.95)^3 \approx 1 - 0.857 = 0.143 \]

Correct answer: \( \approx 0.143 \). The naive approach \( 3 \times 0.05 = 0.15 \) overcounts; the complement method gives the exact value.

Part (b) — Are "zero defectives" and "at least one defective" mutually exclusive?

Yes. Every possible sample of three items either contains zero defectives, or it contains at least one. These two outcomes share no elements — a sample cannot simultaneously have zero and at least one defective item. Together they cover all of \( S \) (they are exhaustive), and they share no outcomes (they are mutually exclusive). This is exactly the complement relationship: \( P(\text{at least one}) + P(\text{none}) = 1 \).

Part (c) — \( P(\text{at least two defective}) \) by direct enumeration:

All arrangements of exactly 2 defective (D) and 1 good (G) among 3 items — there are \( \binom{3}{2} = 3 \) arrangements:

• (D, D, G): \( (0.05)^2 \times (0.95) = 0.002375 \)
• (D, G, D): \( (0.05) \times (0.95) \times (0.05) = 0.002375 \)
• (G, D, D): \( (0.95) \times (0.05)^2 = 0.002375 \)

\[ P(\text{exactly 2 defective}) = 3 \times (0.05)^2 \times (0.95) = 3 \times 0.002375 = 0.007125 \]

\[ P(\text{exactly 3 defective}) = (0.05)^3 = 0.000125 \]

\[ P(\text{at least 2 defective}) = 0.007125 + 0.000125 = \mathbf{0.007250} \]

Comparison with Part (a): Part (a) used one calculation; Part (c) required enumerating 3 arrangements plus a separate "exactly 3" case. As the number of trials grows, the complement strategy becomes far more efficient — for "at least one" with n trials, it is always a single subtraction.

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Challenge 3 — Non-Uniform Probability Space

Setup: Biased die with \( P(6) = 1/3 \). Faces 1–5 equally share the remaining probability.

Part (a) — \( P(\text{each face from 1 to 5}) \):

Step 1 — remaining probability: \( 1 - P(6) = 1 - 1/3 = 2/3 \).

Step 2 — divide equally among 5 faces: \( P(\text{each of 1–5}) = (2/3) / 5 = 2/15 \).

Verify: \( 5 \times (2/15) + 1/3 = 10/15 + 5/15 = 15/15 = 1 \). ✓

Correct answer: \( P(1) = P(2) = P(3) = P(4) = P(5) = 2/15 \).

Part (b) — \( P(\text{even}) = P(2) + P(4) + P(6) \):

\[ P(\text{even}) = \frac{2}{15} + \frac{2}{15} + \frac{1}{3} = \frac{2}{15} + \frac{2}{15} + \frac{5}{15} = \frac{9}{15} = \frac{3}{5} \]

Note: the classical formula \( 3/6 = 1/2 \) for a fair die does not apply here — each face must be weighted by its actual probability, not just counted.

Part (c) — \( P(\text{prime}) = P(2) + P(3) + P(5) \):

\[ P(\text{prime}) = \frac{2}{15} + \frac{2}{15} + \frac{2}{15} = \frac{6}{15} = \frac{2}{5} \]

The prime faces \( \{2, 3, 5\} \) each have probability 2/15. Note 6 is not prime (\( 6 = 2 \times 3 \)).

Part (d) — Are "even" and "prime" mutually exclusive? Compute \( P(\text{even} \cap \text{prime}) \) and \( P(\text{even} \cup \text{prime}) \):

Even faces: \( \{2, 4, 6\} \). Prime faces: \( \{2, 3, 5\} \). Intersection: \( \{2\} \) — the number 2 is both even and prime.

\[ P(\text{even} \cap \text{prime}) = P(2) = \frac{2}{15} \neq 0 \]

Therefore the events are not mutually exclusive.

\[ P(\text{even} \cup \text{prime}) = P(\text{even}) + P(\text{prime}) - P(\text{even} \cap \text{prime}) \]

\[ = \frac{9}{15} + \frac{6}{15} - \frac{2}{15} = \frac{13}{15} \]