L-PR-1: Solutions — Basic Probability Concepts

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — Venn Diagram and Mutual Exclusivity

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

Step 1 — Complete the Venn regions.

• Both:

• Sport only:

• Instrument only:

Step 2 — Count the union.

Step 3 — Count neither.

Step 4 — Find the union probability.

Part (c) — Mutually exclusive? No.

The 12 students in both groups prove that the events can co-occur.

Part (d) — Independence claim. “Different activities” is not a mathematical test for independence. The supplied counts are enough for the formal test introduced in L-PR-2:

Because , the events are dependent.

Problem 2 — Classical Probability and Complement

This is a generated typed-entry problem, so the counts change. For every instance:

1. Count the full equally likely sample space, .
2. Count the outcomes in the event, .
3. Compute

4. Check the result with

The denominator is always the full sample space. A favourable count by itself is not a probability.

Problem 3 — Applying the Complement Rule

This is a generated typed-entry problem. The given event and the requested event are complements, so:

The two probabilities must sum to 1. Two outcomes do not imply probabilities of 0.50 each.

Problem 4 — General Addition Rule

The values and context change, but every generated instance has . The events overlap, so use:

Subtracting the intersection once removes its second count. The generated feedback shows the exact substitution for the current instance.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — Empirical Probability from a Two-Way Table

This generated problem asks for a row marginal, a column marginal, or a joint probability.

• For a marginal, add the matching row or column and divide by the grand total .

• For a joint probability, use the matching interior cell and divide by .

Dividing an interior cell by a row or column subtotal answers a conditional-probability question instead, which is not the target here.

Problem 2 — Addition Rule from a Two-Way Table

Read the three counts from the generated table:

• : the matching row total

• : the matching column total

• : the matching interior cell

Then compute in counts before rounding:

Working in counts avoids rounding the three component probabilities before the final result.

Problem 3 — Complement Strategy: At Least One

The generated scenario contains two or three independent trials. Translate “at least one target event” into its complement, “no target events.”

Step 1 — Compute the probability of none.

Step 2 — Subtract from 1.

The exact base, exponent, and four-decimal answer appear in the feedback for the generated instance.

Problem 4 — Find the Error: Probability Reasoning

The generator selects one of six error families. In every version, identify the first incorrect step rather than the first line whose arithmetic looks unusual.

1. Overlapping die events: 6 belongs to both “even” and “greater than 4,” so the intersection must be subtracted.
2. Two coin flips: proves that “first flip heads” and “second flip heads” can occur together.
3. Copied complement: if , then .
4. Axiom violation: pass and fail are complements, so probabilities 0.70 and 0.70 cannot both be correct.
5. Assumed independence: do not replace a given with .
6. Subset union: the outcome 3 is already odd, so “3 or odd” is just the odd event.

Problem 5 — Multi-Step Synthesis

Given: 120 students; 80 submitted every assignment; 90 passed; 75 both submitted every assignment and passed.

Step 1 — Complete the four interior cells.

• Submitted and passed:

• Submitted and failed:

• Did not submit and passed:

• Did not submit and failed:

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

Step 2 — Compute the union.

Step 3 — Compute neither.

The table confirms the result:

Part (d) — Mutually exclusive? No.

Part (e) — Independence claim. “Different things” does not establish independence. The table is sufficient for the formal test introduced in L-PR-2:

Because , the events are dependent. The comparison with — not the raw size of the overlap by itself — establishes that conclusion.

Mixed Review — Retrieval from Earlier Lessons

Review Problem 1 — Range (L-DS-4). For the generated data, identify the maximum and minimum, then compute:

Review Problem 2 — Empirical Rule (L-DS-5). Match the requested interval to its distance from the mean: about 68% lies within , 95% within , and 99.7% within for an approximately normal distribution.

Both blocks display exact feedback for the generated values and record the attempt as prerequisite evidence for the earlier lesson.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Question 1 — Feynman Test

When two events can both occur, adding counts every shared outcome twice. The General Addition Rule subtracts that second count:

For example, suppose 12 of 30 students play piano, 10 play guitar, and 4 play both.

Count the union.

Convert the count to a probability.

The shorter rule is valid only when .

Question 2 — Apply

This typed generator supplies a new count context. Read , , , and , then compute:

The positive intersection means the mutually exclusive shortcut does not apply.

Question 3 — Error Analysis

The student applied the simplified rule without verifying mutual exclusivity.

The result violates the probability bound and signals that an overlap was not subtracted. Without , the exact union is unknown. Its bounds are:

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Path A — The Pollster

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

Task 1 — Read the three probabilities.

Task 2 — Find and interpret the union.

Therefore, 66% of the 500 surveyed residents are gym members, eat healthily, or both.

Task 3 — Find neither and verify it.

The bottom-right cell confirms the result:

Task 4 — Evaluate the claim.

Mutual exclusivity requires a zero intersection. Here:

The colleague is wrong: 110 residents belong to both groups.

Model reflection. If the events were mutually exclusive, the “Gym Member AND Healthy Eating” cell would be 0. The row and column totals could remain positive, but nobody could appear in both groups.

Path B — The Referee

Task 1 — Identify the fastest warning sign.

A probability cannot exceed 1, so 115% is impossible.

Task 2 — Identify the misused rule.

The commentator used:

That shorter rule requires:

Winning and scoring first can happen in the same game, so the required condition is false.

Task 3 — Supply the missing piece and compute.

The missing probability is .

With :

The correct probability is 75%.

Task 4 — Reject the semantic argument.

“Winning” and “scoring first” are different descriptions, but one game can satisfy both. Different labels do not make events mutually exclusive.

Model reflection. Ask, “Can both events happen on the same trial?” If yes, the intersection can be positive and must be included in the General Addition Rule.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Challenge 1 — Derive the General Addition Rule from Counting

Adding and counts every outcome in twice. Subtract the intersection once:

Divide every term by :

Therefore:

If and are mutually exclusive, , so the formula simplifies to:

Challenge 2 — Non-Uniform Probability Space

Step 1 — Find the probability of each face from 1 through 5.

The five faces share the probability left after assigning :

Split that remainder equally:

Step 2 — Find the probability of an even result.

Step 3 — Find the probability of a prime result.

Step 4 — Find the intersection.

The outcome 2 is both even and prime:

Therefore, the events are not mutually exclusive.

Step 5 — Find the union.

Challenge 3 — A Hidden Simplification

There are 12 red, 8 blue, and 5 yellow blocks, for 25 total. Let be “red” and be “warm-coloured,” meaning red or yellow.

Every red block is warm-coloured, so:

The union adds nothing beyond :

There are warm-coloured blocks. Therefore:

The General Addition Rule gives the same simplification because .