L-PR-5: Solutions — Binomial Distribution

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.

Example 1 — BINS Check and Exact Probability

The light-bulb outcomes are binary, independence is stated, is fixed, and is constant. Therefore .

There is about a 17.6% chance that exactly 2 of the 6 bulbs are defective.

Example 2 — Cumulative Probabilities

For :

Thus, using the unrounded term values:

For “at least 2,” the shorter calculation uses the complement:

Example 3 — Centre and Spread

The seed outcomes satisfy BINS, so .

Across many groups of 20 seeds, an average of 12 germinate. The count typically differs from 12 by about 2.19 seeds.

Example 4 — Transfer: When the Binomial Model Fails

Drawing five cards without replacement gives binary outcomes and fixed , but the draws are dependent and the ace probability changes after every draw. Conditions I and S fail.

Use a hypergeometric count:

The trainee’s binomial result is about 0.0465; it comes from an invalid constant-probability assumption.

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 — BINS Classification (C1 + C6)

• Scenario A: Ten independent coin flips satisfy BINS, so .
• Scenario B: Drawing marbles without replacement makes later draws depend on earlier draws. Use a hypergeometric model.
• Scenario C: Stopping when the third response occurs means the number of trials is not fixed. Use a negative-binomial model.

Problem 2 — Exact Binomial Probability (C1 + C2)

Every generated version follows the same invariant:

1. Verify binary outcomes, independence, fixed , and constant .
2. Set and identify the requested count .
3. Compute

4. Interpret the decimal as the probability of exactly successes in trials.

Problem 3 — Cumulative Binomial Probability (C1 + C3)

Translate the boundary before calculating:

The generated feedback identifies an off-by-one boundary, an exact-versus-cumulative error, or the complementary event.

Problem 4 — Centre, Spread, and Shape (C4 + C5)

For every generated :

For the distribution is right-skewed, for it is symmetric, and for it is left-skewed. At all probability is at ; at all probability is at .

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 — Full Binomial Workflow (C1 + C2 + C3 + C4)

The generated solution verifies BINS, calculates one exact probability, translates one cumulative event, and then calculates and . Its invariant sequence is:

Problem 2 — Cumulative Boundaries (C3)

First decide which counts are included. The generator then requires the correct sum or complement and a probability rounded to four decimals. An answer using only is a point probability, not a cumulative probability.

Problem 3 — Majority of Trials (C2 + C3)

For an odd trial count , a strict majority begins at

Then:

Problem 4 — Diagnose a BINS Error (C1 + C6)

The regenerable error taxonomy contains:

• sampling without replacement: independence and constant fail; use hypergeometric;
• waiting for a fixed number of successes: fixed fails; use negative binomial;
• changing success probability: constant fails; use a general PMF or Poisson-binomial model.

Problem 5 — Centre, Spread, and Shape (C4 + C5)

Use , , and . Interpret the mean as a long-run average count and the SD as typical variation around that count. Determine skew from .

Mixed Review 1 — Combinations in Probability (L-PR-3)

For red and blue tokens, drawing without replacement, the probability of exactly red tokens is:

Mixed Review 2 — Expected Value and Variance (L-PR-4)

Build the PMF from net winnings, then compute:

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

The factor counts all placements of successes among trials. For and , the sequences SSF, SFS, and FSS each have probability . Therefore:

Without the coefficient, the formula counts only one of the three sequences.

Question 2 — Applied Scenario

The seed treatment satisfies BINS, so .

For at least 6, a two-term direct sum is shorter:

Question 3 — Error Analysis

The generator randomly presents one of three authentic errors:

1. using target count as the binomial trial count ;
2. omitting and calculating one sequence only;
3. swapping the exponents on and during evaluation.

The correction always returns to:

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 Analyst

Let and count defects in 10 independent inspections:

For either line:

Line A:

Line B:

Both distributions are right-skewed. Only Line B reaches the investigation threshold of 0.50. Without-replacement sampling from a small lot would invalidate the binomial model.

Reflection model: Translating “at least 2” into the complement of 0 or 1 defect makes the policy calculation short and transparent.

🎲 Path B — The Architect

The plans satisfy BINS:

Compact plan:

Broad plan:

Recommend the Compact plan: it has both the larger target probability and the smaller SD. Both distributions are right-skewed. Changing response probabilities across contacts or allowing responses to influence one another would undermine BINS.

Reflection model: The target probability makes the decision, the equal means explain the plans’ apparent similarity, and the SDs explain why their probability patterns differ.

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 — Deriving

Start with the definition:

The term is zero. Use :

Let :

The sum is the complete PMF of , so it equals 1:

Challenge 2 — Mode of the Binomial Distribution

Each generated version asks for the most likely count or counts and then verifies the prediction with neighboring probabilities. The invariant is:

For :

• if is not a whole number, the mode is ;
• if is a whole number , the two modes are and .

At the endpoints, the distribution has one mode: when , and when .

The generated solution independently calculates the neighboring PMF values and checks that every reported mode attains the maximum probability.