How to use this page: Try each problem in the lesson before checking solutions
here. If your answer doesn't match, read the solution carefully — especially the part that explains
why common wrong answers are wrong. Understanding the error matters more than getting the right
answer the first time.
6 bulbs, each independently defective with probability 0.15; = number defective. Find .
BINS: Binary (defective/not) ✓; Independent ✓; fixed ✓; constant ✓ → . With , :
About a 17.6% chance exactly 2 of 6 bulbs are defective.
Example 2 — Cumulative Probabilities (“At Most” and “At Least”)
For :
(a) :, , , so .
(b) .
Why complement for “at least k”? Summing to 6 needs 5 terms; needs only 2. The complement flips the sum direction, so you work with whichever end has fewer terms.
Example 3 — Computing μ and σ, then Interpreting
Medication effective per patient with probability 0.60; . BINS hold → .
: across many 20-patient trials, the long-run average is 12 effective.
: counts typically deviate from 12 by ~2.19 patients; 9 or 15 would be unusual but not extreme (≈1.4 SD).
Example 4 — Find the Error (Non-Binomial Setting)
Five cards drawn without replacement; the student concludes .
Error — BINS condition I fails (dependent trials). Without replacement, is or , not — conditions I and S fail. Correct model: hypergeometric:
The student’s binomial answer (≈0.0334) underestimates it by treating the changing-deck probability as constant.
Section 5: Guided Practice Solutions
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Problem 1 — BINS Classification (All 5 Variants)
Variant 0: (1) 10 coin flips, = heads → binomial. (2) 10 T/F guesses → binomial. (3) 3 marbles without replacement from 5R+3B → not binomial (I, S fail) → hypergeometric.
Variant 1: (1) 12 operations, → binomial. (2) cards with replacement until first ace → not binomial (N fails) → geometric. (3) 8 tested from 20, removed each → not binomial (I, S fail) → hypergeometric.
Variant 2: (1) 15 volunteers, → binomial. (2) roulette 20 spins, red → binomial. (3) poll without replacement until 5 say “online” → not binomial (N, I, S fail) → negative hypergeometric.
Variant 3: (1) 6 arrows, → binomial. (2) test without replacement until 2 defective → not binomial (N fails) → negative hypergeometric. (3) quiz with rising 0.05 each correct → not binomial (S fails).
Variant 4: (1) 7 free throws, → binomial. (2) 15 tested with replacement from 200 (10 defective) → binomial. (3) admit patients until 3 respond → not binomial (N fails) → negative binomial.
Problem 2 — Exact Binomial Probability (Generator)
Verify BINS → .
Identify , , .
Apply .
Interpret as the chance of exactly successes in trials.
Problem 3 — Cumulative Binomial Probability (Generator)
“At most ”:.
“At least ”: (complement).
“More than ”: ( excluded).
“Fewer than ”:.
Write the formal probability statement first, then compute each before summing.
Problem 4 — Mean and Standard Deviation (Generator)
(a) . (b) . (c) — always take the square root; don’t report the variance as the SD. Interpret as the long-run average and as the typical deviation.
Section 6: Independent Practice Solutions
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Problem 1 — Full Binomial Workflow (Generator)
(a) Verify BINS → . (b) Exact: . (c) Cumulative: sum for “at most,” complement for “at least.” (d) , .
Problem 2 — “More Than” and “Fewer Than” (Generator)
“More than ”:.
“Fewer than ”:.
For “more than ” with :
Compute each term, sum, subtract from 1.
Problem 3 — Majority of Trials (All 5 Variants)
Variant 0 (fair coin, , ): (exactly 0.5 by symmetry).
Variant 1 (shooter, , , ): .
Variant 2 (T/F quiz, , ): .
Variant 3 (treatment, , , ): .
Variant 4 (QC, , , ): .
Problem 4 — Identify BINS Violations (All 5 Variants)
Variant 0 (10 of 25 books without replacement): I, S fail → hypergeometric.
Variant 1 (test until 2 defective): N fails → negative binomial.
Variant 2 (decreasing , averaged to 0.67): S fails — varies; use the Poisson-binomial / general PMF, not the binomial.
Variant 3 (misread: actually with replacement): all BINS hold → ; the student wrongly assumed without-replacement.
Variant 4 (calls until first sale): N fails → geometric.
Problem 5 — Shape and Full Workflow (Generator)
(a) . (b) . (c) . Shape: right-skewed if , symmetric if , left-skewed if . Interpret and in context.
Section 7: Mastery Check Solutions
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Problem 1 — Feynman Test
counts every arrangement of successes among trials, not just one.
Example (, ): the sequences SSF, SFS, FSS each have probability ; there are of them, so . Without , gives only one specific sequence.
Problem 2 — Applied Scenario (Vaccine, , )
(a) BINS hold → .
(b).
(c) — direct sum (2 terms here): ; ; total . (The complement needs 6 terms — direct is simpler here.)
Problem 3 — Error Analysis (, find )
Error — missing combination coefficient. The student computed , the probability of one sequence (SSSFFFFF).
Correct:. The student’s value is exactly of this — confirming the missing arrangements.
Task 4:, , . ; since counts can’t be negative, expect 0–3 defectives in most samples.
Path B — The Architect (Carnival Game Design)
Many satisfy and . Worked with , (others: , , …).
Tasks 1–2: → ✓. ✓.
Task 3 — :, so — players win ~5.8%.
Task 4 — Game with BINS: “Flip a biased coin (heads ) 8 times; win with 5+ heads.” Binary ✓, Independent ✓, ✓, constant ✓.
Section 9: Challenge Problem Solutions
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Challenge 1 — Algebraic Derivation of E(X) = np
For :
Apply the identity :
Re-index with :
The inner sum is the total probability of , which equals 1. The shortcut follows directly from the definition.
Challenge 2 — Mode of the Binomial Distribution
Mode: when is not an integer; two modes otherwise.
(a) : → mode . Check: , , — the peak is at ✓.
(b) : → mode . Consistent with the shape rule ( → left-skewed, mode above ). Symmetry check: mirrors , and ✓.