PR-5 Solutions: Binomial Distribution

GP1 — BINS Classification (All 5 Variants)

Variant 0 — Set A:

1. Coin flipped 10 times, \( X \) = heads. Binomial. B ✓ (H/T), I ✓ (coin flips are independent), N ✓ (\( n = 10 \)), S ✓ (\( p = 0.5 \)).
2. Student guesses on 10 T/F questions. Binomial. B ✓, I ✓ (each guess independent), N ✓ (\( n = 10 \)), S ✓ (\( p = 0.5 \)). So \( X \sim B(10, 0.5) \).
3. 3 marbles drawn without replacement from 5 red + 3 blue. Not binomial. Condition I fails: after drawing a red marble, only 4 red remain among 7, so \( p \) changes. Condition S also fails. Correct model: Hypergeometric.

Variant 1 — Set B:

1. Surgeon performs 12 operations, each independently successful with \( p = 0.92 \). Binomial. All BINS hold. \( X \sim B(12, 0.92) \).
2. Cards drawn with replacement until first ace. Not binomial. Condition N fails: no fixed number of draws. Correct model: Geometric distribution.
3. 8 items tested from run of 20, removing each after test. Not binomial (if population is small). Condition I and S fail: removing tested items from 20 changes \( p \) at each draw. Correct model: Hypergeometric (population size 20, drawing 8 without replacement).

Variant 2 — Set C:

1. 15 volunteers, drug response independent with \( p = 0.45 \). Binomial. \( X \sim B(15, 0.45) \).
2. Roulette wheel spun 20 times, \( X \) = number of red outcomes. Binomial. B ✓, I ✓ (spins are independent), N ✓ (\( n = 20 \)), S ✓ (\( p = 18/38 \approx 0.474 \)). So \( X \sim B(20, 18/38) \).
3. Students polled without replacement until 5 say "online." Not binomial. Condition N fails (no fixed \( n \)); Conditions I and S also fail (sampling without replacement from 30). Correct model: Negative hypergeometric (or geometric if population is treated as large).

Variant 3 — Set D:

1. Archer shoots 6 arrows, each independently hits bullseye with \( p = 0.70 \). Binomial. \( X \sim B(6, 0.70) \).
2. Box of 4 defective + 16 good batteries, tested without replacement until 2 defective found. Not binomial. Condition N fails (no fixed \( n \)); sampling without replacement also violates I and S. Correct model: Negative hypergeometric.
3. Quiz where probability of correct answer increases by 0.05 for each previous correct answer. Not binomial. Condition S fails: \( p \) is not constant across questions.

Variant 4 — Set E:

1. Basketball player shoots 7 free throws, each independent with \( p = 0.82 \). Binomial. \( X \sim B(7, 0.82) \).
2. Shipment of 200 items (10 defective), inspector tests 15 items with replacement (reshuffles after each). Binomial. Because replacement restores the original composition, I and S hold. \( X \sim B(15, 10/200) = B(15, 0.05) \).
3. Treatment works with \( p = 0.50 \); hospital admits patients until 3 respond. Not binomial. Condition N fails — total patients admitted is the random variable. Correct model: Negative Binomial (waiting for the 3rd success).

---

GP2 — Exact Binomial Probability (Generator Template)

This problem is generated randomly each time. The method is identical regardless of the specific values of \( n \), \( p \), and \( k \):

1. Verify BINS: Binary (success/failure) ✓; Independent ✓; Fixed \( n = [n] \) ✓; Constant \( p = [p] \) ✓. So \( X \sim B([n], [p]) \).
2. Identify: \( q = 1 - [p] = [q] \), \( k = [k] \), \( \binom{[n]}{[k]} = [C] \).
3. Apply the formula: \[ P(X = [k]) = \binom{[n]}{[k]} ([p])^{[k]} ([q])^{[n]-[k]} = [C] \times [p^k] \times [q^{n-k}] = \mathbf{[answer]} \]
4. Interpret: "There is a [answer × 100]% chance that exactly [k] out of [n] [context noun] are [successes]."

---

GP3 — Cumulative Binomial Probability (Generator Template)

This problem is generated randomly. The method depends on the phrasing:

• "At most [k]": \( P(X \leq [k]) = \sum_{j=0}^{[k]} P(X = j) \). Sum each term \( \binom{[n]}{j}([p])^j([q])^{[n]-j} \) for \( j = 0, 1, \ldots, [k] \), then add.
• "At least [k]": \( P(X \geq [k]) = 1 - P(X \leq [k]-1) \). Use complement to avoid summing many terms.
• "More than [k]": \( P(X > [k]) = 1 - P(X \leq [k]) \). Note: [k] itself is excluded.
• "Fewer than [k]": \( P(X < [k]) = P(X \leq [k]-1) \). Shift boundary down by 1 — [k] is excluded.

Always write the formal probability statement first, then compute each individual \( P(X = j) \) term in full before summing.

---

GP4 — Mean and Standard Deviation (Generator Template)

This problem is generated randomly. The method is always the same:

1. (a) \( \mu = np = [n]([p]) = [np] \)
2. (b) \( \sigma_X^2 = np(1-p) = [n]([p])([q]) = [npq] \)
3. (c) \( \sigma_X = \sqrt{[npq]} \approx [sigma] \) — always take the square root; do not report the variance as the standard deviation.
4. Interpretation: "In the long run, [context] averages [np] [units] per [period]. The actual count typically deviates from this average by about [sigma] [units]."

IP1 — Full Binomial Workflow (Generator Template)

Generated by generateFullWorkflow(). For any generated \( n \), \( p \), \( k \), and cumulative type, the workflow is:

1. (a) Verify BINS: Binary (success/failure) ✓; Independent ✓; Fixed \( n = [n] \) ✓; Constant \( p = [p] \) ✓. So \( X \sim B([n], [p]) \).
2. (b) Exact probability: \[ P(X = [k]) = \binom{[n]}{[k]} ([p])^{[k]} ([q])^{[n]-[k]} = [C] \times [p^k] \times [q^{n-k}] = \mathbf{[answer_b]} \]
3. (c) Cumulative probability (shown here for "at most [k]"; complement method for "at least [k]"): \[ P(X \leq [k]) = \sum_{j=0}^{[k]} P(X=j) = [term_0] + [term_1] + \cdots + [term_k] = \mathbf{[answer_c]} \] or, for "at least [k]": \( P(X \geq [k]) = 1 - P(X \leq [k]-1) \).
4. (d) Mean and SD: \[ \mu = np = [n]([p]) = [np], \quad \sigma_X = \sqrt{np(1-p)} = \sqrt{[npq]} \approx [sigma] \]

---

IP2 — "More Than" and "Fewer Than" (Generator Template)

Generated by generateBinomialCumulative(). The method for the boundary phrasings most likely to appear here:

• "More than [k]": \( P(X > [k]) = 1 - P(X \leq [k]) \). Exclude [k] by subtracting the CDF at [k].
• "Fewer than [k]": \( P(X < [k]) = P(X \leq [k]-1) \). Shift boundary down by 1.

Template computation (example structure):

For "more than [k]" with \( X \sim B([n], [p]) \):

\[ P(X > [k]) = 1 - P(X \leq [k]) = 1 - \sum_{j=0}^{[k]} \binom{[n]}{j}([p])^j([q])^{[n]-j} \]

Compute each term, sum, then subtract from 1. Each term follows the same formula structure as GP2.

---

IP3 — Majority of Trials (Variant Bank, All 5 Variants)

Variant 0 — Fair coin, \( n = 5 \), \( p = 0.50 \), find \( P(X \geq 3) \):

\[ P(X = 3) = \binom{5}{3}(0.5)^3(0.5)^2 = 10 \times 0.125 \times 0.25 = 0.3125 \] \[ P(X = 4) = \binom{5}{4}(0.5)^4(0.5)^1 = 5 \times 0.0625 \times 0.5 = 0.1563 \] \[ P(X = 5) = \binom{5}{5}(0.5)^5 = 0.0313 \] \[ P(X \geq 3) = 0.3125 + 0.1563 + 0.0313 = \mathbf{0.5000} \]

By symmetry (\( p = 0.5 \)), the probability of a majority is exactly 0.5. ✓

Variant 1 — Free-throw shooter, \( n = 5 \), \( p = 0.60 \), majority means \( X \geq 3 \):

\[ P(X = 3) = \binom{5}{3}(0.6)^3(0.4)^2 = 10 \times 0.216 \times 0.16 = 0.3456 \] \[ P(X = 4) = \binom{5}{4}(0.6)^4(0.4)^1 = 5 \times 0.1296 \times 0.4 = 0.2592 \] \[ P(X = 5) = (0.6)^5 = 0.0778 \] \[ P(X \geq 3) = \mathbf{0.6826} \]

Variant 2 — T/F quiz, \( n = 6 \), \( p = 0.50 \), majority means \( X \geq 4 \) (more than 3):

\[ P(X = 4) = \binom{6}{4}(0.5)^6 = 15 \times 0.015625 = 0.2344 \] \[ P(X = 5) = \binom{6}{5}(0.5)^6 = 6 \times 0.015625 = 0.0938 \] \[ P(X = 6) = (0.5)^6 = 0.0156 \] \[ P(X \geq 4) = \mathbf{0.3438} \]

Variant 3 — Treatment, \( n = 6 \), \( p = 0.70 \), majority means \( X \geq 4 \):

\[ P(X = 4) = \binom{6}{4}(0.7)^4(0.3)^2 = 15 \times 0.2401 \times 0.09 = 0.3241 \] \[ P(X = 5) = \binom{6}{5}(0.7)^5(0.3)^1 = 6 \times 0.16807 \times 0.3 = 0.3025 \] \[ P(X = 6) = (0.7)^6 = 0.1176 \] \[ P(X \geq 4) = \mathbf{0.7443} \]

Variant 4 — Quality control, \( n = 5 \), \( p = 0.80 \), majority means \( X \geq 3 \):

\[ P(X = 3) = \binom{5}{3}(0.8)^3(0.2)^2 = 10 \times 0.512 \times 0.04 = 0.2048 \] \[ P(X = 4) = \binom{5}{4}(0.8)^4(0.2)^1 = 5 \times 0.4096 \times 0.2 = 0.4096 \] \[ P(X = 5) = (0.8)^5 = 0.3277 \] \[ P(X \geq 3) = \mathbf{0.9421} \]

---

IP4 — Identify BINS Violations (Find the Error, All 5 Variants)

Variant 0 — Librarian selects 10 of 25 books without replacement:

Error: Conditions I and S both fail. Sampling without replacement from a finite population (25 books) makes draws dependent. After selecting a fiction book, only 7 of 24 remain fiction — \( p \) changes from 0.32. The binomial assumption of constant \( p = 8/25 \) is violated.

Correct model: Hypergeometric. \( P(X = k) = \dfrac{\binom{8}{k}\binom{17}{10-k}}{\binom{25}{10}} \).

Variant 1 — Circuit boards tested until 2 defective found:

Error: Condition N fails. There is no fixed number of trials — testing continues until 2 defectives are found. The total count of boards tested is itself a random variable.

Correct model: Negative Binomial distribution (waiting for the \( r \)-th success).

Variant 2 — Quiz with decreasing probability (0.90, 0.80, 0.65, 0.55, 0.45); student averages to \( p = 0.67 \):

Error: Condition S fails. The probability of success varies across questions and cannot be validly averaged to a single constant \( p \). Using the average probability in the binomial formula would give incorrect probabilities for each value of \( k \).

Correct approach: Since questions are independent but have different \( p_i \) values, compute the distribution directly (Poisson-Binomial distribution) or use the general PMF approach from PR-4. The binomial formula does not apply.

Variant 3 — Student says "sampling without replacement" but the problem says "with replacement":

Error: The student misread the problem. With replacement, I and S hold (\( p = 3/10 = 0.30 \) on every draw). All BINS conditions are satisfied.

Correct conclusion: \( X \sim B(8, 0.30) \). The student incorrectly assumed without-replacement when the scenario explicitly states the chip is replaced and reshuffled.

Variant 4 — Salesperson calls until first sale; student writes \( X \sim B(n, 0.20) \):

Error: Condition N fails. The salesperson stops at the first sale, so the number of calls is not fixed in advance. Writing "\( n \) = however many calls she makes" treats a random variable as a parameter.

Correct model: Geometric distribution. \( P(X = k) = (0.80)^{k-1}(0.20) \) for \( k = 1, 2, 3, \ldots \)

---

IP5 — Shape and Full Workflow (Generator Template)

Generated by generateMeanSD(). Template for any generated \( n \), \( p \), and context:

1. (a) \( \mu = np = [n]([p]) = [np] \)
2. (b) \( \sigma_X^2 = np(1-p) = [n]([p])([q]) = [npq] \)
3. (c) \( \sigma_X = \sqrt{[npq]} \approx [sigma] \)
4. Shape: Since \( p = [p] \), the distribution is
   • right-skewed if \( [p] < 0.5 \) (tail extends toward large values of \( k \))
   • symmetric if \( [p] = 0.5 \)
   • left-skewed if \( [p] > 0.5 \) (tail extends toward small values of \( k \))
5. Interpretation: "In the long run, [context] averages [np] [units] per [period]. The actual count typically deviates from this by about [sigma] [units]."

Question 1 — Feynman Test (Model Answer)

The combination coefficient \( \binom{n}{k} \) is needed because the binomial formula must account for every possible arrangement of \( k \) successes among \( n \) trials, not just one specific arrangement.

Concrete example (\( n = 3 \), \( k = 2 \)): Suppose each trial independently succeeds with probability \( p \). The three sequences that produce exactly 2 successes are:

• SSF: probability \( p \cdot p \cdot (1-p) = p^2(1-p) \)
• SFS: probability \( p \cdot (1-p) \cdot p = p^2(1-p) \)
• FSS: probability \( (1-p) \cdot p \cdot p = p^2(1-p) \)

Each sequence has the same probability \( p^2(1-p) \). There are \( \binom{3}{2} = 3 \) such sequences. The total probability of exactly 2 successes is \( 3 \cdot p^2(1-p) \).

Without \( \binom{n}{k} \), the expression \( p^k(1-p)^{n-k} \) gives only the probability of one specific sequence (e.g., SSF) — not the probability that exactly \( k \) successes occur somewhere in the \( n \) trials. The coefficient counts all arrangements and aggregates them.

---

Question 2 — Applied Scenario (Vaccine, \( n = 7 \), \( p = 0.85 \))

(a) BINS check: B ✓ (effective/not), I ✓ (stated independent), N ✓ (\( n = 7 \) fixed), S ✓ (\( p = 0.85 \) constant). So \( X \sim B(7, 0.85) \).

(b)

\[ P(X = 5) = \binom{7}{5}(0.85)^5(0.15)^2 = 21 \times 0.4437 \times 0.0225 \approx \mathbf{0.2097} \]

(c) \( P(X \geq 6) \) — two equivalent approaches:

Direct sum (simpler here, only 2 terms):

\[ P(X = 6) = \binom{7}{6}(0.85)^6(0.15)^1 = 7 \times 0.3771 \times 0.15 \approx 0.3960 \] \[ P(X = 7) = (0.85)^7 \approx 0.3206 \] \[ P(X \geq 6) = 0.3960 + 0.3206 = \mathbf{0.7166} \]

Complement form (equivalent): \( P(X \geq 6) = 1 - P(X \leq 5) \) — this requires summing 6 terms (k = 0 through 5), so the direct sum is more efficient for this particular problem.

---

Question 3 — Error Analysis (\( X \sim B(8, 0.30) \), find \( P(X = 3) \))

Error identified: Missing combination coefficient.

The student computed \( (0.30)^3(0.70)^5 \approx 0.00454 \), which is the probability of one specific sequence with 3 successes followed by 5 failures (i.e., exactly SSSFFFFF). This is not \( P(X = 3) \).

Correct calculation:

\[ P(X = 3) = \binom{8}{3}(0.30)^3(0.70)^5 = 56 \times 0.027 \times 0.16807 = 56 \times 0.004538 \approx \mathbf{0.2541} \]

The student's answer (0.00454) is exactly \( 1/56 \) of the correct answer — confirming the missing factor is \( \binom{8}{3} = 56 \). There are 56 different arrangements of 3 successes in 8 trials, each with equal probability. The binomial formula counts and sums all of them.

Path A — The Analyst (Quality Control, \( n = 15 \), \( p = 0.08 \))

Task 1 — Verify BINS:

• B ✓: Each item is defective (success for counting) or not (failure).
• I ✓: Items are sampled independently (stated in scenario).
• N ✓: \( n = 15 \) fixed in advance.
• S ✓: \( p = 0.08 \) constant — same defect rate for every item.

So \( X \sim B(15, 0.08) \), where \( X \) = number of defective items in the sample.

Task 2 — Compute \( P(X = 0) \), \( P(X = 1) \), \( P(X = 2) \):

\[ P(X = 0) = \binom{15}{0}(0.08)^0(0.92)^{15} = 1 \times 1 \times (0.92)^{15} \approx 1 \times 1 \times 0.2863 = \mathbf{0.2863} \] \[ P(X = 1) = \binom{15}{1}(0.08)^1(0.92)^{14} = 15 \times 0.08 \times 0.3112 \approx \mathbf{0.3734} \] \[ P(X = 2) = \binom{15}{2}(0.08)^2(0.92)^{13} = 105 \times 0.0064 \times 0.3383 \approx \mathbf{0.2273} \]

Task 3 — Find \( P(X \leq 2) \) and make the approval decision:

\[ P(X \leq 2) = 0.2863 + 0.3734 + 0.2273 = \mathbf{0.8870} \]

Since \( P(X \leq 2) \approx 0.887 \geq 0.70 \), the company policy threshold is met. Conclusion: Approve the batch. The probability of finding 2 or fewer defective items in a sample of 15 is approximately 88.7%, which exceeds the required 70% threshold.

Task 4 — Compute \( \mu \) and \( \sigma_X \), interpret:

\[ \mu = np = 15(0.08) = \mathbf{1.2 \text{ defective items (on average)}} \] \[ \sigma_X^2 = np(1-p) = 15(0.08)(0.92) = 1.104 \] \[ \sigma_X = \sqrt{1.104} \approx \mathbf{1.05} \]

On average, you expect about 1.2 defective items in a sample of 15. The interval \( \mu \pm 2\sigma_X = 1.2 \pm 2(1.05) = (-0.9,\ 3.3) \). Since counts cannot be negative, this is effectively (0, 3.3): in most samples you would expect 0 to 3 defective items.

---

Path B — The Architect (Carnival Game Design)

Note: There are infinitely many valid \( (n, p) \) combinations satisfying \( \mu = np = 2.4 \) and \( \sigma_X < 1.5 \). The worked example below uses \( n = 8 \), \( p = 0.30 \). Other valid pairs include \( (n=12, p=0.20) \), \( (n=6, p=0.40) \), and others — verify each satisfies both constraints.

Tasks 1 and 2 — Choosing \( (n, p) \) and verifying constraints:

Choose \( n = 8 \). From \( \mu = np = 2.4 \): \( p = 2.4/8 = 0.30 \). Valid since \( 0 < 0.30 < 1 \). ✓

\[ \sigma_X = \sqrt{np(1-p)} = \sqrt{8(0.30)(0.70)} = \sqrt{1.68} \approx 1.296 \]

Since \( 1.296 < 1.5 \), the \( \sigma_X \) constraint is satisfied. ✓

Task 3 — Compute the probability a player wins (majority of 8 = at least 5 heads):

Winning requires \( X \geq 5 \) (more than 4 heads out of 8). Using complement:

\[ P(X \geq 5) = 1 - P(X \leq 4) \] \[ P(X = 0) = (0.70)^8 \approx 0.0576 \] \[ P(X = 1) = \binom{8}{1}(0.30)(0.70)^7 = 8 \times 0.30 \times 0.0824 \approx 0.1977 \] \[ P(X = 2) = \binom{8}{2}(0.30)^2(0.70)^6 = 28 \times 0.09 \times 0.1176 \approx 0.2965 \] \[ P(X = 3) = \binom{8}{3}(0.30)^3(0.70)^5 = 56 \times 0.027 \times 0.1681 \approx 0.2541 \] \[ P(X = 4) = \binom{8}{4}(0.30)^4(0.70)^4 = 70 \times 0.0081 \times 0.2401 \approx 0.1361 \] \[ P(X \leq 4) \approx 0.0576 + 0.1977 + 0.2965 + 0.2541 + 0.1361 = 0.9420 \] \[ P(X \geq 5) = 1 - 0.9420 = \mathbf{0.0580} \]

A player wins approximately 5.8% of the time with \( n = 8 \) flips and \( p = 0.30 \).

Task 4 — Game description with BINS verification:

Example: "Flip a biased coin 8 times. The coin lands heads with probability 0.30. Win if you get 5 or more heads."

• B ✓: Each flip is heads (success) or tails (failure).
• I ✓: Each flip is independent — coin has no memory.
• N ✓: \( n = 8 \) flips fixed in advance.
• S ✓: \( p = 0.30 \) constant on every flip (same biased coin throughout).

Challenge 1 — Algebraic Derivation of \( E(X) = np \)

We show that \( E(X) = \sum_{k=0}^{n} k \cdot P(X = k) = np \) for \( X \sim B(n, p) \).

1. Start from the definition of expected value: \[ E(X) = \sum_{k=0}^{n} k \binom{n}{k} p^k (1-p)^{n-k} \]
2. Drop the \( k = 0 \) term (it contributes 0): \[ E(X) = \sum_{k=1}^{n} k \binom{n}{k} p^k (1-p)^{n-k} \]
3. Apply the identity \( k\binom{n}{k} = n\binom{n-1}{k-1} \) to each term. (Verify: \( k \cdot \dfrac{n!}{k!(n-k)!} = \dfrac{n!}{(k-1)!(n-k)!} = n \cdot \dfrac{(n-1)!}{(k-1)!((n-1)-(k-1))!} = n\binom{n-1}{k-1} \). ✓) \[ E(X) = \sum_{k=1}^{n} n\binom{n-1}{k-1} p^k (1-p)^{n-k} \]
4. Factor out \( np \) (since \( p^k = p \cdot p^{k-1} \)): \[ E(X) = np \sum_{k=1}^{n} \binom{n-1}{k-1} p^{k-1} (1-p)^{(n-1)-(k-1)} \]
5. Re-index with \( j = k - 1 \) (as \( k \) runs from 1 to \( n \), \( j \) runs from 0 to \( n-1 \)): \[ E(X) = np \sum_{j=0}^{n-1} \binom{n-1}{j} p^j (1-p)^{(n-1)-j} \]
6. Recognize the sum as the total probability of \( B(n-1, p) \), which sums to 1 (all PMF values sum to 1 for any valid distribution): \[ E(X) = np \cdot 1 = \mathbf{np} \quad \square \]

This derivation shows that the shortcut formula \( \mu = np \) is not magic — it follows directly from the general expected-value definition via a combinatorial identity.

---

Challenge 2 — Mode of the Binomial Distribution

The mode formula: \( k^* = \lfloor (n+1)p \rfloor \) when \( (n+1)p \) is not a whole number; two modes at \( (n+1)p - 1 \) and \( (n+1)p \) when it is a whole number.

(a) \( B(10, 0.30) \):

\( (n+1)p = 11 \times 0.30 = 3.3 \). Not a whole number, so unique mode at \( k^* = \lfloor 3.3 \rfloor = \mathbf{3} \).

Verification:

\[ P(X = 2) = \binom{10}{2}(0.3)^2(0.7)^8 = 45 \times 0.09 \times 0.05765 \approx 0.2335 \] \[ P(X = 3) = \binom{10}{3}(0.3)^3(0.7)^7 = 120 \times 0.027 \times 0.08235 \approx 0.2668 \] \[ P(X = 4) = \binom{10}{4}(0.3)^4(0.7)^6 = 210 \times 0.0081 \times 0.1176 \approx 0.2001 \]

\( P(X = 3) \approx 0.2668 \) is the largest of the three. ✓ The mode is confirmed at \( k = 3 \).

(b) \( B(10, 0.70) \):

\( (n+1)p = 11 \times 0.70 = 7.7 \). Not a whole number, so unique mode at \( k^* = \lfloor 7.7 \rfloor = \mathbf{7} \).

Consistency with C5 shape rule: \( p = 0.70 > 0.5 \) produces a left-skewed distribution, so the mode lies above the midpoint \( n/2 = 5 \). The mode at \( k = 7 \) is above the center, with probability mass concentrated toward the high end and a long left tail. This confirms C5: \( p > 0.5 \Rightarrow \) left-skewed with mode above center. ✓

Symmetry check: \( B(10, 0.70) \) is the mirror image of \( B(10, 0.30) \). The mode of \( B(10, 0.30) \) is at \( k = 3 \), and \( 10 - 3 = 7 \) — confirming the mode of \( B(10, 0.70) \) is at the "mirrored" position. ✓