Binomial Distributions — Full Worked Solutions

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Full Worked Solutions

1. Question: A multiple-choice test has 8 questions, each with 4 options (one correct). A student guesses every answer randomly. Identify n and p for a Binomial model and verify the BINS conditions.

   Verifying BINS Conditions

   Binary: Each question is either correct (success) or incorrect (failure). ✓

   Independent: Guessing one question does not affect any other. ✓

   Number: Fixed number of questions: n = 8. ✓

   Success probability: Constant p = 1/4 = 0.25 for each question (random guess from 4 options). ✓

   All four conditions are satisfied, so X ~ B(8, 0.25) where X = number of correct answers.

2. Question: X ~ B(8, 0.4). Find P(X = 3).

   Applying the Binomial Formula

   P(X = k) = ⁿC_k × p^k × (1 − p)^{n − k}

   P(X = 3) = ⁸C₃ × (0.4)³ × (0.6)⁵

   ⁸C₃ = 8!/(3! × 5!) = 56

   (0.4)³ = 0.064

   (0.6)⁵ = 0.07776

   P(X = 3) = 56 × 0.064 × 0.07776 = 56 × 0.00497664 = 0.2787 (to 4 d.p.)

3. Question: X ~ B(5, 0.3). Find P(X = 0).

   Finding P(X = 0)

   P(X = 0) = ⁵C₀ × (0.3)⁰ × (0.7)⁵

   ⁵C₀ = 1,   (0.3)⁰ = 1

   (0.7)⁵ = 0.7 × 0.7 × 0.7 × 0.7 × 0.7 = 0.16807

   P(X = 0) = 1 × 1 × 0.16807 = 0.1681 (to 4 d.p.)

   Interpretation: There is about a 16.8% chance of zero successes in 5 trials when p = 0.3.

4. Question: X ~ B(6, 0.25). Find P(X ≤ 2).

   Cumulative Probability: P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2)

   n = 6, p = 0.25, q = 0.75

   P(X = 0): ⁶C₀ × (0.25)⁰ × (0.75)⁶ = 1 × 1 × 0.17798 = 0.17798

   P(X = 1): ⁶C₁ × (0.25)¹ × (0.75)⁵ = 6 × 0.25 × 0.23730 = 0.35596

   P(X = 2): ⁶C₂ × (0.25)² × (0.75)⁴ = 15 × 0.0625 × 0.31641 = 0.29663

   P(X ≤ 2) = 0.17798 + 0.35596 + 0.29663 = 0.8306 (to 4 d.p.)

5. Question: X ~ B(7, 0.5). Find P(X ≥ 4) using the complement rule.

   Using the Complement Rule

   P(X ≥ 4) = 1 − P(X ≤ 3)

   n = 7, p = q = 0.5

   P(X = 0) = (0.5)⁷ = 1/128 ≈ 0.00781

   P(X = 1) = 7 × (0.5)⁷ = 7/128 ≈ 0.05469

   P(X = 2) = 21 × (0.5)⁷ = 21/128 ≈ 0.16406

   P(X = 3) = 35 × (0.5)⁷ = 35/128 ≈ 0.27344

   P(X ≤ 3) = (1 + 7 + 21 + 35)/128 = 64/128 = 0.5

   P(X ≥ 4) = 1 − 0.5 = 0.5

   Note: By symmetry (p = 0.5, n = 7), P(X ≥ 4) = P(X ≤ 3) = 0.5, which confirms our answer.

6. Question: X ~ B(12, 0.35). Find E(X) and Var(X).

   Mean and Variance of a Binomial

   n = 12, p = 0.35, q = 1 − 0.35 = 0.65

   E(X) = np = 12 × 0.35 = 4.2

   Var(X) = np(1 − p) = 12 × 0.35 × 0.65 = 12 × 0.2275 = 2.73

   Interpretation: On average, 4.2 out of every 12 trials will be successes.

7. Question: X ~ B(12, 0.35). Find SD(X), correct to 3 decimal places.

   Standard Deviation

   From Q6: Var(X) = 2.73

   SD(X) = √Var(X) = √2.73 ≈ 1.652 (to 3 d.p.)

   Alternatively: SD(X) = √(np(1−p)) = √(12 × 0.35 × 0.65) = √2.73 ≈ 1.652

8. Question: For X ~ B(n, p), you are given E(X) = 6 and Var(X) = 2.4. Find n and p.

   Setting Up Simultaneous Equations

   E(X) = np = 6   … (1)

   Var(X) = np(1 − p) = 2.4   … (2)

   Solving

   Divide equation (2) by equation (1):

   np(1 − p) / np = 2.4 / 6

   (1 − p) = 0.4 ⇒ p = 0.6

   Substitute into (1): n × 0.6 = 6 ⇒ n = 10

   Therefore X ~ B(10, 0.6).

   Check: np = 10 × 0.6 = 6 ✓, np(1−p) = 10 × 0.6 × 0.4 = 2.4 ✓

9. Question: A factory produces electronic components; 5% are defective. A quality inspector randomly tests 20 components. Let X = number of defective components. Find P(at least 1 defective).

   Setting Up the Model

   X ~ B(20, 0.05), so n = 20, p = 0.05, q = 0.95

   Using the Complement Rule

   P(at least 1 defective) = P(X ≥ 1) = 1 − P(X = 0)

   P(X = 0) = ²⁰C₀ × (0.05)⁰ × (0.95)²⁰

   = 1 × 1 × (0.95)²⁰

   (0.95)²⁰ ≈ 0.3585

   P(X ≥ 1) = 1 − 0.3585 = 0.6415

   Interpretation: There is about a 64.2% chance that the inspector finds at least one defective component in a batch of 20.

10. Question: A factory produces light globes; 8% are defective. A quality inspector randomly tests 15 globes. Let X = number of defective globes. Find P(X = 0) and P(X ≥ 2). Interpret your answers.

    Setting Up the Model

    X ~ B(15, 0.08), so n = 15, p = 0.08, q = 0.92

    P(X = 0): No Defective Globes

    P(X = 0) = (0.92)¹⁵

    (0.92)¹⁵ ≈ 0.2863

    P(X = 0) ≈ 0.2863

    Interpretation: There is about a 28.6% chance of finding no defective globes in a batch of 15. This seems reasonable — with only an 8% defect rate, a clean batch is quite likely.

    P(X ≥ 2): Two or More Defective

    P(X ≥ 2) = 1 − P(X ≤ 1) = 1 − [P(X = 0) + P(X = 1)]

    P(X = 1) = ¹⁵C₁ × (0.08)¹ × (0.92)¹⁴

    = 15 × 0.08 × (0.92)¹⁴

    (0.92)¹⁴ = (0.92)¹⁵ / 0.92 ≈ 0.2863 / 0.92 ≈ 0.3112

    P(X = 1) = 15 × 0.08 × 0.3112 ≈ 0.3734

    P(X ≥ 2) = 1 − 0.2863 − 0.3734 = 0.3403

    Interpretation: There is about a 34.0% chance that the inspector finds 2 or more defective globes. Despite the relatively low defect rate, testing 15 items gives a substantial chance of catching multiple defects.