A lot of 37 precast panels contains 5 defectives. 6 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Start with the thinking

• Sampling without replacement from a small lot is hypergeometric, not binomial.
• Each probability is a ratio of combination counts.

Step-by-step solution

1. Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)

2. Denominator

   $$C(37, 6) = 2,324,784$$

3. x = 0 — C(5,0)·C(32,6) = 906,192, so P(0) = 0.3898

4. x = 1 — C(5,1)·C(32,5) = 1,006,880, so P(1) = 0.4331

5. At least one defective

   $$1 - P(0) = 0.6102$$

Why the other options are there

• 0.4185 (binomial approximation)
• 0.8108 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

A lot of 22 precast panels contains 7 defectives. 5 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Start with the thinking

• Sampling without replacement from a small lot is hypergeometric, not binomial.
• Each probability is a ratio of combination counts.

Step-by-step solution

1. Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)

2. Denominator

   $$C(22, 5) = 26,334$$

3. x = 0 — C(7,0)·C(15,5) = 3,003, so P(0) = 0.1140

4. x = 1 — C(7,1)·C(15,4) = 9,555, so P(1) = 0.3628

5. At least one defective

   $$1 - P(0) = 0.8860$$

Why the other options are there

• 0.1473 (binomial approximation)
• 1.5909 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

A lot of 38 precast panels contains 7 defectives. 8 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Start with the thinking

• Sampling without replacement from a small lot is hypergeometric, not binomial.
• Each probability is a ratio of combination counts.

Step-by-step solution

1. Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)

2. Denominator

   $$C(38, 8) = 48,903,492$$

3. x = 0 — C(7,0)·C(31,8) = 7,888,725, so P(0) = 0.1613

4. x = 1 — C(7,1)·C(31,7) = 18,407,025, so P(1) = 0.3764

5. At least one defective

   $$1 - P(0) = 0.8387$$

Why the other options are there

• 0.1962 (binomial approximation)
• 1.4737 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

A lot of 40 precast panels contains 4 defectives. 7 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Start with the thinking

• Sampling without replacement from a small lot is hypergeometric, not binomial.
• Each probability is a ratio of combination counts.

Step-by-step solution

1. Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)

2. Denominator

   $$C(40, 7) = 18,643,560$$

3. x = 0 — C(4,0)·C(36,7) = 8,347,680, so P(0) = 0.4478

4. x = 1 — C(4,1)·C(36,6) = 7,791,168, so P(1) = 0.4179

5. At least one defective

   $$1 - P(0) = 0.5522$$

Why the other options are there

• 0.4783 (binomial approximation)
• 0.7000 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

A lot of 27 precast panels contains 8 defectives. 5 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Start with the thinking

• Sampling without replacement from a small lot is hypergeometric, not binomial.
• Each probability is a ratio of combination counts.

Step-by-step solution

1. Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)

2. Denominator

   $$C(27, 5) = 80,730$$

3. x = 0 — C(8,0)·C(19,5) = 11,628, so P(0) = 0.1440

4. x = 1 — C(8,1)·C(19,4) = 31,008, so P(1) = 0.3841

5. At least one defective

   $$1 - P(0) = 0.8560$$

Why the other options are there

• 0.1726 (binomial approximation)
• 1.4815 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

A lot of 35 precast panels contains 8 defectives. 6 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Start with the thinking

• Sampling without replacement from a small lot is hypergeometric, not binomial.
• Each probability is a ratio of combination counts.

Step-by-step solution

1. Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)

2. Denominator

   $$C(35, 6) = 1,623,160$$

3. x = 0 — C(8,0)·C(27,6) = 296,010, so P(0) = 0.1824

4. x = 1 — C(8,1)·C(27,5) = 645,840, so P(1) = 0.3979

5. At least one defective

   $$1 - P(0) = 0.8176$$

Why the other options are there

• 0.2108 (binomial approximation)
• 1.3714 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

A lot of 20 precast panels contains 4 defectives. 7 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Start with the thinking

• Sampling without replacement from a small lot is hypergeometric, not binomial.
• Each probability is a ratio of combination counts.

Step-by-step solution

1. Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)

2. Denominator

   $$C(20, 7) = 77,520$$

3. x = 0 — C(4,0)·C(16,7) = 11,440, so P(0) = 0.1476

4. x = 1 — C(4,1)·C(16,6) = 32,032, so P(1) = 0.4132

5. At least one defective

   $$1 - P(0) = 0.8524$$

Why the other options are there

• 0.2097 (binomial approximation)
• 1.4000 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

A lot of 27 precast panels contains 4 defectives. 4 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Start with the thinking

• Sampling without replacement from a small lot is hypergeometric, not binomial.
• Each probability is a ratio of combination counts.

Step-by-step solution

1. Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)

2. Denominator

   $$C(27, 4) = 17,550$$

3. x = 0 — C(4,0)·C(23,4) = 8,855, so P(0) = 0.5046

4. x = 1 — C(4,1)·C(23,3) = 7,084, so P(1) = 0.4036

5. At least one defective

   $$1 - P(0) = 0.4954$$

Why the other options are there

• 0.5266 (binomial approximation)
• 0.5926 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

A lot of 29 precast panels contains 7 defectives. 8 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Start with the thinking

• Sampling without replacement from a small lot is hypergeometric, not binomial.
• Each probability is a ratio of combination counts.

Step-by-step solution

1. Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)

2. Denominator

   $$C(29, 8) = 4,292,145$$

3. x = 0 — C(7,0)·C(22,8) = 319,770, so P(0) = 0.0745

4. x = 1 — C(7,1)·C(22,7) = 1,193,808, so P(1) = 0.2781

5. At least one defective

   $$1 - P(0) = 0.9255$$

Why the other options are there

• 0.1097 (binomial approximation)
• 1.9310 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

A lot of 40 precast panels contains 4 defectives. 6 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Start with the thinking

• Sampling without replacement from a small lot is hypergeometric, not binomial.
• Each probability is a ratio of combination counts.

Step-by-step solution

1. Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)

2. Denominator

   $$C(40, 6) = 3,838,380$$

3. x = 0 — C(4,0)·C(36,6) = 1,947,792, so P(0) = 0.5075

4. x = 1 — C(4,1)·C(36,5) = 1,507,968, so P(1) = 0.3929

5. At least one defective

   $$1 - P(0) = 0.4925$$

Why the other options are there

• 0.5314 (binomial approximation)
• 0.6000 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper