4.4 Discrete Distributions: Binomial, Poisson, and Hypergeometric

4.4 Discrete Distributions: Binomial, Poisson, and Hypergeometric

Fundamentals of Probability for Quality Metrology

Before evaluating discrete probability distributions, quality technicians must master the fundamental mathematical rules governing probability. An experiment is an inspection or testing activity that generates an observable outcome. The sample space ($S$) is the collection of all possible outcomes. An event ($A$) is a specific subset of outcomes within that sample space.

Fundamental Probability Axioms

1. For any event $A$, its probability is bounded between zero and one: $0 \le P(A) \le 1$ where $P(A) = 0$ denotes an impossible outcome, and $P(A) = 1$ denotes an absolute certainty.
2. The sum of probabilities for all mutually exclusive outcomes in the sample space equals 1.0: $P(S) = 1$
3. Complement Rule: The probability that event $A$ does not occur equals one minus the probability that it does occur: $P(A^c) = 1 - P(A)$

The Addition Rule (Union of Events)

The probability that event $A$ or event $B$ occurs:

• Mutually Exclusive (Disjoint) Events: Events that cannot occur simultaneously (e.g., a stamped part cannot be simultaneously conforming and nonconforming): $P(A \cup B) = P(A) + P(B)$
• Non-Mutually Exclusive Events: Events that can occur simultaneously (e.g., a circuit board can have both a soldering defect and a missing component): $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

The Multiplication Rule (Joint Probability)

The probability that both event $A$ and event $B$ occur together:

• Independent Events: The occurrence of event $A$ has no influence on the probability of event $B$: $P(A \cap B) = P(A) \times P(B)$
• Dependent Events (Conditional Probability): The occurrence of event $A$ alters the probability of event $B$: $P(A \cap B) = P(A) \times P(B|A)$ where $P(B|A)$ is the conditional probability of $B$ occurring given that $A$ has already occurred.

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The Binomial Distribution

The Binomial distribution is the foundational model for attribute lot inspection where parts are classified into binary categories (Good vs. Defective, Conforming vs. Nonconforming, Pass vs. Fail).

The Four Mandatory BINS Conditions

For a process to be modeled by the binomial distribution, all four BINS conditions must be satisfied:

1. B — Binary Outcomes: Each trial results in one of only two mutually exclusive states (Success or Failure, Nonconforming or Conforming).
2. I — Independent Trials: The outcome of any given trial does not affect the outcome of any subsequent trial.
3. n — Number of Trials is Fixed: The sample size $n$ is predetermined before testing begins.
4. S — Same Probability: The probability of nonconformance ($p$) remains constant across all trials. This condition is met when sampling from an ongoing continuous production stream or when sampling with replacement from a finite lot.

Binomial Probability Mass Function (PMF)

The probability of finding exactly $x$ nonconforming units in a random sample of $n$ units is: $P(X = x) = \binom{n}{x} p^x (1 - p)^{n - x} = \frac{n!}{x!(n - x)!} p^x (1 - p)^{n - x}$ where:

• $\binom{n}{x} = \frac{n!}{x!(n - x)!}$ is the combination coefficient (the number of distinct ways to choose $x$ items from $n$).
• $p$ is the historical probability of obtaining a nonconforming unit.
• $(1 - p) = q$ is the probability of obtaining a conforming unit.

Parameters and Summary Metrics

• Mean (Expected Value): $\mu = np$
• Variance: $\sigma^2 = np(1 - p)$
• Standard Deviation: $\sigma = \sqrt{np(1 - p)}$

Quality Application and Worked Example

Scenario: A stamping press exhibits a historical nonconformance rate of $p = 0.04$ (4%). A quality technician selects a random sample of $n = 6$ brackets from the outgoing conveyor. What is the probability of finding:

1. Exactly zero nonconforming brackets?
2. Exactly one nonconforming bracket?
3. At least one nonconforming bracket?

Step 1: Exactly 0 Nonconforming Units ($x = 0$) $P(X = 0) = \binom{6}{0} (0.04)^0 (0.96)^{6 - 0} = 1 \times 1 \times (0.96)^6 \approx 0.7828 \quad (78.28\%)$

Step 2: Exactly 1 Nonconforming Unit ($x = 1$) $P(X = 1) = \binom{6}{1} (0.04)^1 (0.96)^{6 - 1} = 6 \times 0.04 \times (0.96)^5 = 0.24 \times 0.8154 \approx 0.1957 \quad (19.57\%)$

Step 3: At Least 1 Nonconforming Unit ($x \ge 1$) Applying the complement rule: $P(X \ge 1) = 1 - P(X = 0) = 1 - 0.7828 = 0.2172 \quad (21.72\%)$

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The Poisson Distribution

While the binomial distribution counts the number of nonconforming units (defectives), the Poisson distribution counts the number of discrete defects or blemishes distributed across a continuous area of opportunity (time, length, surface area, volume, or complex assembly).

Conditions for the Poisson Model

1. The count of events ($x$) occurs in a defined, continuous inspection interval (e.g., defects per 100 meters of extruded wire, voids per square meter of fiberglass sheet, blemishes per car door, solder bridges per circuit board).
2. The events occur independently of each other.
3. The average arrival or defect rate ($\lambda$, lambda) is constant across the inspection unit.
4. The probability of two events occurring at the exact same infinitesimal point in space or time is zero (rare event model).

Poisson Probability Mass Function (PMF)

$P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}$ where:

• $\lambda$ (lambda) is the average expected number of defects per inspection unit.
• $e \approx 2.71828$ is Euler's mathematical constant.
• $x$ is the observed number of defects ($x = 0, 1, 2, \dots$).

The Unique Hallmark of the Poisson Distribution

In a Poisson distribution, the mean and variance are mathematically identical: $\mu = \lambda$ $\sigma^2 = \lambda$ $\sigma = \sqrt{\lambda}$

[!IMPORTANT] Classic ASQ Exam Question: If a quality technician observes that the number of solder blemishes per board follows a Poisson distribution with an average of $\lambda = 4.0$ blemishes, what is the standard deviation? Answer: $\sigma = \sqrt{4.0} = 2.0$ blemishes. Whenever mean equals variance, suspect the Poisson distribution.

Poisson Worked Example: Automotive Paint Finish

Scenario: An automated optical inspection system audits painted truck doors, recording an average of $\lambda = 1.5$ paint pinholes per door. What is the probability that a randomly inspected door has:

1. Zero pinholes (a perfect door)?
2. Exactly two pinholes?

Step 1: Probability of 0 Pinholes ($x = 0$) $P(X = 0) = \frac{e^{-1.5} (1.5)^0}{0!} = \frac{0.22313 \times 1}{1} \approx 0.2231 \quad (22.31\%)$

Step 2: Probability of Exactly 2 Pinholes ($x = 2$) $P(X = 2) = \frac{e^{-1.5} (1.5)^2}{2!} = \frac{0.22313 \times 2.25}{2 \times 1} = \frac{0.50204}{2} \approx 0.2510 \quad (25.10\%)$

Poisson Approximation to the Binomial

When the sample size $n$ is very large ($n \ge 50$ or $100$) and the probability of defect $p$ is very small ($p \le 0.05$ or $0.10$), evaluating factorials in the binomial formula becomes computationally difficult. The Poisson distribution serves as an accurate approximation to the binomial by setting: $\lambda = np$

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The Hypergeometric Distribution

In precision manufacturing, technicians frequently inspect parts drawn from a small, finite lot without replacement (e.g., pulling 5 surgical drill bits from a sterile lot container of 25). Under sampling without replacement, the outcome of each draw directly alters the probability of defect on the next draw. The trials are dependent, violating the constant-probability assumption of the Binomial distribution. In this case, the Hypergeometric distribution must be used.

Conditions for the Hypergeometric Model

1. The population consists of a known, finite lot of size $N$.
2. The lot contains exactly $D$ nonconforming units and $N - D$ conforming units.
3. A sample of $n$ units is drawn without replacement.
4. Each draw is dependent; the probability of drawing a defective shifts after each selection.

Hypergeometric Probability Mass Function (PMF)

$P(X = x) = \frac{\binom{D}{x} \binom{N - D}{n - x}}{\binom{N}{n}}$ where:

• $N$ = total population lot size.
• $D$ = total number of nonconforming units in the lot.
• $n$ = sample size drawn.
• $x$ = number of nonconforming units observed in the sample.
• $\binom{D}{x}$ = ways to choose $x$ nonconforming units from the $D$ available.
• $\binom{N - D}{n - x}$ = ways to choose $(n - x)$ conforming units from the $(N - D)$ available.
• $\binom{N}{n}$ = total possible ways to draw $n$ units from lot $N$.

Mean and Variance of the Hypergeometric Distribution

• Mean: $\mu = n \left(\frac{D}{N}\right) = np$
• Variance: $\sigma^2 = n \left(\frac{D}{N}\right) \left(1 - \frac{D}{N}\right) \left(\frac{N - n}{N - 1}\right)$ Notice that the variance formula is identical to the Binomial variance multiplied by the Finite Population Correction Factor (FPCF): $\frac{N - n}{N - 1}$. As lot size $N$ grows large relative to sample size $n$, the FPCF approaches 1.0.

Hypergeometric Worked Example: Medical Pin Lot Acceptance

Scenario: A receiving lot consists of $N = 20$ precision titanium medical implants. Unknown to the technician, exactly $D = 4$ implants have burrs. The technician randomly selects $n = 3$ implants without replacement. What is the probability of finding exactly $x = 1$ nonconforming implant?

Step 1: Compute Combinations

• Ways to pick 1 defective from 4: $\binom{4}{1} = \frac{4!}{1!3!} = 4$
• Ways to pick 2 good implants from the 16 conforming ($20 - 4 = 16$): $\binom{16}{2} = \frac{16 \times 15}{2 \times 1} = 120$
• Total ways to pick 3 implants from the lot of 20: $\binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1,140$

Step 2: Calculate Probability $P(X = 1) = \frac{\binom{4}{1} \binom{16}{2}}{\binom{20}{3}} = \frac{4 \times 120}{1,140} = \frac{480}{1,140} \approx 0.4211 \quad (42.11\%)$

When Does the Binomial Approximate the Hypergeometric?

Calculating factorials for large lots is tedious. Quality engineering standards establish the following rule of thumb: $\text{If } \frac{n}{N} \le 0.10 \quad (10\% \text{ rule})$ When the sample size $n$ represents 10% or less of the total lot size $N$, sampling without replacement has a negligible impact on lot proportion $p$. In this case, the Binomial distribution provides an excellent, mathematically simple approximation to the Hypergeometric distribution. If $n / N > 0.10$, the Hypergeometric distribution is mandatory.

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Master Comparison Table: Discrete Distributions for Quality Inspection

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Defect vs. Defective: The Critical Metrology Distinction

A frequent source of confusion on the ASQ CQT exam is the operational difference between a defect and a defective:

• Defective (Nonconforming Unit): An entire part, component, or assembly that fails to meet one or more engineering specifications, rendering the part unserviceable or noncompliant. A part is binary: either it is defective, or it is conforming. Modeled by Binomial or Hypergeometric distributions ($p$ and $np$ charts).
• Defect (Nonconformance): A specific instance of a flaw, blemish, or departure from specification on an inspected unit. A single defective part may contain multiple distinct defects. For example, an automotive bumper cover (one unit) might have 3 paint runs, 1 scratch, and 2 unpunched mounting holes—representing 1 defective part containing 6 defects. Modeled by the Poisson distribution ($c$ and $u$ charts).