6.2 Discrete Probability Distributions

Discrete probability distributions model countable random phenomena where outcomes take on isolated integer values. In industrial and systems engineering, discrete distributions form the backbone of statistical quality control, acceptance sampling, inventory stockout evaluations, and service queue analysis. The NCEES FE Reference Handbook provides explicit formulas for the probability mass function (PMF), mean, and variance of several key distributions. Recognizing the physical conditions that dictate which distribution to apply is one of the most critical problem-solving skills on the FE exam.

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1. Discrete Random Variables: PMF, CDF, Expectation, and Variance

Probability Mass Function (PMF)

A discrete random variable $X$ assumes a countable set of distinct values ${x_1, x_2, \dots}$. Its probability mass function, denoted $p(x) = P(X = x)$, must satisfy two foundational criteria:

1. $p(x_i) \ge 0$ for all $x_i$
2. $\sum_{i} p(x_i) = 1$

Cumulative Distribution Function (CDF)

The cumulative distribution function, denoted $F(x)$, gives the probability that the random variable $X$ takes on a value less than or equal to a specified threshold $x$:

$F(x) = P(X \le x) = \sum_{t \le x} p(t)$

Properties of the discrete CDF include:

• $F(x)$ is a step function that is non-decreasing and continuous from the right (right-continuous).
• $\lim_{x \to -\infty} F(x) = 0$ and $\lim_{x \to \infty} F(x) = 1$.
• The probability of an exact outcome is the jump size at that value: $P(X = x) = F(x) - F(x^-)$.
• For any interval $(a, b]$ where $a < b$: $P(a < X \le b) = F(b) - F(a)$

Expected Value and Linear Properties

The expected value (or population mean $\mu$) is the probability-weighted average of all possible values:

$E(X) = \mu = \sum_{i} x_i \cdot p(x_i)$

Expected value is a strictly linear operator. For any constants $a$ and $b$, and any two random variables $X$ and $Y$ (whether independent or dependent):

$E(aX + b) = a E(X) + b$ $E(X + Y) = E(X) + E(Y)$

Variance and Standard Deviation

The variance $\sigma^2$ or $\text{Var}(X)$ measures the expected squared deviation about the mean:

$\text{Var}(X) = \sigma^2 = E[(X - \mu)^2] = \sum_{i} (x_i - \mu)^2 p(x_i)$

Using the computational shortcut formula:

$\text{Var}(X) = E(X^2) - [E(X)]^2 = \left( \sum_{i} x_i^2 p(x_i) \right) - \mu^2$

Properties of variance:

• $\text{Var}(aX + b) = a^2 \text{Var}(X)$ (adding a constant shifts location without altering spread; multiplying scales by $a^2$).
• Standard deviation is the positive square root: $\sigma = \sqrt{\text{Var}(X)}$.
• For independent random variables $X$ and $Y$: $\text{Var}(X \pm Y) = \text{Var}(X) + \text{Var}(Y)$ (Note: Variances always add; they never subtract!)

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2. The Binomial Distribution in Manufacturing Quality Control

Conditions for a Binomial Experiment (The BINS Framework)

A random variable $X$ follows a Binomial distribution, denoted $X \sim \text{Bin}(n, p)$, if and only if the process satisfies four conditions:

1. Binary Outcomes: Each trial results in one of exactly two mutually exclusive outcomes: Success ($S$) or Failure ($F$). In quality control, "success" often denotes finding a defective unit.
2. Independent Trials: The outcome of any individual trial does not influence any other trial.
3. Number of Trials ($n$) is Fixed: The experiment consists of a predetermined number of trials $n$.
4. Same Probability ($p$): The probability of success $p$ remains strictly constant across all trials (sampling with replacement or sampling from an infinite/very large lot).

Probability Mass Function, Mean, and Variance

The probability of obtaining exactly $k$ successes in $n$ independent trials is:

$P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} = \frac{n!}{k! \, (n - k)!} p^k (1 - p)^{n - k}, \quad k = 0, 1, 2, \dots, n$

• Mean: $E(X) = \mu = n p$
• Variance: $\text{Var}(X) = \sigma^2 = n p (1 - p) = n p q$ (where $q = 1 - p$)
• Standard Deviation: $\sigma = \sqrt{n p (1 - p)}$

Binomial Distribution Conditions (BINS):
  [B]inary outcomes ──────> Conforming vs. Non-conforming
  [I]ndependent trials ───> One part's status does not affect the next
  [N]umber of trials ─────> n is fixed in advance
  [S]ame probability ─────> p is constant across all trials

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3. The Poisson Distribution and Poisson Process Modeling

Conditions for a Poisson Process

The Poisson distribution models the number of occurrences of an event over a specified continuous interval (time, area, volume, or length). A Poisson process assumes:

1. Events occur at a constant average rate $\lambda$ per unit interval.
2. Occurrences in non-overlapping intervals are statistically independent.
3. The probability of two or more events occurring at the exact same instantaneous point in time/space is negligible (events occur individually).

Probability Mass Function, Mean, and Variance

If $X$ represents the count of events in an interval where the expected number of occurrences is $\lambda$:

$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k = 0, 1, 2, 3, \dots$

• Mean: $E(X) = \mu = \lambda$
• Variance: $\text{Var}(X) = \sigma^2 = \lambda$
• Standard Deviation: $\sigma = \sqrt{\lambda}$

Defining Characteristic: In a Poisson distribution, the variance is always equal to the mean ($\sigma^2 = \mu = \lambda$). If empirical defect data shows a variance significantly larger than the mean, the process exhibits overdispersion and violates Poisson assumptions.

Scaling the Rate Parameter

If occurrences occur at rate $\alpha$ per unit interval and the evaluated interval length is $t$, then:

$\lambda = \alpha \cdot t$

Example: If customer orders arrive at an automated distribution center at an average rate of $\alpha = 12\text{ orders/hour}$, then over a $20\text{-minute}$ window ($t = 1/3\text{ hour}$), the parameter is $\lambda = 12 \times (1/3) = 4.0\text{ orders}$.

Poisson Approximation to the Binomial

When the number of trials $n$ in a binomial experiment is large and the probability of success $p$ is small (modeling "rare events"), the binomial distribution is closely approximated by a Poisson distribution with:

$\lambda = n p$

Rule of Thumb for NCEES FE Exam: The approximation is acceptable when $n \ge 20$ and $p \le 0.05$, and exceptionally accurate when $n \ge 100$ and $n p \le 10$. This saves candidates from computing large factorials like $\binom{500}{3}$.

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4. The Hypergeometric Distribution: Finite-Lot Sampling Without Replacement

Physical Modeling Conditions

In quality acceptance sampling, parts are drawn from a finite lot of size $N$ without replacement. Because drawn parts are not replaced, the trials are dependent, and the probability of drawing a defective unit changes with every consecutive pick. This violates the binomial assumptions.

Let:

• $N$ = total population (lot) size
• $K$ = total number of successes (defective units) in the lot
• $n$ = sample size drawn without replacement
• $k$ = observed number of successes in the sample

Probability Mass Function, Mean, and Variance

$P(X = k) = \frac{\binom{K}{k} \binom{N - K}{n - k}}{\binom{N}{n}}, \quad \max(0, n - (N - K)) \le k \le \min(n, K)$

• Mean: $E(X) = \mu = n \left( \frac{K}{N} \right) = n p$ (where $p = K/N$ is the initial lot fraction defective)
• Variance: $\text{Var}(X) = \sigma^2 = n \left( \frac{K}{N} \right) \left( 1 - \frac{K}{N} \right) \left( \frac{N - n}{N - 1} \right) = n p (1 - p) \left( \frac{N - n}{N - 1} \right)$

The Finite Population Correction (FPC) Factor

The term $\frac{N - n}{N - 1}$ is known as the finite population correction (FPC) factor. Notice that:

• As the population size $N \to \infty$, the FPC $\frac{N - n}{N - 1} \to 1$, and the hypergeometric variance converges to the binomial variance $n p (1 - p)$.
• General Rule: When the sampling fraction is small ($n / N \le 0.05$ or $5%$), sampling without replacement can be modeled using the simpler Binomial distribution with minimal error.

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5. Geometric and Negative Binomial Distributions

The Geometric Distribution (Waiting for the First Success)

The geometric distribution models the number of Bernoulli trials $X$ required to encounter the first success.

• Assumptions: Independent trials, binary outcomes, constant success probability $p$ per trial.
• PMF (NCEES Trial Formulation): For the first success occurring on trial $k$ ($k - 1$ consecutive failures followed by one success): $P(X = k) = (1 - p)^{k - 1} p, \quad k = 1, 2, 3, \dots$
• Cumulative Distribution: $P(X \le k) = 1 - (1 - p)^k$
• Tail Probability: $P(X > k) = (1 - p)^k$ (probability of failing $k$ consecutive trials)
• Mean: $E(X) = \mu = \frac{1}{p}$
• Variance: $\text{Var}(X) = \sigma^2 = \frac{1 - p}{p^2}$

Memoryless Property of the Geometric Distribution The geometric distribution is the only discrete distribution that possesses the memoryless property: $P(X > s + t \mid X > s) = P(X > t)$ In an inspection process, if an inspector has already examined $s = 20$ parts without finding a defect, the probability of requiring at least $t = 5$ more trials to find the first defect is identical to starting fresh from trial zero. Past non-defective parts provide no accumulated "credit."

The Negative Binomial Distribution (Waiting for $r$ Successes)

The negative binomial distribution generalizes the geometric distribution. It models the total number of trials $X$ required to observe exactly $r$ successes.

• PMF: For the $r$-th success to occur on the $k$-th trial, exactly $r - 1$ successes must occur in the preceding $k - 1$ trials (which can happen in $\binom{k - 1}{r - 1}$ ways), followed by a success on trial $k$: $P(X = k) = \binom{k - 1}{r - 1} p^r (1 - p)^{k - r}, \quad k = r, r + 1, r + 2, \dots$
• Relationship to Geometric: A negative binomial variable is the sum of $r$ independent and identically distributed geometric random variables ($X = G_1 + G_2 + \dots + G_r$).
• Mean: $E(X) = \mu = \frac{r}{p}$
• Variance: $\text{Var}(X) = \sigma^2 = \frac{r(1 - p)}{p^2}$

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6. Comparative Summary of Discrete Probability Distributions

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7. Step-by-Step Worked Engineering Examples

Worked Example 6.2.1: Binomial vs. Poisson Approximation in PCB Solder Defect Screening

Problem: A wave soldering process produces solder shorts on complex circuit boards at an average probability of $p = 0.004$ per solder joint. A single production board contains $n = 500$ solder joints.

1. Compute the exact Binomial probability that a finished board has at most 1 solder defect.
2. Compute the Poisson approximation for the same event and determine the percentage error.

Solution:

1. Exact Binomial Calculation:
   • Let $X \sim \text{Bin}(500, 0.004)$.
   • $P(X \le 1) = P(X = 0) + P(X = 1)$
   • $P(X = 0) = \binom{500}{0} (0.004)^0 (0.996)^{500} = (0.996)^{500} \approx 0.13482$
   • $P(X = 1) = \binom{500}{1} (0.004)^1 (0.996)^{499} = 500 \times 0.004 \times (0.996)^{499} = 2.0 \times 0.13536 \approx 0.27071$
   • Total exact probability: $P(X \le 1) = 0.13482 + 0.27071 = 0.40553 \approx 40.55\%$
2. Poisson Approximation Calculation:
   • Check conditions: $n = 500 \ge 20$, $p = 0.004 \le 0.05$, and $n p = 500(0.004) = 2.0 \le 10$. The approximation is valid.
   • Set parameter $\lambda = n p = 2.0$.
   • $P(Y \le 1) = P(Y = 0) + P(Y = 1)$
   • $P(Y = 0) = \frac{2.0^0 e^{-2.0}}{0!} = e^{-2} \approx 0.13534$
   • $P(Y = 1) = \frac{2.0^1 e^{-2.0}}{1!} = 2.0 e^{-2} \approx 0.27067$
   • Total approximate probability: $P(Y \le 1) = 0.13534 + 0.27067 = 0.40601 \approx 40.60\%$
3. Percentage Error Analysis: $\text{Error} = \frac{|0.40601 - 0.40553|}{0.40553} \times 100\% = \frac{0.00048}{0.40553} \times 100\% \approx 0.12\%$ Engineering Conclusion: The Poisson model provides extreme precision (within $0.2%$) while avoiding tedious binomial expansions.

Worked Example 6.2.2: Hypergeometric Lot Acceptance Sampling for Aerospace Fasteners

Problem: A small incoming shipment of $N = 20$ titanium aerospace bolts contains exactly $K = 4$ non-conforming bolts with defective thread pitch. An incoming receiving inspector selects a random sample of $n = 3$ bolts without replacement. What is the probability that the sample contains at most 1 defective bolt?

Solution:

1. Identify Model Parameters:
   • Population $N = 20$
   • Defective units in population $K = 4$
   • Conforming units in population $N - K = 16$
   • Sample size $n = 3$
   • We require $P(X \le 1) = P(X = 0) + P(X = 1)$.
2. Calculate Total Sample Combinations: $\binom{N}{n} = \binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 20 \times 19 \times 3 = 1{,}140$
3. Compute $P(X = 0)$ (Zero Defects): $\binom{K}{0} = \binom{4}{0} = 1, \quad \binom{N - K}{n - 0} = \binom{16}{3} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560$ $P(X = 0) = \frac{1 \times 560}{1{,}140} = \frac{560}{1{,}140} \approx 0.49123$
4. Compute $P(X = 1)$ (Exactly One Defect): $\binom{K}{1} = \binom{4}{1} = 4, \quad \binom{N - K}{n - 1} = \binom{16}{2} = \frac{16 \times 15}{2 \times 1} = 120$ $P(X = 1) = \frac{4 \times 120}{1{,}140} = \frac{480}{1{,}140} \approx 0.42105$
5. Sum Cumulative Probability: $P(X \le 1) = \frac{560 + 480}{1{,}140} = \frac{1{,}040}{1{,}140} = \frac{52}{57} \approx 0.91228$
6. Final Result: The probability of accepting the lot under a $c = 1$ threshold is $91.23%$.

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8. NCEES Reference Handbook Tips & Realistic Exam Traps

• Binomial vs. Hypergeometric Distinction: Always look for the phrasing "without replacement from a lot of size N". If $N$ is small (e.g., $N = 20$ or $N = 50$), you MUST use the hypergeometric distribution. If sampling occurs with replacement, or from a continuous production flow where $N$ is practically infinite, use the binomial distribution.
• Poisson Rate Scaling Errors: The parameter $\lambda$ must always match the physical unit of exposure in the question. If defects occur at $1.5\text{ defects/meter}$, and the problem asks for the probability of defects on a $4\text{-meter}$ pipe, $\lambda = 1.5 \times 4 = 6.0$. Using $\lambda = 1.5$ directly is an intentional distractor trap on the exam.
• Cumulative Phrasing Traps:
  • "At most $k$" means $P(X \le k) = P(0) + P(1) + \dots + P(k)$.
  • "Fewer than $k$" means $P(X < k) = P(X \le k - 1)$.
  • "At least $k$" means $P(X \ge k) = 1 - P(X \le k - 1)$.
  • "More than $k$" means $P(X > k) = 1 - P(X \le k)$.
• Geometric Mean Formulation: The NCEES Reference Handbook uses the trial formulation $E(X) = 1/p$, where $X$ is the trial on which the first success occurs ($X \in {1, 2, \dots}$). Some outside textbooks define $Y$ as the number of failures before the first success ($Y \in {0, 1, \dots}$), giving $E(Y) = (1 - p)/p$. For the FE exam, always follow the trial formulation ($1/p$).