6.2 Probability Concepts & Distributions

6.2 Probability Concepts & Distributions

Probability theory quantifies uncertainty, providing the mathematical bridge between descriptive statistics and statistical inference. Quality engineers use probability models to evaluate component failure risk, design acceptance sampling plans, and establish process capability under inherent randomness.

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Basic Rules & Axioms of Probability

Probability $P(A)$ measures the likelihood of an event $A$ occurring, bounded by $0 \le P(A) \le 1$.

1. Complement Rule

The probability that event $A$ does not occur is:

$P(A^c) = 1 - P(A)$

2. Addition Rule (Union of Events)

For any two events $A$ and $B$, the probability that at least one occurs is:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

If $A$ and $B$ are mutually exclusive (cannot occur simultaneously, $P(A \cap B) = 0$):

$P(A \cup B) = P(A) + P(B)$

3. Conditional Probability & Multiplication Rule

The conditional probability of event $A$ given event $B$ has occurred is:

$P(A|B) = \frac{P(A \cap B)}{P(B)}, \quad \text{for } P(B) > 0$

Rearranging gives the general Multiplication Rule:

$P(A \cap B) = P(A|B) \cdot P(B)$

If events $A$ and $B$ are independent (the occurrence of $B$ does not alter the likelihood of $A$, $P(A|B) = P(A)$):

$P(A \cap B) = P(A) \cdot P(B)$

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Discrete Probability Distributions

Discrete distributions model countable outcomes, such as number of defects, rejected lots, or customer complaints.

1. Binomial Distribution

Models the number of success occurrences ($X = k$) in $n$ independent trials, each with constant success probability $p$.

• Probability Mass Function (PMF): $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} = \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}$
• Mean & Variance: $\mu = np, \quad \sigma^2 = np(1-p), \quad \sigma = \sqrt{np(1-p)}$
• Assumptions: Fixed number of trials $n$; binary outcomes (Success/Failure); independent trials; constant probability $p$.

2. Poisson Distribution

Models the count of rare events occurring randomly across a continuous interval of time, area, or volume at an average rate $\lambda$.

• Probability Mass Function (PMF): $P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k = 0, 1, 2, \dots$
• Mean & Variance: $\mu = \lambda, \quad \sigma^2 = \lambda$
• Key Characteristic: Equal mean and variance ($\mu = \sigma^2$). Serves as a limiting approximation to the Binomial distribution when $n \ge 100$ and $np \le 10$.

3. Hypergeometric Distribution

Models sampling without replacement from a finite population of size $N$ containing $K$ defective items, where $n$ items are sampled.

• Probability Mass Function (PMF): $P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$
• Distinction: Unlike the Binomial distribution, trial outcomes are dependent because sampling without replacement alters remaining probabilities.

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Continuous Probability Distributions

Continuous distributions model measurable variables such as dimension, weight, temperature, or operating lifetime.

1. Normal (Gaussian) Distribution

The cornerstone of statistical quality control, symmetric and bell-shaped, denoted $N(\mu, \sigma^2)$.

• Standard Normal Transformation (Z-score): $Z = \frac{X - \mu}{\sigma}$
• Empirical Rule (68-95-99.7 Rule):
  • $\mu \pm 1\sigma$ contains $\approx 68.27%$ of data.
  • $\mu \pm 2\sigma$ contains $\approx 95.45%$ of data.
  • $\mu \pm 3\sigma$ contains $\approx 99.73%$ of data.

2. Exponential Distribution

Models continuous time-to-failure or inter-arrival times for components exhibiting a constant failure rate $\lambda$ (memoryless property).

• Probability Density Function (PDF): $f(t) = \lambda e^{-\lambda t}, \quad t \ge 0$
• Reliability Function $R(t)$: $R(t) = P(T > t) = e^{-\lambda t}$
• Mean Time Between Failures (MTBF): $\mu = \text{MTBF} = \frac{1}{\lambda}$

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Distribution Selection Matrix for Quality Engineers

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Comprehensive Worked Numerical Examples

Example 1: Binomial Inspection

A printed circuit board (PCB) assembly line has a historical defect rate of $p = 0.05$. A sample of $n = 10$ boards is inspected. Find the probability of finding exactly 1 defective board.

P(X = 1) &= \binom{10}{1} (0.05)^1 (1 - 0.05)^{10-1} \\ &= 10 \times 0.05 \times (0.95)^9 \\ &= 0.50 \times 0.6302 = 0.3151 \text{ (or } 31.51\%\text{)} \end{aligned}$$ ### Example 2: Poisson Equipment Jam Rate A high-speed labeling machine experiences an average of $\lambda = 2.0$ stoppages per 8-hour shift. Calculate the probability of experiencing at most 1 stoppage during a shift. $$\begin{aligned} P(X \le 1) &= P(X = 0) + P(X = 1) \\ &= \frac{2.0^0 e^{-2.0}}{0!} + \frac{2.0^1 e^{-2.0}}{1!} \\ &= e^{-2.0} (1 + 2.0) = 0.1353 \times 3 = 0.4060 \text{ (or } 40.60\%\text{)} \end{aligned}$$ ### Example 3: Exponential Reliability An avionics power module has a constant failure rate of $\lambda = 0.0005$ failures per hour (MTBF = 2,000 hours). Calculate the reliability $R(t)$ for a 500-hour mission. $$R(500) = e^{-\lambda t} = e^{-(0.0005 \times 500)} = e^{-0.25} \approx 0.7788 \text{ (or } 77.88\%\text{)}$$ --- ## CQE Exam Tips & Strategy - **Distinguish Replacement:** Sampling with replacement or from infinite lot = **Binomial**. Sampling without replacement from small finite lot = **Hypergeometric**. - **Poisson Equality:** If an exam question mentions that mean equals variance ($\mu = \sigma^2$), immediately recognize the **Poisson distribution**. - **Exponential Memoryless:** The exponential distribution assumes no wear-out mechanism. The probability of surviving the next 100 hours is independent of how many hours the part has already operated.