Discrete Probability Distributions: Binomial, Poisson, and Geometric Models

Fundamentals of Probability Distributions

• Definition of a Probability Distribution: A probability distribution is a graph, table, or frequency distribution that displays the probabilities associated with each possible event or outcome, rather than displaying the raw observed frequencies.

• Standard Notational Symbols:

  • Success Probability ($p$): Represents the probability that a specific designated event occurs.

  • Failure Probability ($q$): Represents the probability that the designated event does not occur.

  • Complementarity Relationship: The events are complementary, giving the identities:

    • $q = 1 - p$

    • $p + q = 1$

• Classification of Probability Distributions:

  • Discrete Probability Distributions: Used when data must assume distinct, separate, countable values (for example, non-negative integers: $0, 1, 2, 3, \dots$).

  • Continuous Probability Distributions: Used when the random variable can assume any real value along a continuum or within a specified continuous interval.

• Core Discrete Distributions:

  • Binomial Distribution - most common

  • Poisson Distribution

  • Geometric Distribution

The Binomial Distribution

• Underlying Principles and Properties:

  • One of the most widely used and intuitive discrete distributions. most useful and easiest to use

  • Models mutually exclusive events where each trial produces exactly one of two possible outcomes (dichotomous "yes/no", "success/failure" conditions).

  • Trials must be independent: the outcome of any given trial does not alter or influence the probability of outcomes in subsequent trials.

• The Multiplication Rule for Independent Events:

  • If two events $A$ and $B$ occur independently, the joint probability that both occur simultaneously is the product of their individual probabilities:     $P(AB)=P(A)\cdot P(B)$

  • Example: Tossing two fair coins simultaneously to obtain two heads ($p = 0.5$ for each coin):     $P(\text{Two Heads}) = (0.5)^2 = 0.25$

• Binomial Expansions for Small Sample Sizes ($k$):

  • Expanding the binomial expression $(p + q)^k = 1$, where $k$ represents the total number of events or trials:

  • Two Events ($k = 2$):     $(p + q)^2 = p^2 + 2pq + q^2 = 1$

  • $p^2$: Probability of two successes (e.g., two heads).

  • $2pq$: Probability of one success and one failure (e.g., one head and one tail).

  • $q^2$: Probability of two failures (e.g., two tails).

  • Three Events ($k = 3$):     $(p + q)^3 = p^3 + 3p^2q + 3pq^2 + q^3 = 1$

  • $p^3$: Probability of three successes.

  • $3p^2q$: Probability of two successes and one failure.

  • $3pq^2$: Probability of one success and two failures.

  • $q^3$: Probability of three failures.

• General Mathematical Formulation:

  • Expanding polynomials algebraically becomes impractical as the number of trials increases, necessitating the general binomial formula.

  • Variables Defined:

  • $n$: Total number of trials.

  • $x$: Exact number of successes desired.

  • $n - x$: Number of failures.

  • $p$: Probability of success on a single trial.

  • $q$: Probability of failure on a single trial ($q = 1 - p$).

  • Probability Mass Function:     $p(x) = \frac{n!}{x!(n - x)!} p^x q^{(n - x)}$

  • Mean ($\mu$):     $\mu = np$

  • Variance ($\sigma^2$):     $\sigma^2 = npq$

Binomial Distribution: Worked Examples and Applications

• Example 1: Tossing Four Fair Coins ($n = 4$, $p = 0.5$, $q = 0.5$):

  • Parameters:     $\mu = np = (4)(0.5) = 2$     $\sigma^2 = npq = (4)(0.5)(0.5) = 1$

  • Discrete Probability Distribution Schedule:

  • $X = 0$ Heads: $P(0) = \frac{4!}{0!4!} (0.5)^0 (0.5)^4 = 0.0625$

  • $X = 1$ Head: $P(1) = \frac{4!}{1!3!} (0.5)^1 (0.5)^3 = 0.25$

  • $X = 2$ Heads: $P(2) = \frac{4!}{2!2!} (0.5)^2 (0.5)^2 = 0.375$

  • $X = 3$ Heads: $P(3) = \frac{4!}{3!1!} (0.5)^3 (0.5)^1 = 0.25$

  • $X = 4$ Heads: $P(4) = \frac{4!}{4!0!} (0.5)^4 (0.5)^0 = 0.0625$

  • Visual Distribution:

• Example 2: Basketball Free Throw Shooting:

  • A basketball player shoots $75\%$ from the free-throw line ($p = 0.75$, $q = 0.25$).

  • The player takes $n = 8$ free throws.

  • Question 1: What is the probability that she hits exactly six shots ($x = 6$)?     $P(6) = \frac{8!}{6!2!} (0.75)^6 (0.25)^2 = 28 \times 0.1779785 \times 0.0625 \approx 0.311462$

  • In spreadsheet applications such as Microsoft Excel, this exact point probability is computed via the function:       =BINOM.DIST(6, 8, 0.75, FALSE) yielding $0.311462$.

    • Number_s - the number of successes

    • Trials - number of trials

    • Probability_s - probability of success

    • Cumulative - a logical function

• Question 2: What is the probability that she hits at least six shots ($x \ge 6$)?     $P(X \ge 6) = P(6) + P(7) + P(8)$

  • Example 3: Mendelian Genetics in Garden Peas:

• Progeny seeds are either yellow or green, following a phenotypic segregation ratio of $3\text{ yellow} : 1\text{ green}$.

• Probability of a yellow seed: $p = \frac{3}{4} = 0.75$.

• Probability of a green seed: $q = \frac{1}{4} = 0.25$.

• Sample size: $n = k = 4$ progeny seeds examined.

• Case A: Probability that exactly three are yellow and one is green ($x = 3$):     $P(3) = \frac{4!}{3!1!} (0.75)^3 (0.25)^1 = 4 \times 0.421875 \times 0.25 = 0.421875$

• Case B: Probability that all four are yellow ($x = 4$):     $P(4) = \frac{4!}{4!0!} (0.75)^4 (0.25)^0 = 1 \times 0.31640625 \times 1 \approx 0.316406$

• Case C: Probability that all four progeny have the same color (all yellow or all green):

  • Probability that all four are green ($x = 0$ yellow):       $P(0) = \frac{4!}{0!4!} (0.75)^0 (0.25)^4 = 1 \times 1 \times 0.00390625 \approx 0.003906$

  • Combined probability of identical color:       $P(\text{Same Color}) = P(4) + P(0) = 0.316406 + 0.003906 = 0.320313$

• Example 4: Hookworm Drug Treatment Trials:

  • A clinical drug treatment demonstrates a $90\%$ cure rate for hookworm infections in children ($p = 0.90$, $q = 0.10$).

  • A random cohort of $n = 20$ children is treated.

  • Calculation Objectives:

  • All $20$ cured: $P(20) = (0.90)^{20} \approx 0.121576$

  • All but one cured ($x = 19$): $P(19) = \frac{20!}{19!1!} (0.90)^{19} (0.10)^1 \approx 0.270170$

  • More than $18$ cured ($x > 18$): $P(X > 18) = P(19) + P(20)$

    • This can use the cumulative function, which includes the probability of everything starting from 0 to the selected number

  • Fewer than $16$ cured ($x < 16$): $P(X \le 15) = \sum_{x=0}^{15} P(x)$

The Poisson Distribution

• Conceptual Framework and Properties:

  • Discrete probability distribution

  • Models the occurrence of rare events where the event occurs a small number of times relative to the total number of opportunities for occurrence.

  • Individual occurrences must be mutually independent of one another. If not random, events might be uniform (spread evenly) or aggregated (Occurring more often at some places or times).

  • Predicts the number of expected occurrences ($0, 1, 2, \dots$) per predefined unit of space, area, volume, or time.

• Spatial and Ecological Examples:

  • Number of deer per acre in a forest.

  • Number of parasites on a fish.

  • Distribution of tumors on plant surfaces.

  • Number of ticks infesting dogs.

  • Number of bacteriophages attached to bacterial cells.

• Temporal Examples:

  • Car accidents occurring per hour in a major city.

  • Number of lightning strikes hitting trees per day in a national reserve.

• Spatial/Temporal Dispersion Patterns:

  • Random Dispersion: Occurrences happen entirely independent of one another; strictly follows a Poisson distribution.

  • Uniform Dispersion: Occurrences are spread out evenly over space or time, leading to lower-than-expected variance.

  • Aggregated (Clumped) Dispersion: Occurrences cluster heavily at specific locations or time intervals, leading to higher-than-expected variance ("hit-or-miss" distribution).

• Mathematical Definition and Parameters:

  • Equality of Mean and Variance: In a true Poisson distribution, the population mean and variance are equal:     $\mu = \sigma^2 = \lambda$

  • Probability Mass Function:     $P(x) = \frac{\lambda^x e^{-\lambda}}{x!} = \frac{\mu^x e^{-\mu}}{x!} = \frac{\bar{Y}^x e^{-\bar{Y}}}{x!}$     where $e \approx 2.71828$ is Euler's constant and $\lambda$ (or $\mu$, $\bar{Y}$) represents the mean arrival or occurrence rate per unit interval.

• Worked Example: Lung Cancer Mortality:

  • Average annual mortality from lung cancer in a specified population is observed to be $12$ deaths per year ($\lambda = 12$).

  • Probability of observing exactly ten deaths ($x = 10$) in a given year:     $P(10) = \frac{12^{10} e^{-12}}{10!} \approx 0.1048$

Poisson Model: Forest Lightning Strikes and Goodness-of-Fit

• Scenario Context: In a state forest, an average of $1437$ trees are struck by lightning per year ($365$ days). Lightning strikes are assumed to occur randomly across the calendar year.

• Calculation of Daily Mean Rate:   $\bar{Y} = \frac{1437}{365} \approx 3.94\text{ strikes/day}$

• Probability of Two or Fewer Strikes on a Given Day ($P(X \le 2)$):   $P(X \le 2) = P(0) + P(1) + P(2)$

  • Evaluating $P(0)$:     $P(0) = \frac{3.94^0 e^{-3.94}}{0!} = \frac{1 \times e^{-3.94}}{1} \approx 0.0194$

  • Evaluating $P(1)$:     $P(1) = \frac{3.94^1 e^{-3.94}}{1!} = 3.94 \times 0.019448 \approx 0.0766$

  • Evaluating $P(2)$:     $P(2) = \frac{3.94^2 e^{-3.94}}{2!} = \frac{15.5236 \times 0.019448}{2} \approx 0.1510$

  • Summing the Cumulative Probability:     $P(X \le 2) = 0.0194 + 0.0766 + 0.1510 = 0.247$

  • There is a $24.7\%$ probability that two or fewer lightning strikes occur on any randomly selected day.

• Usually use the poisson distribution to produce an EXPECTED distribution to then compare to the observed data.

• Testing the Assumption of Randomness (Observed vs. Expected):

  • A field study recorded the empirical frequency of tree fires caused by lightning strikes across the $365$ days:

• Expected Day Frequencies under the Poisson Model ($\text{Expected} = 365 \times P(X)$):

  • $X = 0$: $P(0) = 0.019448 \rightarrow \text{Expected} = 7.098598$ vs. $\text{Observed} = 127$

  • $X = 1$: $P(1) = 0.076626 \rightarrow \text{Expected} = 27.96848$ vs. $\text{Observed} = 21$

  • $X = 2$: $P(2) = 0.150953 \rightarrow \text{Expected} = 55.09790$ vs. $\text{Observed} = 4$

  • $X = 3$: $P(3) = 0.198252 \rightarrow \text{Expected} = 72.36191$ vs. $\text{Observed} = 13$

  • $X = 4$: $P(4) = 0.195278 \rightarrow \text{Expected} = 71.27648$ vs. $\text{Observed} = 20$

  • $X = 5$: $P(5) = 0.153879 \rightarrow \text{Expected} = 56.16587$ vs. $\text{Observed} = 23$

  • $X = 6$: $P(6) = 0.101047 \rightarrow \text{Expected} = 36.88225$ vs. $\text{Observed} = 27$

  • $X = 7$: $P(7) = 0.056875 \rightarrow \text{Expected} = 20.75944$ vs. $\text{Observed} = 28$

  • $X = 8$: $P(8) = 0.028011 \rightarrow \text{Expected} = 10.22402$ vs. $\text{Observed} = 102$

  • $X > 8$: $P(>8) = 0.019630 \rightarrow \text{Expected} = 7.165049$

  • Variance-to-Mean Ratio ($V/M$) as an Index of Dispersion:

• Diagnostic Thresholds:

  • $\frac{s^2}{\bar{X}} \approx 1$: Indicates a Random distribution (Poisson assumption holds).

  • $\frac{s^2}{\bar{X}} < 1$ (approaching $0$): Indicates a Uniform distribution (underdispersion).

  • $\frac{s^2}{\bar{X}} > 1$ (substantially greater than $1$): Indicates an Aggregated / Clumped distribution (overdispersion).

• Sample Variance Computation for Lightning Data:     $s^2 = \frac{\sum X^2 - \frac{(\sum X)^2}{n}}{n - 1} = \frac{9921 - \frac{(1437)^2}{365}}{364} = 11.71$

• Variance-to-Mean Ratio Evaluation:     $\frac{s^2}{\bar{X}} = \frac{11.71}{3.94} = 2.97$

• Diagnostic Conclusion: Because the index of dispersion ($2.97$) is markedly greater than $1$, the occurrence of lightning strikes across days is not random. Instead, strikes are highly clumped or aggregated on specific severe-weather days, with long intervening stretches having zero strikes.

The Geometric Distribution

• Conceptual Framework:

  • Formulated from a sequence of independent Bernoulli trials with a constant probability of success $p$.

  • Rather than evaluating the number of successes in a fixed number of trials (as in the binomial distribution), the geometric distribution evaluates the trial index $x$ on which the first success occurs.

  • The random variable $x$ is defined as the total number of trials up to and including the first success.

• Mathematical Formulations:

  • Probability Mass Function:     $p(x) = q^{(x - 1)} p = (1 - p)^{(x - 1)} p$     where $x \in \{1, 2, 3, \dots\}$.

  • Mean (Expected Value, $\mu$):     $\mu = \frac{1}{p}$

  • Variance ($\sigma^2$):     $\sigma^2 = \frac{1 - p}{p^2}$

• Worked Example: Angling in Lake LU:

  • In Lake LU, the probability of catching a fish on any single cast is $p = 0.04$.

  • Complementary failure rate: $q = 1 - 0.04 = 0.96$.

  • Question: What is the probability that the first fish is caught exactly on the fourth cast ($x = 4$)?     $p(4) = q^{4 - 1} p = (0.96)^3 (0.04)$     $p(4) = (0.884736) \times (0.04) = 0.03538944 \approx 0.035$

  • There is a $3.5\%$ probability of achieving the first successful catch on the fourth cast.

• Practical Applications:

  • Modeling failure rates in engineering and electrical reliability (e.g., number of duty cycles before an electronic component fails for the first time).

  • Modeling ecological encounters that conclude upon a single occurrence, such as predator-prey pursuit attempts before a successful capture.