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Probability Formulas

Basic Principles
Poisson Distribution
Multinomial Distribution
Conditional Probability
Disjoint Events
Negative Binomial Distribution
Negative Hypergeometric Distribution
Advanced Probabilities
Binomial Distribution
Hypergeometric Distribution
Logarithmic Distribution
Independent Events
Geometric Distribution
Discrete Uniform Distribution

Basic Principles






Simple Probability



Formula:

P(A)=Number of favorable outcomesTotal number of possible outcomesP(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}
Calculates the probability of event A happening when all outcomes are equally likely, quantifying how likely a specific event is to occur.
Probability of event A occurring.
Event whose probability is being calculated.
Count of outcomes that qualify as event A occurring.
Count of all possible outcomes in the sample space.
Flipping a fair coin, probability of heads: P(Heads)=12P(Heads) = \frac{1}{2}
Simple games of chance, basic statistical calculations.





Probability Range Of An Event



Formula:

0≤P(A)≤10 \leq P(A) \leq 1
The probability of any event is quantified on a scale from 0 to 1, inclusive. A probability of 0 indicates an event is impossible and will not occur under any circumstances. A probability of 1 indicates an event is certain to occur. Probabilities between 0 and 1 indicate the likelihood of the event's occurrence, with values closer to 1 suggesting a higher probability and values closer to 0 suggesting a lower probability.
Event
Prabability of AA.
0 - Rolling a 7 on a standard six-sided die; 1 - Drawing any card from a full deck of cards; Between 0 and 1 - Flipping a coin and it landing on heads, which has a probability of 0.5.
This fundamental concept is used in all fields involving probability calculations, including statistics, risk assessment, finance, science, and engineering.





Complement Rule



Formula:

P(A′)+P(A)=1P(A') + P(A) = 1
States that the probability of an event occurring plus the probability of its complement (the event not occurring) equals one. This is because the event and its complement together encompass all possible outcomes.
Probability of the complement of event A, which is the event not occurring.
Probability of event A occurring.
If there is a 30% chance it will rain tomorrow (P(A) = 0.30), then there is a 70% chance it will not rain (P(A') = 0.70).
Derived from the definition of probability space, where the total probability is 1. Thus, P(A′)=1−P(A)P(A') = 1 - P(A).
Frequently used to simplify calculations where it is easier to calculate the chance of something not happening rather than it happening, such as in quality control or risk assessment scenarios.

Conditional Probability






Conditional Probability Basic Formula



Formula:

P(A∣B)=P(A∩B)P(B)P(A \mid B) = \frac{P(A \cap B)}{P(B)}
Measures the probability of event A occurring given that event B has already occurred, assessing how B influences A.
Probability of both events A and B happening together.
Probabilities of A and B where B's occurrence affects the likelihood of A.
Drawing two cards, probability second card is an ace given the first was an ace: P(Ace2∣Ace1)=351P(Ace_2 \mid Ace_1) = \frac{3}{51}
Situations where prior information about one event affects the outcome of another.

Advanced Probabilities






Bayes' Theorem



Formula:

P(A∣B)=P(B∣A)×P(A)P(B)P(A \mid B) = \frac{P(B \mid A) \times P(A)}{P(B)}
Updates the probability of event A occurring based on the occurrence of event B, integrating prior belief (prior probability) with new evidence.
Initial or prior probability of A occurring, overall probability of B, and probability of B given A.
Events where B's occurrence provides evidence about A's likelihood.
Disease with 1% prevalence, 99% sensitive test, probability of disease given a positive test: Calculated using Bayes' Theorem.
Medical testing, machine learning algorithms, updating predictions with new information.

Independent Events






Probability Of Both Events Occurring (Multiplication Rule)



Formula:

P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B)
Calculates the probability that both events A and B occur simultaneously. It assumes that A and B are independent, meaning the occurrence of A does not affect the occurrence of B, and vice versa.
Probability of both events A and B occurring together.
Probability of event A occurring independently, and probability of event B occurring independently.
Flipping two fair coins, the probability of both being heads: P(Heads∩Heads)=P(Heads)×P(Heads)=12×12=14P(Heads \cap Heads) = P(Heads) \times P(Heads) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
Used in scenarios involving multiple independent trials or events.





Probability Of Either Event Occurring (Addition Rule)



Formula:

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)
This rule calculates the probability that at least one of the events A or B occurs, adjusting for the overlap by subtracting the probability of both occurring together.
Probability of either event A or event B occurring.
P(A∪B)=P(A)+P(B)−P(A)×P(B)P(A \cup B) = P(A) + P(B) - P(A) \times P(B)
For rolling a die, the probability of rolling a 2 or a 5: P(2∪5)=P(2)+P(5)−P(2)×P(5)=16+16−16×16=1136P(2 \cup 5) = P(2) + P(5) - P(2) \times P(5) = \frac{1}{6} + \frac{1}{6} - \frac{1}{6} \times \frac{1}{6} = \frac{11}{36}
Useful in evaluating the likelihood of multiple possible outcomes where events are independent.





Probability Of At Least One Event Not Occurring



Formula:

P(¬A∩¬B)=P(¬A)×P(¬B)P(\neg A \cap \neg B) = P(\neg A) \times P(\neg B)
Calculates the probability that neither event A nor event B occurs, using the complement rule for each event.
Probability of neither event A nor event B occurring.
Complements of probabilities of A and B occurring, respectively.
If the probability of rain on any given day is 0.20, then the probability of no rain for two independent days is: P(¬Rain1∩¬Rain2)=(1−0.20)×(1−0.20)=0.64P(\neg Rain1 \cap \neg Rain2) = (1 - 0.20) \times (1 - 0.20) = 0.64
Important in scenarios where the non-occurrence of multiple independent events is critical, such as in reliability engineering.





Probability Of Exactly One Event Occurring



Formula:

P(exactly one of A or B)=P(A∩¬B)+P(¬A∩B)P(\text{exactly one of } A \text{ or } B) = P(A \cap \neg B) + P(\neg A \cap B)
Determines the probability of exactly one of the two events occurring, not both.
For independent events: P(exactly one of A or B)=P(A)×(1−P(B))+(1−P(A))×P(B)P(\text{exactly one of } A \text{ or } B) = P(A) \times (1 - P(B)) + (1 - P(A)) \times P(B)
Using a deck of cards, the probability that one draws an ace or a king, but not both: P(Ace∩¬King)+P(¬Ace∩King)P(Ace \cap \neg King) + P(\neg Ace \cap King)
Used in analytical models where exclusive outcomes are analyzed.





General Formula For Multiple Independent Events



Formula:

P(A∩B∩C)=P(A)×P(B)×P(C)P(A \cap B \cap C) = P(A) \times P(B) \times P(C)
Extends the multiplication rule to any number of events, calculating the probability that all specified independent events occur.
Probability of events A, B, and C all occurring together.
Individual Probabilities of AA,BB and CC .
For three independent events each with a 50% chance of occurring, the probability of all occurring is: 0.5×0.5×0.5=0.1250.5 \times 0.5 \times 0.5 = 0.125
Applicable in complex systems where multiple independent events must be simultaneously considered, such as in network reliability.

Disjoint Events






Probability Of Both Disjoint Events Occurring



Formula:

P(A∩B)=0P(A \cap B) = 0
Since disjoint events cannot occur simultaneously, the probability of both occurring at the same time is zero.
This formula pretty much holds the essense of disjoint events definition : chance that both of them happen is equal to zero.
Probability of both disjoint events A and B occurring together.
If event A is drawing a heart from a deck of cards and event B is drawing a spade, since one draw cannot be both a heart and a spade, P(A∩B)=0P(A \cap B) = 0.
Applicable in any scenario where two or more outcomes are mutually exclusive.





Probability Of Either Disjoint Event Occurring (Addition Rule)



Formula:

P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B)
The probability of either disjoint event occurring is the sum of their individual probabilities, because there is no overlap between them.
Probability of either disjoint event A or event B occurring.
If event A is drawing a heart (25% chance) and event B is drawing a spade (25% chance) from a deck, P(A∪B)=25P(A \cup B) = 25% + 25% = 50%.
Useful in determining the likelihood of one of several exclusive events occurring.





Probability Of Neither Disjoint Event Occurring



Formula:

P(¬A∩¬B)=P(¬A)×P(¬B)P(\neg A \cap \neg B) = P(\neg A) \times P(\neg B)
Calculates the probability that neither disjoint event occurs, assuming each event’s non-occurrence is independent of the other.
Probability of neither disjoint event A nor event B occurring.
If the probability of not drawing a heart is 75% and not drawing a spade is 75%, then P(¬A∩¬B)=75P(\neg A \cap \neg B) = 75% \times 75% = 56.25%.
Important for risk assessment and contingency planning in mutually exclusive scenarios.





Conditional Probability For Disjoint Events



Formula:

P(A∣B)=0P(A \mid B) = 0 and P(B∣A)=0P(B \mid A) = 0
If one disjoint event occurs, the other cannot possibly happen, thus the conditional probability is zero.
Probability of A given B and probability of B given A, respectively, for disjoint events.
If event A has already occurred (e.g., a heart was drawn), the probability of event B (drawing a spade) occurring is 0.
Used in scenarios where the occurrence of one outcome precludes the occurrence of others.





Generalization To Multiple Disjoint Events



Formula:

P(A∪B∪C∪…)=P(A)+P(B)+P(C)+…P(A \cup B \cup C \cup \ldots) = P(A) + P(B) + P(C) + \ldots
For multiple disjoint events, the probability of any event occurring is the sum of the probabilities of each event.
General formula for the union of multiple disjoint events.
The probability of drawing a heart, a spade, or a diamond from a deck of cards.
Useful in complex scenarios involving several mutually exclusive outcomes.

Binomial Distribution






Probability Mass Function (PMF)



Formula:

P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^{k} (1 - p)^{n - k}
Calculates the probability of exactly kk successes in nn independent Bernoulli trials, each with success probability pp.
Probability of getting exactly kk successes.
Binomial coefficient representing the number of ways to choose kk successes from nn trials.
Total number of trials.
Number of successful trials.
Probability of success on a single trial.
Probability of failure on a single trial.
If n=5n = 5, p=0.6p = 0.6, and k=3k = 3, then P(X=3)=(53)(0.6)3(0.4)2≈0.3456P(X = 3) = \binom{5}{3} (0.6)^3 (0.4)^2 \approx 0.3456.
Used to find the likelihood of a specific number of successes in processes like quality control, clinical trials, and survey analysis.





Cumulative Distribution Function (CDF)



Formula:

P(X≤k)=∑i=0k(ni)pi(1−p)n−iP(X \leq k) = \sum_{i=0}^{k} \binom{n}{i} p^{i} (1 - p)^{n - i}
Calculates the probability of obtaining at most kk successes in nn trials.
Probability of getting kk or fewer successes.
Summation from i=0i = 0 to i=ki = k.
All other variables are as previously defined in the PMF.
For n=5n = 5, p=0.6p = 0.6, and k=3k = 3, P(X≤3)P(X \leq 3) sums the probabilities for k=0k = 0 to k=3k = 3.
Assessing cumulative probabilities in risk assessment and reliability engineering.





Mean (Expected Value)



Formula:

μ=E[X]=np\mu = E[X] = n p
Provides the average number of successes expected over nn trials.
Mean or expected number of successes.
Expected value of the random variable XX.
Total number of trials.
Probability of success on a single trial.
If n=10n = 10 and p=0.5p = 0.5, then μ=10×0.5=5\mu = 10 \times 0.5 = 5.
Predicting average outcomes in repeated experiments like manufacturing defects.





Variance



Formula:

σ2=Var⁡(X)=np(1−p)\sigma^2 = \operatorname{Var}(X) = n p (1 - p)
Measures the dispersion of the number of successes around the mean.
Variance of the random variable XX.
Variance notation.
Total number of trials.
Probability of success on a single trial.
Probability of failure on a single trial.
For n=10n = 10 and p=0.5p = 0.5, σ2=10×0.5×0.5=2.5\sigma^2 = 10 \times 0.5 \times 0.5 = 2.5.
Assessing variability in outcomes for quality assurance and forecasting.





Standard Deviation



Formula:

σ=np(1−p)\sigma = \sqrt{n p (1 - p)}
Represents the average distance of the data from the mean.
Standard deviation of the random variable XX.
As previously defined.
With n=10n = 10 and p=0.5p = 0.5, σ=10×0.5×0.5≈1.5811\sigma = \sqrt{10 \times 0.5 \times 0.5} \approx 1.5811.
Used to understand the spread of binomial outcomes in fields like finance and epidemiology.

Poisson Distribution






Probability Mass Function (PMF)



Formula:

P(X=k)=e−λλkk!P(X = k) = \frac{e^{-\lambda} \lambda^{k}}{k!}
Calculates the probability of observing exactly kk events in a fixed interval when events occur independently at a constant average rate λ\lambda.
Probability of exactly kk occurrences in the interval.
The probability of zero occurrences.
Mean number of occurrences raised to the kk-th power.
Factorial of kk, the number of occurrences.
Average rate of occurrence (mean number of events per interval).
Number of occurrences (non-negative integer).
If λ=4\lambda = 4 and k=2k = 2, then P(X=2)=e−4×422!≈0.1465P(X = 2) = \frac{e^{-4} \times 4^{2}}{2!} \approx 0.1465.
Modeling the number of rare events over a fixed period, such as the number of incoming calls to a call center in an hour.





Cumulative Distribution Function (CDF)



Formula:

P(X≤k)=e−λ∑i=0kλii!P(X \leq k) = e^{-\lambda} \sum_{i=0}^{k} \frac{\lambda^{i}}{i!}
Calculates the probability of observing up to kk events in a fixed interval.
Probability of kk or fewer occurrences.
Summation from i=0i = 0 to i=ki = k.
Average rate of occurrence.
Factorial of ii, each term in the sum.
For λ=3\lambda = 3 and k=2k = 2, P(X≤2)=e−3(300!+311!+322!)≈0.4232P(X \leq 2) = e^{-3} \left( \frac{3^{0}}{0!} + \frac{3^{1}}{1!} + \frac{3^{2}}{2!} \right) \approx 0.4232.
Assessing cumulative probabilities, such as the probability of observing a certain number of defects in a batch.





Mean (Expected Value)



Formula:

μ=E[X]=λ\mu = E[X] = \lambda
Represents the average number of events occurring in the given interval.
Mean or expected value of the distribution.
Expected value operator.
Average rate of occurrence.
If λ=5\lambda = 5, then μ=5\mu = 5.
Predicting average outcomes in processes like arrival rates, such as customers entering a store per hour.





Variance



Formula:

σ2=Var⁡(X)=λ\sigma^2 = \operatorname{Var}(X) = \lambda
Measures the dispersion of the number of events around the mean.
Variance of the distribution.
Variance operator.
Average rate of occurrence.
If λ=5\lambda = 5, then σ2=5\sigma^2 = 5.
Assessing variability in event counts for planning and resource allocation.





Standard Deviation



Formula:

σ=λ\sigma = \sqrt{\lambda}
Represents the average distance of the data from the mean.
Standard deviation of the distribution.
Average rate of occurrence.
If λ=5\lambda = 5, then σ=5≈2.2361\sigma = \sqrt{5} \approx 2.2361.
Understanding the spread of Poisson-distributed outcomes in fields like telecommunications and biology.

Geometric Distribution






Probability Mass Function (PMF)



Formula:

P(X=k)=(1−p)k−1pP(X = k) = (1 - p)^{k - 1} p
Calculates the probability that the first success occurs on the kk-th trial.
Probability that the first success occurs on trial kk.
Probability of success on each trial.
Probability of failure on each trial.
Trial number on which the first success occurs (k=1,2,3,…k = 1, 2, 3, \dots).
If p=0.3p = 0.3 and k=4k = 4, then P(X=4)=(1−0.3)4−1×0.3=(0.7)3×0.3≈0.1029P(X = 4) = (1 - 0.3)^{4 - 1} \times 0.3 = (0.7)^{3} \times 0.3 \approx 0.1029.
Used to model the number of trials needed to get the first success, such as in quality control testing or reliability analysis.





Cumulative Distribution Function (CDF)



Formula:

P(X≤k)=1−(1−p)kP(X \leq k) = 1 - (1 - p)^{k}
Calculates the probability that the first success occurs on or before the kk-th trial.
Probability that the first success occurs on or before trial kk.
Probability of success on each trial.
Probability of failure on each trial.
Trial number (k=1,2,3,…k = 1, 2, 3, \dots).
For p=0.3p = 0.3 and k=4k = 4, P(X≤4)=1−(1−0.3)4=1−(0.7)4≈0.7599P(X \leq 4) = 1 - (1 - 0.3)^{4} = 1 - (0.7)^{4} \approx 0.7599.
Assessing the likelihood of achieving the first success within a certain number of trials.





Mean (Expected Value)



Formula:

μ=E[X]=1p\mu = E[X] = \frac{1}{p}
Represents the average number of trials required to achieve the first success.
Mean or expected value of the distribution.
Probability of success on each trial.
If p=0.3p = 0.3, then μ=10.3≈3.3333\mu = \frac{1}{0.3} \approx 3.3333.
Predicting average trial counts in processes like quality inspections or repeated experiments.





Variance



Formula:

σ2=Var⁡(X)=1−pp2\sigma^2 = \operatorname{Var}(X) = \frac{1 - p}{p^{2}}
Measures the dispersion of the number of trials around the mean.
Variance of the distribution.
Probability of success on each trial.
Probability of failure on each trial.
If p=0.3p = 0.3, then σ2=1−0.3(0.3)2=0.70.09≈7.7778\sigma^2 = \frac{1 - 0.3}{(0.3)^{2}} = \frac{0.7}{0.09} \approx 7.7778.
Assessing variability in the number of trials needed for the first success.





Standard Deviation



Formula:

σ=1−pp2\sigma = \sqrt{\frac{1 - p}{p^{2}}}
Represents the average distance of the number of trials from the mean.
Standard deviation of the distribution.
Probability of success on each trial.
Probability of failure on each trial.
If p=0.3p = 0.3, then σ=1−0.3(0.3)2≈7.7778≈2.7889\sigma = \sqrt{\frac{1 - 0.3}{(0.3)^{2}}} \approx \sqrt{7.7778} \approx 2.7889.
Understanding the spread of trials needed to achieve the first success.

Negative Binomial Distribution






Probability Mass Function (PMF)



Formula:

P(X=k)=(k−1r−1)pr(1−p)k−rP(X = k) = \binom{k - 1}{r - 1} p^{r} (1 - p)^{k - r}
Calculates the probability that the rr-th success occurs on the kk-th trial.
Probability that the rr-th success occurs on trial kk.
Binomial coefficient representing the number of ways to arrange r−1r - 1 successes in k−1k - 1 trials.
Probability of success on each trial.
Probability of failure on each trial.
Trial number on which the rr-th success occurs (k=r,r+1,r+2,…k = r, r+1, r+2, \dots).
Number of successes to be achieved.
If p=0.4p = 0.4, r=3r = 3, and k=7k = 7, then P(X=7)=(62)(0.4)3(0.6)4≈0.0595P(X = 7) = \binom{6}{2} (0.4)^{3} (0.6)^{4} \approx 0.0595.
Used to model the number of trials needed to achieve a specified number of successes, such as in reliability testing or sequential analysis.





Cumulative Distribution Function (CDF)



Formula:

P(X≤k)=Ip(r,k−r+1)P(X \leq k) = I_{p}(r, k - r + 1)
Calculates the probability that the rr-th success occurs on or before the kk-th trial using the regularized incomplete beta function.
Probability that the rr-th success occurs on or before trial kk.
Regularized incomplete beta function.
Probability of success on each trial.
Trial number.
Number of successes.
For p=0.4p = 0.4, r=3r = 3, and k=7k = 7, P(X≤7)P(X \leq 7) can be calculated using the cumulative sum of PMF values or the incomplete beta function.
Assessing the likelihood of achieving rr successes within a certain number of trials.





Mean (Expected Value)



Formula:

μ=E[X]=rp\mu = E[X] = \frac{r}{p}
Represents the average number of trials required to achieve rr successes.
Mean or expected value of the distribution.
Number of successes.
Probability of success on each trial.
If r=3r = 3 and p=0.4p = 0.4, then μ=30.4=7.5\mu = \frac{3}{0.4} = 7.5.
Predicting average trial counts in processes like quality control or repeated experiments until a certain number of successes are achieved.





Variance



Formula:

σ2=Var⁡(X)=r(1−p)p2\sigma^2 = \operatorname{Var}(X) = \frac{r (1 - p)}{p^{2}}
Measures the dispersion of the number of trials around the mean.
Variance of the distribution.
Number of successes.
Probability of success on each trial.
Probability of failure on each trial.
If r=3r = 3 and p=0.4p = 0.4, then σ2=3×0.6(0.4)2=11.25\sigma^2 = \frac{3 \times 0.6}{(0.4)^{2}} = 11.25.
Assessing variability in the number of trials needed for achieving rr successes.





Standard Deviation



Formula:

σ=r(1−p)p2\sigma = \sqrt{\frac{r (1 - p)}{p^{2}}}
Represents the average distance of the number of trials from the mean.
Standard deviation of the distribution.
Number of successes.
Probability of success on each trial.
Probability of failure on each trial.
If r=3r = 3 and p=0.4p = 0.4, then σ=11.25≈3.3541\sigma = \sqrt{11.25} \approx 3.3541.
Understanding the spread of trials needed to achieve rr successes.

Hypergeometric Distribution






Probability Mass Function (PMF)



Formula:

P(X=k)=(Kk)(N−Kn−k)(Nn)P(X = k) = \frac{\binom{K}{k} \binom{N - K}{n - k}}{\binom{N}{n}}
Calculates the probability of observing exactly kk successes in nn draws without replacement from a finite population.
Probability of getting exactly kk successes.
Number of ways to choose kk successes from KK total successes in the population.
Number of ways to choose n−kn - k failures from N−KN - K total failures.
Total number of ways to choose nn items from NN total items.
Total number of items in the population.
Total number of successes in the population.
Number of draws or trials.
Number of observed successes.
If N=50N = 50, K=10K = 10, n=5n = 5, and k=2k = 2, then P(X=2)=(102)(403)(505)P(X = 2) = \frac{\binom{10}{2} \binom{40}{3}}{\binom{50}{5}}.
Used in scenarios involving sampling without replacement, such as quality control inspection or lotteries.





Mean (Expected Value)



Formula:

μ=E[X]=n(KN)\mu = E[X] = n \left( \frac{K}{N} \right)
Represents the average number of successes expected in nn draws.
Mean or expected value of the distribution.
Number of draws.
Total number of successes in the population.
Total number of items in the population.
If N=50N = 50, K=10K = 10, and n=5n = 5, then μ=5(1050)=1\mu = 5 \left( \frac{10}{50} \right) = 1.
Predicting the average number of successes in random sampling without replacement.





Variance



Formula:

σ2=Var⁡(X)=n(KN)(N−KN)(N−nN−1)\sigma^2 = \operatorname{Var}(X) = n \left( \frac{K}{N} \right) \left( \frac{N - K}{N} \right) \left( \frac{N - n}{N - 1} \right)
Measures the dispersion of the number of successes around the mean.
Variance of the distribution.
Number of draws.
Total number of successes in the population.
Total number of items in the population.
Total number of failures in the population.
Number of items not drawn.
Adjustment for finite population.
If N=50N = 50, K=10K = 10, and n=5n = 5, then σ2=5(1050)(4050)(4549)≈0.9184\sigma^2 = 5 \left( \frac{10}{50} \right) \left( \frac{40}{50} \right) \left( \frac{45}{49} \right) \approx 0.9184.
Assessing variability in outcomes when sampling without replacement.





Standard Deviation



Formula:

σ=n(KN)(N−KN)(N−nN−1)\sigma = \sqrt{n \left( \frac{K}{N} \right) \left( \frac{N - K}{N} \right) \left( \frac{N - n}{N - 1} \right)}
Represents the average distance of the number of successes from the mean.
Standard deviation of the distribution.
All other variables are as previously defined.
Using the previous variance example, σ=0.9184≈0.9583\sigma = \sqrt{0.9184} \approx 0.9583.
Understanding the spread of the number of successes in sampling without replacement.





Probability Of At Least $k$ Successes



Formula:

P(X≥k)=1−∑i=0k−1P(X=i)P(X \geq k) = 1 - \sum_{i=0}^{k - 1} P(X = i)
Calculates the probability of observing at least kk successes in nn draws.
Probability of kk or more successes.
Sum of probabilities of observing fewer than kk successes.
PMF evaluated at ii successes.
To find P(X≥2)P(X \geq 2), compute 1−[P(X=0)+P(X=1)]1 - [P(X = 0) + P(X = 1)].
Assessing the likelihood of meeting or exceeding a certain number of successes.

Multinomial Distribution






Probability Mass Function (PMF)



Formula:

P(X1=x1,…,Xk=xk)=n!x1!x2!…xk!p1x1p2x2…pkxkP(X_1 = x_1, \dots, X_k = x_k) = \frac{n!}{x_1! x_2! \dots x_k!} p_1^{x_1} p_2^{x_2} \dots p_k^{x_k}
Calculates the probability of obtaining a specific combination of counts (x1,x2,…,xk)(x_1, x_2, \dots, x_k) in nn independent trials, where each trial can result in one of kk possible outcomes.
Probability of observing counts x1,x2,…,xkx_1, x_2, \dots, x_k for outcomes 11 to kk respectively.
Total number of independent trials.
Number of times outcome ii occurs, where ∑i=1kxi=n\sum_{i=1}^{k} x_i = n.
Probability of outcome ii on a single trial, where ∑i=1kpi=1\sum_{i=1}^{k} p_i = 1.
Factorial of nn.
Factorial of xix_i.
If n=5n = 5, k=3k = 3, p1=0.2p_1 = 0.2, p2=0.5p_2 = 0.5, p3=0.3p_3 = 0.3, and we observe counts x1=1x_1 = 1, x2=2x_2 = 2, x3=2x_3 = 2, then P=5!1!2!2!(0.2)1(0.5)2(0.3)2≈0.135P = \frac{5!}{1! 2! 2!} (0.2)^1 (0.5)^2 (0.3)^2 \approx 0.135.
Used in experiments with more than two possible outcomes per trial, such as dice rolls, categorical data analysis, or genetic allele frequencies.





Mean (Expected Value) Of Each Outcome



Formula:

E[Xi]=npiE[X_i] = n p_i
Represents the average number of times outcome ii is expected to occur in nn trials.
Expected value of the count for outcome ii.
Total number of trials.
Probability of outcome ii.
For n=10n = 10 and p2=0.4p_2 = 0.4, E[X2]=10×0.4=4E[X_2] = 10 \times 0.4 = 4.
Predicting average counts of each category in repeated experiments.





Variance Of Each Outcome



Formula:

Var⁡(Xi)=npi(1−pi)\operatorname{Var}(X_i) = n p_i (1 - p_i)
Measures the dispersion of the count for outcome ii around its mean.
Variance of the count for outcome ii.
Total number of trials.
Probability of outcome ii.
Probability of not observing outcome ii.
For n=10n = 10 and p2=0.4p_2 = 0.4, Var⁡(X2)=10×0.4×0.6=2.4\operatorname{Var}(X_2) = 10 \times 0.4 \times 0.6 = 2.4.
Assessing variability in counts for each category.





Covariance Between Outcomes



Formula:

Cov⁡(Xi,Xj)=−npipj\operatorname{Cov}(X_i, X_j) = -n p_i p_j
Measures the degree to which counts of outcomes ii and jj vary together.
Covariance between counts of outcomes ii and jj.
Total number of trials.
Probabilities of outcomes ii and jj, respectively.
For n=10n = 10, p1=0.2p_1 = 0.2, p2=0.5p_2 = 0.5, Cov⁡(X1,X2)=−10×0.2×0.5=−1\operatorname{Cov}(X_1, X_2) = -10 \times 0.2 \times 0.5 = -1.
Understanding the negative dependence between category counts in multinomial experiments.





Correlation Coefficient Between Outcomes



Formula:

ρij=Cov⁡(Xi,Xj)Var⁡(Xi)Var⁡(Xj)=−pipjpi(1−pi)pj(1−pj)\rho_{ij} = \frac{\operatorname{Cov}(X_i, X_j)}{\sqrt{\operatorname{Var}(X_i) \operatorname{Var}(X_j)}} = \frac{-p_i p_j}{\sqrt{p_i (1 - p_i) p_j (1 - p_j)}}
Quantifies the linear relationship between counts of outcomes ii and jj.
Correlation coefficient between counts of outcomes ii and jj.
Covariance between counts of outcomes ii and jj.
Variances of counts of outcomes ii and jj, respectively.
Probabilities of outcomes ii and jj.
Using previous values, ρ12=−0.2×0.50.2×0.8×0.5×0.5≈−0.577\rho_{12} = \frac{-0.2 \times 0.5}{\sqrt{0.2 \times 0.8 \times 0.5 \times 0.5}} \approx -0.577.
Assessing the strength of the relationship between category counts.

Discrete Uniform Distribution






Probability Mass Function (PMF)



Formula:

P(X=k)=1b−a+1P(X = k) = \frac{1}{b - a + 1}
Calculates the probability of the discrete random variable XX taking on a specific integer value kk within the range from aa to bb, where all outcomes are equally likely.
Probability that XX equals kk.
Minimum integer value in the distribution's range.
Maximum integer value in the distribution's range.
An integer within the range a≤k≤ba \leq k \leq b.
Total number of possible integer values in the range.
If a=1a = 1 and b=6b = 6 (like a fair die), then P(X=3)=16P(X = 3) = \frac{1}{6}.
Modeling situations where a finite number of outcomes are equally likely, such as rolling a fair die or selecting a random card from a deck.





Mean (Expected Value)



Formula:

μ=E[X]=a+b2\mu = E[X] = \frac{a + b}{2}
Represents the average value of the distribution, calculated as the midpoint of the range.
Mean or expected value of the distribution.
Minimum integer value.
Maximum integer value.
For a=1a = 1 and b=6b = 6, μ=1+62=3.5\mu = \frac{1 + 6}{2} = 3.5.
Determining the average expected outcome in uniformly random scenarios.





Variance



Formula:

σ2=Var⁡(X)=(b−a+1)2−112\sigma^2 = \operatorname{Var}(X) = \frac{(b - a + 1)^2 - 1}{12}
Measures the dispersion of the values around the mean.
Variance of the distribution.
Minimum integer value.
Maximum integer value.
Total number of possible integer values.
For a=1a = 1 and b=6b = 6, σ2=(6−1+1)2−112=36−112=3512≈2.9167\sigma^2 = \frac{(6 - 1 + 1)^2 - 1}{12} = \frac{36 - 1}{12} = \frac{35}{12} \approx 2.9167.
Assessing the spread of outcomes in uniformly random scenarios.





Standard Deviation



Formula:

σ=(b−a+1)2−112\sigma = \sqrt{\frac{(b - a + 1)^2 - 1}{12}}
Represents the average distance of the values from the mean.
Standard deviation of the distribution.
All other variables are as previously defined.
Using the previous variance example, σ=2.9167≈1.7078\sigma = \sqrt{2.9167} \approx 1.7078.
Understanding the spread of uniformly distributed discrete outcomes.





Cumulative Distribution Function (CDF)



Formula:

P(X≤k)=k−a+1b−a+1P(X \leq k) = \frac{k - a + 1}{b - a + 1} for k=a,a+1,…,bk = a, a+1, \dots, b
Calculates the probability that the random variable XX is less than or equal to a specific value kk.
Cumulative probability up to and including kk.
Minimum integer value.
Maximum integer value.
An integer within the range a≤k≤ba \leq k \leq b.
For a=1a = 1, b=6b = 6, and k=3k = 3, P(X≤3)=3−1+16−1+1=36=0.5P(X \leq 3) = \frac{3 - 1 + 1}{6 - 1 + 1} = \frac{3}{6} = 0.5.
Determining cumulative probabilities in uniformly random discrete events.

Negative Hypergeometric Distribution






Probability Mass Function (PMF)



Formula:

P(X=k)=(k−1r−1)(N−kK−r)(NK)P(X = k) = \frac{\binom{k - 1}{r - 1} \binom{N - k}{K - r}}{\binom{N}{K}}
Calculates the probability that the rr-th success occurs on the kk-th trial when sampling without replacement from a finite population.
Probability that the rr-th success occurs on trial kk.
Number of ways to choose r−1r - 1 successes in the first k−1k - 1 trials.
Number of ways to choose the remaining K−rK - r successes from the remaining N−kN - k items.
Total number of ways to choose KK successes from NN items.
Total number of items in the population.
Total number of successes in the population.
Trial number on which the rr-th success occurs (k=r,r+1,…,N−K+rk = r, r+1, \dots, N - K + r).
Number of successes to be achieved.
If N=20N = 20, K=5K = 5, r=2r = 2, and k=4k = 4, then P(X=4)=(31)(163)(205)≈0.2381P(X = 4) = \frac{\binom{3}{1} \binom{16}{3}}{\binom{20}{5}} \approx 0.2381.
Used to model the number of trials needed to achieve rr successes without replacement, such as in card games or quality inspections.





Mean (Expected Value)



Formula:

μ=E[X]=r(N+1)K+1\mu = E[X] = \frac{r(N + 1)}{K + 1}
Represents the average number of trials required to achieve rr successes when sampling without replacement.
Mean or expected value of the distribution.
Number of successes to be achieved.
Total number of items in the population.
Total number of successes in the population.
If N=20N = 20, K=5K = 5, and r=2r = 2, then μ=2×215+1=7\mu = \frac{2 \times 21}{5 + 1} = 7.
Predicting average trial counts in processes like drawing specific cards from a deck without replacement.





Variance



Formula:

σ2=r(N+1)(N−K)(N−r)(K+1)2(K+2)\sigma^2 = \frac{r (N + 1)(N - K)(N - r)}{(K + 1)^{2} (K + 2)}
Measures the dispersion of the number of trials around the mean.
Variance of the distribution.
Number of successes to be achieved.
Total number of items in the population.
Total number of successes in the population.
Number of failures in the population.
Number of items not yet drawn after achieving rr successes.
If N=20N = 20, K=5K = 5, and r=2r = 2, then σ2=2×21×15×18(5+1)2×7≈5.7143\sigma^2 = \frac{2 \times 21 \times 15 \times 18}{(5 + 1)^{2} \times 7} \approx 5.7143.
Assessing variability in the number of trials needed for achieving rr successes without replacement.





Standard Deviation



Formula:

σ=r(N+1)(N−K)(N−r)(K+1)2(K+2)\sigma = \sqrt{\frac{r (N + 1)(N - K)(N - r)}{(K + 1)^{2} (K + 2)}}
Represents the average distance of the number of trials from the mean.
Standard deviation of the distribution.
All other variables are as previously defined.
Using the previous variance example, σ=5.7143≈2.3917\sigma = \sqrt{5.7143} \approx 2.3917.
Understanding the spread of trials needed to achieve rr successes without replacement.





Cumulative Distribution Function (CDF)



Formula:

P(X≤k)=1−(N−rk−r)(r−1r−1)(Nk)P(X \leq k) = 1 - \frac{\binom{N - r}{k - r} \binom{r - 1}{r - 1}}{\binom{N}{k}}
Calculates the probability that the rr-th success occurs on or before the kk-th trial.
Probability that the rr-th success occurs on or before trial kk.
Number of ways to choose k−rk - r failures from N−rN - r failures.
Total number of ways to choose kk items from NN items.
Trial number.
For N=20N = 20, K=5K = 5, r=2r = 2, and k=7k = 7, compute P(X≤7)P(X \leq 7) accordingly.
Assessing the likelihood of achieving rr successes within a certain number of trials when sampling without replacement.

Logarithmic Distribution






Probability Mass Function (PMF)



Formula:

P(X=k)=−1ln⁡(1−p)pkkP(X = k) = -\frac{1}{\ln(1 - p)} \frac{p^{k}}{k}
Calculates the probability of the discrete random variable XX taking on the value kk, where kk represents the number of failures required to get the first success in a logarithmic series.
Probability that XX equals kk.
Parameter of the distribution, such that 0<p<10 < p < 1.
A positive integer (k=1,2,3,…k = 1, 2, 3, \dots).
Natural logarithm of 1−p1 - p, normalizing constant.
If p=0.5p = 0.5 and k=3k = 3, then P(X=3)=−1ln⁡(1−0.5)(0.5)33≈0.0625P(X = 3) = -\frac{1}{\ln(1 - 0.5)} \frac{(0.5)^{3}}{3} \approx 0.0625.
Modeling species abundance in ecology, or the number of times a rare event occurs.





Mean (Expected Value)



Formula:

μ=E[X]=−p(1−p)ln⁡(1−p)\mu = E[X] = \frac{-p}{(1 - p) \ln(1 - p)}
Represents the average value of the distribution.
Mean or expected value of the distribution.
Parameter of the distribution.
Natural logarithm of 1−p1 - p.
Complement of parameter pp.
If p=0.5p = 0.5, then μ=−0.5(0.5)(−0.6931)≈1.4427\mu = \frac{-0.5}{(0.5)(-0.6931)} \approx 1.4427.
Determining the average number of occurrences in phenomena following a logarithmic distribution.





Variance



Formula:

σ2=Var⁡(X)=−p(p+ln⁡(1−p))(1−p)2[ln⁡(1−p)]2\sigma^2 = \operatorname{Var}(X) = \frac{-p (p + \ln(1 - p))}{(1 - p)^{2} [\ln(1 - p)]^{2}}
Measures the dispersion of the values around the mean.
Variance of the distribution.
Parameter of the distribution.
Natural logarithm of 1−p1 - p.
Complement of parameter pp.
If p=0.5p = 0.5, calculate σ2\sigma^2 accordingly.
Assessing the spread of outcomes in logarithmically distributed data.





Standard Deviation



Formula:

σ=Var⁡(X)\sigma = \sqrt{\operatorname{Var}(X)}
Represents the average distance of the values from the mean.
Standard deviation of the distribution.
Variance of the distribution.
Calculate σ\sigma using the variance from the previous example.
Understanding the spread of logarithmically distributed discrete outcomes.





Generating Function



Formula:

GX(s)=ln⁡(1−ps)ln⁡(1−p)G_X(s) = \frac{\ln(1 - p s)}{\ln(1 - p)}
Generates the probabilities of the distribution and can be used to find moments.
Probability generating function of the distribution.
Parameter of the distribution.
Dummy variable used in generating functions.
Natural logarithm of 1−ps1 - p s.
Natural logarithm of 1−p1 - p.
For p=0.5p = 0.5, compute GX(s)G_X(s) accordingly.
Used in advanced probability to derive properties of the distribution.