class: center, middle, inverse, title-slide .title[ # Binomial Distributions <br> 🐫 ] .author[ ### S. Mason Garrison ] --- layout: true <div class="my-footer"> <span> <a href="https://psychmethods.github.io/coursenotes/" target="_blank">Methods in Psychological Research</a> </span> </div> --- class: middle # Binomial Distributions --- class: middle # Road Map - Binomial distributions - Binomial requirements - Sampling distribution of a count --- # Motivation - So far, we have used a combination of specific rules, such as the - Addition Rule, Multiplication Rule, Bayes Rule, etc, - to calculate probabilities and expected values. -- - But, it seems a bit out of place, - given this class’s emphasis on application... -- - These specific rules can be transformed into probability models, which, in turn, generate distributions – not unlike the normal distribution. --- ## Binomial Distributions - Just like the normal distribution describes bell-shaped data, the binomial distribution describes count data. - Specifically, it models the number of successful outcomes, - when the total number of trials is known beforehand. - (The distribution of a count depends on how the data are produced. The binomial setting is a common situation.) --- ## Binomial Distribution - What does a binomial distribution look like? - It depends on two parameters: - Probability of Success (P) - Sample Size/Number of Trials (N) - (Recall that the normal distribution depends on μ and σ) --- # Ten Trials, Probability of Success is 50% .pull-left-narrow[ <table class="table" style="color: black; width: auto !important; margin-left: auto; margin-right: auto;"> <thead> <tr> <th style="text-align:center;"> Successes </th> <th style="text-align:center;"> Probability </th> </tr> </thead> <tbody> <tr> <td style="text-align:center;"> 0 </td> <td style="text-align:center;"> 0.0009766 </td> </tr> <tr> <td style="text-align:center;"> 1 </td> <td style="text-align:center;"> 0.0097656 </td> </tr> <tr> <td style="text-align:center;"> 2 </td> <td style="text-align:center;"> 0.0439453 </td> </tr> <tr> <td style="text-align:center;"> 3 </td> <td style="text-align:center;"> 0.1171875 </td> </tr> <tr> <td style="text-align:center;"> 4 </td> <td style="text-align:center;"> 0.2050781 </td> </tr> <tr> <td style="text-align:center;"> 5 </td> <td style="text-align:center;"> 0.2460938 </td> </tr> <tr> <td style="text-align:center;"> 6 </td> <td style="text-align:center;"> 0.2050781 </td> </tr> <tr> <td style="text-align:center;"> 7 </td> <td style="text-align:center;"> 0.1171875 </td> </tr> <tr> <td style="text-align:center;"> 8 </td> <td style="text-align:center;"> 0.0439453 </td> </tr> <tr> <td style="text-align:center;"> 9 </td> <td style="text-align:center;"> 0.0097656 </td> </tr> <tr> <td style="text-align:center;"> 10 </td> <td style="text-align:center;"> 0.0009766 </td> </tr> </tbody> </table> ] .pull-right-wide[ <img src="data:image/png;base64,#binomial_files/figure-html/unnamed-chunk-3-1.png" width="90%" style="display: block; margin: auto;" /> ] --- # Ten Trials, Probability of Success is 25% .pull-left-narrow[ <table class="table" style="color: black; width: auto !important; margin-left: auto; margin-right: auto;"> <thead> <tr> <th style="text-align:center;"> Successes </th> <th style="text-align:center;"> Probability </th> </tr> </thead> <tbody> <tr> <td style="text-align:center;"> 0 </td> <td style="text-align:center;"> 0.0563135 </td> </tr> <tr> <td style="text-align:center;"> 1 </td> <td style="text-align:center;"> 0.1877117 </td> </tr> <tr> <td style="text-align:center;"> 2 </td> <td style="text-align:center;"> 0.2815676 </td> </tr> <tr> <td style="text-align:center;"> 3 </td> <td style="text-align:center;"> 0.2502823 </td> </tr> <tr> <td style="text-align:center;"> 4 </td> <td style="text-align:center;"> 0.1459980 </td> </tr> <tr> <td style="text-align:center;"> 5 </td> <td style="text-align:center;"> 0.0583992 </td> </tr> <tr> <td style="text-align:center;"> 6 </td> <td style="text-align:center;"> 0.0162220 </td> </tr> <tr> <td style="text-align:center;"> 7 </td> <td style="text-align:center;"> 0.0030899 </td> </tr> <tr> <td style="text-align:center;"> 8 </td> <td style="text-align:center;"> 0.0003862 </td> </tr> <tr> <td style="text-align:center;"> 9 </td> <td style="text-align:center;"> 0.0000286 </td> </tr> <tr> <td style="text-align:center;"> 10 </td> <td style="text-align:center;"> 0.0000010 </td> </tr> </tbody> </table> ] .pull-right-wide[ <img src="data:image/png;base64,#binomial_files/figure-html/unnamed-chunk-5-1.png" width="90%" style="display: block; margin: auto;" /> ] --- # Ten Trials, Probability of Success is 5% .pull-left-narrow[ <table class="table" style="color: black; width: auto !important; margin-left: auto; margin-right: auto;"> <thead> <tr> <th style="text-align:center;"> Successes </th> <th style="text-align:center;"> Probability </th> </tr> </thead> <tbody> <tr> <td style="text-align:center;"> 0 </td> <td style="text-align:center;"> 0.5987369 </td> </tr> <tr> <td style="text-align:center;"> 1 </td> <td style="text-align:center;"> 0.3151247 </td> </tr> <tr> <td style="text-align:center;"> 2 </td> <td style="text-align:center;"> 0.0746348 </td> </tr> <tr> <td style="text-align:center;"> 3 </td> <td style="text-align:center;"> 0.0104751 </td> </tr> <tr> <td style="text-align:center;"> 4 </td> <td style="text-align:center;"> 0.0009648 </td> </tr> <tr> <td style="text-align:center;"> 5 </td> <td style="text-align:center;"> 0.0000609 </td> </tr> <tr> <td style="text-align:center;"> 6 </td> <td style="text-align:center;"> 0.0000027 </td> </tr> <tr> <td style="text-align:center;"> 7 </td> <td style="text-align:center;"> 0.0000001 </td> </tr> <tr> <td style="text-align:center;"> 8 </td> <td style="text-align:center;"> 0.0000000 </td> </tr> <tr> <td style="text-align:center;"> 9 </td> <td style="text-align:center;"> 0.0000000 </td> </tr> <tr> <td style="text-align:center;"> 10 </td> <td style="text-align:center;"> 0.0000000 </td> </tr> </tbody> </table> ] .pull-right-wide[ <img src="data:image/png;base64,#binomial_files/figure-html/unnamed-chunk-7-1.png" width="90%" style="display: block; margin: auto;" /> ] --- ## Binomial Distribution - Note: This is a discrete probability model - The distribution isn’t smooth because every outcome can fall into a bin - i.e, There are a finite number of values to measure. - In contrast, a continuous probability model has a smooth range of values. - Even when N is large it is still discrete. - See https://shiny.rit.albany.edu/stat/binomial/ --- ## Binomial Requirements A binomial distribution requires: - a fixed number, `\(n\)`, of observations. - The `\(n\)` observations are all independent. That is, knowing the result of one observation does not change the probabilities we assign to other observations. - Each observation falls into one of just two categories, which, for convenience, we call “success” and “failure.” - The probability of a success, call it `\(p\)`, is the same for each observation. --- # Wrapping Up... --- # This Time… N chose K --- # Formula for Binomial Probabilities - The probability of getting exactly k successes in n independent Bernoulli trials is: `$$P(X = k) = \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}$$` - where - n is the number of trials, - k is the number of successful trials, - p is the probability of success on an individual trial, and - (1-p) is the probability of failure on an individual trial. - Sometimes this equation is written as: `$$P(X = k|pN) = \binom{n}{k} p^k q^{n-k}$$` --- # Formula for Binomial Probabilities `$$P(X = k|pN) = \binom{n}{k} p^k q^{n-k}$$` - The term `\(\binom{n}{k}\)` is read "n choose k" and is calculated as: `$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$` - where n! (n factorial) is the product of all positive integers up to n. --- ## Binomial Formula Breakdown - The binomial coefficient counts the number of different ways in which `\(k\)` successes can be arranged among `\(n\)` trials. - The binomial probability `\(P(X=k)\)` is this count multiplied by the probability of any one specific arrangement of the `\(k\)` successes. - The probability of any one specific arrangement is `\(p^k(1-p)^{n-k}\)`, the probability of `\(k\)` successes and `\(n-k\)` failures. --- # Binomial Formula Breakdown - If `\(X\)` has the binomial distribution with `\(n\)` trials and probability `\(p\)` of success on each trial, the possible values of `\(X\)` are 0, 1, 2,…, `\(n\)`. If `\(k\)` is any one of these values: `$$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$` - Number of arrangements of k successes = `\(\binom{n}{k}\)` - Probability of any one arrangement = `\(p^k (1-p)^{n-k}\)` - Probability of k successes: `\(p^k\)` - Probability of n-k failures: `\((1-p)^{n-k}\)` --- # N chose K - Chose K numbers from N options: `\(\binom{n}{k}\)` - `\(\frac{n!}{k!(n-k)!}\)` - Note: 0! =1 - n! = n × (n − 1) × (n − 2) × ... × 3 × 2 × 1 - 6!/4!= 6*5 --- # Example: N chose K - Example: How many ways can we choose 3 people from a group of 5? - Answer: `\(\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3!}{3! \times 2!} = \frac{5 \times 4}{2 \times 1} = 10\)` - The 10 combinations are: - ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE - Note: Order does not matter, so ABC is the same as ACB, BAC, BCA, CAB, CBA --- # Draw a card, guess its color - P=1/2; 1-p=1/2 - Let's do n = 3 - `$$\binom{n}{k} p^k (1-p)^{n-k}$$` --- # Draw a card, guess its color (0 correct) `$$\binom{n}{k} p^k (1-p)^{n-k}$$` `$$P(x=0|1/2,3) = \binom{3}{0} (.5)^0 (1-.5)^{3-0}$$` `$$\frac{3!}{0!(3-0)!} (.5)^0 (1-.5)^{3-0}$$` `$$\frac{3!}{1(3!)} (.5)^0 (.5)^{3}$$` `$$1 (1) (.5)^{3} = .125$$` --- # Draw a card, guess its color (1 correct) `$$\binom{n}{k} p^k (1-p)^{n-k}$$` `$$P(x=1|1/2,3) = \binom{3}{1} (.5)^1 (1-.5)^{3-1}$$` `$$\frac{3!}{1!(3-1)!} (.5)^1 (1-.5)^{3-1}$$` `$$\frac{3!}{1!(3-1)!} (.5)^1 (.5)^{2}$$` `$$3 (.5)^1 (.5)^{2} = 3/8 = .375$$` --- # Draw a card, guess its color (2 correct) `$$\binom{n}{k} p^k (1-p)^{n-k}$$` `$$P(x=2|1/2,3) = \binom{3}{2} (.5)^2 (1-.5)^{3-2}$$` `$$\frac{3!}{2!(3-2)!} (.5)^2 (1-.5)^{3-2}$$` `$$\frac{3!}{2!(3-2)!} (.5)^2 (.5)^{1}$$` `$$3 (.5)^2 (.5)^{1} = 3/8 = .375$$` --- # Draw a card, guess its color (3 correct) `$$\binom{n}{k} p^k (1-p)^{n-k}$$` `$$P(x=3|1/2,3) = \binom{3}{3} (.5)^3 (1-.5)^{3-3}$$` `$$\frac{3!}{3!(3-3)!} (.5)^3 (1-.5)^{3-3}$$` `$$\frac{3!}{3!(3-3)!} (.5)^3 (.5)^{0}$$` `$$1 (.5)^3 (1) = 1/8 = .125$$` --- class: middle # Wrapping Up... --- class: middle # Unmarketable tomatoes `🍅` --- # Binomial distributions in statistical sampling .pull-left[ - The binomial distributions are important in statistics when we want to make inferences about the proportion p of successes in a population. - Suppose 11% of one producer’s tomatoes are unmarketable. At the warehouse, an official inspects a simple random sample of 10 tomatoes from a shipment of 10,000. Let `\(X\)` = number of unmarketable tomatoes. - What is `\(P(X=0)\)`? ] .pull-right[ <img src="data:image/png;base64,#https://upload.wikimedia.org/wikipedia/commons/f/fe/Solanum_aethiopicum_for_seed_production.JPG" width="80%" style="display: block; margin: auto;" /> Image source: [Michael Hermann](https://commons.wikimedia.org/wiki/User:NusHub) and http://www.cropsforthefuture.org/ ] --- # Example: Unmarketable tomatoes - Suppose 11% of one producer’s tomatoes are unmarketable. At the warehouse, an official inspects a simple random sample of 10 tomatoes from a shipment of 10,000. Let `\(X\)` = number of unmarketable tomatoes. - What is `\(P(X=0)\)`? - Without the binomial, we could calculate the probability like this: `$$P(X=0) = P(\text{first is good}) \times P(\text{second is good}) \times ... \times P(\text{tenth is good})$$` `$$= \frac{8900}{10000} \times \frac{8899}{9999} \times \frac{8898}{9998} ... \times 8891/9991 = 0.3116$$` But this is tedious. --- # Example: Unmarketable tomatoes - Suppose 11% of one producer’s tomatoes are unmarketable. At the warehouse, an official inspects a simple random sample of 10 tomatoes from a shipment of 10,000. Let `\(X\)` = number of unmarketable tomatoes. - What is `\(P(X=0)\)`? - With the binomial, we could calculate the probability like this: `$$P(X=0) \approx \binom{10}{0} (0.11)^0 (1-0.11)^{10-0}$$` `$$= 1 \times 1 \times (0.89)^{10} = 0.3138$$` - This is much easier, and the approximation is pretty good because the population is much larger than the sample. - Note: The binomial distribution is exact if the sample is drawn with replacement. --- # Sampling Distribution of a Count - Choose an simple random sample (SRS) of size `\(n\)` from a population with proportion p of successes. - When the sample size is less than 5% of the population size, the count `\(X\)` of successes in the sample has approximately the binomial distribution with parameters `\(n\)` and `\(p\)` - Conditions for the Binomial Approximation - The sample is an SRS from the population. - The sample size is no more than 5% of the population size. - The population has a proportion `\(p\)` of successes. - The count `\(X\)` of successes in the sample is the variable of interest. --- # Binomial mean and standard deviation - If a count `\(X\)` has the binomial distribution based on `\(n\)` observations with probability `\(p\)` of success, what is its mean, `\(\mu\)` - If a count `\(X\)` has the binomial distribution based on `\(n\)` observations with probability `\(p\)` , the mean and standard deviation of 𝑋 are: `$$\mu = np$$` `$$\sigma = \sqrt{np(1-p)}$$` --- # Recall our tomatoes 🍅 - Suppose 11% of one producer’s tomatoes are unmarketable. At the warehouse, an official inspects a simple random sample of 10 tomatoes from a shipment of 10,000. Let `\(X\)` = number of unmarketable tomatoes. - What would the mean and standard deviation be for the binomial distribution of these 10 tomatoes. - Mean: `\(\mu = np = 10(0.11) = 1.1\)` - Standard Deviation: `\(\sigma = \sqrt{np(1-p)} = \sqrt{10(0.11)(0.89)} = 0.989\)` --- # Wrapping Up... --- # Normal Approximation to the Binomial - As `\(n\)` gets larger, something interesting happens to the shape of a binomial distribution. - The distribution becomes more symmetric and bell-shaped. <img src="data:image/png;base64,#binomial_files/figure-html/unnamed-chunk-9-1.png" width="60%" style="display: block; margin: auto;" /> --- ## Normal Approximation to the Binomial - When n is large, the binomial distribution can be approximated by a normal distribution with mean `\(\mu = np\)` and standard deviation `\(\sigma = \sqrt{np(1-p)}\)`. - As a rule of thumb, we will use the Normal approximation when `\(n\)` is so large that `\(n p\)`≥10 and `\(n\)`(1 – `\(p\)`)≥10. - This rule ensures that the binomial distribution is not too skewed. --- # Wrapping Up… --- # Functions in R - `dbinom`(x, size, prob) - gives the probability of x successes in size trials with probability prob of success on each trial. - `pbinom(q, size, prob…)` - gives the cumulative density function, computes the probability that a binomially distributed random number will be at that value or less than that number - `qbinom()` gives the quantile function, and, is the inverse of `pbinom`, give it a probability, - it produces the number whose cumulative distribution matches the probability - `rbinom()` generates random deviates. --- # Examples using R - Question: Suppose widgets produced at Acme Widget Works have probability 0.005 of being defective. Suppose widgets are shipped in cartons containing 25 widgets. What is the probability that a randomly chosen carton contains exactly one defective widget? - Question Rephrased: What is P(X = 1) when X has the Bin(25, 0.005) distribution? ``` r dbinom(1, 25, 0.005) ``` ``` ## [1] 0.1108317 ``` - In case you're curious, here's what Github Copilot guessed was the answer: > [1] 0.07461825 --- # Examples using R - Question: Suppose widgets produced at Acme Widget Works have probability 0.005 of being defective. Suppose widgets are shipped in cartons containing 25 widgets. What is the probability that a randomly chosen carton contains no more than one defective widget? - Question Rephrased: What is P(X ≤ 1) when X has the Bin(25, 0.005) distribution? ``` r pbinom(1, 25, 0.005) ``` ``` ## [1] 0.9930519 ``` - In case you're curious, here's what Github Copilot guessed was the answer: > [1] 0.9743589 --- # Examples using R - Question: What are the 10th, 20th, and so forth quantiles of the Bin(10, 1/3) distribution? ``` r qbinom(0.1, 10, 1/3) ``` ``` ## [1] 1 ``` ``` r qbinom(0.2, 10, 1/3) ``` ``` ## [1] 2 ``` - and so forth, or all at once with ``` r qbinom(seq(0.1, 0.9, 0.1), 10, 1/3) ``` ``` ## [1] 1 2 3 3 3 4 4 5 5 ``` --- # Wrapping Up…