#set document(title: "5.11 Hypergeometric Distribution", author: "OpenStax") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 5.11#h(0.6em)Hypergeometric Distribution #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Prerequisites] Binomial Distribution, Permutations and Combinations #linebreak() #linebreak() ] The hypergeometric distribution is used to calculate probabilities when sampling without replacement. For example, suppose you first randomly sample one card from a deck of 52. Then, without putting the card back in the deck you sample a second and then (again without replacing cards) a third. Given this sampling procedure, what is the probability that exactly two of the sampled cards will be aces (4 of the 52 cards in the deck are aces). You can calculate this probability using the following formula based on the hypergeometric distribution: #math.equation(block: false, alt: "p equals the fraction open parenthesis k C sub x close parenthesis open parenthesis open parenthesis N minus k close parenthesis C sub open parenthesis n minus x close parenthesis close parenthesis over N C sub n")[$p = frac(( k C_(x) ) ( ( N − k ) C_(( n − x )) ), N C_(n))$] where k is the number of "successes" in the population #linebreak() x is the number of "successes" in the sample #linebreak() N is the size of the population #linebreak() n is the number sampled #linebreak() p is the probability of obtaining exactly x successes #linebreak() #sub[k]C#sub[x] is the number of combinations of k things taken x at a time In this example, k = 4 because there are four aces in the deck, x = 2 because the problem asks about the probability of getting two aces, N = 52 because there are 52 cards in a deck, and n = 3 because 3 cards were sampled. Therefore, #math.equation(block: false, alt: "p equals the fraction open parenthesis 4 C sub 2 close parenthesis open parenthesis open parenthesis 52 minus 4 close parenthesis C sub open parenthesis 3 minus 2 close parenthesis close parenthesis over 52 C sub 3")[$p = frac(( 4 C_(2) ) ( ( 52 − 4 ) C_(( 3 − 2 )) ), 52 C_(3))$] = #math.equation(block: true, alt: "p equals the fraction the fraction 4 ! over 2 ! 2 ! the fraction 48 ! over 47 ! 1 ! over the fraction 52 ! over 49 ! 3 ! equals 0.013")[$p = frac(frac(4 !, 2 ! #h(0.167em) 2 !) frac(48 !, 47 ! #h(0.167em) 1 !), frac(52 !, 49 ! #h(0.167em) 3 !)) = 0.013$] The mean and standard deviation of the hypergeometric distribution are: #math.equation(block: true, alt: "mean equals the fraction open parenthesis n close parenthesis open parenthesis k close parenthesis over N")[$"mean" = frac(( n ) ( k ), N)$] #math.equation(block: true, alt: "sd equals the square root of the fraction open parenthesis n close parenthesis open parenthesis k close parenthesis open parenthesis N minus k close parenthesis open parenthesis N minus n close parenthesis over N squared open parenthesis N minus 1 close parenthesis")[$"sd" = sqrt(frac(( n ) ( k ) ( N − k ) ( N − n ), N^(2) ( N − 1 )))$]