#set document(title: "4.5 Independent Events", author: "Rachel Webb") #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")) == 4.5#h(0.6em)Independent Events Two trials (or events or results of a random experiment) are independent trials if the outcome of one trial does not influence the outcome of the second trial. If two events are not independent, they are dependent events. For instance, if two coins are flipped they are independent since flipping one coin does not affect the outcome of the second coin. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #strong[Independent Events:] If A and B are independent events, then P(A ∩ B) = P(A) ‧ P(B). ] Be careful with this rule. You cannot just multiply probabilities to find an intersection unless you know they are independent. Also, do not confuse independent events with mutually exclusive events. Two events are mutually exclusive when the P(A ∩ B) = 0. #examplebox("Example 1")[][ If a random experiment consists of flipping a coin twice, find the probability of getting heads twice in a row. #solutionbox[ The event of getting a head on the first flip is independent of getting a head on the second flip since the probability does not change with each flip of the coin. Thus, using the multiplication rule of independent events, P(Both coins are heads) = P(1#super[st] coin is a head)·P(2#super[nd] coin is a head) = #math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$] #linebreak() ∙ #math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$] = #math.equation(block: false, alt: "the fraction 1 over 4")[$frac(1, 4)$] = 0.25. ] ] #examplebox("Example 2")[][ The probability of Apple stock rising is 0.3, the probability of Boeing stock rising is 0.4. Assume Apple and Boeing stocks are independent. What is the probability that neither stock rises? #solutionbox[ Let A = Apple stock and B = Boeing stock. Since A and B are independent the probability of both stocks rising at the same time is P(A ∩ B) = 0.3 ‧ 0.4 = 0.12. Neither is the complement to either. P(Not Either) = 1 – P(A U B) = 1 – (P(A) + P(B) – P(A ∩ B)) = 1 – (0.3 + 0.4 + 0.12) = 1 – 0.58 = 0.42. ] ] #examplebox("Example 3")[][ The probability that a student has their own laptop is 0.78. If three students are randomly selected, what is the probability that at least one owns a laptop? #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[At least one, via the complement] Opens a distribution that counts how many of the 3 independent students own a laptop (a preview of Chapter 5). It loads already showing P(X \<= 0) = P(none own one) = 0.22^3 = 0.0106; subtract from 1 for the book's P(at least one) = 0.9894. Change n to 5 students and watch "at least one" get even more likely. - P(none of 3 own a laptop) = 0.0106 ] #solutionbox[ There is an assumption that the three students are not related and that the probability of one owning a laptop is independent of the other people owning a laptop. The probability of none owning a laptop is (1 – 0.78)3 = 0.0106. The probability of at least one is the same as 1 – P(None) = 1 – 0.0106 = 0.9894. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ When two events are dependent, you cannot simply multiply their corresponding probabilities to find their intersection. You will need to use the General Multiplication Rule discussed in the next section. ]