```{r} #| label: setup #| include: false set.seed(2026) ``` ## ๐ฌ The Idea in 3 Minutes ::: {style="text-align:center"} [Watch this short intuition video before (or after) the slides. Captions: CC button.]{style="font-size:22px"} ::: --- ## ๐ฏ Learning Objectives ::: {style="font-size: 32px"} By the end of this lecture, you will be able to: - **Follow** the four steps of the event-composition method (Wackerly ยง2.9) - **Write** a compound event as unions, intersections and complements of events whose probabilities are known - **Test** a composition against the sample points before trusting it - **Compute** the reliability of series and parallel systems, and of a chain that mixes them - **Handle** "at least one" and sequential draws without replacement by composition ::: --- ## ๐บ๏ธ Where We Are ::: {style="font-size: 32px"} **Wackerly ยง2.9** Last class ended on a promise: *the event-composition method โ how to decide, for a compound event with no obvious decomposition, which events to name in the first place.* ::: {.fragment} We now own the tools: Theorem 2.5 for $\cap$, Theorem 2.6 for $\cup$, Theorem 2.7 for complements. ::: ::: {.fragment} Today is about the step **before** the tools โ choosing the events so that every probability the laws ask for is one you actually know. ::: ::: --- ## ๐ณ Motivating Question ::: {style="font-size: 30px"} ::: {.callout-important} ## A card payment in a Baku supermarket The payment passes a **terminal**, the **acquirer's switch**, the **card network** (run from two redundant data centres) and the **issuing bank**. Each piece has a published availability. What is the probability that the payment is authorised? ::: ::: {.fragment} Nobody publishes "the probability the payment goes through". It has to be **built** from the pieces โ and the building is the method. ::: ::: --- ## ๐ The Event-Composition Method ::: {style="font-size: 30px"} ::: {.callout-note} ## Wackerly ยง2.9: the four steps 1. Define the experiment. 2. Visualize the nature of the sample points. Identify a few to clarify your thinking. 3. Write an equation expressing the event of interest, $A$, as a composition of two or more events, using unions, intersections and/or complements. Make certain that $A$ and the composition represent the **same set of sample points**. 4. Apply the additive and multiplicative laws to the composition to find $P(A)$. ::: Unlike the sample-point method, no list of $S$ is needed โ only a clear picture of a typical sample point. ::: --- ## ๐ Step 3 Is the Hard One ::: {style="font-size: 30px"} Many compositions equal $A$. **Pick the one in which every probability in step 4 is known.** If one is unknown, the method fails. ::: {.fragment} Two kinds of pieces make step 4 easy: - **mutually exclusive** pieces, so the additive law has no subtraction - **independent** pieces, so the multiplicative law has no conditionals ::: ::: {.fragment} The most common error is a composition that is simply not equal to $A$. **Test the equality**: take a sample point in $A$ and check it lands in the composition, and the other way round. ::: ::: --- ## ๐ข Example 2.17: Export Shipments ::: {style="font-size: 28px"} An exporter sends 40% of its shipments to Europe through **Alat port** ($R$) and 60% overland through the **Red Bridge** crossing ($D$). Customs clears 70% of Alat shipments and 80% of Red Bridge shipments within 48 hours ($F$). A shipment is picked at random. ::: {.fragment} **Step 3.** Every shipment takes exactly one route, so $$F = (F \cap R) \cup (F \cap D), \qquad \text{two mutually exclusive pieces.}$$ ::: ::: {.fragment} **Step 4.** $P(F \cap R) = P(F \mid R)P(R) = 0.7 \times 0.4 = 0.28$ and $P(F \cap D) = 0.8 \times 0.6 = 0.48$, so $$P(F) = 0.28 + 0.48 = 0.76$$ ::: ::: --- ## ๐ฆ Example 2.19: A Clearing Mandate ::: {style="font-size: 28px"} Five correspondent banks bid to clear a company's euro payments; two quote the **lowest fees**. The treasurer shortlists **two at random**. What is $P(A)$, $A$ = *exactly one of the two cheapest is shortlisted*? ::: {.fragment} Let $B_1$: cheapest bank on the first pick, $B_2$: one of the three dearer banks on the second, and $B_3, B_4$ the same the other way round. Define $C_1, \ldots, C_4$ likewise for the **second-cheapest** bank. Then $$A = (B_1 \cap B_2) \cup (B_3 \cap B_4) \cup (C_1 \cap C_2) \cup (C_3 \cap C_4)$$ ::: ::: {.fragment} **Test it:** every shortlist with exactly one cheap bank lies in exactly one of the four pieces, and no other shortlist lies in any. ::: ::: --- ## ๐งฎ Example 2.19: Step 4, and a Check ::: {style="font-size: 28px"} The four pieces are mutually exclusive, and each is a two-stage draw: $$P(B_1 \cap B_2) = P(B_1)P(B_2 \mid B_1) = \tfrac{1}{5} \times \tfrac{3}{4} = \tfrac{3}{20}, \qquad P(A) = 4 \times \tfrac{3}{20} = \tfrac{3}{5}$$ ::: ```{r} #| label: clearing-check # Sample-point check: list all C(5,2) = 10 shortlists; banks 1 and 2 are cheapest shortlists <- combn(5, 2) n_cheap <- colSums(shortlists <= 2) c(shortlists = ncol(shortlists), exactly_one = sum(n_cheap == 1), P_A = mean(n_cheap == 1)) ``` ::: {style="font-size: 28px"} Same answer as the sample-point method, reached without writing $S$ down. ::: --- ## โก Example 2.20: Power Feeds ::: {style="font-size: 28px"} A Baku data centre hosting a bank's core system has **three independent power feeds**, each live during a grid event with probability 0.9. What is $P(A)$, $A$ = *at least one feed is live*? ::: {.fragment} Let $B_i$: feed $i$ fails, so $P(B_i) = 0.1$. Then $\bar{A} = B_1 \cap B_2 \cap B_3$, and by Theorem 2.7 and independence $$P(A) = 1 - P(B_1)P(B_2)P(B_3) = 1 - (0.1)^3 = 0.999$$ ::: ::: {.fragment} As a union, $A$ needs seven terms of inclusionโexclusion. As a complement, it is **one intersection**. Frequently $P(\bar{A})$ is the easy one. ::: ::: --- ## ๐ Series and Parallel Systems ::: {style="font-size: 28px"} $n$ independent components, each working with probability $p$. :::: {.columns} ::: {.column width="50%"} ::: {.callout-note} ## Series: all must work A payment settlement chain. $$W = W_1 \cap W_2 \cap \cdots \cap W_n$$ $$P(W) = p^n$$ ::: ::: ::: {.column width="50%"} ::: {.callout-note} ## Parallel: one is enough Redundant power feeds. $$\bar{W} = \bar{W}_1 \cap \bar{W}_2 \cap \cdots \cap \bar{W}_n$$ $$P(W) = 1 - (1-p)^n$$ ::: ::: :::: ::: {.fragment} Same components, opposite compositions: every link added to a series **costs** reliability, every feed added in parallel **buys** it. ::: ::: --- ## ๐ป Series vs Parallel in R ```{r} #| label: series-parallel p <- 0.95 # each component works 95% of the time n <- c(1, 2, 4, 8, 12, 20) data.frame(components = n, series = round(p^n, 4), parallel = round(1 - (1 - p)^n, 4), parallel_fails = signif((1 - p)^n, 3)) ``` ::: {style="font-size: 28px"} Twelve 95% links in series work only 54% of the time; two 95% feeds in parallel already give 99.75%. ::: --- ## ๐ฌ Interactive: Series Versus Parallel {.smaller} ```{ojs} //| echo: false viewof p_comp = { const input = Inputs.range([0.50, 0.99], {value: 0.95, step: 0.01, label: "Reliability of each component p:"}); ['pointerdown','touchstart','mousedown','click','wheel','pointermove','touchmove'] .forEach(e => input.addEventListener(e, ev => ev.stopPropagation())); return input; } ``` ```{ojs} //| echo: false rel = Array.from({length: 20}, (_, i) => i + 1).flatMap(n => [ {n, R: Math.pow(p_comp, n), system: "series"}, {n, R: 1 - Math.pow(1 - p_comp, n), system: "parallel"} ]) halfN = (rel.find(d => d.system === "series" && d.R < 0.5) || {n: "more than 20"}).n md`At **p = ${p_comp.toFixed(2)}**, a series chain drops below 50% at **${halfN}** components.` Plot.plot({ width: 1150, height: 300, marginLeft: 78, marginBottom: 58, marginRight: 110, marginTop: 40, style: {fontSize: "18px"}, x: {label: "Number of components n", domain: [1, 20], ticks: [1, 5, 10, 15, 20]}, y: {label: "P(system works)", domain: [0, 1], tickFormat: ".1f"}, color: {domain: ["series", "parallel"], range: ["#8b2635", "#2f5d8a"]}, marks: [ Plot.ruleY([0.5], {stroke: "#cbb8a9", strokeDasharray: "4 4"}), Plot.line(rel, {x: "n", y: "R", stroke: "system", strokeWidth: 3}), Plot.text(rel.filter(d => d.n === 20), {x: "n", y: "R", text: "system", fill: "system", dx: 12, textAnchor: "start", fontSize: 20}), Plot.ruleY([0]) ] }) ``` --- ## ๐ณ Back to the Card Payment ::: {style="font-size: 28px"} Terminal $T$ (0.995), acquirer switch $Q$ (0.99), network data centres $N_1, N_2$ (0.97 each), issuer $I$ (0.99), all independent. Authorisation needs every stage, and the network needs **either** centre: $$\text{Auth} = T \cap Q \cap (N_1 \cup N_2) \cap I$$ ::: {.fragment} $$P(N_1 \cup N_2) = 1 - (0.03)^2 = 0.9991$$ $$P(\text{Auth}) = 0.995 \times 0.99 \times 0.9991 \times 0.99 = 0.9743$$ ::: ::: {.fragment} With a single data centre it would be $0.995 \times 0.99 \times 0.97 \times 0.99 = 0.9459$. The second centre halves the failure rate, from 5.4% to 2.6%. ::: ::: --- ## ๐ Example 2.21: Mortgage Pipeline ::: {style="font-size: 28px"} Each application arriving at a bank's mortgage desk needs **manual underwriting** with probability $p = 1/6$, independently. $A_i$: application $i$ does **not**. The $r$th is the first manual case: $$E_r = A_1 \cap \cdots \cap A_{r-1} \cap \bar{A}_r, \qquad P(E_r) = (5/6)^{r-1}(1/6)$$ ::: ```{r} #| label: first-manual r <- 1:6 P_r <- (5/6)^(r - 1) * (1/6) round(c(setNames(P_r, paste0("r=", r)), within_6 = sum(P_r)), 4) ``` ::: {style="font-size: 28px"} Over all $r$ the geometric series sums to $\frac{1/6}{1 - 5/6} = 1$. ::: --- ## ๐ฆ Draws Without Replacement ::: {style="font-size: 28px"} A customs post holds **8 containers**, of which **2** carry undeclared goods. Inspectors open them one at a time in random order. $N_i$: the $i$th container opened is clean. ::: {.fragment} **(a) At least one flagged container among the first three.** The complement is "the first three are all clean", a chain in which each factor conditions on what came before: $$P = 1 - P(N_1)P(N_2 \mid N_1)P(N_3 \mid N_1 \cap N_2) = 1 - \tfrac{6}{8}\cdot\tfrac{5}{7}\cdot\tfrac{4}{6} = 1 - \tfrac{5}{14} = 0.6429$$ ::: ::: {.fragment} The factors are **not** $\tfrac{6}{8}$ three times: a clean container that has been opened is not put back. ::: ::: --- ## ๐ฆ Found on the Fourth Opening ::: {style="font-size: 28px"} **(b)** Exactly one flagged container in the first three, then a flagged one: three **mutually exclusive** orders, FNNF, NFNF, NNFF, each with the probability of FNNF: $$\tfrac{2}{8}\cdot\tfrac{6}{7}\cdot\tfrac{5}{6}\cdot\tfrac{1}{5} = \tfrac{1}{28}, \qquad P = 3 \times \tfrac{1}{28} = \tfrac{3}{28} = 0.1071$$ ::: ```{r} #| label: containers # The two flagged containers occupy 2 of 8 opening positions, all C(8,2) = 28 equally likely pos <- combn(8, 2) c(at_least_one_in_first_3 = mean(pos[1, ] <= 3), second_found_on_4th = mean(pos[2, ] == 4)) ``` --- ## ๐ง Think-Pair-Share ```{r} #| label: tps-timer #| echo: false # The timer is the only thing in this deck that needs a package beyond base R. # Guarded so a machine without it renders the deck anyway, with a static # figure in the same corner, rather than halting the whole build. if (requireNamespace("countdown", quietly = TRUE)) { countdown::countdown(minutes = 4, seconds = 0, top = 0, right = 0, font_size = "2em", warn_when = 30) } else { htmltools::HTML(paste0( '4:00
')) } ``` ::: {style="font-size: 30px"} A cotton shipment from Baku to Poti leaves on time only if **customs** clears it (0.92), a **rail slot** is booked through either of two forwarders (0.75 each), and the **port** accepts it (0.96). All independent. **Four minutes, in pairs:** 1. Write the event "leaves on time" as a composition. Then find its probability. 2. Which is worth more: a **third forwarder**, or raising customs to **0.97**? ::: --- ## โ
Think-Pair-Share: Solution ::: {style="font-size: 28px"} 1. $\text{OnTime} = C \cap (F_1 \cup F_2) \cap P$, and $P(F_1 \cup F_2) = 1 - 0.25^2 = 0.9375$: $$P(\text{OnTime}) = 0.92 \times 0.9375 \times 0.96 = 0.828$$ ::: {.fragment} 2. Third forwarder: $0.92 \times (1 - 0.25^3) \times 0.96 = 0.8694$. Better customs: $0.97 \times 0.9375 \times 0.96 = 0.8730$. ::: ::: {.fragment} Customs wins, narrowly. The parallel block is already at 94%; the weakest **series** link is where the reliability leaks. The composition tells you where to spend money. ::: ::: --- ## ๐ Quiz #1: A Settlement Chain {.quiz-question} A cross-border payment passes three independent links, each working with probability 0.98. What is the probability the payment settles? - [$0.98^3 = 0.9412$]{.correct data-explanation="โ
Settlement needs every link, an intersection of independent events: 0.98 ร 0.98 ร 0.98 = 0.9412."} - $1 - 0.02^3 = 0.999992$ - $0.98$ - $3 \times 0.02 = 0.06$ --- ## ๐ Quiz #2: Test the Composition {.quiz-question} $A$: the export licence is approved; $B$: the import licence is approved. Which composition is the event **exactly one** licence is approved? - $A \cup B$ - [$(A \cap \bar{B}) \cup (\bar{A} \cap B)$]{.correct data-explanation="โ
A sample point with exactly one approval lies in exactly one of these two disjoint pieces. A โช B also contains the point where both are approved, so it fails the test."} - $\overline{A \cap B}$ - $\bar{A} \cap \bar{B}$ --- ## ๐ Quiz #3: Without Replacement {.quiz-question} Of 5 letters of credit, 2 contain discrepancies. They are checked one at a time in random order. What is the probability the first two checked are both clean? - [$\tfrac{3}{5} \times \tfrac{2}{4} = 0.30$]{.correct data-explanation="โ
Theorem 2.5 in sequence: 3/5 clean on the first check, then 2 clean left among 4. The second factor conditions on the first."} - $\left(\tfrac{3}{5}\right)^2 = 0.36$ - $\tfrac{3}{5} = 0.60$ - $\tfrac{2}{5} \times \tfrac{1}{4} = 0.10$ --- ## ๐ Key Formulas ::: {style="font-size: 30px"} | | Statement | |---|---| | partition by route | $P(F) = P(F \mid R)P(R) + P(F \mid D)P(D)$ | | series, $n$ components | $P(W) = p^n$ | | parallel, $n$ components | $P(W) = 1 - (1-p)^n$ | | at least one | $P(A) = 1 - P(\bar{A}_1 \cap \cdots \cap \bar{A}_n)$ | | first success on trial $r$ | $P(E_r) = (1-p)^{r-1}p$ | | sequential, no replacement | $P(A_1 \cap A_2 \cap A_3) = P(A_1)P(A_2 \mid A_1)P(A_3 \mid A_1 \cap A_2)$ | ::: --- ## ๐ Summary ::: {style="font-size: 30px"} - The method has four steps; step 3, **writing the composition**, is where the work is - Choose pieces whose probabilities you know, preferably mutually exclusive or independent - Test the composition against sample points before applying any law - Series is an intersection, parallel is the complement of an intersection - "At least one" goes through the complement; draws without replacement go through the chain of conditionals ::: --- ## ๐ Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Exercises at the end of ยง2.9 โ start with 2.110, 2.112, 2.116, then 2.120 and 2.121 - For 2.116, draw the three lines as a series or parallel diagram **before** writing any probability **Week 4, Problem Set 1** is open now and closes **Sunday 11 October at 23:59** on WeBWorK, covering ยง2.9. **Next class:** what to do when the event depends on which state the world is in โ the law of total probability and Bayes' rule. ::: --- ## ๐ Thank You ::: {style="font-size: 34px"} **Dr. Samir Orujov** ๐ง sorujov@ada.edu.az ๐ข Building D, Room D325 ๐ Office hours: Wednesday, 16:00 โ 18:00 Slides and readings: **sorujov.net/teaching** ::: --- ## โ Questions ::: {style="font-size: 32px"} - Example 2.17 split $F$ by route. What goes wrong if some shipments could use both routes? - A system of four 0.9 components can be wired as two parallel pairs in series, or two series pairs in parallel. Which is more reliable? - Why does the method fail, rather than approximate, when one probability in step 4 is unknown? ::: 