```{r} #| label: setup #| include: false set.seed(2026) ``` ## π¬ 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"} ::: --- ## π― Learning Objectives ::: {style="font-size: 32px"} By the end of this lecture, you will be able to: - **Apply** the multiplicative law $P(A \cap B) = P(A)P(B \mid A)$, in either order - **Extend** it to a chain of three or more dependent stages - **Apply** the additive law $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, and say what the subtraction repairs - **Use** the complement $P(A) = 1 - P(\bar{A})$ to convert an "at least one" event into an easy one - **Choose** between the two laws by reading whether the event is an intersection or a union ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 32px"} **Wackerly Β§2.8** Wednesday: $P(A \mid B) = P(A \cap B)/P(B)$. We used it to *read information out of* a table. ::: {.fragment} Today we read it the other way. Multiply both sides by $P(B)$ and the definition stops being a measurement and becomes a **construction**: a way to build the probability of an intersection out of pieces. ::: ::: {.fragment} Two laws, one for $\cap$ and one for $\cup$. Between them they handle every compound event in this chapter. ::: ::: --- ## π Theorem 2.5: The Multiplicative Law ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 2.5 For any two events $A$ and $B$, $$P(A \cap B) = P(A)P(B \mid A) = P(B)P(A \mid B)$$ and if $A$ and $B$ are independent, $$P(A \cap B) = P(A)P(B)$$ ::: ::: {.fragment} The proof is Definition 2.9 with the denominator moved across. There is no new content β only a new use. ::: ::: --- ## π Two Forms, One of Which You Can Evaluate ::: {style="font-size: 30px"} $$P(A \cap B) = P(A)P(B \mid A) = P(B)P(A \mid B)$$ ::: {.fragment} The two expressions are always equal, but a problem rarely hands you both. It gives you one marginal and the conditional that goes with it. ::: ::: {.fragment} **So the first question is which conditioning you were given, not which formula you prefer.** Choosing the other form leaves you with two unknowns and one equation. ::: ::: --- ## π Worked Example: Two Loans From a Pool ::: {style="font-size: 30px"} A securitisation pool holds 12 mortgages, of which 4 are in arrears. Two are drawn without replacement for a file review. What is the probability that **both** are in arrears? ::: {.fragment} $$P(A_1 \cap A_2) = P(A_1)\,P(A_2 \mid A_1) = \frac{4}{12} \times \frac{3}{11} = \frac{12}{132} = 0.0909$$ ::: ::: {.fragment} Note what the law did: it turned a question about a **pair** into two questions about a **single draw**, and the second one was easy because the first had already been answered. ::: ::: --- ## βοΈ The Extension to Three ::: {style="font-size: 30px"} ::: {.callout-important} ## Three events $$P(A \cap B \cap C) = P(A)\,P(B \mid A)\,P(C \mid A \cap B)$$ ::: ::: {.fragment} Each factor conditions on **everything already assumed**. Drawing a third mortgage from the same pool: $$P(A_1 \cap A_2 \cap A_3) = \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} = 0.0182$$ ::: ::: {.fragment} The pattern continues for any number of stages, and it is how sampling without replacement is handled throughout the course. ::: ::: --- ## π€ The Independent Case ::: {style="font-size: 30px"} Azerbaijani and Georgian trade receivables are insured under separate policies, and a claim on one is taken to be independent of a claim on the other. With $P(A) = 0.07$ and $P(G) = 0.05$, $$P(A \cap G) = 0.07 \times 0.05 = 0.0035$$ ::: {.fragment} Independence is what licenses the shortcut β it is not the default. **When a problem does not say "independent", the conditional does not disappear**, and assuming it does is the most common way to arrive confidently at the wrong number. ::: ::: --- ## π Theorem 2.6: The Additive Law ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 2.6 For any two events $A$ and $B$, $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ and if $A$ and $B$ are mutually exclusive, $$P(A \cup B) = P(A) + P(B)$$ ::: ::: {.fragment} The second line is Axiom 3 itself. The first is what Axiom 3 becomes when the events overlap. ::: ::: --- ## β What the Subtraction Repairs ::: {style="font-size: 30px"} Write $A \cup B$ as a union of three **disjoint** pieces: $A \cap \bar{B}$, $\bar{A} \cap B$, and $A \cap B$. Axiom 3 applies to those, and adding them gives the law. ::: {.fragment} Adding $P(A)$ and $P(B)$ counts the overlap twice β once inside $A$, once inside $B$. Subtracting $P(A \cap B)$ removes one of the two copies. ::: ::: {.fragment} So the subtraction is not a correction for an error. It is **bookkeeping**, and it disappears exactly when there is nothing to double count. ::: ::: --- ## π¦ Worked Example: A Customer Relationship ::: {style="font-size: 28px"} Of a bank's retail customers, 62% hold a current account ($C$), 45% hold a card ($K$), and 31% hold both. ::: {.fragment} $$P(C \cup K) = 0.62 + 0.45 - 0.31 = 0.76$$ ::: ::: {.fragment} Drop the subtraction and you get 1.07 β a probability above 1, which Axiom 2 forbids. **An answer above 1 is almost always a forgotten intersection.** ::: ::: {.fragment} And $P(\bar{C} \cap \bar{K}) = 1 - 0.76 = 0.24$ hold neither: 24% of the customer base carries no product at all. ::: ::: --- ## π Theorem 2.7: The Complement ::: {style="font-size: 32px"} ::: {.callout-important} ## Theorem 2.7 $$P(A) = 1 - P(\bar{A})$$ ::: ::: {.fragment} $A$ and $\bar{A}$ are mutually exclusive and their union is $S$, so $P(A) + P(\bar{A}) = P(S) = 1$ by Axiom 2 and Axiom 3. ::: ::: {.fragment} Three lines of proof, and it is the most useful result in the chapter. ::: ::: --- ## π‘ "At Least One" Is an Intersection in Disguise ::: {style="font-size: 30px"} A bank guarantees 8 SME loans. Each is called with probability 0.06, independently. What is the probability that **at least one** is called? ::: {.fragment} As a union, this is eight events overlapping in every combination β 255 terms of inclusion and exclusion. ::: ::: {.fragment} As a complement, it is one line. "At least one is called" fails exactly when **none** is called, and *none* is an intersection: $$P(\text{at least one}) = 1 - (1 - 0.06)^8 = 1 - 0.6096 = 0.3904$$ ::: ::: --- ## π» At Least One, Across Portfolio Sizes ```{r} #| label: at-least-one p_call <- 0.06 n <- c(1, 4, 8, 12, 20, 40) data.frame( guarantees = n, none_called = round((1 - p_call)^n, 4), at_least_one = round(1 - (1 - p_call)^n, 4) ) ``` ::: {style="font-size: 28px"} A 6% chance on one guarantee is better than even odds by twelve. Small independent risks accumulate, and the complement is what makes that visible in one line. ::: --- ## π¬ Interactive: The Accumulation {.smaller} ```{ojs} //| echo: false viewof p_each = { const input = Inputs.range([0.01, 0.30], {value: 0.06, step: 0.01, label: "Probability each guarantee is called:"}); ['pointerdown','touchstart','mousedown','click','wheel','pointermove','touchmove'] .forEach(e => input.addEventListener(e, ev => ev.stopPropagation())); return input; } ``` ```{ojs} //| echo: false atLeastOne = (n, p) => 1 - Math.pow(1 - p, n) curve = Array.from({length: 40}, (_, i) => ({n: i + 1, p: atLeastOne(i + 1, p_each)})) half = curve.find(d => d.p >= 0.5) md`At **${(100 * p_each).toFixed(0)}%** each, the portfolio passes a 50-50 chance of at least one call at **${half ? half.n : "more than 40"}** guarantees.` Plot.plot({ width: 1150, height: 320, marginLeft: 78, marginBottom: 58, style: {fontSize: "18px"}, x: {label: "Number of guarantees", domain: [1, 40]}, y: {label: "P(at least one called)", domain: [0, 1], tickFormat: ".1f"}, marks: [ Plot.ruleY([0.5], {stroke: "#cbb8a9", strokeDasharray: "4 4"}), Plot.line(curve, {x: "n", y: "p", stroke: "#14130f", strokeWidth: 2}), Plot.dot(half ? [half] : [], {x: "n", y: "p", r: 8, fill: "#8b2635"}), Plot.ruleY([0]) ] }) ``` --- ## π§ 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 borrower must clear two hurdles to draw on a credit line: an **affordability test** ($T$) and a **collateral valuation** ($V$). $P(T) = 0.80$, and among those who pass the affordability test, 70% also pass the valuation. **Four minutes, in pairs:** 1. What is $P(T \cap V)$? 2. If additionally $P(V) = 0.62$, what is $P(T \cup V)$? 3. What is the probability the borrower fails **both**? ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 30px"} 1. The 70% is a **conditional**, $P(V \mid T) = 0.70$. Theorem 2.5: $$P(T \cap V) = 0.80 \times 0.70 = 0.56$$ 2. Theorem 2.6: $P(T \cup V) = 0.80 + 0.62 - 0.56 = 0.86$. ::: {.fragment} 3. "Fails both" is $\bar{T} \cap \bar{V}$, which by De Morgan is $\overline{T \cup V}$. Theorem 2.7: $$P(\bar{T} \cap \bar{V}) = 1 - 0.86 = 0.14$$ ::: ::: --- ## β
Think-Pair-Share: The Habit ::: {style="font-size: 30px"} Part 3 is where the three results become one tool: De Morgan turned an **intersection of complements** into the **complement of a union**, and the union was already computed. ::: {.fragment} **Before reaching for a formula, write the event in set notation.** "Neither", "both", "at least one", "only one" each have exactly one reading, and the reading picks the law. ::: ::: {.fragment} It is the same discipline as Wednesday's *order matters or not* β the mathematics is easy once the event is named correctly. ::: ::: --- ## π§ Which Law? A Reading Guide ::: {style="font-size: 28px"} | The event says | In set notation | Reach for | |---|---|---| | "both", "and", "all of" | $A \cap B$ | Theorem 2.5 | | "either", "or", "at least one of two" | $A \cup B$ | Theorem 2.6 | | "at least one of many" | $\overline{\bar{A_1} \cap \cdots \cap \bar{A_n}}$ | 2.7, then 2.5 | | "neither", "none" | $\bar{A} \cap \bar{B}$ | De Morgan, then 2.7 | | "exactly one" | $(A \cap \bar{B}) \cup (\bar{A} \cap B)$ | disjoint, so Axiom 3 | ::: {.fragment} The last row is worth remembering: those two pieces never overlap, so no subtraction is needed. ::: ::: --- ## π Quiz #1: Which Law? {.quiz-question} $P(A) = 0.5$, $P(B) = 0.4$, $P(A \cap B) = 0.2$. What is $P(A \cup B)$? - [$0.7$]{.correct data-explanation="β
Theorem 2.6: 0.5 + 0.4 β 0.2 = 0.7. The intersection is subtracted once, because adding the marginals counted it twice."} - $0.9$ - $0.2$ - $1.1$ --- ## π Quiz #2: The Conditional Is Not Optional {.quiz-question} $P(A) = 0.6$ and $P(B \mid A) = 0.25$. What is $P(A \cap B)$? - [$0.15$]{.correct data-explanation="β
Theorem 2.5: P(A β© B) = P(A)P(B | A) = 0.6 Γ 0.25 = 0.15. No independence is needed or claimed."} - $0.85$ - $0.25$ - Cannot be found without $P(B)$ --- ## π Quiz #3: At Least One {.quiz-question} Three independent suppliers each deliver late with probability 0.2. What is the probability that at least one is late? - [$1 - 0.8^3 = 0.488$]{.correct data-explanation="β
'At least one' fails only when none is late, and 'none' is an intersection of independent events: 1 β (0.8)Β³."} - $3 \times 0.2 = 0.6$ - $0.2^3 = 0.008$ - $1 - 0.2^3 = 0.992$ --- ## π Quiz #4: Neither {.quiz-question} $P(A \cup B) = 0.75$. What is $P(\bar{A} \cap \bar{B})$? - [$0.25$]{.correct data-explanation="β
De Morgan gives Δ β© BΜ = complement of (A βͺ B), so Theorem 2.7 gives 1 β 0.75 = 0.25."} - $0.75$ - $0.5625$ - Cannot be found from $P(A \cup B)$ alone --- ## π Key Formulas ::: {style="font-size: 30px"} | | Statement | |---|---| | Theorem 2.5 | $P(A \cap B) = P(A)P(B \mid A) = P(B)P(A \mid B)$ | | independent case | $P(A \cap B) = P(A)P(B)$ | | Theorem 2.5, three events | $P(A \cap B \cap C) = P(A)P(B \mid A)P(C \mid A \cap B)$ | | Theorem 2.6 | $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ | | mutually exclusive case | $P(A \cup B) = P(A) + P(B)$ | | Theorem 2.7 | $P(A) = 1 - P(\bar{A})$ | ::: --- ## π Summary ::: {style="font-size: 30px"} - The multiplicative law is Definition 2.9 rearranged β it **builds** an intersection from a marginal and a conditional - Use the form whose conditioning the problem actually gave you - The additive law subtracts the intersection because adding the marginals counts it twice - The complement turns "at least one" from a 255-term union into a single line - De Morgan converts between "neither" and "not either", and the laws do the rest - Name the event in set notation **first**; the law follows from the name ::: --- ## π Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Exercises at the end of Β§2.8 β start with 2.85 β 2.91, then 2.95 and 2.97 - Return to Wednesday's bond table and verify Theorem 2.5 on it directly, computing $P(S \cap D)$ both ways round **Week 3, Problem Set 2** opens today and closes **3 October at 23:59** on WeBWorK, covering Β§2.8. The TA-led tutorial runs this week, before both deadlines. **Next class:** the event-composition method β how to decide, for a compound event with no obvious decomposition, which events to name in the first place. ::: --- ## π Thank You ::: {style="font-size: 34px"} **Dr. Samir Orujov** π§ sorujov@ada.edu.az π’ Building D, Room D325 π Office hours: Wednesday, 16:00 β 18:00 Slides and readings: **sorujov.net/teaching** ::: --- ## β Questions ::: {style="font-size: 32px"} - Given only $P(A)$ and $P(B)$, what are the largest and smallest values $P(A \cap B)$ can take? - The additive law subtracts one term for two events. How many terms does it need for three? - Mutually exclusive events add, independent events multiply. Which of the two is the stronger assumption to make without evidence? ::: 