Mathematical Statistics

🎯 Learning Objectives

By the end of this lecture, you will be able to:

Follow the four steps of the event-composition method (Wackerly §2.9)
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

🗺️ Where We Are

Wackerly §2.9

Last class ended on a promise: the event-composition method — how to decide, for a compound event with no obvious decomposition, which events to name in the first place.

We now own the tools: Theorem 2.5 for \(\cap\), Theorem 2.6 for \(\cup\), Theorem 2.7 for complements.

Today is about the step before the tools — choosing the events so that every probability the laws ask for is one you actually know.

💳 Motivating Question

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?

Nobody publishes “the probability the payment goes through”. It has to be built from the pieces — and the building is the method.

📐 The Event-Composition Method

Wackerly §2.9: the four steps

Define the experiment.
Visualize the nature of the sample points. Identify a few to clarify your thinking.
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.
Apply the additive and multiplicative laws to the composition to find \(P(A)\).

Unlike the sample-point method, no list of \(S\) is needed — only a clear picture of a typical sample point.

🔑 Step 3 Is the Hard One

Many compositions equal \(A\). Pick the one in which every probability in step 4 is known. If one is unknown, the method fails.

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

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.

🚢 Example 2.17: Export Shipments

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.

Step 3. Every shipment takes exactly one route, so \[F = (F \cap R) \cup (F \cap D), \qquad \text{two mutually exclusive pieces.}\]

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 \[P(F) = 0.28 + 0.48 = 0.76\]

🏦 Example 2.19: A Clearing Mandate

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?

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 \[A = (B_1 \cap B_2) \cup (B_3 \cap B_4) \cup (C_1 \cap C_2) \cup (C_3 \cap C_4)\]

Test it: every shortlist with exactly one cheap bank lies in exactly one of the four pieces, and no other shortlist lies in any.

🧮 Example 2.19: Step 4, and a Check

The four pieces are mutually exclusive, and each is a two-stage draw: \[P(B_1 \cap B_2) = P(B_1)P(B_2 \mid B_1) = \tfrac{1}{5} \times \tfrac{3}{4} = \tfrac{3}{20}, \qquad P(A) = 4 \times \tfrac{3}{20} = \tfrac{3}{5}\]

Code

# 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))

 shortlists exactly_one         P_A 
       10.0         6.0         0.6

Same answer as the sample-point method, reached without writing \(S\) down.

⚡ Example 2.20: Power Feeds

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?

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 \[P(A) = 1 - P(B_1)P(B_2)P(B_3) = 1 - (0.1)^3 = 0.999\]

As a union, \(A\) needs seven terms of inclusion–exclusion. As a complement, it is one intersection. Frequently \(P(\bar{A})\) is the easy one.

🔗 Series and Parallel Systems

\(n\) independent components, each working with probability \(p\).

Series: all must work

A payment settlement chain. \[W = W_1 \cap W_2 \cap \cdots \cap W_n\] \[P(W) = p^n\]

Parallel: one is enough

Redundant power feeds. \[\bar{W} = \bar{W}_1 \cap \bar{W}_2 \cap \cdots \cap \bar{W}_n\] \[P(W) = 1 - (1-p)^n\]

Same components, opposite compositions: every link added to a series costs reliability, every feed added in parallel buys it.

💻 Series vs Parallel in R

Code

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))

  components series parallel parallel_fails
1          1 0.9500   0.9500       5.00e-02
2          2 0.9025   0.9975       2.50e-03
3          4 0.8145   1.0000       6.25e-06
4          8 0.6634   1.0000       3.91e-11
5         12 0.5404   1.0000       2.44e-16
6         20 0.3585   1.0000       9.54e-27

Twelve 95% links in series work only 54% of the time; two 95% feeds in parallel already give 99.75%.

🔬 Interactive: Series Versus Parallel

Code

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;
}

Code

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])
  ]
})

💳 Back to the Card Payment

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: \[\text{Auth} = T \cap Q \cap (N_1 \cup N_2) \cap I\]

\[P(N_1 \cup N_2) = 1 - (0.03)^2 = 0.9991\] \[P(\text{Auth}) = 0.995 \times 0.99 \times 0.9991 \times 0.99 = 0.9743\]

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%.

🏠 Example 2.21: Mortgage Pipeline

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: \[E_r = A_1 \cap \cdots \cap A_{r-1} \cap \bar{A}_r, \qquad P(E_r) = (5/6)^{r-1}(1/6)\]

Code

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)

     r=1      r=2      r=3      r=4      r=5      r=6 within_6 
  0.1667   0.1389   0.1157   0.0965   0.0804   0.0670   0.6651

Over all \(r\) the geometric series sums to \(\frac{1/6}{1 - 5/6} = 1\).

📦 Draws Without Replacement

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.

(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: \[P = 1 - P(N_1)P(N_2 \mid N_1)P(N_3 \mid N_1 \cap N_2) = 1 - \tfrac{6}{8}\cdot\tfrac{5}{7}\cdot\tfrac{4}{6} = 1 - \tfrac{5}{14} = 0.6429\]

The factors are not \(\tfrac{6}{8}\) three times: a clean container that has been opened is not put back.

📦 Found on the Fourth Opening

(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: \[\tfrac{2}{8}\cdot\tfrac{6}{7}\cdot\tfrac{5}{6}\cdot\tfrac{1}{5} = \tfrac{1}{28}, \qquad P = 3 \times \tfrac{1}{28} = \tfrac{3}{28} = 0.1071\]

Code

# 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))

at_least_one_in_first_3     second_found_on_4th 
              0.6428571               0.1071429

🧠 Think-Pair-Share

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:

Write the event “leaves on time” as a composition. Then find its probability.
Which is worth more: a third forwarder, or raising customs to 0.97?

✅ Think-Pair-Share: Solution

\(\text{OnTime} = C \cap (F_1 \cup F_2) \cap P\), and \(P(F_1 \cup F_2) = 1 - 0.25^2 = 0.9375\): \[P(\text{OnTime}) = 0.92 \times 0.9375 \times 0.96 = 0.828\]

Third forwarder: \(0.92 \times (1 - 0.25^3) \times 0.96 = 0.8694\). Better customs: \(0.97 \times 0.9375 \times 0.96 = 0.8730\).

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.

📝 Quiz #1: A Settlement Chain

A cross-border payment passes three independent links, each working with probability 0.98. What is the probability the payment settles?

\(0.98^3 = 0.9412\)
\(1 - 0.02^3 = 0.999992\)
\(0.98\)
\(3 \times 0.02 = 0.06\)

📝 Quiz #2: Test the Composition

\(A\): the export licence is approved; \(B\): the import licence is approved. Which composition is the event exactly one licence is approved?

\(A \cup B\)
\((A \cap \bar{B}) \cup (\bar{A} \cap B)\)
\(\overline{A \cap B}\)
\(\bar{A} \cap \bar{B}\)

📝 Quiz #3: Without Replacement

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?

\(\tfrac{3}{5} \times \tfrac{2}{4} = 0.30\)
\(\left(\tfrac{3}{5}\right)^2 = 0.36\)
\(\tfrac{3}{5} = 0.60\)
\(\tfrac{2}{5} \times \tfrac{1}{4} = 0.10\)

📋 Key Formulas

	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	\(P(A_1 \cap A_2 \cap A_3) = P(A_1)P(A_2 \mid A_1)P(A_3 \mid A_1 \cap A_2)\)

📋 Summary

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

📚 Practice Problems

Wackerly, 7th edition

Exercises at the end of §2.9 — 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 opens today and closes 10 October at 23:59 on WeBWorK, covering §2.9.

Next class: what to do when the event depends on which state the world is in — the law of total probability and Bayes’ rule.

🙏 Thank You

Dr. Samir Orujov

📧 sorujov@ada.edu.az 🏢 Building D, Room D325 🕓 Office hours: Wednesday, 16:00 – 18:00

Slides and readings: sorujov.net/teaching

❓ Questions

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?