Hyper

Probability and Statistics · FE Reference Handbook section

Probability and Statistics

1 formulas

10 exam-style examples

~47 min

All Probability and Statistics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Hyper within Probability and Statistics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what hyper describes physically and when it applies.
State every one of the 1 relation the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: probabilities are dimensionless and must land in [0, 1].

Lecture

Why this section exists. Hyper is the part of Probability and Statistics that lets you connect a sample of measurements from a construction or materials process to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as one distribution or one counting rule, then a single probability or interval. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. probabilities are dimensionless and must land in [0, 1]. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Three engineers in hard hats and safety vests reviewing drawings on a truck tailgate. — Photo 1. Where this shows up in practice: hyper.
Capstone Studio instructional photograph

Probability and Statistics — Hyper: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a sample of measurements from a construction or materials process. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 1 relation on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

() N n−x nr

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Hypergeometric sampling from a finite lot — Hyper

A lot of 37 precast panels contains 5 defectives. 6 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Given

N = 37
K = 5 defectives
n = 6 sampled

Find

P(x = 0) and P(x = 1) from the hypergeometric distribution

Start with the thinking

Sampling without replacement from a small lot is hypergeometric, not binomial.
Each probability is a ratio of combination counts.

Step-by-step solution

Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)
Denominator
$C (37, 6) = 2, 324, 784$
x = 0 — C(5,0)·C(32,6) = 906,192, so P(0) = 0.3898
x = 1 — C(5,1)·C(32,5) = 1,006,880, so P(1) = 0.4331
At least one defective
$1 - P (0) = 0.6102$

Answer: P(0) = 0.3898, P(1) = 0.4331

Why the other options are there

0.4185 (binomial approximation)
0.8108 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

Example 2

Hypergeometric sampling from a finite lot — Hyper (2)

A lot of 22 precast panels contains 7 defectives. 5 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Given

N = 22
K = 7 defectives
n = 5 sampled

Find

P(x = 0) and P(x = 1) from the hypergeometric distribution

Start with the thinking

Sampling without replacement from a small lot is hypergeometric, not binomial.
Each probability is a ratio of combination counts.

Step-by-step solution

Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)
Denominator
$C (22, 5) = 26, 334$
x = 0 — C(7,0)·C(15,5) = 3,003, so P(0) = 0.1140
x = 1 — C(7,1)·C(15,4) = 9,555, so P(1) = 0.3628
At least one defective
$1 - P (0) = 0.8860$

Answer: P(0) = 0.1140, P(1) = 0.3628

Why the other options are there

0.1473 (binomial approximation)
1.5909 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

Example 3

Hypergeometric sampling from a finite lot — Hyper (3)

A lot of 38 precast panels contains 7 defectives. 8 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Given

N = 38
K = 7 defectives
n = 8 sampled

Find

P(x = 0) and P(x = 1) from the hypergeometric distribution

Start with the thinking

Sampling without replacement from a small lot is hypergeometric, not binomial.
Each probability is a ratio of combination counts.

Step-by-step solution

Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)
Denominator
$C (38, 8) = 48, 903, 492$
x = 0 — C(7,0)·C(31,8) = 7,888,725, so P(0) = 0.1613
x = 1 — C(7,1)·C(31,7) = 18,407,025, so P(1) = 0.3764
At least one defective
$1 - P (0) = 0.8387$

Answer: P(0) = 0.1613, P(1) = 0.3764

Why the other options are there

0.1962 (binomial approximation)
1.4737 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

Example 4

Hypergeometric sampling from a finite lot — Hyper (4)

A lot of 40 precast panels contains 4 defectives. 7 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Given

N = 40
K = 4 defectives
n = 7 sampled

Find

P(x = 0) and P(x = 1) from the hypergeometric distribution

Start with the thinking

Sampling without replacement from a small lot is hypergeometric, not binomial.
Each probability is a ratio of combination counts.

Step-by-step solution

Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)
Denominator
$C (40, 7) = 18, 643, 560$
x = 0 — C(4,0)·C(36,7) = 8,347,680, so P(0) = 0.4478
x = 1 — C(4,1)·C(36,6) = 7,791,168, so P(1) = 0.4179
At least one defective
$1 - P (0) = 0.5522$

Answer: P(0) = 0.4478, P(1) = 0.4179

Why the other options are there

0.4783 (binomial approximation)
0.7000 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

Example 5

Hypergeometric sampling from a finite lot — Hyper (5)

A lot of 27 precast panels contains 8 defectives. 5 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Given

N = 27
K = 8 defectives
n = 5 sampled

Find

P(x = 0) and P(x = 1) from the hypergeometric distribution

Start with the thinking

Sampling without replacement from a small lot is hypergeometric, not binomial.
Each probability is a ratio of combination counts.

Step-by-step solution

Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)
Denominator
$C (27, 5) = 80, 730$
x = 0 — C(8,0)·C(19,5) = 11,628, so P(0) = 0.1440
x = 1 — C(8,1)·C(19,4) = 31,008, so P(1) = 0.3841
At least one defective
$1 - P (0) = 0.8560$

Answer: P(0) = 0.1440, P(1) = 0.3841

Why the other options are there

0.1726 (binomial approximation)
1.4815 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

Example 6

Hypergeometric sampling from a finite lot — Hyper (6)

A lot of 35 precast panels contains 8 defectives. 6 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Given

N = 35
K = 8 defectives
n = 6 sampled

Find

P(x = 0) and P(x = 1) from the hypergeometric distribution

Start with the thinking

Sampling without replacement from a small lot is hypergeometric, not binomial.
Each probability is a ratio of combination counts.

Step-by-step solution

Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)
Denominator
$C (35, 6) = 1, 623, 160$
x = 0 — C(8,0)·C(27,6) = 296,010, so P(0) = 0.1824
x = 1 — C(8,1)·C(27,5) = 645,840, so P(1) = 0.3979
At least one defective
$1 - P (0) = 0.8176$

Answer: P(0) = 0.1824, P(1) = 0.3979

Why the other options are there

0.2108 (binomial approximation)
1.3714 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

Example 7

Hypergeometric sampling from a finite lot — Hyper (7)

A lot of 20 precast panels contains 4 defectives. 7 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Given

N = 20
K = 4 defectives
n = 7 sampled

Find

P(x = 0) and P(x = 1) from the hypergeometric distribution

Start with the thinking

Sampling without replacement from a small lot is hypergeometric, not binomial.
Each probability is a ratio of combination counts.

Step-by-step solution

Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)
Denominator
$C (20, 7) = 77, 520$
x = 0 — C(4,0)·C(16,7) = 11,440, so P(0) = 0.1476
x = 1 — C(4,1)·C(16,6) = 32,032, so P(1) = 0.4132
At least one defective
$1 - P (0) = 0.8524$

Answer: P(0) = 0.1476, P(1) = 0.4132

Why the other options are there

0.2097 (binomial approximation)
1.4000 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

Example 8

Hypergeometric sampling from a finite lot — Hyper (8)

A lot of 27 precast panels contains 4 defectives. 4 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Given

N = 27
K = 4 defectives
n = 4 sampled

Find

P(x = 0) and P(x = 1) from the hypergeometric distribution

Start with the thinking

Sampling without replacement from a small lot is hypergeometric, not binomial.
Each probability is a ratio of combination counts.

Step-by-step solution

Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)
Denominator
$C (27, 4) = 17, 550$
x = 0 — C(4,0)·C(23,4) = 8,855, so P(0) = 0.5046
x = 1 — C(4,1)·C(23,3) = 7,084, so P(1) = 0.4036
At least one defective
$1 - P (0) = 0.4954$

Answer: P(0) = 0.5046, P(1) = 0.4036

Why the other options are there

0.5266 (binomial approximation)
0.5926 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

Example 9

Hypergeometric sampling from a finite lot — Hyper (9)

A lot of 29 precast panels contains 7 defectives. 8 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Given

N = 29
K = 7 defectives
n = 8 sampled

Find

P(x = 0) and P(x = 1) from the hypergeometric distribution

Start with the thinking

Sampling without replacement from a small lot is hypergeometric, not binomial.
Each probability is a ratio of combination counts.

Step-by-step solution

Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)
Denominator
$C (29, 8) = 4, 292, 145$
x = 0 — C(7,0)·C(22,8) = 319,770, so P(0) = 0.0745
x = 1 — C(7,1)·C(22,7) = 1,193,808, so P(1) = 0.2781
At least one defective
$1 - P (0) = 0.9255$

Answer: P(0) = 0.0745, P(1) = 0.2781

Why the other options are there

0.1097 (binomial approximation)
1.9310 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

Example 10

Hypergeometric sampling from a finite lot — Hyper (10)

A lot of 40 precast panels contains 4 defectives. 6 panels are inspected without replacement. Find the probability of finding no defectives and of finding exactly one.

Given

N = 40
K = 4 defectives
n = 6 sampled

Find

P(x = 0) and P(x = 1) from the hypergeometric distribution

Start with the thinking

Sampling without replacement from a small lot is hypergeometric, not binomial.
Each probability is a ratio of combination counts.

Step-by-step solution

Formula — P(x) = C(K,x)·C(N−K, n−x) / C(N, n)
Denominator
$C (40, 6) = 3, 838, 380$
x = 0 — C(4,0)·C(36,6) = 1,947,792, so P(0) = 0.5075
x = 1 — C(4,1)·C(36,5) = 1,507,968, so P(1) = 0.3929
At least one defective
$1 - P (0) = 0.4925$

Answer: P(0) = 0.5075, P(1) = 0.3929

Why the other options are there

0.5314 (binomial approximation)
0.6000 (expected count reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Hyper

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a sample of measurements from a construction or materials process, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Hyper contains 1 relation; you must be able to find this page in under 15 seconds.
Exam style: one distribution or one counting rule, then a single probability or interval.
Unit rule: probabilities are dimensionless and must land in [0, 1].
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

probabilities are dimensionless and must land in [0, 1]
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Hyper

Learning objectives

Lecture

Theory, developed

Handbook notes for this section

Core formulas for this FE topic

h(x; n, r, N ) =Needs technical review

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section