De Morgan's Law
Probability and Statistics · FE Reference Handbook section
Learning objectives
What you must be able to do before leaving this section.
This chapter section covers De Morgan's Law 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 de morgan's law describes physically and when it applies.
- State every one of the 2 relations 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. De Morgan's Law 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.
Photo 1. Where this shows up in practice: de morgan's law.
Capstone Studio instructional photograph
Probability and Statistics — De Morgan's Law: 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 2 relations 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.
Photo 2. Probability and Statistics: the physical system the theory above idealises.
Capstone Studio instructional photograph
Notation used in this section
| A,B | Quantity produced by "A,B= A+B" — read its definition and unit from the handbook line directly above the equation. |
|---|---|
| A+B | Quantity produced by "A+B= A,B" — read its definition and unit from the handbook line directly above the equation. |
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.
Event A (a weld fails inspection) has P(A) = 0.40 and event B (a bolt fails inspection) has P(B) = 0.35, with P(A∩B) = 0.06. Use De Morgan's law to compute the probability that neither event occurs.
Given
- P(A) = 0.40
- P(B) = 0.35
- P(A∩B) = 0.06
Find
P(Ā ∩ B̄) using De Morgan's law
Start with the thinking
- De Morgan's law states that the complement of a union is the intersection of the complements.
- So 'neither fails' is one minus the probability that at least one fails.
Step-by-step solution
Formula — P(A∪B) = P(A) + P(B) − P(A∩B)
Substituting — P(A∪B) = 0.40 + 0.35 − 0.06 = 0.6900
De Morgan's law
Substituting — P(Ā∩B̄) = 1 − 0.6900 = 0.3100
Answer: P(neither) = 0.3100
Why the other options are there
- 0.3900 (assumed independence)
- 0.2500 (ignored the overlap)
Reference: FE Reference Handbook — Probability and Statistics → De Morgan's Law
Event A (a weld fails inspection) has P(A) = 0.30 and event B (a bolt fails inspection) has P(B) = 0.30, with P(A∩B) = 0.08. Use De Morgan's law to compute the probability that neither event occurs.
Given
- P(A) = 0.30
- P(B) = 0.30
- P(A∩B) = 0.08
Find
P(Ā ∩ B̄) using De Morgan's law
Start with the thinking
- De Morgan's law states that the complement of a union is the intersection of the complements.
- So 'neither fails' is one minus the probability that at least one fails.
Step-by-step solution
Formula — P(A∪B) = P(A) + P(B) − P(A∩B)
Substituting — P(A∪B) = 0.30 + 0.30 − 0.08 = 0.5200
De Morgan's law
Substituting — P(Ā∩B̄) = 1 − 0.5200 = 0.4800
Answer: P(neither) = 0.4800
Why the other options are there
- 0.4900 (assumed independence)
- 0.4000 (ignored the overlap)
Reference: FE Reference Handbook — Probability and Statistics → De Morgan's Law
Event A (a weld fails inspection) has P(A) = 0.40 and event B (a bolt fails inspection) has P(B) = 0.30, with P(A∩B) = 0.09. Use De Morgan's law to compute the probability that neither event occurs.
Given
- P(A) = 0.40
- P(B) = 0.30
- P(A∩B) = 0.09
Find
P(Ā ∩ B̄) using De Morgan's law
Start with the thinking
- De Morgan's law states that the complement of a union is the intersection of the complements.
- So 'neither fails' is one minus the probability that at least one fails.
Step-by-step solution
Formula — P(A∪B) = P(A) + P(B) − P(A∩B)
Substituting — P(A∪B) = 0.40 + 0.30 − 0.09 = 0.6100
De Morgan's law
Substituting — P(Ā∩B̄) = 1 − 0.6100 = 0.3900
Answer: P(neither) = 0.3900
Why the other options are there
- 0.4200 (assumed independence)
- 0.3000 (ignored the overlap)
Reference: FE Reference Handbook — Probability and Statistics → De Morgan's Law
Event A (a weld fails inspection) has P(A) = 0.25 and event B (a bolt fails inspection) has P(B) = 0.30, with P(A∩B) = 0.11. Use De Morgan's law to compute the probability that neither event occurs.
Given
- P(A) = 0.25
- P(B) = 0.30
- P(A∩B) = 0.11
Find
P(Ā ∩ B̄) using De Morgan's law
Start with the thinking
- De Morgan's law states that the complement of a union is the intersection of the complements.
- So 'neither fails' is one minus the probability that at least one fails.
Step-by-step solution
Formula — P(A∪B) = P(A) + P(B) − P(A∩B)
Substituting — P(A∪B) = 0.25 + 0.30 − 0.11 = 0.4400
De Morgan's law
Substituting — P(Ā∩B̄) = 1 − 0.4400 = 0.5600
Answer: P(neither) = 0.5600
Why the other options are there
- 0.5250 (assumed independence)
- 0.4500 (ignored the overlap)
Reference: FE Reference Handbook — Probability and Statistics → De Morgan's Law
Event A (a weld fails inspection) has P(A) = 0.45 and event B (a bolt fails inspection) has P(B) = 0.50, with P(A∩B) = 0.07. Use De Morgan's law to compute the probability that neither event occurs.
Given
- P(A) = 0.45
- P(B) = 0.50
- P(A∩B) = 0.07
Find
P(Ā ∩ B̄) using De Morgan's law
Start with the thinking
- De Morgan's law states that the complement of a union is the intersection of the complements.
- So 'neither fails' is one minus the probability that at least one fails.
Step-by-step solution
Formula — P(A∪B) = P(A) + P(B) − P(A∩B)
Substituting — P(A∪B) = 0.45 + 0.50 − 0.07 = 0.8800
De Morgan's law
Substituting — P(Ā∩B̄) = 1 − 0.8800 = 0.1200
Answer: P(neither) = 0.1200
Why the other options are there
- 0.2750 (assumed independence)
- 0.0500 (ignored the overlap)
Reference: FE Reference Handbook — Probability and Statistics → De Morgan's Law
Event A (a weld fails inspection) has P(A) = 0.45 and event B (a bolt fails inspection) has P(B) = 0.55, with P(A∩B) = 0.08. Use De Morgan's law to compute the probability that neither event occurs.
Given
- P(A) = 0.45
- P(B) = 0.55
- P(A∩B) = 0.08
Find
P(Ā ∩ B̄) using De Morgan's law
Start with the thinking
- De Morgan's law states that the complement of a union is the intersection of the complements.
- So 'neither fails' is one minus the probability that at least one fails.
Step-by-step solution
Formula — P(A∪B) = P(A) + P(B) − P(A∩B)
Substituting — P(A∪B) = 0.45 + 0.55 − 0.08 = 0.9200
De Morgan's law
Substituting — P(Ā∩B̄) = 1 − 0.9200 = 0.0800
Answer: P(neither) = 0.0800
Why the other options are there
- 0.2475 (assumed independence)
- 0.0000 (ignored the overlap)
Reference: FE Reference Handbook — Probability and Statistics → De Morgan's Law
Event A (a weld fails inspection) has P(A) = 0.35 and event B (a bolt fails inspection) has P(B) = 0.55, with P(A∩B) = 0.08. Use De Morgan's law to compute the probability that neither event occurs.
Given
- P(A) = 0.35
- P(B) = 0.55
- P(A∩B) = 0.08
Find
P(Ā ∩ B̄) using De Morgan's law
Start with the thinking
- De Morgan's law states that the complement of a union is the intersection of the complements.
- So 'neither fails' is one minus the probability that at least one fails.
Step-by-step solution
Formula — P(A∪B) = P(A) + P(B) − P(A∩B)
Substituting — P(A∪B) = 0.35 + 0.55 − 0.08 = 0.8200
De Morgan's law
Substituting — P(Ā∩B̄) = 1 − 0.8200 = 0.1800
Answer: P(neither) = 0.1800
Why the other options are there
- 0.2925 (assumed independence)
- 0.1000 (ignored the overlap)
Reference: FE Reference Handbook — Probability and Statistics → De Morgan's Law
Event A (a weld fails inspection) has P(A) = 0.35 and event B (a bolt fails inspection) has P(B) = 0.50, with P(A∩B) = 0.11. Use De Morgan's law to compute the probability that neither event occurs.
Given
- P(A) = 0.35
- P(B) = 0.50
- P(A∩B) = 0.11
Find
P(Ā ∩ B̄) using De Morgan's law
Start with the thinking
- De Morgan's law states that the complement of a union is the intersection of the complements.
- So 'neither fails' is one minus the probability that at least one fails.
Step-by-step solution
Formula — P(A∪B) = P(A) + P(B) − P(A∩B)
Substituting — P(A∪B) = 0.35 + 0.50 − 0.11 = 0.7400
De Morgan's law
Substituting — P(Ā∩B̄) = 1 − 0.7400 = 0.2600
Answer: P(neither) = 0.2600
Why the other options are there
- 0.3250 (assumed independence)
- 0.1500 (ignored the overlap)
Reference: FE Reference Handbook — Probability and Statistics → De Morgan's Law
Event A (a weld fails inspection) has P(A) = 0.40 and event B (a bolt fails inspection) has P(B) = 0.45, with P(A∩B) = 0.08. Use De Morgan's law to compute the probability that neither event occurs.
Given
- P(A) = 0.40
- P(B) = 0.45
- P(A∩B) = 0.08
Find
P(Ā ∩ B̄) using De Morgan's law
Start with the thinking
- De Morgan's law states that the complement of a union is the intersection of the complements.
- So 'neither fails' is one minus the probability that at least one fails.
Step-by-step solution
Formula — P(A∪B) = P(A) + P(B) − P(A∩B)
Substituting — P(A∪B) = 0.40 + 0.45 − 0.08 = 0.7700
De Morgan's law
Substituting — P(Ā∩B̄) = 1 − 0.7700 = 0.2300
Answer: P(neither) = 0.2300
Why the other options are there
- 0.3300 (assumed independence)
- 0.1500 (ignored the overlap)
Reference: FE Reference Handbook — Probability and Statistics → De Morgan's Law
Event A (a weld fails inspection) has P(A) = 0.30 and event B (a bolt fails inspection) has P(B) = 0.30, with P(A∩B) = 0.07. Use De Morgan's law to compute the probability that neither event occurs.
Given
- P(A) = 0.30
- P(B) = 0.30
- P(A∩B) = 0.07
Find
P(Ā ∩ B̄) using De Morgan's law
Start with the thinking
- De Morgan's law states that the complement of a union is the intersection of the complements.
- So 'neither fails' is one minus the probability that at least one fails.
Step-by-step solution
Formula — P(A∪B) = P(A) + P(B) − P(A∩B)
Substituting — P(A∪B) = 0.30 + 0.30 − 0.07 = 0.5300
De Morgan's law
Substituting — P(Ā∩B̄) = 1 − 0.5300 = 0.4700
Answer: P(neither) = 0.4700
Why the other options are there
- 0.4900 (assumed independence)
- 0.4000 (ignored the overlap)
Reference: FE Reference Handbook — Probability and Statistics → De Morgan's Law
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
- De Morgan's Law contains 2 relations; 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.