De Morgan's Law

Probability and Statistics · FE Reference Handbook section

Probability and Statistics

2 formulas

10 exam-style examples

~49 min

All Probability and Statistics lectures

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

De Morgan's law applied to two inspection events — 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) = 0.35, 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
$Ā \cap B̄ = (A \cup B)‾$
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

Example 2

De Morgan's law applied to two inspection events — De Morgan's Law (2)

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
$Ā \cap B̄ = (A \cup B)‾$
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

Example 3

De Morgan's law applied to two inspection events — De Morgan's Law (3)

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
$Ā \cap B̄ = (A \cup B)‾$
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

Example 4

De Morgan's law applied to two inspection events — De Morgan's Law (4)

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
$Ā \cap B̄ = (A \cup B)‾$
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

Example 5

De Morgan's law applied to two inspection events — De Morgan's Law (5)

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
$Ā \cap B̄ = (A \cup B)‾$
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

Example 6

De Morgan's law applied to two inspection events — De Morgan's Law (6)

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
$Ā \cap B̄ = (A \cup B)‾$
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

Example 7

De Morgan's law applied to two inspection events — De Morgan's Law (7)

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
$Ā \cap B̄ = (A \cup B)‾$
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

Example 8

De Morgan's law applied to two inspection events — De Morgan's Law (8)

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
$Ā \cap B̄ = (A \cup B)‾$
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

Example 9

De Morgan's law applied to two inspection events — De Morgan's Law (9)

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
$Ā \cap B̄ = (A \cup B)‾$
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

Example 10

De Morgan's law applied to two inspection events — De Morgan's Law (10)

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
$Ā \cap B̄ = (A \cup B)‾$
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

De Morgan's Law

Core formulas for this FE topic

A,B=A+BA,B= A+BA,B=A+B

A+B=A,BA+B= A,BA+B=A,B

Worked exam-style examples