Associative 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 — Associative Law

Event A (a weld fails inspection) has P(A) = 0.30 and event B (a bolt fails inspection) has P(B) = 0.50, with P(A∩B) = 0.06. Use De Morgan's law to compute the probability that neither event occurs.

Given

$P(A) = 0.30$
$P(B) = 0.50$
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) = 0.30 + 0.50 − 0.06 = 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

0.3500 (assumed independence)
0.2000 (ignored the overlap)

Reference: FE Reference Handbook — Probability and Statistics → Associative Law

Example 2

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

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) = 0.12. 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) = 0.12

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 − 0.12 = 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

0.2475 (assumed independence)
0.0000 (ignored the overlap)

Reference: FE Reference Handbook — Probability and Statistics → Associative Law

Example 3

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

Event A (a weld fails inspection) has P(A) = 0.60 and event B (a bolt fails inspection) has P(B) = 0.60, with P(A∩B) = 0.12. Use De Morgan's law to compute the probability that neither event occurs.

Given

$P(A) = 0.60$
$P(B) = 0.60$
P(A∩B) = 0.12

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) = 0.60 + 0.60 − 0.12 = 1.0800
De Morgan's law
$Ā \cap B̄ = (A \cup B)‾$
Substituting — P(Ā∩B̄) = 1 − 1.0800 = -0.0800

Answer:

P(neither) = -0.0800

Why the other options are there

0.1600 (assumed independence)
-0.2000 (ignored the overlap)

Reference: FE Reference Handbook — Probability and Statistics → Associative Law

Example 4

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

Event A (a weld fails inspection) has P(A) = 0.25 and event B (a bolt fails inspection) has P(B) = 0.35, with P(A∩B) = 0.09. Use De Morgan's law to compute the probability that neither event occurs.

Given

$P(A) = 0.25$
$P(B) = 0.35$
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) = 0.25 + 0.35 − 0.09 = 0.5100
De Morgan's law
$Ā \cap B̄ = (A \cup B)‾$
Substituting — P(Ā∩B̄) = 1 − 0.5100 = 0.4900

Answer:

P(neither) = 0.4900

Why the other options are there

0.4875 (assumed independence)
0.4000 (ignored the overlap)

Reference: FE Reference Handbook — Probability and Statistics → Associative Law

Example 5

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

Event A (a weld fails inspection) has P(A) = 0.40 and event B (a bolt fails inspection) has P(B) = 0.25, with P(A∩B) = 0.15. Use De Morgan's law to compute the probability that neither event occurs.

Given

$P(A) = 0.40$
$P(B) = 0.25$
P(A∩B) = 0.15

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 + 0.25 − 0.15 = 0.5000
De Morgan's law
$Ā \cap B̄ = (A \cup B)‾$
Substituting — P(Ā∩B̄) = 1 − 0.5000 = 0.5000

Answer:

P(neither) = 0.5000

Why the other options are there

0.4500 (assumed independence)
0.3500 (ignored the overlap)

Reference: FE Reference Handbook — Probability and Statistics → Associative Law

Example 6

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

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) = 0.08. 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) = 0.08

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 + 0.50 − 0.08 = 0.8700
De Morgan's law
$Ā \cap B̄ = (A \cup B)‾$
Substituting — P(Ā∩B̄) = 1 − 0.8700 = 0.1300

Answer:

P(neither) = 0.1300

Why the other options are there

0.2750 (assumed independence)
0.0500 (ignored the overlap)

Reference: FE Reference Handbook — Probability and Statistics → Associative Law

Example 7

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

Event A (a weld fails inspection) has P(A) = 0.35 and event B (a bolt fails inspection) has P(B) = 0.35, with P(A∩B) = 0.07. Use De Morgan's law to compute the probability that neither event occurs.

Given

$P(A) = 0.35$
$P(B) = 0.35$
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) = 0.35 + 0.35 − 0.07 = 0.6300
De Morgan's law
$Ā \cap B̄ = (A \cup B)‾$
Substituting — P(Ā∩B̄) = 1 − 0.6300 = 0.3700

Answer:

P(neither) = 0.3700

Why the other options are there

0.4225 (assumed independence)
0.3000 (ignored the overlap)

Reference: FE Reference Handbook — Probability and Statistics → Associative Law

Example 8

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

Event A (a weld fails inspection) has P(A) = 0.35 and event B (a bolt fails inspection) has P(B) = 0.45, with P(A∩B) = 0.07. Use De Morgan's law to compute the probability that neither event occurs.

Given

$P(A) = 0.35$
$P(B) = 0.45$
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) = 0.35 + 0.45 − 0.07 = 0.7300
De Morgan's law
$Ā \cap B̄ = (A \cup B)‾$
Substituting — P(Ā∩B̄) = 1 − 0.7300 = 0.2700

Answer:

P(neither) = 0.2700

Why the other options are there

0.3575 (assumed independence)
0.2000 (ignored the overlap)

Reference: FE Reference Handbook — Probability and Statistics → Associative Law

Example 9

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

Given

$P(A) = 0.25$
$P(B) = 0.35$
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) = 0.25 + 0.35 − 0.09 = 0.5100
De Morgan's law
$Ā \cap B̄ = (A \cup B)‾$
Substituting — P(Ā∩B̄) = 1 − 0.5100 = 0.4900

Answer:

P(neither) = 0.4900

Why the other options are there

0.4875 (assumed independence)
0.4000 (ignored the overlap)

Reference: FE Reference Handbook — Probability and Statistics → Associative Law

Example 10

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

Event A (a weld fails inspection) has P(A) = 0.45 and event B (a bolt fails inspection) has P(B) = 0.35, with P(A∩B) = 0.13. Use De Morgan's law to compute the probability that neither event occurs.

Given

$P(A) = 0.45$
$P(B) = 0.35$
P(A∩B) = 0.13

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 + 0.35 − 0.13 = 0.6700
De Morgan's law
$Ā \cap B̄ = (A \cup B)‾$
Substituting — P(Ā∩B̄) = 1 − 0.6700 = 0.3300

Answer:

P(neither) = 0.3300

Why the other options are there

0.3575 (assumed independence)
0.2000 (ignored the overlap)

Reference: FE Reference Handbook — Probability and Statistics → Associative Law

Associative Law

Core formulas for this FE topic

A,B,Ch=]A,Bg,CA , ^ B , C h = ] A , Bg , CA,B,Ch=]A,Bg,C

A+B+Ch=]A+Bg+CA + ^ B + C h = ] A + Bg + CA+B+Ch=]A+Bg+C

Worked exam-style examples