Distributive 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 — Distributive Law

Event A (a weld fails inspection) has P(A) = 0.45 and event B (a bolt fails inspection) has P(B) = 0.30, with P(A∩B) = 0.05. Use De Morgan's law to compute the probability that neither event occurs.

Given

$P(A) = 0.45$
$P(B) = 0.30$
P(A∩B) = 0.05

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 + 0.30 − 0.05 = 0.7000
De Morgan's law
$Ā \cap B̄ = (A \cup B)‾$
Substituting — P(Ā∩B̄) = 1 − 0.7000 = 0.3000

Answer:

P(neither) = 0.3000

Why the other options are there

0.3850 (assumed independence)
0.2500 (ignored the overlap)

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

Example 2

De Morgan's law applied to two inspection events — Distributive 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) = 0.06. 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) = 0.06

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) = 0.45 + 0.55 − 0.06 = 0.9400
De Morgan's law
$Ā \cap B̄ = (A \cup B)‾$
Substituting — P(Ā∩B̄) = 1 − 0.9400 = 0.0600

Answer:

P(neither) = 0.0600

Why the other options are there

0.2475 (assumed independence)
0.0000 (ignored the overlap)

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

Example 3

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

Event A (a weld fails inspection) has P(A) = 0.25 and event B (a bolt fails inspection) has P(B) = 0.25, with P(A∩B) = 0.10. Use De Morgan's law to compute the probability that neither event occurs.

Given

$P(A) = 0.25$
$P(B) = 0.25$
P(A∩B) = 0.10

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 + 0.25 − 0.10 = 0.4000
De Morgan's law
$Ā \cap B̄ = (A \cup B)‾$
Substituting — P(Ā∩B̄) = 1 − 0.4000 = 0.6000

Answer:

P(neither) = 0.6000

Why the other options are there

0.5625 (assumed independence)
0.5000 (ignored the overlap)

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

Example 4

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

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 → Distributive Law

Example 5

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

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) = 0.10. 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) = 0.10

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 − 0.10 = 0.8000
De Morgan's law
$Ā \cap B̄ = (A \cup B)‾$
Substituting — P(Ā∩B̄) = 1 − 0.8000 = 0.2000

Answer:

P(neither) = 0.2000

Why the other options are there

0.2925 (assumed independence)
0.1000 (ignored the overlap)

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

Example 6

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

Event A (a weld fails inspection) has P(A) = 0.55 and event B (a bolt fails inspection) has P(B) = 0.30, with P(A∩B) = 0.13. Use De Morgan's law to compute the probability that neither event occurs.

Given

$P(A) = 0.55$
$P(B) = 0.30$
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) = 0.55 + 0.30 − 0.13 = 0.7200
De Morgan's law
$Ā \cap B̄ = (A \cup B)‾$
Substituting — P(Ā∩B̄) = 1 − 0.7200 = 0.2800

Answer:

P(neither) = 0.2800

Why the other options are there

0.3150 (assumed independence)
0.1500 (ignored the overlap)

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

Example 7

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

Event A (a weld fails inspection) has P(A) = 0.40 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

$P(A) = 0.40$
$P(B) = 0.60$
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) = 0.40 + 0.60 − 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

0.2400 (assumed independence)
0.0000 (ignored the overlap)

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

Example 8

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

Event A (a weld fails inspection) has P(A) = 0.45 and event B (a bolt fails inspection) has P(B) = 0.45, with P(A∩B) = 0.09. Use De Morgan's law to compute the probability that neither event occurs.

Given

$P(A) = 0.45$
$P(B) = 0.45$
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) = 0.45 + 0.45 − 0.09 = 0.8100
De Morgan's law
$Ā \cap B̄ = (A \cup B)‾$
Substituting — P(Ā∩B̄) = 1 − 0.8100 = 0.1900

Answer:

P(neither) = 0.1900

Why the other options are there

0.3025 (assumed independence)
0.1000 (ignored the overlap)

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

Example 9

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

Event A (a weld fails inspection) has P(A) = 0.55 and event B (a bolt fails inspection) has P(B) = 0.25, with P(A∩B) = 0.06. Use De Morgan's law to compute the probability that neither event occurs.

Given

$P(A) = 0.55$
$P(B) = 0.25$
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) = 0.55 + 0.25 − 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

0.3375 (assumed independence)
0.2000 (ignored the overlap)

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

Example 10

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

Event A (a weld fails inspection) has P(A) = 0.30 and event B (a bolt fails inspection) has P(B) = 0.55, with P(A∩B) = 0.09. Use De Morgan's law to compute the probability that neither event occurs.

Given

$P(A) = 0.30$
$P(B) = 0.55$
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) = 0.30 + 0.55 − 0.09 = 0.7600
De Morgan's law
$Ā \cap B̄ = (A \cup B)‾$
Substituting — P(Ā∩B̄) = 1 − 0.7600 = 0.2400

Answer:

P(neither) = 0.2400

Why the other options are there

0.3150 (assumed independence)
0.1500 (ignored the overlap)

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