Study guide
5.1 Probability Theory, Counting Techniques & Bayes' Theorem
PRC Registered Electrical Engineer REE
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1 Chapter 1: Overview, RA 7920 & Electrical Engineering Ethics
1.1 PRC Board of Electrical Engineering & REE Examination Overview 1.2 Republic Act No. 7920 (New Electrical Engineering Law) & License Classes 1.3 Code of Ethics & Professional Practice for Electrical Engineers
2 Chapter 2: Math — Algebra, Trig & Geometry
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3 Chapter 3: Engineering Mathematics — Calculus & Differential Equations
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4 Chapter 4: Engineering Mathematics — Advanced Math & Transform Methods
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5 Chapter 5: Engineering Mathematics — Probability, Statistics & Engineering Data Analysis
5.1 Probability Theory, Counting Techniques & Bayes' Theorem 5.2 Engineering Data Analysis I — Descriptive Statistics & Probability Distributions 5.3 Engineering Data Analysis II — Confidence Intervals, Hypothesis Testing & Regression
6 Chapter 6: Engineering Mathematics — Numerical Methods & Analysis
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9 Chapter 9: ESAS — Engineering Economics, Management & Contracts
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10 Chapter 10: PEC 1 — Safety, Sizing & Protection
10.1 PEC Fundamentals, Safety Standards & Grounding/Bonding 10.2 Conductor Sizing, Ampacity Adjustment & Voltage Drop 10.3 Overcurrent Protection, Branch Circuit & Feeder Sizing 10.4 Basic Occupational Safety & Health (BOSH) and Electrical Safety Standards
11 Chapter 11: PEC 1 & 2 — Equipment & Utility Systems
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12 Chapter 12: Electrical Professional — DC & AC Circuit Fundamentals
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15.1 Single-Phase & Three-Phase Transformers 15.2 Transformer Connections & Special Transformers 15.3 DC Generators & DC Motors
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16.1 Synchronous Generators (Alternators) 16.2 Synchronous Motors & Special AC Machines 16.3 Three-Phase & Single-Phase Induction Motors
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18.1 Transmission Line Parameters & Performance Models 18.2 Mechanical Overhead Line Design, Underground Cables & Insulators 18.3 Distribution Systems, Substation Layouts & Power Flow Control
19 Chapter 19: Electrical Professional — Power System Analysis
19.1 The Per-Unit System, Base Conversion & Impedance Diagrams 19.2 Symmetrical Components & Unbalanced Fault Analysis 19.3 Load Flow, Economic Dispatch & Power System Stability
20 Chapter 20: Electrical Professional — Protection, Control, Instrumentation & Illumination
20.1 Power System Protection & Protective Relaying 20.2 Industrial Motor Control, Logic Circuits & PLC Automation 20.3 Electrical Instrumentation, Transducers & Measurement Systems 20.4 Illumination Engineering, Lighting Design & Photometrics 20.5 Feedback Control Systems & Research Methods
5.1 Probability Theory, Counting Techniques & Bayes' Theorem
Key Takeaways Explain This Topic AI
Probability & Statistics is its own topic in the PRC Enhanced Table of Specifications (Res. 40, s. 2024), carrying 1.25% of the total exam and 5 of the 100 Mathematics items.
Permutations count ordered arrangements, nPr = n!/(n − r)!; combinations count unordered selections, nCr = n!/[r!(n − r)!]. The word 'committee' signals combinations; 'ranking' or 'seating' signals permutations.
The addition rule is P(A ∪ B) = P(A) + P(B) − P(A ∩ B); the subtraction term vanishes only when the events are mutually exclusive.
Independence means P(A ∩ B) = P(A)·P(B); mutual exclusivity means P(A ∩ B) = 0. Two events with non-zero probability cannot be both independent and mutually exclusive.
Bayes' theorem, P(A|B) = P(B|A)P(A)/P(B), is the standard REE reliability item: given a defective relay, find the probability it came from a particular supplier.
Last updated: August 2026
5.1 Probability Theory, Counting Techniques & Bayes' Theorem
The PRC Enhanced Table of Specifications for the Registered Electrical Engineer Licensure Examination (Professional Regulatory Board of Electrical Engineering Resolution No. 40, series of 2024 , which superseded Board Resolution No. 23, s. 2011) lists Probability & Statistics as topic D of the Mathematics subject, weighted 1.25% of the whole examination — 5 of the 100 Mathematics items . Probability also underpins every reliability, availability and quality-control calculation an electrical engineer performs on protection schemes, redundant feeders and equipment procurement.
1. Sample Spaces, Events & the Axioms of Probability
A random experiment is any process whose outcome cannot be predicted with certainty. The sample space S S S is the set of all possible outcomes; an event A A A is any subset of S S S .
For equally likely outcomes, the classical definition applies: P ( A ) = n ( A ) n ( S ) = number of favourable outcomes total number of outcomes P(A) = \frac{n(A)}{n(S)} = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}} P ( A ) = n ( S ) n ( A ) = total number of outcomes number of favourable outcomes
The three Kolmogorov axioms constrain every valid probability:
- 0 ≤ P ( A ) ≤ 1 0 \le P(A) \le 1 0 ≤ P ( A ) ≤ 1 for every event A A A ;
- P ( S ) = 1 P(S) = 1 P ( S ) = 1 (some outcome must occur);
For mutually exclusive events, P ( A 1 ∪ A 2 ∪ … ) = P ( A 1 ) + P ( A 2 ) + … P(A_1 \cup A_2 \cup \dots) = P(A_1) + P(A_2) + \dots P ( A 1 ∪ A 2 ∪ … ) = P ( A 1 ) + P ( A 2 ) + …
The complement rule follows immediately and is the fastest route through most "at least one" board problems: P ( A ′ ) = 1 − P ( A ) P(A') = 1 - P(A) P ( A ′ ) = 1 − P ( A )
2. Counting Techniques
Because the classical definition needs n ( A ) n(A) n ( A ) and n ( S ) n(S) n ( S ) , counting is half the work.
Fundamental Counting Principle
If a first task can be done in m m m ways and a second independently in n n n ways, the pair can be done in m × n m \times n m × n ways.
Permutations (order matters) n P r = n ! ( n − r ) ! _{n}P_{r} = \frac{n!}{(n-r)!} n P r = ( n − r )! n !
Circular permutations of n n n distinct objects: ( n − 1 ) ! (n-1)! ( n − 1 )! — fix one object to remove rotational duplicates.
Permutations with repeated objects : n ! n 1 ! n 2 ! ⋯ n k ! \dfrac{n!}{n_1!\,n_2!\cdots n_k!} n 1 ! n 2 ! ⋯ n k ! n !
Combinations (order does not matter) n C r = ( n r ) = n ! r ! ( n − r ) ! _{n}C_{r} = \binom{n}{r} = \frac{n!}{r!\,(n-r)!} n C r = ( r n ) = r ! ( n − r )! n !
Board-Exam Signal Word Technique Formula
"arrangement", "ranking", "seating order", "password" Permutation n ! ( n − r ) ! \dfrac{n!}{(n-r)!} ( n − r )! n !
"committee", "sample", "selection", "handshake" Combination n ! r ! ( n − r ) ! \dfrac{n!}{r!(n-r)!} r ! ( n − r )! n !
"seated around a round table" Circular permutation ( n − 1 ) ! (n-1)! ( n − 1 )!
"letters of the word ENGINEER" Permutation with repeats n ! n 1 ! n 2 ! ⋯ \dfrac{n!}{n_1!n_2!\cdots} n 1 ! n 2 ! ⋯ n !
3. Addition, Multiplication & Conditional Probability
Addition Rule (union — "A or B") P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) P(A \cup B) = P(A) + P(B) - P(A \cap B) P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B )
The overlap term P ( A ∩ B ) P(A \cap B) P ( A ∩ B ) disappears only when A A A and B B B are mutually exclusive. Forgetting to subtract it is the single most common REE probability error.
Conditional Probability & the Multiplication Rule P ( A ∣ B ) = P ( A ∩ B ) P ( B ) , P ( B ) ≠ 0 ⟹ P ( A ∩ B ) = P ( B ) P ( A ∣ B ) P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) \neq 0 \qquad \Longrightarrow \qquad P(A \cap B) = P(B)\,P(A \mid B) P ( A ∣ B ) = P ( B ) P ( A ∩ B ) , P ( B ) = 0 ⟹ P ( A ∩ B ) = P ( B ) P ( A ∣ B )
Independence vs. Mutual Exclusivity
Property Definition Consequence
Independent P ( A ∩ B ) = P ( A ) P ( B ) P(A \cap B) = P(A)P(B) P ( A ∩ B ) = P ( A ) P ( B ) P ( A ∣ B ) = P ( A ) P(A \mid B) = P(A) P ( A ∣ B ) = P ( A ) — knowing B B B tells you nothing about A A A
Mutually exclusive P ( A ∩ B ) = 0 P(A \cap B) = 0 P ( A ∩ B ) = 0 P ( A ∣ B ) = 0 P(A \mid B) = 0 P ( A ∣ B ) = 0 — the events cannot co-occur
These are different ideas, and two events with non-zero probability can never be both. Examiners exploit the confusion routinely.
Series and Parallel Reliability
For n n n independent components with individual reliabilities R i R_i R i : R series = ∏ i = 1 n R i R parallel = 1 − ∏ i = 1 n ( 1 − R i ) R_{\text{series}} = \prod_{i=1}^{n} R_i \qquad\qquad R_{\text{parallel}} = 1 - \prod_{i=1}^{n} (1 - R_i) R series = i = 1 ∏ n R i R parallel = 1 − i = 1 ∏ n ( 1 − R i )
A protection scheme with two redundant relays is a parallel block: it fails only if both relays fail.
4. Total Probability & Bayes' Theorem
When the sample space is partitioned into mutually exclusive, exhaustive causes B 1 , B 2 , … , B n B_1, B_2, \dots, B_n B 1 , B 2 , … , B n : P ( A ) = ∑ i = 1 n P ( B i ) P ( A ∣ B i ) (Law of Total Probability) P(A) = \sum_{i=1}^{n} P(B_i)\,P(A \mid B_i) \qquad \text{(Law of Total Probability)} P ( A ) = i = 1 ∑ n P ( B i ) P ( A ∣ B i ) (Law of Total Probability)
Reversing the conditioning gives Bayes' theorem : P ( B k ∣ A ) = P ( B k ) P ( A ∣ B k ) ∑ i = 1 n P ( B i ) P ( A ∣ B i ) P(B_k \mid A) = \frac{P(B_k)\,P(A \mid B_k)}{\displaystyle\sum_{i=1}^{n} P(B_i)\,P(A \mid B_i)} P ( B k ∣ A ) = i = 1 ∑ n P ( B i ) P ( A ∣ B i ) P ( B k ) P ( A ∣ B k )
In REE practice this answers questions of the form: a randomly pulled relay is defective — what is the probability it came from Supplier B?
- Solved Board Exam Examples
- Example 1: Parallel-Redundant Protection Reliability
A 69 kV feeder is protected by a primary relay of reliability 0.95 0.95 0.95 and an independent backup relay of reliability 0.88 0.88 0.88 . Find the probability that the feeder is protected (at least one relay operates).
Solution. The relays form a parallel reliability block: R parallel = 1 − ( 1 − 0.95 ) ( 1 − 0.88 ) = 1 − ( 0.05 ) ( 0.12 ) = 1 − 0.006 = 0.994 R_{\text{parallel}} = 1 - (1 - 0.95)(1 - 0.88) = 1 - (0.05)(0.12) = 1 - 0.006 = \boxed{0.994} R parallel = 1 − ( 1 − 0.95 ) ( 1 − 0.88 ) = 1 − ( 0.05 ) ( 0.12 ) = 1 − 0.006 = 0.994
Redundancy raises protection reliability from 95% to 99.4%, and the probability of an unprotected fault falls from 5 % 5\% 5% to 0.6 % 0.6\% 0.6% .
Example 2: Bayes' Theorem on Contactor Supply
A panel builder sources contactors from three suppliers: A supplies 50% with a 1% defect rate, B supplies 30% with a 2% defect rate, and C supplies 20% with a 4% defect rate. A contactor drawn at random is defective. What is the probability it came from Supplier C?
Solution. First the total defect probability: P ( D ) = ( 0.50 ) ( 0.01 ) + ( 0.30 ) ( 0.02 ) + ( 0.20 ) ( 0.04 ) = 0.005 + 0.006 + 0.008 = 0.019 P(D) = (0.50)(0.01) + (0.30)(0.02) + (0.20)(0.04) = 0.005 + 0.006 + 0.008 = 0.019 P ( D ) = ( 0.50 ) ( 0.01 ) + ( 0.30 ) ( 0.02 ) + ( 0.20 ) ( 0.04 ) = 0.005 + 0.006 + 0.008 = 0.019
Then Bayes' theorem: P ( C ∣ D ) = ( 0.20 ) ( 0.04 ) 0.019 = 0.008 0.019 = 0.421 or 42.1 % P(C \mid D) = \frac{(0.20)(0.04)}{0.019} = \frac{0.008}{0.019} = \boxed{0.421 \text{ or } 42.1\%} P ( C ∣ D ) = 0.019 ( 0.20 ) ( 0.04 ) = 0.019 0.008 = 0.421 or 42.1%
Supplier C ships only 20% of the units but accounts for 42.1% of the defects — the practical lesson behind the arithmetic.
Example 3: Committee Selection
From 7 registered electrical engineers and 5 registered master electricians, a 4-member safety committee is formed. What is the probability it contains exactly 2 of each?
Solution. Order is irrelevant, so use combinations: n ( S ) = ( 12 4 ) = 495 , n ( A ) = ( 7 2 ) ( 5 2 ) = ( 21 ) ( 10 ) = 210 n(S) = \binom{12}{4} = 495, \qquad n(A) = \binom{7}{2}\binom{5}{2} = (21)(10) = 210 n ( S ) = ( 4 12 ) = 495 , n ( A ) = ( 2 7 ) ( 2 5 ) = ( 21 ) ( 10 ) = 210 P ( A ) = 210 495 = 0.4242 P(A) = \frac{210}{495} = \boxed{0.4242} P ( A ) = 495 210 = 0.4242