5.1 Probability Theory, Counting Techniques & Bayes' Theorem

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Licensure Examination (Professional Regulatory Board of Electrical Engineering Resolution No. 40, series of 2024, which superseded Board Resolution No. 23, s. 2011) lists Probability & Statistics as topic D of the Mathematics subject, weighted 1.25% of the whole examination — 5 of the 100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 $S$ is the set of all possible outcomes; an event $A$ is any subset of $S$.

For equally likely outcomes, the classical definition applies:

$P(A) = \frac{n(A)}{n(S)} = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$

The three Kolmogorov axioms constrain every valid probability:

1. $0 \le P(A) \le 1$ for every event $A$;
2. $P(S) = 1$ (some outcome must occur);
3. For mutually exclusive events, $P(A_1 \cup A_2 \cup \dots) = P(A_1) + P(A_2) + \dots$

The complement rule follows immediately and is the fastest route through most "at least one" board problems:

$P(A') = 1 - P(A)$

---

2. Counting Techniques

Because the classical definition needs $n(A)$ and $n(S)$, counting is half the work.

Fundamental Counting Principle

If a first task can be done in $m$ ways and a second independently in $n$ ways, the pair can be done in $m \times n$ ways.

Permutations (order matters)

$_{n}P_{r} = \frac{n!}{(n-r)!}$

• Circular permutations of $n$ distinct objects: $(n-1)!$ — fix one object to remove rotational duplicates.
• Permutations with repeated objects: $\dfrac{n!}{n_1!,n_2!\cdots n_k!}$

Combinations (order does not matter)

$_{n}C_{r} = \binom{n}{r} = \frac{n!}{r!\,(n-r)!}$

---

3. Addition, Multiplication & Conditional Probability

Addition Rule (union — "A or B")

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

The overlap term $P(A \cap B)$ disappears only when $A$ and $B$ are mutually exclusive. Forgetting to subtract it is the single most common REE probability error.

Conditional Probability & the Multiplication Rule

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) \neq 0 \qquad \Longrightarrow \qquad P(A \cap B) = P(B)\,P(A \mid B)$

Independence vs. Mutual Exclusivity

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 $n$ independent components with individual reliabilities $R_i$:

$R_{\text{series}} = \prod_{i=1}^{n} R_i \qquad\qquad R_{\text{parallel}} = 1 - \prod_{i=1}^{n} (1 - R_i)$

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 $B_1, B_2, \dots, B_n$:

$P(A) = \sum_{i=1}^{n} P(B_i)\,P(A \mid B_i) \qquad \text{(Law of Total Probability)}$

Reversing the conditioning gives Bayes' theorem:

$P(B_k \mid A) = \frac{P(B_k)\,P(A \mid B_k)}{\displaystyle\sum_{i=1}^{n} P(B_i)\,P(A \mid B_i)}$

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 $0.95$ and an independent backup relay of reliability $0.88$. Find the probability that the feeder is protected (at least one relay operates).

Solution. The relays form a parallel reliability block:

$R_{\text{parallel}} = 1 - (1 - 0.95)(1 - 0.88) = 1 - (0.05)(0.12) = 1 - 0.006 = \boxed{0.994}$

Redundancy raises protection reliability from 95% to 99.4%, and the probability of an unprotected fault falls from $5%$ to $0.6%$.

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:

$P(D) = (0.50)(0.01) + (0.30)(0.02) + (0.20)(0.04) = 0.005 + 0.006 + 0.008 = 0.019$

Then Bayes' theorem:

$P(C \mid D) = \frac{(0.20)(0.04)}{0.019} = \frac{0.008}{0.019} = \boxed{0.421 \text{ or } 42.1\%}$

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:

$n(S) = \binom{12}{4} = 495, \qquad n(A) = \binom{7}{2}\binom{5}{2} = (21)(10) = 210$

$P(A) = \frac{210}{495} = \boxed{0.4242}$