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3-Phase Servo AVR (AC Voltage Stabilizer) — Troubleshooting, Repair & Maintenance By Edwin Ogie • December 18, 2025 • -- AC Voltage Stabilizer — 3-phase servo control type (example from user photo) A practical, step-by-step guide to diagnose, repair and maintain 3-phase servo Automatic Voltage Regulators (AVR) / servo voltage stabilizers. Written in simple terms for technicians and maintenance teams working with generators, UPS rooms and factories. Includes videos, spare-parts list, safety checklist, troubleshooting flow and links to internal/external resources. Contents Why this matters In environments with unstable mains (frequent sags, surges or phase imbalance) a servo AVR protects sensitive equipment by continuously adjusting an autotransformer tap via a small servo motor. A well-maintained stabilizer saves equipment, reduces downtime and prevents costly damage. ...
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Probability E-Note for Senior Secondary School | Edwin Ogie Library 2
Definition:
Mutually inclusive probability involves two or more events where the occurrence of one event has no effect on the occurrence of the others. In other words, events A and B are mutually inclusive (or independent) if the probability of A and B occurring together is the same as the probability of A occurring alone multiplied by the probability of B occurring alone. In these cases, the term “and” is used, and the events are multiplied.
Example:
When tossing two coins, the occurrence of a head (H) on one coin does not affect the occurrence of a tail (T) on the other. It is possible to get one head and one tail.
Formula:
Pr(A and B) = Pr(A) × Pr(B)
2. Mutually Exclusive Probability
Definition:
Mutually exclusive probability involves two or more events where the occurrence of one event prevents any of the others from occurring. That is, if event X occurs, event Y cannot occur at the same time. These are sometimes referred to as dependent events, and the term “or” (or “either”) is used. In this case, the probabilities of the individual events are added.
Example:
If events X and Y are mutually exclusive, then
Pr(X or Y) = Pr(X) + Pr(Y)
Example 6.7 (With Replacement)
A bag contains some identical balls: 10 red, 12 white, and 8 yellow. Three balls are drawn at random with replacement. Find the probability that:
Again, this can occur in three orders: (R, W, Y), (W, Y, R), or (Y, R, W).
Calculation:
For one order: (10/30) × (12/29) × (8/28)
Multiply by 3 for all orders:
(10/30 × 12/29 × 8/28) × 3
Alternatively written as: (1/3 × 12/29 × 2/7) = (4/29 × 2/7) = 8/203
(c) Two white and one red ball
The arrangements possible are: (W, W, R), (W, R, W), or (R, W, W).
Calculation:
For one order: (12/30) × (11/29) × (8/28)
Similarly, also compute for the other orders:
(12/30 × 8/29 × 12/28) and (10/30 × 12/29 × 11/28)
Summing these probabilities:
(44/1015) × 3 = 132/1015
Evaluation (Brain Match)
The probability of picking a letter T from the word OBSTRUCTION is:
A. 2/11
B. 3/11
C. 2/8
D. 1/11
What is the probability of obtaining a factor of 3 in a single toss of a fair die?
A. 1/3
B. 1/3
C. 2/3
D. 1/2
What is the probability of obtaining two heads in a single toss of two coins?
A. 1/4
B. 2/5
C. 1/2
D. 3/4
The following table shows the distribution of recharge cards for four major GSM operators:
What is the probability that a recharge card selected at random is Airtel?
A. 3/10
B. 1/10
C. 3/20
D. 5/17
What is the probability that a recharge card selected at random is MTN?
A. 1/4
B. 3/10
C. 7/10
D. 1/20
Theory
A basket contains 10 identical balls that differ only in colour. There are 5 blue balls, 3 green balls, and the rest are yellow. If a ball is chosen at random, what is the probability that it is:
(a) Blue
(b) Yellow
(c) Green
(d) Yellow and black
(e) Red and green
(f) Not blue
(g) Not green
The following table shows the number of limes and apples of the same size in a bag:
If two fruits are picked at random, one at a time, without replacement, find the probability that:
(i) Both are good limes
(ii) Both are bad fruits
(iii) One is a good apple and the other is a bad lime
1. What is a Sample Space?
In probability theory, a sample space is the set of all possible outcomes of a random experiment or process. It is denoted by the symbol S or Ω. For example, when rolling a six-sided die, the sample space is
S = {1, 2, 3, 4, 5, 6}
Each outcome in the sample space is called an element or sample point. The sample space represents the complete set of possible results that can occur in the experiment. (For a fair coin, the sample space is often given as {H, T}.)
2. Sample Space for Two Dice
When two dice are tossed, it is useful to organize the outcomes in a table. Since each die has 6 outcomes, the complete sample space consists of 36 ordered pairs. A table can list the results with the first die’s outcome along one axis and the second die’s outcome along the other.
3. Example 7.1: Tossing Three Coins
Problem:
What is the probability of obtaining at least two heads when three coins are tossed?
Sample Space for Three Coins:
HHH, HHT, HTH, THH, TTT, TTH, THT, HTT
There are 8 possible outcomes.
Calculations:
Probability of at least two heads:
Favorable outcomes: HHH, HHT, HTH, THH (4 outcomes)
Pr(at least two heads) = 4/8 = ½
Probability of at least one tail:
Only one outcome (HHH) has no tail, so
Pr(at least one tail) = 7/8
Probability of exactly two heads (two heads and one tail):
Pr(two heads and one tail) = 3/8
4. Example 7.2: Tossing Two Fair Dice
Problem:
Two fair dice are tossed once. Find the probability of obtaining:
(a) a sum of 7,
(b) the same digits on both dice, and
(c) a sum of at least 9.
Solution:
Sample Space:
The complete sample space is the set of 36 ordered pairs. For example, the outcomes when the first die shows 1 are:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
and so on for 2 through 6.
(a) Sum of 7:
Favorable outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
Pr(sum of 7) = 6/36 = 1/6
(c) Sum of at least 9:
There are 10 outcomes that yield a sum of 9 or more.
Pr(sum of at least 9) = 10/36 = 5/18
5. Probability of Special Cases
Sometimes a given probability is used to determine the probability of related events.
Example 7.3: Civil Servants and Car Ownership
Problem:
If the probability that a civil servant owns a car is 1/6, find the probability that:
(i) Two civil servants, A and B, selected at random, each own a car.
(ii) Of two civil servants, C and D selected at random, only one owns a car.
(iii) Of three civil servants, X, Y, and Z selected at random, only one owns a car.
Solution:
Step 1:
Pr(owns a car) = 1/6
Pr(does not own a car) = 1 – 1/6 = 5/6
(i) Both A and B own a car:
Pr = (1/6) × (1/6) = 1/36
(ii) Only one of C and D owns a car:
Two cases:
- C owns and D does not: (1/6) × (5/6)
- D owns and C does not: (5/6) × (1/6)
Total Pr = (1/6 × 5/6) + (5/6 × 1/6) = 5/36 + 5/36 = 10/36 = 5/18
(iii) Only one of X, Y, and Z owns a car:
For one specific arrangement (e.g., X owns, Y and Z do not):
Pr = (1/6) × (5/6) × (5/6)
Since any one of the three could be the sole owner, multiply by 3:
Pr = 3 × [(1/6) × (5/6) × (5/6)] = 3 × (25/216) = 75/216 = 25/72
6. Evaluation – Brain Match Questions
Question:
From the sample space {a, b, d, f, g, k}, what is the probability of picking a consonant if a letter is chosen at random? Options:
A. 1/6 B. 5/6 C. 3/5 D. 6/5 Answer: Since only “a” is a vowel, 5 out of 6 letters are consonants. Probability = 5/6
Question:
If the probability of hitting the target is 2/3, what is the probability of not hitting the target? Options:
A. 1/3 B. 2/3 C. 4/3 D. None Answer:
Pr(not hitting) = 1 – 2/3 = 1/3
Question:
If three unbiased coins are tossed, what is the probability that they all land heads? Options:
A. ½ B. 1/3 C. 1/9 D. 1/8 Answer:
Pr(all heads) = (½)³ = 1/8
Question:
A bag contains 5 black, 4 white, and x red marbles. If the probability of picking a red marble is 2/5, find the value of x. Options:
A. 8 B. 10 C. 4 D. 6 Solution:
Total marbles = 5 + 4 + x = 9 + x
Pr(red) = x/(9 + x) = 2/5
Cross-multiply: 5x = 2(9 + x) → 5x = 18 + 2x → 3x = 18 → x = 6
Question:
Find the probability that a number selected at random from 41 to 56 (inclusive) is a multiple of 9. Options:
A. 1/8 B. 2/15 C. 3/16 D. 7/8 Solution:
Total numbers = 56 – 41 + 1 = 16
Multiples of 9 between 41 and 56: 45 and 54 (2 numbers)
Probability = 2/16 = 1/8
7. Theory Problems
Problem:
The probabilities that Ade, Kujo, and Fati will pass an examination are 2/3, 5/8, and 3/4 respectively. Find the probability that:
(a) All three pass.
(b) None of them pass.
(c) Only Ade and Kujo pass (i.e., Fati fails).
Solution:
(a) All three pass:
Pr = (2/3) × (5/8) × (3/4) = 30/96 = 5/16
Boys 1 and 2 pass, Boy 3 fails: (2/3) × (5/8) × (1/4) = 10/96
Boys 1 and 3 pass, Boy 2 fails: (2/3) × (3/8) × (1/4) = 6/96 (Note: Alternatively, if Boy 2’s failure probability is used correctly, recalculate as: (2/3) × (1 - 5/8 = 3/8) × (3/4) = 18/96)
Boys 2 and 3 pass, Boy 1 fails: (1/3) × (5/8) × (3/4) = 15/96
Total Pr = (10 + 18 + 15)/96 = 43/96
Problem:
Two fair dice are thrown once.
(a) Draw a sample space for the possible outcomes.
(b) What is the probability of obtaining two prime numbers?
Solution:
(a) Sample Space:
The sample space is the set of 36 ordered pairs:
{ (1,1), (1,2), …, (1,6), (2,1), …, (6,6) }.
(b) Two prime numbers:
The prime numbers on a die are 2, 3, and 5.
Number of favorable outcomes = 3 (choices for die 1) × 3 (choices for die 2) = 9
Pr = 9/36 = 1/4
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