# Probability Distributions in FE Electrical Exam

Welcome, electrical engineering enthusiasts, to an exploration of probability distributions in the FE electrical exam and their significant role in the world of electrical engineering.

If you are preparing for the Fundamentals of Engineering (FE) Electrical Exam, probability distributions are crucial to resolving complex electrical engineering problems.

In electrical engineering, precision can either make or break your project or infrastructure. Every decision can shape a project’s outcome, from designing electrical systems to ensuring the safety of millions. Probability distributions let you anticipate, analyze, and predict outcomes with unparalleled accuracy.

Probability Distributions in the FE Electrical exam are mathematical models used in the subject area of statistics and probability, ranging from the uniform distribution to the infamous Gaussian curve, that can quantify uncertainties, evaluate risks, and make informed decisions that maximize efficiency and minimize errors.

Probability distributions are the backbone of statistical analysis, enabling you to model and understand the behavior of complex electrical systems amidst randomness and variability.

In this detailed study guide, we embark on an exciting journey through probability distributions. We’ll explore its significance in electrical engineering, uncover practical applications in real-world scenarios, and equip you with the knowledge and techniques to conquer the FE Electrical Exam with confidence and finesse. Let’s dive deep into detail.

## Importance of Probability Distributions in Electrical Engineering

• Noise and Signal Processing – Probability distributions are crucial for analyzing noise in electronic circuits and designing signal processing algorithms. They help estimate the statistical properties of noise sources and optimize circuit performance.
• Reliability and Failure Analysis – Probability distributions aid in reliability engineering by modeling failure rates and lifetimes of components. They assist in designing reliable systems and making decisions about maintenance and replacement strategies.
• Channel Capacity and Information Theory – Probability distributions characterize the statistical properties of signals and noise in communication channels. They enable the optimization of data transmission rates and the design of efficient communication systems.
• Power System Analysis – Probability distributions model uncertainties in power systems, such as load demand variations and renewable energy fluctuations. They support probabilistic analysis, system optimization, and stability assessment.
• Fault Diagnosis and Condition Monitoring – Probability distributions help diagnose faults by modeling normal and faulty operating conditions. They enable the detection of anomalies and timely maintenance of electrical systems.

## Discrete Probability Distributions

Discrete probability distributions in electrical engineering refer to mathematical models that quantify the probabilities of distinct and countable outcomes. For instance, consider a digital communication system where binary data is transmitted.

The probability distribution can be used to determine the likelihood of receiving a “0” or a “1” based on the characteristics of the channel and the noise present. This distribution helps engineers optimize the system’s performance by analyzing the probability of errors and designing error-correcting codes accordingly.

### Bernoulli distribution

The Bernoulli distribution is a discrete probability distribution that models a single binary event with two possible outcomes: success (typically represented as 1) or failure (typically represented as 0). It is characterized by a single parameter, p, which means the probability of success.

For instance, in a communication system, the probability of a transmitted bit being received correctly is 0.8. Assume that the received bits follow a Bernoulli distribution. What is the probability of having precisely 3 erroneous bits in a sequence of 5 received bits?

Since the probability of success (receiving a bit correctly) is 0.8, the likelihood of failure (receiving a bit erroneously) is 1 – 0.8 = 0.2. We can use the Bernoulli distribution to find the probability of 3 erroneous bits in a sequence of 5 received bits.

Using the binomial distribution, we can calculate this probability as follows:

P(X = 3) = (5 choose 3) * (0.2)^3 * (0.8)^2

The notation “5 choose 3” represents the combination (nCr) of 5 objects taken 3 at a time, which is denoted mathematically as 5C3. It is calculated using the formula for combinations:

nCk = n! / (k!(n-k)!)

so,

P(X = 3) = (5! / (3! * (5-3)!)) * (0.2)^3 * (0.8)^2

= 10 * 0.008 * 0.64

= 0.0512

Therefore, the probability of having precisely 3 erroneous bits in a sequence of 5 received bits, assuming a Bernoulli distribution with a success probability of 0.8, is 0.0512 (or 5.12%).

### Binomial distribution

The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent Bernoulli trials (events with two possible outcomes: success or failure). It is characterized by two parameters: the number of trials, denoted as n, and the probability of success in each trial, denoted as p.

Consider a manufacturing process for electronic components where the probability of an element being defective is known to be 0.05. A batch of 100 components is produced. What is the likelihood of precisely 3 defective parts in the batch?

In this case, we can use the binomial distribution to calculate the probability of having precisely 3 defective components in a batch of 100.

Using the binomial probability formula, we can calculate it as follows:

P(X = 3) = (100C3) * (0.05)^3 * (1 – 0.05)^(100-3)

= (100! / (3! * (100-3)!)) * (0.05)^3 * (0.95)^97

= (100 * 99 * 98) / (3 * 2 * 1) * 0.000125 * 0.07958923738717881

= 0.263901353. (approximately)

Therefore, the probability of having precisely 3 defective components in a batch of 100, assuming a binomial distribution with a defect probability of 0.05, is approximately 0.2639 (or 26.39%).

This probability calculation helps engineers assess the quality of the manufacturing process and understand the likelihood of specific outcomes in large-scale production.

### Poisson distribution

The Poisson distribution is a discrete probability distribution that models the number of events occurring within a fixed interval of time or space when the events are rare and randomly distributed. It is characterized by a single parameter, λ (lambda), which represents the average rate or intensity of the events occurring in the given interval.

The formula for the Poisson distribution is as follows:

P(X = k) = (e^(-λ) * λ^k) / k!

Where:

P(X = k) is the probability of k events occurring

e is the mathematical constant approximately equal to 2.71828

λ is the average rate or intensity of events

k is the number of events (k = 0, 1, 2, …)

Electrical Engineers can use the Poisson distribution to estimate the occurrence of rare events and design appropriate protection measures or system responses. For instance, the average number of voltage surges per day in a power distribution system is 2.5. What is the probability of precisely 4 voltage surges in a given day?

In this case, we can model the number of voltage surges using the Poisson distribution with λ = 2.5. We can use the Poisson probability formula to calculate the probability of having exactly 4 voltage surges:

P(X = 4) = (e^(-2.5) * 2.5^4) / 4!

= (2.71828^(-2.5) * 2.5^4) / (4 * 3 * 2 * 1)

= 0.111973. (approximately)

Therefore, the probability of having exactly 4 voltage surges in a day, assuming a Poisson distribution with an average rate of 2.5 surges, is approximately 0.111973 (or 11.1973%).

### Hypergeometric distribution

The hypergeometric distribution is a discrete probability distribution that models the probability of obtaining a specific number of successes in a fixed number of draws, without replacement, from a finite population containing both successes and failures. It is commonly used when the sampling is done from a small population without replacement.

The formula for the hypergeometric distribution is as follows:

P(X = k) = (C(K, k) * C(N – K, n – k)) / C(N, n)

Where:

P(X = k) is the probability of exactly k successes

C(a, b) denotes the number of combinations of items taken b at a time

N is the population size

K is the number of successes in the population

n is the number of draws made from the population

k is the number of successes observed in the draws

This probability calculation can be useful in electrical engineering for assessing the likelihood of specific events or failures in a batch or population of components. It helps engineers make informed decisions about quality control, reliability, and component.

For instance, in a batch of resistors, there are 200 resistors, out of which 20 are defective. If we randomly select 10 resistors from the batch, what is the probability that exactly 3 of them will be defective?

In this case, we can use the hypergeometric distribution to calculate the probability of having exactly 3 defective resistors in a sample of 10 resistors, without replacement.

Using the hypergeometric probability formula, we can calculate it as follows:

P(X = 3) = (C(20, 3) * C(200 – 20, 10 – 3)) / C(200, 10)

= (C(20, 3) * C(180, 7)) / C(200, 10)

= (1140 * 142506) / 19167973160

= 0.000084. (approximately)

Therefore, the probability of randomly selecting exactly 3 defective resistors from a sample of 10, assuming a hypergeometric distribution with 20 defective resistors in a batch of 200, is approximately 0.000084 (or 0.0084%).

### Geometric distribution

The geometric distribution is a discrete probability distribution that models the number of trials needed to achieve the first success in a sequence of independent Bernoulli trials, where each trial has a constant probability of success, denoted as p.

The formula for the geometric distribution is as follows:

P(X = k) = (1 – p)^(k-1) * p

Where:

P(X = k) is the probability of the first success occurring on the kth trial.

p is the probability of success in a single trial.

k is the number of trials needed to achieve the first success (k = 1, 2, 3, …)

Consider a communication system, the probability of successfully transmitting a data packet without errors is 0.9. What is the probability that the first error occurs on the 4th transmission attempt?

In this case, we can use the geometric distribution to calculate the probability of the first error occurring on the 4th transmission attempt, given a success probability of 0.9.

Using the geometric probability formula, we can calculate it as follows:

P(X = 4) = (1 – 0.9)^(4-1) * 0.9

= (0.1)^3 * 0.9

= 0.001 * 0.9

= 0.0009

Therefore, the probability of the first error occurring on the 4th transmission attempt, assuming a geometric distribution with a success probability of 0.9, is 0.0009 (or 0.09%).

### Negative Binomial distribution

The negative binomial distribution is a discrete probability distribution that models the number of trials needed to achieve a fixed number of successes in a sequence of independent Bernoulli trials. It is often used when we are interested in the number of failures that occur before achieving a certain number of successes.

The formula for the negative binomial distribution is as follows:

P(X = k) = (k + r – 1)C(k) * p^r * (1 – p)^(k)

Where:

P(X = k) is the probability of achieving the kth success on the (k + r)th trial.

(k + r – 1)C(k) represents the number of combinations of (k + r – 1) items taken k at a time.

p is the probability of success in a single trial.

r is the number of successes needed.

Consider a reliability test for a power supply unit, the probability of failure in a given hour is 0.01. What is the likelihood that the power supply unit fails precisely on the 10th hour of operation if it requires at least 3 failures to deem it unreliable?

In this case, we can use the negative binomial distribution to calculate the probability of the 10th failure occurring precisely on the 10th hour of operation, given a failure probability of 0.01 and a requirement of at least 3 failures.

Using the negative binomial probability formula, we can calculate it as follows:

P(X = 10) = (10 + 3 – 1)C(10) * 0.01^3 * (1 – 0.01)^(10)

= (12C10) * 0.000001 * 0.99045817185

= (66) * 0.000001 * 0.99045817185

= 0.0000653234

Therefore, the probability of the power supply unit failing precisely on the 10th hour of operation, assuming a negative binomial distribution with a failure probability of 0.01 and a requirement of at least 3 failures, is approximately 0.0000653234 (or 0.00653234%).

## Continuous Probability Distributions

Continuous probability distributions are mathematical models used to describe the probability of a random variable taking on a range of values within a continuous interval.

Unlike discrete probability distributions, which deal with variables that can only take on specific values, continuous probability distributions deal with variables that can take on any value within a given range.

In simple terms, continuous probability distributions describe the likelihood of a variable falling within a particular range or interval. Smooth curves, such as the standard or exponential distribution, often represent them.

The main difference between continuous and discrete probability distributions lies in the variables they model. Continuous probability distributions are used when dealing with variables with infinite possible values within a given range, such as time, distance, or temperature.

On the other hand, discrete probability distributions are used when dealing with variables that can only take on a finite or countable set of values, such as the number of defects in a product or the number of customers in a queue.

### Uniform Distribution

The uniform distribution is a continuous probability distribution where all outcomes in a given interval are equally likely.

Formula: The probability density function (PDF) of the uniform distribution is defined as:

f(x) = 1 / (b – a), for a ≤ x ≤ b

where a and b are the lower and upper bounds of the interval, respectively.

Consider the voltage output of a power supply unit is known to follow a uniform distribution between 110V and 120V. What is the probability that the voltage output will be between 115V and 117V?

In this case, the lower bound (a) is 115V, and the upper bound (b) is 117V. Since the uniform distribution has a constant probability density over the interval, the probability of the voltage output falling within this range is:

P(115 ≤ X ≤ 117) = 1 / (b – a) = 1 / (117 – 115) = 1 / 2 = 0.5 = 50%

Therefore, the probability that the voltage output will be between 115V and 117V, assuming a uniform distribution between 110V and 120V, is 50%.

### Normal Distribution

The normal distribution, also known as the Gaussian distribution, is a continuous, symmetric, bell-shaped probability distribution. It is widely used to model various natural phenomena.

The probability density function (PDF) of the normal distribution is defined as:

f(x) = (1 / (σ * √(2π))) * e^(-((x – μ)^2) / (2σ^2))

where μ is mean, and σ is the standard deviation of the distribution.

Consider the lifetimes of electronic components produced by a particular manufacturer are normally distributed with a mean of 500 hours and a standard deviation of 20 hours. What is the probability that an element will have a lifetime between 480 and 520 hours?

In this case, we need to calculate the probability of a component’s lifetime falling within 480 to 520 hours, assuming a normal distribution with a mean (μ) of 500 hours and a standard deviation (σ) of 20 hours.

To solve this problem, we can utilize the properties of the standard normal distribution and standardize the values using z-scores. We can find the corresponding probabilities using the standard normal distribution table.

P(480 ≤ X ≤ 520) = P((480 – 500) / 20 ≤ Z ≤ (520 – 500) / 20)

= P(-1 ≤ Z ≤ 1)

Looking up the standard normal distribution table, the probability corresponding to Z = -1 is 0.1587, and the probability corresponding to Z = 1 is 0.8413.

Therefore, the probability that a component will have a lifetime between 480 and 520 hours, assuming a normal distribution with a mean of 500 hours and a standard deviation of 20 hours, is:

P(480 ≤ X ≤ 520) = 0.8413 – 0.1587 = 0.6826 = 68.26%

### Exponential Distribution

The exponential distribution is a continuous probability distribution that models the time between events in a Poisson process, where events occur continuously and independently at a constant average rate.

The probability density function (PDF) of the exponential distribution is defined as

f(x) = λ * e^(-λx), for x ≥ 0

where λ is the rate parameter, representing the average number of events per unit of time.

Consider the time between consecutive power outages in a particular area follows an exponential distribution with an average rate of 0.1 outages per day. What is the probability that the time between two consecutive power outages is at least 8 days?

In this case, we need to calculate the probability of the time between consecutive power outages being at least 8 days, assuming an exponential distribution with a rate (λ) of 0.1 outages per day.

Using the exponential distribution formula, we can calculate it as follows:

P(X ≥ 8) = ∫[8, ∞] λ * e^(-λx) dx

Integrating the above expression from 8 to infinity, we can calculate the probability using calculus techniques or consult exponential distribution tables.

P(X ≥ 8) = e^(-λx) | [8, ∞]

= e^(-0.1 * 8)

= e^(-0.8)

≈ 0.4493

Therefore, the probability that the time between two consecutive power outages is at least 8 days, assuming an exponential distribution with an average rate of 0.1 outages per day, is approximately 0.4493 (or 44.93%).

### Gamma distribution

Definition: The gamma distribution is a continuous probability distribution often used to model the time until an event occurs. It is a versatile distribution that can take on different shapes depending on its parameters.

The probability density function (PDF) of the gamma distribution is defined as:

f(x) = (1 / (Γ(α) * β^α)) * x^(α-1) * e^(-x/β)

where x ≥ 0, α > 0, β > 0, and Γ(α) is the gamma function.

Consider the time until a power transformer fails follows a gamma distribution with shape parameter α = 3 and rate parameter β = 0.2. What is the probability that the transformer fails within the first 10 hours?

The gamma distribution has a probability density function (PDF) given by:

f(x) = (1 / (Γ(α) * β^α)) * x^(α-1) * e^(-x/β)

In this case, the shape parameter α = 3 and the rate parameter β = 0.2.

To find the probability that the transformer fails within the first 10 hours, we need to calculate the cumulative distribution function (CDF) of the gamma distribution up to 10 hours.

The CDF is given by the integral of the PDF from 0 to the desired time (in this case, 10 hours):

CDF(x) = ∫[0, x] f(t) dt

Using the provided PDF formula, we can calculate the CDF as follows:

CDF(x) = ∫[0, x] (1 / (Γ(3) * 0.2^3)) * t^(3-1) * e^(-t/0.2) dt

To solve this integral, we can use numerical integration techniques. At the same time, the value of Γ(3) is 2. The gamma function is defined as the extension of the factorial function to real and complex numbers (excluding negative integers). Specifically, Γ(n) = (n-1)! where “!” denotes the factorial of a non-negative integer. In this case, Γ(3) = 2, as 2 is the factorial of (3-1).

Calculating the integral from 0 to 10 hours, we find:

CDF(10) ≈ 0.9917

Therefore, the probability that the transformer fails within the first 10 hours, assuming a gamma distribution with shape parameter α = 3 and rate parameter β = 0.2, is approximately 0.9917 (or 99.17%).

This probability calculation helps assess electrical components’ reliability and maintenance scheduling, enabling engineers to make informed decisions regarding maintenance planning, system design, and risk analysis.

### Weibull distribution

The Weibull distribution is a continuous probability distribution commonly used to model the time to failure or the lifetime of objects. It can describe many failure patterns, including early failures, constant failure rates, and wear-out failures.

The probability density function (PDF) of the Weibull distribution is defined as:

f(x) = (β/λ) * (x/λ)^(β-1) * e^(-(x/λ)^β)

where x ≥ 0, β > 0, and λ > 0.

Consider the time to failure of a particular electrical component follows a Weibull distribution with shape parameter β = 2.5 and scale parameter λ = 1000 hours. What is the probability that the component fails before 1500 hours?

The Weibull distribution has a probability density function (PDF) given by:

f(x) = (β/λ) * (x/λ)^(β-1) * e^(-(x/λ)^β)

In this case, the shape parameter β = 2.5 and the scale parameter λ = 1000 hours.

To find the probability that the component fails before 1500 hours, we need to calculate the cumulative distribution function (CDF) of the Weibull distribution up to 1500 hours.

The CDF is given by the integral of the PDF from 0 to the desired time (in this case, 1500 hours):

CDF(x) = ∫[0, x] f(t) dt

Using the provided PDF formula, we can calculate the CDF as follows:

CDF(x) = ∫[0, x] (2.5/1000) * (t/1000)^(2.5-1) * e^(-(t/1000)^2.5) dt

To solve this integral, we can use numerical integration techniques.

Calculating the integral from 0 to x=1500 hours, we find:

CDF(1500) ≈ 0.8631

Therefore, the probability that the component fails before 1500 hours, assuming a Weibull distribution with shape parameter β = 2.5 and scale parameter λ = 1000 hours, is approximately 0.8631 (or 86.31%).

## Joint Probability Distributions

Joint Probability Distributions refer t probability distributions that involve multiple random variables simultaneously. They provide a way to model and analyze the collective behavior of multiple variables in a system or experiment.

By examining the joint probabilities of these variables, engineers can gain insights into the relationships and dependencies between different electrical parameters, facilitating system design, optimization, and decision-making processes.

### Joint Probability Density Function

A joint probability density function (PDF) is a function that describes the probability distribution of multiple random variables. It provides a mathematical representation of the likelihood that the variables take specific values simultaneously.

Consider a circuit with two resistors, R1 and R2, connected in series. The resistance values are normally distributed with mean values of 10 ohms and 15 ohms and standard deviations of 1 ohm and 2 ohms, respectively. The probability that the circuit’s total resistance, R_total, is less than 25 ohms.

To solve this problem, we need to find the joint PDF of R1 and R2 and then determine the probability that R_total is less than 25 ohms using the joint PDF.

#### Step 1: Define the joint PDF

Let’s assume that R1 and R2 are independent random variables. The joint PDF of R1 and R2, denoted as f(R1, R2), can be defined as the product of their individual PDFs, f1(R1) and f2(R2), respectively.

f(R1, R2) = f1(R1) * f2(R2)

Given that R1 follows a normal distribution with mean μ1 = 10 ohms and standard deviation σ1 = 1 ohm, its PDF f1(R1) can be expressed as:

f1(R1) = (1 / (σ1 * sqrt(2π))) * exp(-(R1 – μ1)^2 / (2σ1^2))

Similarly, for R2, with mean μ2 = 15 ohms and standard deviation σ2 = 2 ohms, its PDF f2(R2) can be expressed as:

f2(R2) = (1 / (σ2 * sqrt(2π))) * exp(-(R2 – μ2)^2 / (2σ2^2))

#### Step 2: Calculate the probability

We need to integrate the joint PDF over the desired range to find the probability that R_total is less than 25 ohms.

P(R_total < 25) = ∬[f(R1, R2) dR1 dR2]

Since R1 and R2 are independent, the joint PDF can be written as the product of their individual PDFs:

P(R_total < 25) = ∬[f1(R1) * f2(R2) dR1 dR2]

Now, let’s substitute the f1(R1) and f2(R2) expressions into the integral and solve it.

P(R_total < 25) = ∫[∫[(1 / (σ1 * sqrt(2π))) * exp(-(R1 – μ1)^2 / (2σ1^2)) * (1 / (σ2 * sqrt(2π))) * exp(-(R2 – μ2)^2 / (2σ2^2)) dR1] dR2]

Simplifying the expression, we get:

P(R_total < 25) = ∫[∫[(1 / (σ1 * σ2 * 2π)) * exp(-(R1 – μ1)^2 / (2σ1^2)) * exp(-(R2 – μ2)^2 / (2σ2^2)) dR1] dR2]

Now, we can evaluate this double integral numerically using the appropriate numerical integration method.

The final result will give us the probability that the circuit’s total resistance, R_total, is less than 25 ohms based on the given joint PDF and the specified resistance distributions.

*Please note that I assumed that R1 and R2 are independent and normally distributed random variables in this example.

### Marginal and Conditional Probability Distributions

#### Marginal Probability Distribution

The marginal probability distribution refers to the probability distribution of a single random variable in a joint probability distribution. It is obtained by summing or integrating the joint probability distribution over all possible values of the other variables.

#### Conditional Probability Distribution

The conditional probability distribution describes the probability distribution of one random variable, given that another random variable has a specific value. It is obtained by fixing one variable’s value and calculating the other variable’s probabilities.

Let’s consider an example related to electrical engineering where a circuit consists of two components, Component A and Component B. The lifetimes of these components, denoted by X and Y, respectively, are random variables. The joint probability density function (PDF) of X and Y is given by:

f(X, Y) = 3e^(-3X) * 2e^(-2Y) for X > 0 and Y > 0

Let’s find the marginal probability distribution of Component A’s lifetime and the conditional probability distribution of Component B’s lifetime, given that Component A’s lifetime is 1.

Marginal Probability Distribution of Component A’s Lifetime (X):

To find the marginal probability distribution of Component A’s lifetime, we integrate the joint PDF over all possible values of Component B’s lifetime (Y).

P(X = x) = ∫[f(X, Y) dy] from 0 to ∞

P(X = x) = ∫[3e^(-3x) * 2e^(-2y) dy] from 0 to ∞

P(X = x) = 3e^(-3x) * ∫[2e^(-2y) dy] from 0 to ∞

P(X = x) = 3e^(-3x) * (-e^(-2y)) / 2 from 0 to ∞

P(X = x) = 3e^(-3x) * (0 – (-1/2)) = 3/2 * e^(-3x) for x > 0

Therefore, the marginal probability distribution of Component A’s lifetime (X) is:

P(X = x) = (3/2) * e^(-3x) for x > 0

Conditional Probability Distribution of Component B’s Lifetime (Y) given X = 1:

To find the conditional probability distribution of Component B’s lifetime, given that Component A’s lifetime is 1, we use the joint PDF and normalize it by dividing it by the marginal probability of X = 1.

P(Y = y | X = 1) = f(X = 1, Y = y) / P(X = 1)

P(Y = y | X = 1) = (3e^(-31) * 2e^(-2y)) / ((3/2) * e^(-31))

P(Y = y | X = 1) = 2e^(-2y)

Therefore, the conditional probability distribution of Component B’s lifetime (Y) given that Component A’s lifetime is 1 is:

P(Y = y | X = 1) = 2e^(-2y) for y > 0

This means that given Component A’s lifetime is 1, the lifetime of Component B follows an exponential distribution compounded with a rate parameter of 2.

### Conclusion

As you have seen throughout the blog, Probability distributions are vital in electrical engineering and the importance of probability distributions in the FE Electrical exam. They help us understand and predict system behavior by considering uncertainties and random variables.

Engineers can design reliable systems, optimize performance, and make informed decisions using probability distributions. Whether assessing component lifetimes or analyzing signal variability, probability distributions provide the insights to tackle real-world challenges confidently.

For more detailed and dedicated assistance concerning any aspect of the FE exam, consult Study for FE – Your go-to place for FE Electrical exam preparation. Licensed Professional Engineer in Texas (PE), Florida (PE) and Ontario (P. Eng) with consulting experience in design, commissioning and plant engineering for clients in Energy, Mining and Infrastructure.