FE section 2 of 16 · free theory
Probability & Statistics
Probability and statistics is a modest slice of the FE Civil exam — typically 4–6 questions — but the questions are high-value. They test whether you can set up a probability model, read a distribution, and interpret data without getting tangled in arithmetic.
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Probability laws and counting
Most probability questions are won or lost at the setup: which events, which rule, and whether order matters. Sort that out before touching the calculator:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
| P(A ∪ B) | probability that A or B (or both) occurs |
| P(A ∩ B) | probability that both occur |
P(B|A) = P(A ∩ B)/P(A) if independent: P(A ∩ B) = P(A)·P(B)
P(not A) = 1 − P(A) (the complement — the exam's favourite shortcut)
nPk = n!/(n − k)! nCk = n!/(k!(n − k)!)
Permutations (nPk) when the order of selection matters; combinations (nCk) when it does not. "Choose a panel of 4 from 9" is a combination — dividing by 4! is the step most people forget.
Discrete distributions
The binomial distribution is the workhorse: a fixed number of independent trials, each with the same success probability. The Poisson covers rare events over time or space:
P(X = k) = nCk pk(1 − p)n−k (binomial)
| μ = np | binomial mean (expected number of successes) |
| σ² = np(1 − p) | binomial variance |
P(X = k) = λke−λ/k! (Poisson; mean = variance = λ)
"At most k" means sum P(0) through P(k); "at least 1" is fastest via the complement, 1 − P(0). Read the quantifier before you start.
The normal distribution
The normal distribution is the continuous counterpart — and the exam almost always routes it through the z-score and a standard normal table or calculator:
z = (x − μ)/σ
| z | number of standard deviations the value lies from the mean |
| μ, σ | population mean and standard deviation |
f(x) = 1/(σ√(2π)) e−(x−μ)²/(2σ²)
Empirical rule: ≈68% within ±1σ, ≈95% within ±2σ, ≈99.7% within ±3σ
A standard normal table gives P(Z < z), the area to the left. For P(X > x), compute 1 minus the table value. Try it hands-on with the free normal-distribution calculator.
Mean, variance, and standard deviation
Descriptive statistics questions hinge on one distinction: the sample formulas (divide by n − 1) versus the population formulas (divide by N):
x̄ = Σxi/n median = middle ordered value mode = most frequent value
s² = Σ(xi − x̄)²/(n − 1) (sample variance)
σ² = Σ(xi − μ)²/N (population variance)
standard deviation = √variance
The n − 1 in the sample variance (Bessel's correction) makes it an unbiased estimate. If the question says "sample", use n − 1; if it describes the whole population, use N.
Confidence intervals
A confidence interval says how precisely the sample mean estimates the true mean. The key insight for the exam is how the width behaves:
x̄ ± z* · s/√n
| z* | critical value: 1.645 for 90%, 1.96 for 95%, 2.576 for 99% |
| s/√n | standard error of the mean |
The margin of error shrinks with 1/√n, not 1/n. To halve the interval width you must quadruple the sample size.
Linear regression basics
Regression fits the line that minimises the squared errors. The exam tests interpretation far more often than the fitting arithmetic:
least squares: minimise Σ(yi − (mxi + b))²
R² = fraction of the variation in y explained by the model (0 ≤ R² ≤ 1)
correlation r: −1 ≤ r ≤ +1 (sign matches the slope)
R² measures how well the line fits the data — it never proves causation. That distinction is exactly the kind of conceptual question the exam likes to ask.
Worked example Binomial probability
Given: A precast plant's defect rate is 15%. In a random sample of 10 units, what is the probability that exactly 2 are defective?
Solution:
- Binomial with n = 10, p = 0.15, k = 2. First the combinations: 10C2 = 10!/(2!·8!) = 45.
- pk = (0.15)² = 0.0225. (1 − p)n−k = (0.85)8 = 0.27249.
- Multiply: 45 × 0.0225 × 0.27249 = 0.27590.
Answer: ≈ 0.276 (about a 28% chance).
Worked example Normal probability via z-score
Given: 28-day concrete strengths are normally distributed with μ = 30 MPa and σ = 3.5 MPa. What fraction of cylinders exceed 35 MPa?
Solution:
- z = (35 − 30)/3.5 = 5/3.5 = 1.4286.
- The standard normal table gives P(Z < 1.43) ≈ 0.9236.
- We want the upper tail: P(X > 35) = 1 − 0.9236 = 0.0764.
Answer: ≈ 0.076 (about 7.6% of cylinders).
Worked example 95% confidence interval
Given: 25 soil-moisture readings have a sample mean of 48.2 and a sample standard deviation of 6.0. Give the 95% confidence interval for the true mean.
Solution:
- Standard error: s/√n = 6.0/√25 = 6.0/5 = 1.2.
- 95% critical value z* = 1.96. Margin = 1.96 × 1.2 = 2.352.
- Interval: 48.2 ± 2.352 → (45.848, 50.552).
Answer: 45.8 to 50.6 (rounded).