Probability Fundamentals

FE Exam Weight: Probability and Statistics is 6–9 questions (~7% of the 110-question FE Other Disciplines exam). These items are usually formula-driven and fast points if you recognize the right rule — the formulas live in the Engineering Probability and Statistics pages of the NCEES FE Reference Handbook.

Sample Spaces and the Axioms

The sample space S is the set of all possible outcomes; an event A is a subset of S. For equally likely outcomes, P(A) = (favorable outcomes)/(total outcomes). Three axioms govern every probability:

1. 0 ≤ P(A) ≤ 1 for any event.
2. P(S) = 1 — something in the sample space must occur.
3. Complement: P(A′) = 1 − P(A). The complement rule is the single biggest time-saver: "at least one" problems are almost always faster as 1 − P(none).

Combining Events

Mutually exclusive means the events cannot both happen (A∩B = ∅), so you subtract nothing. Independent means one event's occurrence does not change the other's probability. These two ideas are distinct and the FE deliberately mixes them up in distractors.

Worked example — independent AND. If P(A) = 0.3 and P(B) = 0.4 and the events are independent, then P(A and B) = (0.3)(0.4) = 0.12. If instead they were mutually exclusive, P(A and B) would be 0, and P(A or B) would be 0.7 — showing why you must read which relationship the problem states.

Conditional Probability and Bayes' Theorem

The probability of A given B has occurred is P(A|B) = P(A∩B)/P(B). Bayes' theorem reverses the conditioning when you know P(B|A) but want P(A|B):

$P(A\mid B) = \frac{P(B\mid A)\,P(A)}{P(B)}, \qquad P(B) = P(B\mid A)P(A) + P(B\mid A')P(A')$

Worked example — without replacement. A box holds 5 red and 3 blue marbles. Draw two without replacement; find P(both red). The first draw is red with probability 5/8; given that, 4 reds remain of 7, so the second is 4/7. By the multiplication rule, P(both red) = (5/8)(4/7) = 20/56 = 5/14 ≈ 0.357. The key is that the second probability is conditional — the counts shrink after the first draw.

Worked example — Bayes (defect testing). A process yields 2% defective parts (P(D) = 0.02). A test flags a true defective 95% of the time (P(+|D) = 0.95) and falsely flags a good part 3% of the time (P(+|D′) = 0.03). If a part tests positive, how likely is it actually defective?

P(D|+) = (0.95)(0.02) / [(0.95)(0.02) + (0.03)(0.98)] = 0.019 / (0.019 + 0.0294) = 0.019 / 0.0484 ≈ 0.393.

Despite a 95%-accurate test, a positive part is only ~39% likely to be defective. This base-rate result surprises test-takers: because defectives are rare, the many false positives from the large good population dominate. The same logic underlies medical-screening and reliability-inspection FE questions.

Counting Methods

Many probability problems reduce to counting the favorable and total arrangements. The decisive question is does order matter?

Worked example — permutation. Choosing a President, Vice-President, and Treasurer from 10 candidates is ordered (the roles differ), so P(10,3) = 10!/7! = 10·9·8 = 720 ways.

Worked example — combination. Choosing a 3-person committee (no roles) from 10 people is unordered, so C(10,3) = 10!/(3!·7!) = (10·9·8)/(3·2·1) = 720/6 = 120 ways. The factor of 3! = 6 difference between 720 and 120 is exactly the number of orderings of any chosen trio — that ratio is the heart of the permutation-vs-combination distinction.

Worked example — multiplication principle. A 4-character password using digits 0–9 with repetition allowed has 10⁴ = 10,000 possibilities; without repetition it is the permutation 10·9·8·7 = 5,040. Deciding whether repetition is allowed is the trap that separates these two counts. When a probability question gives "how many ways," map it to one of these three tools before computing any ratio.

Random Variables and Expected Value

A random variable X assigns a number to each outcome. Its expected value (mean) is the probability-weighted average E[X] = Σ xᵢ·P(xᵢ) for a discrete variable. The variance is Var(X) = E[X²] − (E[X])², measuring spread about the mean. Two handy linearity rules the FE uses: E[aX + b] = a·E[X] + b, and Var(aX + b) = a²·Var(X) (adding a constant b shifts the mean but never changes the variance).

Worked example — expected value. A game pays $10 with probability 0.2, $2 with probability 0.5, and $0 with probability 0.3. The expected payout is E[X] = 10(0.2) + 2(0.5) + 0(0.3) = 2 + 1 + 0 = $3. If a ticket costs $4, the expected net is 3 − 4 = −$1, so the game is unfavorable. Expected-value reasoning like this also underlies decision and risk problems where you weight each outcome's cost or benefit by its probability — a recurring theme that ties probability to the Engineering Economics portion of the exam.