FE Other Disciplines Probability and Statistics Practice Problems

Work 3 free FE Other Disciplines probability and statistics practice problems with complete solutions, governing equations and checks. Attempt the setup before reading the answer.

Free sample problem

A conduit plant fills orders from three extrusion lines. Line 1 supplies 50% of output at a 2% defect rate, Line 2 supplies 30% at a 4% defect rate, and Line 3 supplies 20% at a 5% defect rate. A unit drawn at random from finished stock is found to be defective. The probability that it was produced on Line 2 is most nearly:

(A)0.040
(B)0.300
(C)0.313
(D)0.375

Show worked solution

Answer: (D)

Line 2 accounts for 0.375 of all defective units, more than its 30% share of production because its defect rate runs above the plant average.

Total probability applies because the three lines partition the finished stock, so the plant-wide defect rate is the share-weighted sum of the line rates.

P(D) = 0.50(0.02) + 0.30(0.04) + 0.20(0.05) = 0.032

Bayes' theorem then rescales the 30% prior share of Line 2 by how readily that line generates the observed defect.

P(L_2 \mid D) = \frac{0.30(0.04)}{0.032} = \frac{0.012}{0.032} = 0.375

FE Reference Handbook — Engineering Probability and Statistics: Bayes' Theorem

Why the other choices appear

(A)Reports the conditional defect rate of Line 2 itself, $P(D \mid L_2) = 0.04$ , rather than the reverse conditional probability.
(B)Reports the prior output share $P(L_2) = 0.30$ , ignoring the defect evidence entirely.
(C)Uses the Line 1 (or Line 3) numerator by mistake, $0.010/0.032 = 0.313$ .

Two more free problem types

Try a different setup before reading the solution. These additional original questions are separate from the sold practice forms, so solving them does not reveal another question from a full practice form.

At least two defects in a five-unit sample

Each unit independently has a 0.10 probability of being defective. Five units are sampled. What is the probability that at least two are defective?

(A)0.01000
(B)0.08146
(C)0.32805
(D)0.40951

Show worked solution

Answer: (B)

The probability is 0.08146, or about 8.15%.

Independent identical defect events give a binomial count with n = 5 and p = 0.10.

X\sim\operatorname{Binomial}(5,0.10)

At least two is the complement of zero or one, so only two terms are needed.

P(X\ge2)=1-(0.9)^5-5(0.1)(0.9)^4=0.08146

Summing the four direct terms for k = 2, 3, 4, 5 independently gives 0.0729 + 0.0081 + 0.00045 + 0.00001 = 0.08146. At least is not the same as exactly.

FE Reference Handbook: Probability and Statistics — binomial distribution and complementary events.

Why the other choices appear

(A)This uses the chance of two particular units being defective, ignoring other combinations.
(C)This is the probability of exactly one defective unit.
(D)This is the probability of at least one defective unit.

A positive inspection and the base rate

Four percent of incoming parts are defective. An inspection flags 95% of defective parts and falsely flags 8% of nondefective parts. A randomly selected part is flagged. What is the probability that it is defective?

(A)0.0400
(B)0.3310
(C)0.8700
(D)0.9500

Show worked solution

Answer: (B)

About 33.10% of flagged parts are defective under the stated model.

A flag can come from a defective part or a false positive. Weight both branches by their population fractions.

P(F)=0.04(0.95)+0.96(0.08)=0.1148

Restrict the denominator to all flagged parts, not to all parts or all defective parts.

P(D\mid F)=\frac{0.04(0.95)}{0.1148}=0.331010

In a hypothetical 10,000 parts, expect 380 true flags and 768 false flags. The share 380/(380 + 768) reproduces the conditional probability. Sensitivity alone does not answer the inverse question.

FE Reference Handbook: Probability and Statistics — total probability and Bayes theorem.

Why the other choices appear

(A)This is the prior defect probability before inspection.
(C)This does not use the correctly weighted false-positive branch.
(D)This is sensitivity P(flag given defect), reversing the conditioning.

Download all 14 additional questions, diagrams and complete solutions (free PDF). Download the editable practice review sheet (CSV).

Original questions and diagrams by EngineerProof, September 30, 2026. Personal study and unchanged redistribution with attribution are permitted. The questions are independent practice, not official NCEES material. Report a correction.

The complete FE Other Disciplines practice exam contains 7 problems in this area and 110 problems overall, with an answer key, full solutions, and a score sheet by knowledge area.

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