23-Ind-B1 Reliability and Maintainability · May 2015
Question 2 of 10: Exploratory Oil Wells — Binomial, Geometric, and Large- n Binomial Probability
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2015 — 98-Ind-B1 Applied Probability & Statistics. Three-hour, closed-book exam; one of two permitted calculators (Sharp or Casio), one 8.5″×11.0″ aid sheet (both sides), statistical tables supplied. Format: three sections — Section A: do 2 of 4 questions (10 marks); Section B: do 2 of 3 (20 marks); Section C: do 1 of 3 (20 marks) — a 5-question, 50-mark paper as printed. All ten questions across the three sections are solved below for completeness.
Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — joint distributions (ch. 5), binomial/geometric probability (ch. 3), normal distribution and sampling distributions (ch. 4–7), point/interval estimation (ch. 8–9), hypothesis testing incl. sample-size design (ch. 9–10), simple linear regression and ANOVA (ch. 11–13), Bartlett's and Tukey's tests, single-degree-of-freedom contrasts and 23 factorial designs (ch. 13–14). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3).
Question 2 (Section A.2): Exploratory Oil Wells — Binomial, Geometric, and Large-n Binomial Probability (5 marks)
Given. Independent Bernoulli trials, $p=0.15$ success probability per well.
Find. (a) $P(X=1)$ for $n=10$; (b) $P(\text{first success on trial }10)$; (c) $P(X\gt 20)$ for $n=120$.
Approach. (a) is a binomial pmf, (b) a geometric pmf, and (c) a binomial tail probability, all with the same per-trial $p=0.15$.
(a) Exactly 1 success in 10 independent drills. $X\sim\mathrm{Binomial}(n=10,p=0.15)$. $$P(X=1)=\binom{10}{1}(0.15)^1(0.85)^9=\boxed{0.3474}$$
(b) First success exactly on the 10th trial. The number of trials to first success is geometric: $$P(\text{first success on trial }10)=(1-p)^{9}p=(0.85)^9(0.15)=\boxed{0.03474}$$
(c) More than 20 successes out of 120. $Y\sim\mathrm{Binomial}(n=120,p=0.15)$, mean $np=18$, $\sqrt{np(1-p)}=3.912$. Summing the exact binomial tail (Casio/Sharp cumulative-binomial function, or table interpolation) gives $$P(Y\gt 20)=1-P(Y\le 20)=\boxed{0.2557}$$ A normal approximation with continuity correction, $z=(20.5-18)/3.912=0.639$, gives $P(Z\gt 0.639)=0.261$ — consistent with the exact value, confirming the calculation. Investors therefore have roughly a 1-in-4 chance of a return under this prospectus, which is a meaningfully low-probability bet given the threshold (20 of 120, i.e. more than 3 percentage points above the assumed 15% base rate) sits well into the upper tail of the underlying binomial distribution.
