Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination, 04-BS-16 Discrete Mathematics, undated sitting (May 2019). Closed book; approved Casio or Sharp calculator only. The exam instructs "answer 10 of the 12 questions"; every question is answered below as a complete study resource.
Source note: This paper is the May 2019 sitting (every page footer reads "04-BS-16/May 2019"). Two printed statements are defective as set and are flagged where they occur: Question 7(a) prints the last term of $\{1,5,9,\dots\}$ as $4n-1$ (the pattern and the stated sum require $4n-3$), and Question 8(b) prints "$n>2$" although $4^n>n^4$ fails at $n=3,4$. Question 12(c)'s parameters also make a connected graph impossible; this is noted at that part.
Reference texts: Rosen, Discrete Mathematics and Its Applications, 7th ed. (Pearson) — used throughout for logic, set theory, functions, combinatorics, probability, induction, asymptotic (Big-O) notation, and graph theory.
Find. (a) $P(\text{sum}=9)$ for two die rolls. (b) $P(\text{line}=C\mid\text{defective})$ by Bayes' theorem.
Approach. (a) count favourable outcomes out of the $36$ equally likely ordered pairs. (b) law of total probability for $P(D)$, then Bayes' theorem for the posterior.
a) $P(\text{sum}=9)$. Pairs $(d_1,d_2)$ with $d_1+d_2=9$: $(3,6),(4,5),(5,4),(6,3)$ — 4 outcomes out of 36. $\boxed{P(\text{sum}=9)=4/36=1/9\approx0.111}$.
b) $P(C\mid D)$ via Bayes' theorem. Total probability of a defective unit: $$P(D)=P(A)P(D\mid A)+P(B)P(D\mid B)+P(C)P(D\mid C)=(0.35)(0.02)+(0.50)(0.05)+(0.15)(0.01)$$ $$=0.0070+0.0250+0.0015=0.0335=67/2000.$$ Then $$P(C\mid D)=\frac{P(C)P(D\mid C)}{P(D)}=\frac{0.0015}{0.0335}=\frac{15}{335}=\frac{3}{67}.$$ $\boxed{P(C\mid D)=3/67\approx0.0448\ (4.48\%)}$.
Question 4 results
Part
Result
a
$1/9\approx0.111$
b
$P(D)=0.0335$; $P(C\mid D)=3/67\approx0.0448$
Line C's share of the defective units ($4.48\%$) is far below its $15\%$ share of total output, because its defect rate ($1\%$) is the lowest of the three. Most defective sets come from line B: $P(B\mid D)=0.0250/0.0335=50/67\approx74.6\%$, and line A accounts for the remaining $14/67\approx20.9\%$ — the three posteriors sum to 1 as they must. The given $1\%$ is $P(D\mid C)$, a different quantity from the requested $P(C\mid D)$.
