04-BS-16, May 2016, Q4 | Nivaar Exam Prep

Question 4 (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given data
Quantity	Value
Class size	60 (20 girls, 40 boys)
Girls who passed	16
Line output shares $P(A),P(B),P(C)$	0.30, 0.50, 0.20
Line defect rates	4%, 8%, 3%

Find. (a) The number of boys who passed. (b) $P(C\mid\text{Defective})$.

Approach. (a) use $P(E_1),P(E_2)$ independence to find the class-wide fail rate from the girls-only fail rate, then back out the boys' fail/pass counts. (b) apply Bayes' theorem with the law of total probability over the three lines.

(a) Boys who passed. $16$ of $20$ girls passed, so $4$ girls failed: $P(E_2\mid E_1)=4/20=0.20$. Independence of $E_1,E_2$ means $P(E_2\mid E_1)=P(E_2)$, so the class-wide fail rate is also $0.20$: $$P(E_2)=0.20 \;\Rightarrow\; \text{total failed} = 0.20\times 60 = 12$$ Of these 12 failures, 4 are girls, so boys who failed $=12-4=8$. Boys who passed: $$\boxed{40-8=32}$$
(b) $P(C\mid D)$ via Bayes' theorem. Let $D$ = "lamp is defective". By the law of total probability: $$P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)=0.04(0.30)+0.08(0.50)+0.03(0.20)=0.012+0.040+0.006=0.058$$ Bayes' theorem then gives: $$P(C\mid D)=\frac{P(D\mid C)P(C)}{P(D)}=\frac{0.006}{0.058}\approx\boxed{0.1034\ (10.34\%)}$$

Question 4 – results
Part	Result
a	32 boys passed (8 boys failed)
b	$P(C\mid D)\approx 0.1034$

Topic: Conditional independence and Bayes' theorem.

Key relations: Independence: $P(E_2\mid E_1)=P(E_2)$; law of total probability $P(D)=\sum_i P(D\mid X_i)P(X_i)$; Bayes' theorem $P(X_i\mid D)=P(D\mid X_i)P(X_i)/P(D)$.

Why this works: Independence lets a conditional rate observed on one subgroup (girls) be treated as the population-wide rate, converting a subgroup count into a class-wide count; Bayes' theorem inverts a "given the cause, probability of the effect" statement into "given the effect, probability of the cause" using the full partition of causes.

Common pitfall: Confusing $P(E_2\mid E_1)$ (fail rate among girls) with $P(E_1\mid E_2)$ (fraction of failures who are girls); forgetting to renormalize by the total $P(D)$ in Bayes' theorem rather than using the numerator alone.

Source: Rosen, Discrete Mathematics and Its Applications, 7th ed., Ch. 7 (Discrete Probability, Bayes' Theorem).

Question 4 of 12

Question 4 (10 marks)