6 A company buys \( p 0 \) of its electronic components from supplier \( A \) and the remaining \( \frac{(100-p) \% \text { fromsupplier } B \text {. The probability that a randomly chosen component supplied }}{\text { by } A \text { is faulty (s } 0.05 \text {. The probability that a randomly chosen component supplied by } B \text { is }} \) faulty is 0.03 . (i) Given that \( p=25 \), find the probability that a randomly chosen component is faulty. (ii) For a general value of \( p \), the probability that a randomly chosen component that is faulty was supplied by \( A \) is denoted by \( \mathrm{f}(p) \). Show that \( \mathrm{f}(p)=\frac{0.05 p}{0.02 p+3} \). Prove by differentiation that f is an increasing function for \( 0 \leq p \leq 100 \), and explain what this statement means in the context of the question.

We are given that the company buys p% of its components from supplier A and (100 – p)% from supplier B. The probability that a component from A is faulty is 0.05 and from B is faulty is 0.03. ────────────────────────────── Part (i): p = 25 1. The probability a component comes from A is 25/100 = 0.25 and from B is 75/100 = 0.75. 2. The overall probability that a component is faulty is found using the law of total probability:   P(faulty) = P(faulty | A) × P(A) + P(faulty | B) × P(B)         = 0.05 × 0.25 + 0.03 × 0.75         = 0.0125 + 0.0225         = 0.035 Thus, when p = 25, the probability that a randomly chosen component is faulty is 0.035. ────────────────────────────── Part (ii): General p and f(p) Let f(p) denote the probability that a randomly chosen faulty component was supplied by A. Step 1. Determine the overall probability that a component is faulty for a general p:   P(faulty) = 0.05 (p/100) + 0.03 ((100 – p)/100)         = [0.05p + 0.03(100 – p)]/100         = [0.05p + 3 – 0.03p]/100         = (3 + 0.02p)/100 Step 2. Find the probability that a component was supplied by A and is faulty:   P(A and faulty) = 0.05 (p/100) Step 3. Use Bayes’ Theorem to get f(p):   f(p) = P(A | faulty) = (P(A and faulty))/(P(faulty))       = (0.05 (p/100))/((3 + 0.02p)/100)       = (0.05p)/(3 + 0.02p) Thus, we have shown that   f(p) = (0.05p)/(0.02p + 3). ────────────────────────────── Prove by Differentiation That f is Increasing We now differentiate f(p) with respect to p. Let:   N(p) = 0.05p   (numerator)   D(p) = 3 + 0.02p   (denominator) Using the quotient rule:   f'(p) = [D(p) × N'(p) – N(p) × D'(p)] / [D(p)]² 1. Compute the derivatives:   N'(p) = 0.05   D'(p) = 0.02 2. Substitute into the quotient rule:   f'(p) = [(3 + 0.02p)(0.05) – (0.05p)(0.02)] / (3 + 0.02p)² 3. Simplify the numerator:   = [0.05 × 3 + 0.05 × 0.02p – 0.05p × 0.02] / (3 + 0.02p)²   Notice that 0.05 × 0.02p = 0.001p and 0.05p × 0.02 = 0.001p, so:   = (0.15 + 0.001p – 0.001p) / (3 + 0.02p)²   = 0.15 / (3 + 0.02p)² Since 0.15 > 0 and (3 + 0.02p)² > 0 for all p (with 0 ≤ p ≤ 100), we have:   f'(p) > 0 Thus, f(p) is an increasing function for 0 ≤ p ≤ 100. ────────────────────────────── Interpretation in Context The result that f(p) is increasing means that as the percentage p increases (i.e., as the company purchases a larger proportion of its components from supplier A), the probability that a faulty component comes from A also increases. In other words, if the company relies more on supplier A, then among the faulty components produced, a greater proportion will be attributed to supplier A.