Q.(a) A company has three machines A, B, C which produces 25%, 30% and 50% of the product respectively. Their respective defective percentages are 5, 4 and 2. From these products one is chosen and inspected. If it is defective, what is the probability that it has been made by the machine B.

(a) By Bayes' theorem, $P(B\mid D) = \dfrac{8}{23} \approx 0.348$. (b) The circle through $(1,0),(0,1)$ with centre on $x+y=1$ is $x^2+y^2-x-y=0$.

Part (a): Bayes' theorem. Let $A, B, C$ be the machines and $D$ = item is defective.

$P(A)=0.25,\ P(B)=0.30,\ P(C)=0.50;\quad P(D\mid A)=0.05,\ P(D\mid B)=0.04,\ P(D\mid C)=0.02.$

Step 1 — Total probability of a defective item.

$P(D) = P(A)P(D\mid A)+P(B)P(D\mid B)+P(C)P(D\mid C).$

$= 0.25(0.05)+0.30(0.04)+0.50(0.02) = 0.0125+0.012+0.010 = 0.0345.$

Step 2 — Posterior probability (Bayes).

$P(B\mid D) = \frac{P(B)P(D\mid B)}{P(D)} = \frac{0.012}{0.0345} = \frac{120}{345} = \frac{8}{23} \approx 0.348.$

Part (b): Circle through $(1,0),(0,1)$ with centre on $x+y=1$. Let the circle be $x^2+y^2+2gx+2fy+c=0$, centre $(-g,-f)$.

Passes through $(1,0)$: $1 + 2g + c = 0$.

Passes through $(0,1)$: $1 + 2f + c = 0$.

Subtracting: $2g = 2f \Rightarrow g = f$. …