Q.(a) Three boxes $B_1$, $B_2$, $B_3$ contain Lamp bulbs some of which are defective. The defective proportions in box $B_1$, box $B_2$ and box $B_3$ are respectively $\dfrac{1}{2}$, $\dfrac{1}{8}$ and $\dfrac{3}{4}$. A box is selected at random and a bulb drawn from it. If the selected bulb is found to be defective, what is the probability that the selected bulb is from the box $B_1$ ?

(a) Bayes' theorem gives $P(B_1\mid D)=\dfrac{4}{11}$. (b) Induction: true for $n=1$; assume for $n=k$; then it holds for $n=k+1$.

Part (a) — Bayes' theorem

Given: $P(B_1)=P(B_2)=P(B_3)=\dfrac13$; $P(D\mid B_1)=\dfrac12,\ P(D\mid B_2)=\dfrac18,\ P(D\mid B_3)=\dfrac34$.

Step 1 — total probability of drawing a defective bulb.

$P(D)=\frac13\cdot\frac12+\frac13\cdot\frac18+\frac13\cdot\frac34=\frac13\left(\frac{4+1+6}{8}\right)=\frac13\cdot\frac{11}{8}=\frac{11}{24}.$

Step 2 — apply Bayes' theorem.

$P(B_1\mid D)=\frac{P(B_1)P(D\mid B_1)}{P(D)}=\frac{\frac13\cdot\frac12}{\frac{11}{24}}=\frac{\frac16}{\frac{11}{24}}=\frac16\times\frac{24}{11}=\frac{4}{11}.$

So $P(B_1\mid D)=\dfrac{4}{11}\approx0.364$.

Part (b) — Mathematical induction

Claim: $P(n):\ 1+2+3+\cdots+n=\dfrac{n(n+1)}{2}$ for all $n\in\mathbb N$.

Step 1 — base case $n=1$. LHS $=1$; RHS $=\dfrac{1(2)}{2}=1$. So $P(1)$ is true.

Step 2 — inductive hypothesis. Assume $P(k)$ is true:

$1+2+\cdots+k=\frac{k(k+1)}{2}.$

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