Question 11

Let $P(M_1) = 2/5$, $P(M_2) = 2/5$, $P(M_3) = 1/5$. Let $D$ be the event that a bulb is defective. Given $P(D) = 0.20$ and $P(D|M_1) = 0.15$. Also given $P(M_2|D) = 2/5$. By Bayes' Theorem: $P(M_2|D) = \frac{P(M_2)P(D|M_2)}{P(D)} \implies \frac{2}{5} = \frac{(2/5)P(D|M_2)}{0.20} \implies P(D|M_2) = 0.20$. By the law of total probability: $P(D) = P(M_1)P(D|M_1) + P(M_2)P(D|M_2) + P(M_3)P(D|M_3)$. $0.20 = (2/5)(0.15) + (2/5)(0.20) + (1/5)P(D|M_3)$. $0.20 = 0.06 + 0.08 + 0.2 P(D|M_3) \implies 0.06 = 0.2 P(D|M_3) \implies P(D|M_3) = 0.3$.