Question

In a certain assembly plant, three machines B1, B2, and B3, make 35%, 40%, and 25%, respectively. It is known from past experience that 2%, 3%, and 4% of the products made by each machine, respectively, are defective. A finished product is randomly selected and found to be defective, what is the probability that it was made by machine B3? Enter your answer as a decimal and use 3 significant digits.

Answer 1

I suppose the part about

"B1, B2, and B3, make 35%, 40%, and 25%"

should be taken to mean $B_1$ produces 35% of some product, $B_2$ produces 40%, and $B_3$ produces 25%. Then

$\begin{cases}P(B_1)=0.35\\P(B_2)=0.4\\P(B_3)=0.25\end{cases}$

Each machine has some probability of making a defective product - denote this event by $D$. Then

$\begin{cases}P(D\mid B_1)=0.02\\P(D\mid B_2)=0.03\\P(D\mid B_3)=0.04\end{cases}$

We want to find the probability that a defective product was made by machine $B_3$, i.e. $P(B_3\mid D)$. By definition of conditional probability and the law of total probability, we have

$P(B_3\mid D)=\dfrac{P(B_3\cap D)}{P(D)}$

(def. of conditional probability)

$P(B_3\mid D)=\dfrac{P(D\mid B_3)P(B_3)}{P(D)}$

(def. of conditional probability)

$P(B_3\mid D)=\dfrac{P(D\mid B_3)P(B_3)}{P(D\cap B_1)+P(D\cap B_2)+P(D\cap B_3)}$

(law of total probability)

$P(B_3\mid D)=\dfrac{P(D\mid B_3)P(B_3)}{P(D\mid B_1)P(B_1)+P(D\mid B_2)P(B_2)+P(D\mid B_3)P(B_3)}$

(def. of conditional probability; this result is also known as Bayes' theorem)

So we have

$P(B_3\mid D)=\dfrac{0.04\cdot0.25}{0.02\cdot0.35+0.03\cdot0.04+0.04\cdot0.25}\approx\boxed{0.345}$