In a manufacturing plant, three machines A,B, and C produce 40 %, 35 %, and 25 %, respectively, of the total production. The company's quality control department determined that 1 % of the items produced by machine A, 1.5 % of the items produced by machine B, and 2 % of the items produced by machine C are defective. If an item is selected at random and found to be defective, what is the probability that it was produced by machine B

Let [tex]X[/tex] denote the machine that produced the item. Let [tex]Y[/tex] denote whether the item was defective or not. (Let [tex]Y = 1[/tex] if the item is defective and [tex]Y = 0[/tex] otherwise.)

The question is asking for the probability of the event [tex]X = B[/tex] (the selected item was produced by machine [tex]B[/tex]) given that [tex]Y = 1[/tex] (the selected item was defective.) This probability could be denoted as: [tex]P(X = B \; | \; Y = 1)[/tex].

Quantities given in this question are:

[tex]P(Y = 1\; | \; X = A)[/tex], [tex]P(Y = 1\; | \; X = B)[/tex], and [tex]P(Y = 1\; | \; X = C)[/tex], as well as
[tex]P(X = A)[/tex], [tex]P(X = B)[/tex], and [tex]P(X = C)[/tex].

Apply Bayes' Theorem to express [tex]P(X = B \; | \; Y = 1)[/tex] in terms of these quantities. By Bayes' Theorem:

[tex]\begin{aligned} & P(X = B\; | \; Y = 1) \\ =\; & \frac{P(X = B\; \cap Y = 1)}{P(Y = 1)}\\ =\; & \frac{P(Y = 1\; |\; X = B)\, P(X = B)}{P(Y = 1)}\end{aligned}[/tex].

Find the value of [tex]P(Y = 1)[/tex] (probability that a random item from this plant is defective) by marginalizing over all possible values of [tex]X[/tex]. This value is like a weighted average of the defective rate of each machine. The weights here would be the percentage of items produced by each machine.

[tex]\begin{aligned}& P(Y = 1) \\=\; & P(Y = 1\; | \; X=A)\, P(X = A) \\ &+ P(Y = 1\; |\; X = B)\, P(X = B) \\ & + P(Y = 1\; |\; X = C)\, P(X = C) \\ =\; & 0.01 \times 0.40 \\ &+ 0.015 \times 0.35 \\ &+ 0.02 \times 0.25 \\ =\; & 0.01425\end{aligned}[/tex].

Substitute in the given quantities [tex]P(Y = 1\; |\; X = B) = 0.015[/tex] and [tex]P(X = B) 0.35[/tex] into the expression from Bayes' Theorem:

[tex]\begin{aligned} & P(X = B\; | \; Y = 1) \\ =\; & \frac{P(Y = 1\; |\; X = B)\, P(X = B)}{P(Y = 1)} \\=\; & \frac{0.015 \times 0.35}{0.01425} \\\approx\; & 0.37 = 37\%\end{aligned}[/tex].