Question

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

Answer 1

Approximately
`37\%`.

Explanation:

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

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

Quantities given in this question are:

• 
  `P(Y = 1\; | \; X = A)`,
  `P(Y = 1\; | \; X = B)`, and
  `P(Y = 1\; | \; X = C)`, as well as
• 
  `P(X = A)`,
  `P(X = B)`, and
  `P(X = C)`.

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

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

Find the value of
`P(Y = 1)` (probability that a random item from this plant is defective) by marginalizing over all possible values of
`X`. 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.

`\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 * 0.40 \\ &+ 0.015 * 0.35 \\ &+ 0.02 * 0.25 \\ =\; & 0.01425\end{aligned}`.

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

`\begin{aligned} & P(X = B\; | \; Y = 1) \\ =\; & (P(Y = 1\; |\; X = B)\, P(X = B))/(P(Y = 1)) \\=\; & (0.015 * 0.35)/(0.01425) \\\approx\; & 0.37 = 37\%\end{aligned}`.