# Probability of defective machines question

[solved]probability question

## Homework Statement

Three machines A, B, and C produce 50%, 30% and 20% respectively of the total
number of items at a factory. Each produces a number of defective: 3%, 4% and 5%
respectively.
(a) If an item produced by one of these machines is selected, find the probability that
it is defective.
(b) Now suppose that a manufactured item is selected at random and is found to be
defective. Find the probability that this item was produced by machine A.

N/A

## The Attempt at a Solution

is part (a)
Pr = (3%+4%+5%)/(50%*30%*20%)?

and no idea on part b

thanks a lot!

Last edited:

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Use Bayes' theorem.

Let D be the event that a selected item is defective

Let A, B, and C be the events that that an item is produced by machine A, B, and C respectively

$$P(D) = P(D\cap A) + P(D \cap B) +P(D \cap C)$$

$$= P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)$$

$$P(A|D) = \frac {P(A \cap D)}{P(D)}$$

$$= \frac {P(D|A)P(A)}{P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)}$$

HallsofIvy
Homework Helper
How I would do it: Imagine that 1000 of the items are produced. Then 50% of 1000= 500 come from machine A, 30% of 1000= 300 from machine B, and 20% of 1000= 200 from machine C.

Of the 500 from machine A, 3% of 500= 15 are defective, 4% of 300= 12 are from machine B, and 2% of 200= 4 are from machine C.

A total of 15+ 12+ 4= 41 defective items are produced of a total of 1000 items. With that information (a) is easy.

Of a total of 41 defective items, 15 of them were from A. With that information, (b) is easy.

thank you guys!!