Question

A manufacturing process has two assembly lines, A and B. Suppose that line A produces 60% of the product. and line B produces the rest. We are told that 5% of the products produced by line A are defective in some way, and 8% of the line B products are defective. it may be helpful to construct a tree diagram with first and second-generation branches to answer the following:

C) if the end product is defective, what is the probability that it was produced by line B

Answer 1

Solution:

We are given that:

A manufacturing process has two assembly lines, A and B.

Line A produces 60% of the product.Thus P(A) = 0.60

and line B produces the rest. That is: P(B) = 1 - 0.60 = 0.40

We are told that 5% of the products produced by line A are defective in some way and 8% of the line B products are defective.

Let D = Defective and ND = Non-defective then we have:

P(D|A) =5% = 0.05 and P(ND |A) = 1 - 0.05 = 0.95

and

P(D|B) =8% = 0.08 and P(ND |B) = 1 - 0.08 = 0.92

Thus we have following tree diagram:

We have to find:
P( Product is produced by line B given that it is defective)=........?

That is:

P( B | D) =........?

where

and

Thus

P( Product is produced by line B given that it is defective)=