Question

A factory has two machines A and B. Past records show that the machine A produced 60% of the items of output and machine B produce 40% of the items. Further 2% of the items produced by machine A were defective and 1% produced by machine B were defective. If an item is drawn at random, what is the probability that it is defective?

Answer 1

Given:

Machine A produced 60% of the total items.

Machine B produced 40% of the total items.

2% of the items produced by machine A were defective.

1% of the items produced by machine B were defective.

There are two mutually exclusive ways to draw a defective item produced by one of the two machines – 

a. Item was produced by machine A, and then, the item is defective

b. Item was produced by machine B, and then, the item is defective

Let E₁ be the event that the item was produced by machine A and E2 be the event that the item was produced by machine B.

As 60% of the total items were produced by machine A, we have

Similarly, as 40% of the total items were produced by machine B, we have

Let E₃ denote the event that the item is defective.

Hence, we have

We also have

Using the theorem of total probability, we get

P(E₃) = P(E₁)P(E₃|E₁) + P(E₂)P(E₃|E₂)

Thus, the probability of the item being defective is \(\cfrac2{125}.\)