# A factory h

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

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

We also have

1% of the items produced by machine A are defective.

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.

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