Q. 294.5( 2 Votes )

# 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 :

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 E_{1} be the event that the item was produced by machine A and E_{2} 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 E_{3} 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(E_{3}) = P(E_{1})P(E_{3}|E_{1}) + P(E_{2})P(E_{3}|E_{2})

Thus, the probability of the item being defective is.

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