Question

A factory has two machines A and B. Past record shows that machine Aproduced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

Answer 1

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Answer:

Step-by-step explanation:

Let :

A = items produced by machine A

B =items produced by machine B

D =items is defective

P(A)=60%

=60/100

=0.6

P(B) = 40%

=0.4

P(D/A) =2%

=0.02

P(D/B)=1%

=0.01

Now we need to find the probability that item was produced by machine B,if it found to be defective

So P(B/D) =P(B)*P(B/D)÷P(B)*P(D/B)+P(A)*P(D/A)

= 0.4*0.01÷0.6*0.02+0.4*0.01

=0.4÷1.6

=4÷16

=1÷4

So the probability of the items was produced by machine B is 1/4