Question

A company setup three manufacturing unit in Nagpur, all the manufacturing unit producing
the large batch of similar item but the manufacturing capacity is different of all the manufacturing
unit. Let manufacturing unit P1 produced 20 % item, manufacturing unit P2 produce 30 % and
manufacturing unit P3 produced 50 %. Suppose the item produced by the Manufacturing unit P1 is
2 % defective similarly the item produced by P2 is 3 % defective and item produced by p3 is 4%
defective. Suppose one item is randomly selected from the entire batch and it found to be defective.
What is the probability that this item is produced by the manufacturing unit P3?

Answer 1

Using Baye's Theorem:

	Manufacturing Unit P1	Manufacturing Unit P2	Manufacturing Unit P3
% Produced	20%	30%	50%
	2%	3%	4%
Probability of selected item comes from Manufacturing Unit P1 = P(P1)	0.2	Probability that defective comes from Manucaturing Unit P 1 = P(D|P1)	0.02
Probability of selected item comes from Manufacturing Unit P2 = P(P2)	0.3	Probability that defective comes from Manucaturing Unit P 2 = P(D|P2)	0.03
Probability of selected item comes from Manufacturing Unit P3 = P(P3)	0.5	Probability that defective comes from Manucaturing Unit P 3 = P(D|P3)	0.04

Given the randomly selected item is defected, the probability that this item is produced by the manufacturing unit P3 is,

P( P3|D) = P(P3)*P(D|P3) / [ P(P1)*P(D|P1) +P(P2)*P(D|P2) +P(P3)*P(D|P3)]

= 0.5*0.04 / [ .5*.04 + .3*.03 + .2*.02]

=0.02/ [ 0.004 + 0.009 + 0.02]

=.02/.033

=0.606