A contractor buys white paint from two different suppliers. Sixty percent of the

Question

A contractor buys white paint from two different suppliers. Sixty percent of the paint comes from supplier 1. The rest supplier 2.
The paint is supplied in unmarked five gallon cans. When the contractor receives a shipment of paint, no record is kept of its origin. Usually, a considerable amount of time goes by before the paint is used, so that no one can remember for sure where it came from, even if they try, Sometimes, supplier 1’s cans don’t contain five full gallons. One half of a percent of supplier 1’s cans are not completely full.
Supplier 2 has the same problem but with a different percentage. If a can came from Supplier 2, it is not completely full with probability 0.0075 ( three quarters of a percent of Supplier 2’s cans.
A. A worker is about to open a can of paint. What is the probability that the can will not be completely full AND that it came from supplier 1?
B. Calculate the probability that the worker will discover that the can of paint is not completely full
C. Given that the can of paint was not completely full when opened, what is the probability that it came from supplier 1?

Explanation / Answer

P(1 & 'not full') = P(1)*P('not full'|1) = 0.60*0.005 = 0.003 or 0.3% P(1 & 'not full') + P(2 & 'not full') = P(1)*P('not full'|1) + P(2)*P('not full'|2) = 0.60*0.005 + 0.40*0.0075 = 0.003 + 0.003 = 0.006 or 0.6% C . Since the probability P(1 & 'not full') is equal to P(2 & 'not full') (as seen in answer B.) the probability that it came from either supplier is equal to 50% The notation would be: P(1 & 'not full') / [P(1 & 'not full') + P(2 & 'not full')] = 0.003 / [0.003 + 0.003] = 1/2 or 50% P(1 & 'not full') / [P(1 & 'not full') + P(2 & 'not full')] = 0.003 / [0.003 + 0.003] = 1/2 or 50% P(1 & 'not full') / [P(1 & 'not full') + P(2 & 'not full')] = 0.003 / [0.003 + 0.003] = 1/2 or 50%

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