Factories A and B produce all of the computers sold at Cheapo Discount Stores. F

Question

Factories A and B produce all of the computers sold at Cheapo Discount Stores. Factory A produces 4 times as many computers as factory B. The probability that a computer produced by factory A is defective is 0.4 and the probability that one produced by factory B is defective is 0.3. If a Cheapo computer is selected at random and it is found to be defective, what is the probability it came from factory B? If a Cheapo computer is selected at random and it is found not to be defective, what is the probability it came from factory A?

Explanation / Answer

Let E be any event and probability of event E occuring be denoted by P(E) .

Let A be the event that computer has been produced by factory A and B be the event that computer has been produced by factory B.

Since factory A produces 4 times as many computer as factory B thus P(A) is 4 times P(B) and we know P(A) + P(B) = 1 or 5 * P(B) = 1 therefore P(B) = 0.2 and P(A) = 0.8

Let D be the event that the computer chosen at random is defective.

Given to us P(D|A) = 0.4 and P(D|B) = 0.3

First question required us to calculate the conditional probability that the computer is produced by factory B given it is defective which is P(B|D)

Now P(B|D) = P(B and D) / P(D) = P(D|B) * P(B) / ( P(D|A) * P(A) + P(D|B) * P(B) )

since P(D) = ( P(D|A) * P(A) + P(D|B) * P(B) ) = 0.4 * 0.8 + 0.3 * 0.2 = 0.38 and P(B and D) = P(D|B) * P(B) = 0.3 * 0.2 = 0.06 (Using Bayes Theorem)

Therefore P(B|D) = 6/38 =  0.158 approximately

For the second part we need firstly define ND is the event that the selected computer is not a defective computer

So basically we need to calculate P(A| ND) = P(ND and A) / P(ND)

P(ND and A) = P(ND|A) * P(A) = ( 1 - P(D|A) ) * P(A) = 0.6 * 0.8 = 0.48

and P(ND) = P(ND|A) * P(A) + P(ND|B)*P(B) = ( 1 - P(D|A) ) * P(A) + ( 1 - P(D|B) ) * P(B) = 0.6 * 0.8 + 0.7 * 0.2 = 0.62

So the required probability is P(A| ND) = 0.48 / 0.62 = 0.7742 approximately

Please upvote the answer if found helpful and do comment for further doubts

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