Probability and Stats problem. Please help

Question

A large chain retailer purchases 45% of a certain kind of electronic device from Company A and the rest from Company B.  Suppose that the devices from Company A are defective 2% of the time, while the devices from Company B are defective 4% of the time.  The retailer selects 10 devices at random from one of the two companies andfinds that 3 of the 10 are defective.  What is the probability that the shipment is from Caompany A?

 

The answer is .1057, but I have no clue how they got that answer.

Answer 1

We need conditional probability for this. Let X be the event that the devices are from company A, and let Y be the probability that exactly 3 out of 10 randomly selected devices are defective.

 

You are asked to find P(X|Y). Note that:

 

P(X|Y)P(Y)=P(Y|X)P(X)

 

So we compute all of the other probabilities first.

 

P(X)=45/100, because 45% of devices come from company A.

 

P(Y|X)=(10!/3!7!)(2/100)^3(98/100)^7, the probability that exactly 3 out of 10 randomly chosen devices are defective given that they are from company A.

 

P(Y)=(45/100)(10!/3!7!)(2/100)^3(98/100)^7+(55/100)(10!/3!7!)(4/100)^3(96/100)^7

 

Now you can compute P(X|Y). Your answer is correct and here is the answer in more decimal places: 0.105667786.