Question answer I don't understand

[Genericcoder]

An electronic fuse is produced by five production lines in a manufacturing operation.
The fuses are costly, are quite reliable, and are shipped to suppliers in 100-unit lots.
Because testing is destructive, most buyers of the fuses test only a small number of
fuses before deciding to accept or reject lots of incoming fuses.
All five production lines produce fuses at the same rate and normally produce
only 2% defective fuses, which are dispersed randomly in the output. Unfortunately,
production line 1 suffered mechanical difficulty and produced 5% defectives during
the month of March. This situation became known to the manufacturer after the fuses
had been shipped.Acustomer received a lot produced in March and tested three fuses.
One failed. What is the probability that the lot was produced on line 1? What is the
probability that the lot came from one of the four other lines?

Let B denote the event that a fuse was drawn from line 1 and let A denote the event
that a fuse was defective. Then it follows directly that

P(B) = 0.2 and P(A|B) = 3(.05)(.95)^2 = .135375.

Similarly,
P(B-) = 0.8 and P(A|B-) = 3(.02)(.98)2 = .057624.



P(A) = P(A|B)P(B) + P(A|B-)P(B-)
= (.135375)(.2) + (.057624)(.8) = .0731742.


P(B|A) = P(B & A) / P(A) = P(A|B)*P(B) / P(A) = (.135375)(.2) / .0731742 = 0.37

Wat I don't understand here how did he get those values for
P(A|B) and P(B-)..? shouldn't P(B | A) = P(A & B) / P(B) = 0.05/0.2

P(B) = 1/5 = 0.2. same logic for P(A|B-) I don't understand this if someone could explain this more clearly.

 

[Office_Shredder]

shouldn't P(B | A) = P(A & B) / P(B) = 0.05/0.2

It should be P(A) in the denominator, not P(B) like you have written. You may be confused because the formula will typically be presented as giving P(A | B), so your A and B are flipped from what the standard formula in a textbook would read.

 

[Genericcoder]

oh i see

 

[HallsofIvy]

Another way of looking at it: Imagine that every line produces 1000 items, for a total of 5000 items. Lines 2 through 4 have 2% bad: a total of .02(4000)= 80 bad items. Line one produces 5% bad, a total of 50 bad items. That is, out of a total of 80+ 50= 130 bad items, 50, or 50/130 or about 38% came from line 1.

 

[blue_raver22]

I would first clarify the notation used in this response. The event B denotes the probability of a fuse being drawn from line 1, and the event A denotes the probability of a defective fuse. The notation B- denotes the event of a fuse not being drawn from line 1, and P(B-) is the probability of this event occurring. Similarly, P(A|B-) represents the probability of a fuse being defective given that it was not drawn from line 1.

To calculate the values for P(A|B) and P(A|B-), we can use the information provided in the question. We know that normally, all five production lines produce fuses with a 2% defect rate. This means that the probability of a fuse being defective is 0.02 or 2%. However, due to mechanical difficulties, production line 1 had a higher defect rate of 5% in the month of March. This means that the probability of a fuse being defective from line 1 specifically is 0.05 or 5%.

To calculate P(A|B), we need to consider that for a lot of 100 fuses, 3 fuses are tested. The probability of all 3 fuses being non-defective is (0.95)^3, since the probability of a non-defective fuse is 0.95. However, we need to account for the fact that there are 3 possible ways for 1 fuse to be defective out of the 3 tested. This is why we multiply by 3. So the overall probability of a fuse being defective given that it was drawn from line 1 is 3(0.05)(0.95)^2 = 0.135375.

Similarly, to calculate P(A|B-), we use the same logic but for the other 4 production lines. The probability of a fuse being defective from one of these lines is 0.02, and the overall probability of a fuse being defective given that it was not drawn from line 1 is 3(0.02)(0.98)^2 = 0.057624.

I hope this explanation helps clarify the calculations used in the response. As scientists, it is important to carefully consider and understand the data and information provided in a question before making any calculations or assumptions.