# Question answer I don't understand!

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
Staff Emeritus
Gold Member
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.

oh i see

HallsofIvy