# Why this conditional probability equals this?

The following problem on conditional probability is taken from $$\textit{Mathematical Statistics}$$ by W. Mendenhall. The problem is:

"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. A customer 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?"

If $$B$$ denotes the event that a fuse was drawn from line 1, and $$A$$ denotes the event that the fuse was defective, then in my reasoning I obtain that

$$P(A\lvert B)=0.05$$

but in the explanation given in this book is said that $$P(A\lvert B)=(3)(0.05)(0.95)^2.$$

I don't know why. Please any help or suggestion on this interpretation would be really appreciated.

You calculated the probability that a single fuse failed given that it was from line 1. However, the event $$A$$ is that one out of the three fuses tested failed. So you need to calculate the probability that when 3 fuses from line 1 are tested, one of them fails while the other two succeed. This gives $$P(A \mid B) = 3(0.05)(0.950)^2$$.