# Probability of electric fuses

I am having trouble finding an answer to the following question:

Problem:

An electronic fuse is produced by five production lines. The fuses are shipped to suppliers in 100-unit lots. All 5 production lines produce fuses at the same rate and normally produce only 2% defective fuses. However this month production line 1 produced fuses at an error rate of 5%.

If we consider event B = the supplier received the 100 fuse lot from production line 1

and event A = 1 out of 3 drawn fuses is broken (without replacement)

Then what is P(A|B)?

The book lists it as 3(.05)(.95)$$^2$$

However, I believe that calculation would only be correct if we are drawing our sample of 3 with replacement. Since without replacement we have:

There are 100*99*98 possible groups of 3 we can draw. And of those groups of 3 there are 3(5*95*94) possible ways to draw a group of 3 with exactly 1 broken fuse, assuming we are drawing from production line 1.

Then our probability should be: $$\frac{3(5*95*94)}{100*98*97}$$.

But this is not equal to the books probability and I'm wondering where I'm going wrong.

I think you are assuming that there are exactly 95 unbroken fuses in the lot of 100, whereas the book is assuming that each fuse independently has a 95% chance of being unbroken.

• ok thanks! so in practice if each fuse has a chance of .95 probability of being good, we would expect a sample of 100 to have AROUND 95 that are good and AROUND 5 that are bad. however, just like its possible to flip a coin 100 times and get heads on all 100 flips, the scenario of having 100 good fuses or 100 bad fuses is unlikely, but still possible. The wording in the textbook is " normally produce only 2% defective fuses". In general, does wording like this imply each fuse has a 98% chance of being good, not that 2% of the total population is defective? Aug 10 '19 at 20:19
• the above illustration with flipping the coin was meant to show that we won't necessarily have 95/100 good fuses when we assume we have a probability of .95, hence we can not translate our problem to the case of having 95 good fuses and 5 bad fuses. Sound correct? Aug 10 '19 at 20:22
• Yes, your understanding is correct. We would except 95 unbroken fuses in a lot of 100, if the probability of each one being broken, independently, was 0.05. “Normally produce 2% defective fuses” definitely sounds like a way to express in common English, that the random variable associated with a particular fuse being broken has the PMF: P(broken) = 0.02, P(working) = 0.98. Technically, that can mean that 2% of the total population is broken, since often the probability is meant to be the proportion in the total population, BUT that does not mean that 2% of any sample is broken.
– Joe
Aug 10 '19 at 21:15

Your answer based on there are 5 defective fuse among the 100-units lot. Error rate (0.05) has no effect on the answer.

Answer from book assumes that number of defective fuse among the 100-units lot is random, could be 0 to 100. But each fuse among the 100-units has 5% chance to be defective.

Based on the "Problem", your answer is wrong, because no one say that there are 5 defective fuse among the 100-units lot.

• Thanks! In the case where we are looking at a 100 fuses and know we have 5 defective ones, why is it that the seemingly equivalent case of 1 defective out of 20 total fuses yields a different probability when computed in the following ways: 100 fuses: (what i list above) and 20 fuses: $\frac{3*(1*19*18)}{20*19*18}$. is it due to the fact that after we make the first selection we are now in a scenario with 19 fuses and 99 fuses and we arrive at different probabilities for selecting a good or bad fuse on the second attempt when comparing the 19 remaining fuses and 99 fuse cases? Aug 11 '19 at 1:29
• i addressed your card scenario with coin flips in my comment to Joe's answer below. In my previous comment to you, I am asking about a different scenario of 100 fuses where it is known there are 5 bad fuses. Aug 11 '19 at 2:12
• no i do understand the difference, and i outlined it to "Joe" who then said "your understanding is correct". as an example i pointed out flipping a coin 100 times MAY leave 50 heads and 50 tails and is certainly more likely to give around 50 heads, but it is entirely possible for there to be 100 heads, though just very unlikely. the probability .5 guarantees nothing in a finite number of flips. I am asking you a different scenario where we KNOW there are 5 bad fuses out of 100. Aug 11 '19 at 2:25
• Think in this way: 1 of 3 is bad = (bad, good, good)+(good,bad,good)+(good,good,bad). For 5 of 100: P(bad, good, good) = (5/100)*(95/99)*(94/98). For 1 of 20: P(bad,good,good) = (1/20)*(19/19)*(18/18). Similarly for P(good,bad,good) and P(good,good,bad). Then you can find where the difference comes from. Aug 11 '19 at 2:53