Straight Dope Message Board > Main Probability question about aircraft rivets
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#1
10-15-2009, 01:12 PM
 statsman1982 Guest Join Date: Feb 2009

A student emailed me a question on homework this morning, and after working out the solution for myself and checking the answer in the back of the book and finding it different from mine, I am worried that I am missing something. I hope someone can confirm my suspicion that the book's answer is wrong. Here goes:

Quote:
 An aircraft seam requires 25 rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of one another, each with the same probability p. a). If 20% of all seams need reworking, what is the probability that a rivet is defective?
My solution:

P(seam needs rework) = P(>= 1 rivet is defective) = 1 - P(all rivets are fine) = 1 - (1 - p)^25.

Then set
1 - (1 - p)^25 = 0.2 and solve for p. One obtains p = 0.00889.

I should note that Googling this problem results in solutions that lead to my answer, and that the problem is from a well-known textbook for engineering students that is now in its 7th edition. I can't find any errata for the textbook, so I have no idea if the solution in the back of the book is just in error.
#2
10-15-2009, 01:25 PM
 Baracus Guest Join Date: Dec 2006
It appears to me that you did it correctly. The answer the book gives seems to be if 20% of the seams did NOT need reworking.

(1-0.06235)^25 = 0.20 = probability that all the rivets are fine if p=0.06235
#3
10-15-2009, 01:25 PM
 MikeS Charter Member Join Date: Oct 2001 Location: Williamstown, MA Posts: 3,051
If you plug p = 0.06235 into your formula (which I think is correct), you get a result of 80% of seams needing reworking rather than 20%. I'm going to go for a simple error on the publisher's or author's part.

ETA: <Maxwell Smart> Missed it by that much! </Maxwell Smart>

Last edited by MikeS; 10-15-2009 at 01:26 PM.

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