
#1
Oct2706, 06:34 PM

P: 1,629

Hello everyone. I think I have part a of this problem right but part b must be wrong becuase when I did the probabily it was like 99.994% chance that at least 1 board would be defective.
Heres the problem: Suppose that three computer boards in a production run of forty are defective. A Sample of 5 is to be selected to be checked for defects. Part a also might be wrong though, becuase I used 40 possible boards might be defective...but in the beginning they said 3 out of the 40 already where found to be defective, so would I use 38 instead of 40? a. How many different samples can be chosen? b. How many samples will contain at least one defective board? c. What is the probability that a randomly chosen sample of 5 contains at least one defective board? Heres my work for a, b, and c. For part, b, i used 403, because at the beginning of the problem they said 3 out of the 40 boards were defective. So i thought, well you know 38 of the boards may or may not be defective, but u already used 3 out of the 40, so you only have 38 to choose from now. Hm..it seems they disabled embeded images? Just click the link to dispaly the image. Thanks! PS: I used the difference rule. I was modeling my problem after another problem the book did. There problem was the following: Suppose the group of 12 consist of 5 men and 7 women. b. How many 5 person teams contain at least one man? There solution to this was: Observe that the set of 5 person teams containing at least one man equals the set difference between the set of all 5 person teams and the set of 5 person teams that do not contain any men. now a team with no men consist entirely of 5 women chosen from the seven women in the group, so there are 7 choose 5 such teams. The total number of 5 person teams is 12 choose 5 = 792, so: [# of teams with at least 1 man] = [total number of teams of 5]  [# of teams of 5 that do not contain any men] = (12 choose 5)  (7 choose 5) = 792  7!/(5!2!) = 792 12 = 771. 



#2
Oct2706, 11:07 PM

P: 1,119

I think you are right, 99% seems way too high. Your part (a) is fine. You have the right idea for part (b) but you computed the number of samples that have no defects incorrectly. You said there are 403 = 37 good boards. So how many samples are there from these good boards? Once you have that you should get the right answer for (b) and then use your same idea for (c) and you should get the probability to be ~ .337




#3
Oct2806, 01:24 AM

P: 1,629

Aw3some thanks matt!
I got your answer. 37 choose 5 = 435897 222111/658008 = .3376 = 33.76% that sounds more like it! 


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