- #1

mr_coffee

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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:

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?

Heres my work for a, b, and c.

For part, b, i used 40-3, 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.

http://suprfile.com/src/1/40tbojq/lastscan.jpg

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.

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?

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 40-3, 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.

http://suprfile.com/src/1/40tbojq/lastscan.jpg

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.

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