Math Is Fun Forum

lindah

Member

Registered: 2010-01-25

Posts: 121

Defective parts

http://imageshack.us/photo/my-images/688/68986489.png/

Hi guys,

Please ignore part a)
Sorry for the stream of questions, I'm completing a past paper that does not have answers - so trying the bets I can

With part b) may I confirm if my work is correct:

I've decided to take the complement (1 - sample of 3 having no defective items)

I calculated this as

If this is correct, what would be the method using binomial probability?

Thanks
Lin

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Defective parts

Hi lindah;

I am getting 61 / 125 using the binomial distribution.

Something you can do to make your questions a little easier is to
encase your addresses like this.

[url]http://imageshack.us/photo/my-images/688/68986489.png/[/url]

Then they will post as live links.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

lindah

Member

Registered: 2010-01-25

Posts: 121

Re: Defective parts

Hi bobbym,

Thanks for the tip, I've updated all my posts.

I initially did it per your answer, but it seems with probability questions I over think it.

So is the rationale behind this:
p=1 out of 5 chance of obtaining a defective part from the sample

For the 3 parts chosen, the possible combinations with at least one defective part is:

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Defective parts

Yes, it seems they are saying there are 5 defective in the 25.

Whoa! Hold on here! You should be able to see why my answer is wrong!

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

lindah

Member

Registered: 2010-01-25

Posts: 121

Re: Defective parts

Should the combination be from 5 defective parts rather than 3?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Defective parts

Actually your answer was correct and mine is wrong due to the fact that this is not a binomial distribution problem! When you pick a part to test, you do not replace it. Therefore this is a hypergeomtric distribution and your answer in post #1, although different than the way I would calculate it, is correct.

The easiest way to calculate your answer is

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

lindah

Member

Registered: 2010-01-25

Posts: 121

Re: Defective parts

bobbym,

Unless its too much work, could you possibly share with me how you would calculate it?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Defective parts

Hi lindah;

It is all there in post #6. Can you follow that?

Is it clear, because I think that is the easiest.

By the way, very good for having the right answer!

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

lindah

Member

Registered: 2010-01-25

Posts: 121

Re: Defective parts

Hi bobby,

Yes, that's very clear! I knew I'd learn a better way off you.
Thanks very much!!!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Defective parts

Your welcome. Sorry for the confusion but at first I thought it was a binomial problem too.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.