# Thread: very interesting problem!!!

1. ## very interesting problem!!!

Hello all, I hope some of you can help me with this problem...

There are 30 total components in the bag. 21 are good and 9 are bad. It takes 6 components to buid one assembly (we can build 5 total assemblies), and components will be pulled randomly out of bag. What is probability for each assembly to have all 6 good components, and therefore be a good assembly (good assebmly has to have all 6 good components).

I tried using (0.7)^6 formula, but that formula accounts for any 6 good components in a row, which is not correct. If we pulled 6 good components in a row, but ther are pulled 3rd to 9th we will
not have a good assembly. Having a good assembly will require pulling good components from 1 to 6, or 7 to 12 and so on.

I tried formula mentioned above, where there is 70% chance to pull a good component out of bag, but this formula works only for infinite sample where cases are indenpendent from eachother. In our case we have 30 components total.

Thanks

2. Originally Posted by Put
There are 30 total components in the bag. 21 are good and 9 are bad. It takes 6 components to buid one assembly (we can build 5 total assemblies), and components will be pulled randomly out of bag. What is probability for each assembly to have all 6 good components, and therefore be a good assembly (good assebmly has to have all 6 good components).

I tried using (0.7)^6 formula, but that formula accounts for any 6 good components in a row, which is not correct. If we pulled 6 good components in a row, but ther are pulled 3rd to 9th we will
not have a good assembly. Having a good assembly will require pulling good components from 1 to 6, or 7 to 12 and so on.

I tried formula mentioned above, where there is 70% chance to pull a good component out of bag, but this formula works only for infinite sample where cases are indenpendent from eachother. In our case we have 30 components total.

Thanks
You are right. Your formula only works if you were somehow taking the components "with replacement" (like some magic leprechaun replacing each part as you removed it. ).

We need to do this without replacement. This is where good old combinations come in. I assume that order doesn't matter.

The probability of having a good assembly will be the total number of ways of having a good assembly (Q1 below), divided by the total number of ways of making an assembly (Q2 below).

Q1: How many ways are there of making a good assembly? More specifically. how many ways are there to take 6 good parts? You need to take 6 good ones from the 21 good ones.
$\mathbf{C}^{21}_6 = \frac{21!}{6!\cdot 15!}=\ldots$

Q2: How many total ways are there to take 6 parts from all 30?
$\mathbf{C}^{30}_6 = \frac{30!}{6!\cdot 24!}=\ldots$

Can you finish it?

3. I calculated the results, but it looks like pretty high.

"12C6" = 924
"9C3" = 84

Q1 = "12C6 * 9C3" = 77616

Q2 = "21C9" = 293930

Q1/Q2 = .264 = 26.4 %

If I used (0.70)^6 formula we get 16.81%, so I'm not sure if 24.4% is right. We should not have more chance to make a good assembly for my case.

Let me know what you think....

4. Well if you were scratching your head at what I did, bravo. I don't know what drugs I was on. I'm sorry for being confusing.

I edited my above answer to be right. When I do out the calculation, I get 0.091 = 9.1%

Good job on being both analytical and skeptical!!!

5. This is a good answer, and thank you very much. Now it looks pretty simple....

I just caclulated a few more cases and it look like that as we increase sample size (bag size), the result will be exponentially aproaching (0.7)^6 = 11.765%, which is infinite sample size where every part is "replaced".

For example, if I take one easier case, where there are 6 total components, 2 bad, and we need 3 good ones to make a good assembly. I'll represent it as 6, 2, 3, and the result is 20%.

If I increase the sample size to 12, 4, 3 (12 total component, 4 bad, 3 to make a good assembly), the result is 25.45%.

if I keep going...

6, 2, 3, = 20%
12, 4, 3, = 25.45%
24, 8, 3, = 27.67%
48, 16, 3, = 28.68%

So the answer is aproaching (0.6667)^3=29,63%, which is case for infinite big sample size where picking components does not change probability any more.

Thanks again for your help.

6. No worries. It's good to see someone using their intuition.