# math fancy



## Inside (Jan 31, 2010)

Hi, so I read someone replying to another thread about them not minding math so much. Well, I was wondering if someone wanted to be my math buddy so to speak. I've had all the way up to college algebra - completed with an A but I'm in statistics now and get really frustrated. I enjoy it once I get it but it takes days for me to get some of the stuff so, I was just wondering if someone who doesn't mind math would want to chat with me, maybe do a skype thing, don't have to show faces or anything like that... was just wondering... its an online class and most of the homeworks I can get copies of... sometimes it helps to look at the problem and the books has most of the info in powerpoint slides I could send.. :-/ :roll


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## marenubium87 (Jan 11, 2009)

Responded on the other thread. Feel free to skype me - same s/n.


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## Inside (Jan 31, 2010)

Never used Skype before and not sure how to get the questions to you.. is it okay if I send you a pm with my email address?


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## Inside (Jan 31, 2010)

*Probability question*

So I'm able to save the questions I take quizzes on and am able to retake the quizzes. Reason being is most of these questions will be on the exam which is a one time shot. 
This is one question I'm not able to get through with logic. I'll post it along with the answer. Hopefully I'm not breaking some SAS rule or thread rule...

*Find the indicated probability.

*A IRS auditor randomly selects 3 tax returns from 57 returns of which 6 contain errors. What is the probability that she selects none of those containing errors? 

Selected Answer:
0.7117 
That's the correct answer, just not sure how to get there.


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## coldmorning (Jul 4, 2007)

The images you're referencing are local files on your computer so we can't see them.

But for your example, the probability of getting a non-error on the:
1st file = 51 out of 57 = 89.47 % (divide 51 by 57 = .8947)
2nd file =50 out of 56 = 89.28 % (now you have 1 less good file)
3rd file = 49 out of 55 = 89.09 % (yet another good file gone from pile)

You multiply the 3 separate probabilities to get the chance of all 3 being non-error files which is 71.17%

The multiplication is with 1 being 100% so, you get:
.8947 X .8928 X .8909 = .7117 (which is 71.17%)


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## Inside (Jan 31, 2010)

Thats exactly how I have been working the other problems!! I can't believe I didn't try that process on this. Somehow I got caught up in the wording on taking 3 returns from 57...
Seeing it worked out like that makes complete sense. Thank you for taking the time to work through that, I have an entire semester to work through.

I went back and tried to edit the first post that had the unreadable images in it. Hopefully it doesn't try and display that any more.

Thanks again for answering... 
Before I close this thank you post, let me show you one I was using that process on successfully... just as food for thought 

A bin contains 71 light bulbs of which 9 are defective. If 6 light bulbs are randomly selected from the bin with replacement, find the probability that all the bulbs selected are good ones. and the answer is.... drum roll.........
0.443!!!


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## coldmorning (Jul 4, 2007)

greenfuzz said:


> The first solution uses Baye's rule, you can also do it with combinatorics
> 
> C(51,3) / C(57,3) = .712


That's true. But sometimes when you're learning this stuff it helps to understand the little parts that make up the formula.

If a beginner just memorizes formulas, they can become black boxes that hide what's happening inside them. But they're certainly useful once you have it down well.


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## Inside (Jan 31, 2010)

greenfuzz said:


> The first solution uses Baye's rule, you can also do it with combinatorics
> 
> C(51,3) / C(57,3) = .712


I'm not going to even pretend like I know what combinatorics is. :clapThat's right, my brain is full so I'm happy to say - I don't know what that is, means or is used for.


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## Inside (Jan 31, 2010)

coldmorning said:


> That's true. But sometimes when you're learning this stuff it helps to understand the little parts that make up the formula.
> 
> If you just memorize formulas, they can become black boxes that hide what's happening inside them. But they're certainly useful once you have it down well.


:yes Thank you coldmorning


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## coldmorning (Jul 4, 2007)

No problem Inside...


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## Inside (Jan 31, 2010)

Ya know, I may not need to say this but I do enjoy learning and I do like smart people and I want to learn the rest of my life. It's a really hard thing to ask for help, at least for me anyway. If I hadn't spent three days trying to get through 20 problems, well add 4 to it because there are still problems I haven't made it through, I probably wouldn't have asked.
I just wanted to say I like smart peeps I guess and appreciate other people too who help smart people get smarter 

So can I post my questions here as long as I put my logic with it so you know I'm not trying to get out of the work or take an easy route? Hate to make it public and tie up a thread... :-/


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