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February 16th, 2010, 03:46 AM   #1
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Probability formula help please

I have no idea where to start. I think this is joint probability because the computer warranty and the need for repair occur at the same time.

If 10% of all computers are returned for repair with guarantee still in effect and I purchase nine for my business. What is the probability that 3 or more will be returned for repair whilst guarantee is in effect?

What assumptions (if I know the correct formula I can research this)
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February 16th, 2010, 05:33 AM   #2
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Re: Probability formula help please

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Originally Posted by paul111
I think this is joint probability because the computer warranty and the need for repair occur at the same time.
No, it's a simple probability problem (not joint) because you're given the probability of that combined event as a single number.

The probability is the chance that 3 need to be returned + the chance that 4 need to be returned + ... + the chance that 10 need to be returned. Alternatively, it's 1 - the chance that 0 need to be returned - the chance that 1 needs to be returned - the chance that 2 need to be returned.
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February 16th, 2010, 11:44 AM   #3
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Re: Probability formula help please

Oh my~ really! Thank you so much for some direction.

We newbies over complicate things don't we.

Ok, so I divide 3 by nine and add this to 4/9 + 5/9 + 6/9 +7/9 +8/9 ...?
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February 16th, 2010, 12:03 PM   #4
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Re: Probability formula help please

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Originally Posted by paul111
Ok, so I divide 3 by nine and add this to 4/9 + 5/9 + 6/9 +7/9 +8/9 ...?
Er, no. The probability that the first three computers (and no others) need to be returned is
1/10 * 1/10 * 1/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 = (1/10)^3 * (9/10)^7.
This corresponds to the order FFFooooooo (with F = fail and o = OK). But other orders are possible, like FFoFoooooo and FooFoooooF. Since there are 10 spaces, 3 of which are F, there are (10 choose 3) ways to arrange them. You may have seen this written or . So in total the probability that exactly 3 fail is
.

Do you know how to calculate ?
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February 16th, 2010, 01:16 PM   #5
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Re: Probability formula help please

Thank you so much for walking me through this, I divide 10 by 3...multiply by 1/10 to the power of 3 multiply by 9/10 to the power of 7...?

or

1/10 * 1/10 * 1/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 = (1/10)^3 * (9/10)^7 Do my divisions and multiply as I go along...?

We did no probability in my first stats course, and now I have had a couple of weeks introduction crammed into me. Lost~ Give me hypothesis testing anytime.

If there were only standard formulas!

Am reviewing the chapter on classical method of probability, the formula given is P(E)=ne/N

But this doesn't make sense for the 3/ more...I would rather learn to read your formula.

I am so hiring a tutor when this course is over ~
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February 16th, 2010, 01:48 PM   #6
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Re: Probability formula help please

Quote:
Originally Posted by paul111
1/10 * 1/10 * 1/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 = (1/10)^3 * (9/10)^7
This part is fine, but is not 10 divided by 3. Try this article:
http://en.wikipedia.org/wiki/Binomial_coefficient
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February 16th, 2010, 02:05 PM   #7
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Re: Probability formula help please

thank you so much!

Do you tutor online using skype?
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February 16th, 2010, 02:23 PM   #8
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Re: Probability formula help please

I don't have skype installed at the moment, and you couldn't afford me anyway. :P
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February 16th, 2010, 03:32 PM   #9
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