My Math Forum Probability formula help please

 February 16th, 2010, 04:46 AM #1 Newbie   Joined: Feb 2010 Posts: 5 Thanks: 0 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)
February 16th, 2010, 06:33 AM   #2
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Quote:
 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.

 February 16th, 2010, 12:44 PM #3 Newbie   Joined: Feb 2010 Posts: 5 Thanks: 0 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 ...?
February 16th, 2010, 01:03 PM   #4
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Quote:
 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 ${}_{10}\text{C}_3$ or ${10\choose3}$. So in total the probability that exactly 3 fail is
${10\choose3}(1/10)^3(9/10)^7$.

Do you know how to calculate ${10\choose3}$?

 February 16th, 2010, 02:16 PM #5 Newbie   Joined: Feb 2010 Posts: 5 Thanks: 0 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 ~
February 16th, 2010, 02:48 PM   #6
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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 ${10\choose 3}$ is not 10 divided by 3. Try this article:
http://en.wikipedia.org/wiki/Binomial_coefficient

 February 16th, 2010, 03:05 PM #7 Newbie   Joined: Feb 2010 Posts: 5 Thanks: 0 Re: Probability formula help please thank you so much! Do you tutor online using skype?
 February 16th, 2010, 03:23 PM #8 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Probability formula help please I don't have skype installed at the moment, and you couldn't afford me anyway. :P
 February 16th, 2010, 04:32 PM #9 Newbie   Joined: Feb 2010 Posts: 5 Thanks: 0 Re: Probability formula help please

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