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February 16th, 2010, 03:46 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, 05:33 AM  #2  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Probability formula help please Quote:
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, 11:44 AM  #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, 12:03 PM  #4  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Probability formula help please Quote:
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 ?  
February 16th, 2010, 01:16 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, 01:48 PM  #6  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Probability formula help please Quote:
http://en.wikipedia.org/wiki/Binomial_coefficient  
February 16th, 2010, 02:05 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, 02:23 PM  #8 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 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, 03:32 PM  #9 
Newbie Joined: Feb 2010 Posts: 5 Thanks: 0  Re: Probability formula help please 

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