The Odds Of....

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Anonymous Poster	The Odds Of.... 11/26/2007 2:22 PM
	if you are given the choice between 5 doors and a prize is located behind one door and you choose the correct door the probability of doing so is 20% correct ? now if you repeat this process 10 times and you choose correctly 5 times or 50% what is the probability or chance of that occurring ? thanx
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TVP45 Guru Join Date: Jul 2007 Posts: 4448 Good Answers: 143	#1 Re: The Odds Of.... 11/26/2007 7:46 PM
	Isn't there a rule somewhere about not doing statistics where children can see it? __________________ "Well, I've wrestled with reality for 35 years, Doctor, and I'm happy to state I finally won out over it." Elwood P. Dowd
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U V Guru Join Date: Jun 2007 Location: East of Seattle, Washington state Republic of the 50 states of America Posts: 2045 Good Answers: 36	#2 Re: The Odds Of.... 11/27/2007 12:46 AM
	Only when it's their homework Brad __________________ (Larrabee's Law) Half of everything you hear in a classroom is crap. Education is figuring out which half is which.
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Anonymous Poster	#3 Re: The Odds Of.... 11/27/2007 4:39 AM
	not enough information. And you're mixing up between probability and odds.
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Codemaster Guru Join Date: Jul 2005 Location: Stoke-on-Trent, UK Posts: 4496 Good Answers: 137	#4 Re: The Odds Of.... 11/27/2007 4:44 AM
	1st para - 20 % is correct. 2nd para 2.642 %. It's the binomial distribution, b(x) = p^x_(1-p)^N-x_^NC_x where b(x) = probability of x successes, x = number of successes (5), p = probability of success in each trial (0.2), N = number of trials (10), ^NC_x is combinations formula = N!/(x!_*(N-x)!) Cheers....Codey __________________ Give masochists a fair crack of the whip
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Randall Guru Join Date: Aug 2005 Location: Hemel Hempstead, UK Posts: 5826 Good Answers: 322	#5 Re: The Odds Of.... 11/27/2007 9:36 AM
	This is the same answer as Codey, but slightly more intuitive. Suppose you choose correctly the first 5 times and wrongly the last 5 times: the probability of this is:- (1/5)⁵x(4/5)⁵ But actually the correct and wrong selections could have been any of the 10 choices. So choose any 5 from 10. First one can be any one of 10 second any one of 9 etc. = 10x9x8x7x6 , but these can be in any order so divide by 5x4x3x2x1 . So total solution is (1/5)⁵x(4/5)⁵x(10x9x8x7x6)/(5x4x3x2) = 0.026424 __________________ If you spend all your time looking for people and things to complain about: trust me, you will find plenty to complain about.
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Anonymous Poster (1); Codemaster (1); Randall (1); TVP45 (1); U V (1)

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