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Topic: YAHOO! questions (Read 2330 times) |
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sanny
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YAHOO! questions
« on: Dec 3rd, 2008, 1:43am » |
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1. You have 'n' number of balls and 'r' number of boxes. Find the probability that that first 'r1' boxes contains k balls.
2. you have a special crystall ball that breaks only when it is dropped from (or above) a certain storey 'n' in a 64 storeyed building where 'n' can be 1<=n<=64. You have two crystall balls, how many (minimum) throws you have to make to find 'n' in the worst situation.
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towr
wu::riddles Moderator
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Re: YAHOO! questions
« Reply #1 on: Dec 3rd, 2008, 2:16am » |
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For 1) I would think C(n, k) (r1/r)k([r-r1]/r)n-k
For 2) there are previous threads of much the same problem here and here.
A more general problem (with more than two balls/eggs) is here.
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dullhead
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Re: YAHOO! questions
« Reply #2 on: Dec 17th, 2008, 9:19pm » |
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on Dec 3rd, 2008, 1:43am, sanny wrote:1. You have 'n' number of balls and 'r' number of boxes. Find the probability that that first 'r1' boxes contains k balls.
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Can one box contain more that one ball?
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dullhead
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Re: YAHOO! questions
« Reply #3 on: Dec 18th, 2008, 12:52pm » |
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on Dec 3rd, 2008, 2:16am, towr wrote:For 1) I would think C(n, k) (r1/r)k([r-r1]/r)n-k
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Can you explain how you got this answer. Choosing k balls out of n, takes c(n,k). after that how did you get the remaining term?
Thanks in advance.
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towr
wu::riddles Moderator
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Re: YAHOO! questions
« Reply #4 on: Dec 18th, 2008, 12:56pm » |
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on Dec 18th, 2008, 12:52pm, dullhead wrote:| Can you explain how you got this answer. Choosing k balls out of n, takes c(n,k). after that how did you get the remaining term? |
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The probability that a given ball is put in the first r1 boxes is r1/r; so the probability that k given balls are put in the first r1 boxes is (r1/r)k
The other factor is the same reasoning for putting the rest of the balls in the rest of the boxes.
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River Phoenix
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Re: YAHOO! questions
« Reply #5 on: Dec 21st, 2008, 3:32pm » |
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FYI Yahoo! is no longer a real company. Be careful!
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nks
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Re: YAHOO! questions
« Reply #6 on: Dec 22nd, 2008, 12:21am » |
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Quote:| FYI Yahoo! is no longer a real company. Be careful |
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!
How is it so ?
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River Phoenix
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Re: YAHOO! questions
« Reply #7 on: Dec 22nd, 2008, 2:36am » |
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on Dec 22nd, 2008, 12:21am, nks wrote:
semi joking, but they've somewhat fallen apart, laid off a huge number of people, switched leadership around, and fallen catastrophically in value. they may be purchased by microsoft
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tpraja
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Re: YAHOO! questions
« Reply #8 on: Jan 13th, 2009, 7:14am » |
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Its not easy to get yahoo because yahoo now improving so many features.
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| « Last Edit: Jan 13th, 2009, 7:15am by tpraja » |
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darklord
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Re: YAHOO! questions
« Reply #9 on: Apr 7th, 2009, 4:04am » |
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didnt the first answer follow binomial distribution????
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