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probability problem


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Initial post: Jun 7, 2007 7:08:20 AM PDT
the probability of a 5/1 chance occuring at least once in five trials i come up with .598 5/6 to the fifth power how would you compute the probability of a 4k/1 occurring at least once in 4k trials thank you Harold Thomas

In reply to an earlier post on May 20, 2008 7:16:37 PM PDT
Actually, .598 is 1-(5/6)^5 but you have the right idea.

Now, for your question, the answer is 1-(4000/4001)^4000 = 0.36792542131209053968048500693842

This is equal to e^(-1) where e is the exponential growth constant. because
the limit as n goes to infinity of (1+(1/n))^n = e.

In the examples that you gave, you have (1+(1/(n+1)))^n which is close enough.

In reply to an earlier post on Feb 23, 2009 6:55:19 AM PST
Your answer is incorrect. You have computed the probability of exactly 5 occurrences in five trials. The answer is a simple binomial probability. In general let p be the probability of success (in your case p=5/6). Then the probability of at least 1 success in n trials (in your case n=5) is 1- probability of zero success.
This is 1-(1-p)^n, in your case 1-(1-5/6)^5=1-(1/6)^5=1-(1/7776)=1-0.00013=.99987. For the other case you mention p=4k/(4k+1) so the result is
1-(1-[4k/(4k+1)])^4k for k=1 it is .984 for k>=2 it is bigger than .999.

In reply to an earlier post on Feb 23, 2009 7:02:09 AM PST
You are right about the limit but not the answer to his problem.
It should be 1-(1/6)^5 which is a lot bigger than .598! It is 1- the probability that it never comes up in 5 trials not 1-the probability that it come up all 5 times!
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Discussion in:  Probability forum
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Initial post:  Jun 7, 2007
Latest post:  Feb 23, 2009

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