LogFAQs > #934538694

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Topic	SaveEstelle & LeonhartFour in Different Houses: Eclectic Edition [SELF]
LordoftheMorons 02/18/20 5:57:30 AM #285:	If you have probability p per roll of getting X, the odds that exactly k out of n rolls give you X is p^k(1-p)^(n-k)(n choose k) Here (n choose k) = n!/(k!(n-k)!) is a binomial coefficient, which is the number of ways to choose k items out of n without replacement (and where order doesn't matter). --- Congrats to Advokaiser* for winning the CBX Guru Challenge! <cite>LordoftheMorons posted [p:934538694] in SaveEstell... [t:78363536]...</cite> <quote>If you have probability p per roll of getting X, the odds that exactly k out of n rolls give you X is p^k(1-p)^(n-k)(n choose k) Here (n choose k) = n!/(k!*(n-k)!) is a binomial coefficient, which is the number of ways to choose k items out of n without replacement (and where order doesn't matter). --- Congrats to Advokaiser for winning the CBX Guru Challenge!</quote> ... Copied to Clipboard!
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