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for part a: would the probability that a 6 comes up on the nth try be (5/6)^(n-1) *(1/6)? Brandy | for part a: would the probability that a 6 comes up on the nth try be (5/6)^(n-1) *(1/6)? Brandy | ||
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| + | [[Category:MA375Spring2009Walther]] | ||
| + | That makes sense to me JRHaynie | ||
Revision as of 09:48, 4 March 2009
for part a: would the probability that a 6 comes up on the nth try be (5/6)^(n-1) *(1/6)? Brandy That makes sense to me JRHaynie
