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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]]
 
[[Category:MA375Spring2009Walther]]
 
That makes sense to me  JRHaynie
 
That makes sense to me  JRHaynie

Revision as of 09:48, 4 March 2009

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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


That makes sense to me JRHaynie

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin