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Re: question from may 2005 exam

From: giggle
Remote Name: 138.25.13.239
Date: 05 Jun 2005
Time: 21:23:06 -0600

Comments

I think the number of rolls should follow a negative binormial distribution. Say the N stand for the number of rolls, the Prob(N=n)=n!/(n-3)!3!(1/6)^3(5/6)^5.

Last changed: 02/05/07