| http://www.w3.org/ns/prov#value | - But you cannot get more than 1000 sixes from 1000 dice, so $\Pr( ext{at most 1000 sixes}) =1$, and you can rewrite this more briefly as $$\Pr( ext{at least 150 sixes)} = 1 - \Pr( ext{at most 149 sixes}).$$ In other words, the method in you first case is a particular of the method in your second case.
|
