
#19
Jun911, 07:41 AM

Mentor
P: 16,518

(B,BN,NN,B) (N,NN,BB,B) (N,NB,NB,B) (N,NB,BN,B) (B,NN,NB,B) (N,BN,NB,B) And you see from that that the probability is 2/3. 



#20
Jun911, 07:49 AM

Mentor
P: 16,518





#21
Jun1711, 03:34 PM

P: 75

Let's name the pancakes as
[itex]B_{0}[/itex]= Pancake with no burnt sides [itex]B_{1}[/itex]= Pancake with one burnt side [itex]B_{2}[/itex]= Pancake with two burnt sides The probability a priori, for a pancake of being at the top after stacking them is 1/3 for each one. I think there is no doubt about this. Now, let's take into account that the top side is burnt. This let us with only [itex]B_{1}[/itex] and [itex]B_{2}[/itex] as being the pancake on the top. But they don't have the same probability at all. If the pancake at the top is pancake [itex]B_{2}[/itex], it can appear in 2 possible ways, depending on which of its sides (both burnt) is up. On the other hand, if the pancake on the top side is pancake [itex]B_{1}[/itex], then it can appear in only 1 way: its burnt side up. So, the probability that the pancake in the top is [itex]B_{2}[/itex], knowing that the top side is burnt, is clearly 2/3. 



#22
Jun1711, 07:32 PM

P: 67

there are only 3 burnt sides and 2 sides belong to the completely burnt pancake, so the probability is 2/3.



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