Question 348560
the sum 7 is given by 1&6 2&5 3&4 and reverse order, hence 6 possibilities over 6*6=36
 
so P(x)=(1/6)^x*(5/6)^(5-x)
 
Verif : we find 
P(5)=1/6^5=0.0001286
P(0)=(5/6)^5=0.4018
 
however P(2) gives other result (0.01607 ?)
 
P(1)=1/6*(5/6)^4=0.0803