1 . If a fair die is to be tossed 12 times , the probability of getting 1,2,3,4,5
and 6 points exactly twice each is
(apologies cant find decent latex software)
0.166 = fraction 1 in 6
(12! / 2!2!2!2!2!2! ) (0.166*2 ) (0.166*2) (0.166*2) (0.166*2) (0.166*2) (0.166*2) = 0.00344
question 1
does the formula used account for any order of the 2 of each ,meaning
there is no order
3 3 , 4 4 , 6 6 , 5 5 ,1 1, 2 2 is the same as
1 1, 2 2, 3 3, 4 4, 5 5 , 6 6 and so on , it accounts for all the possible combinations of the 2
of each ?.........
if so ...
question 2
what would be the formula for the exact order
the first is 1 , second is 1 , the third is 2 , the fourth is 2 etc in perfect running order?