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I'm trying to find the probabilty of consecutive numbers on tossing three dice. eg 1 2 3 , 4 5 6 etc

My workings so is

nPr = 3P3 (how many permutaions of 3 numbers in order)

= n!/(n-r)! = 3!/0! = 6

What

I did next is calulate all possible outcomes of three dice = 6 x 6 x 6 = 216

Therfore the probabilty of consecutive numbers on three dice = 6/216 = 1/36

Does this look right ?

I'm not quite sure if the permutaion calculation is right or if i have to multiply it by six...

regards

Brendan