Assuming all three die are identical...
ways to get 30
40, any two others
30, and any two others
20, and any two others
10, and any two others, one of which is greater than 10.
So the probablity of getting greater than 30 is
Pr(>30)=1-pr(three 10s)=1-1/4 ^3
ways of getting less than 35
three 10's
Well, that sort of messes things up for getting over 30..
so Pr ( 30<total<35)=0
The faces of a green tetrahedron die are marked 10,20,30 and 40. All three die are thrown together and the scores added. Find the probability that the total is more than 30 but less than 35.
1 answer