Posted by Diana on Tuesday, February 15, 2011 at 8:35am.
even sum 2, 4, 6, 8 , 10, 12
ways 1,1 one way
3,1;2,2; 1,3 three ways
5,1;4,2;3,3;3,3;1,5;2;4 six ways
and continue...
So for the pr (2)=1/6*1/6
pr(4)=1/6*1/6*3
pr(6)=1/6*1/6*6
and continue
To have an even sum
1. all four must be even (E)
2. all four must be odd (O)
3 two must be even, two must be odd.
prob that one die lands even = 3/6 = 1/2
the same as for odd.
prob of EEEE = (1/2)^4
prob of OOOO = (1/2)^4
prob of EEOO = 4!/(2!2!) (1/2)^4
sum = 1/16 + 6(1/16) + 1/16 = 8/16 = 1/2