Assume we have 2 cubes as described, one blue, the other red.
The Event E: "one is a 2, and the sum is < or equal to 6" is:
E={(Red 2, Blue 1), (Red 2, Blue 2), (Red 2, Blue 3), (Red 2, Blue 4), (Red 1, Blue 2), (Red 2, Blue 2), (Red 3, Blue 2), (Red 4, Blue 2)}
thus, n(E)=8,
The sample space S (the set of all the possible accounts) is:
{(Red x, Blue y | x ∈ 1, 2, 3, 4, 5, 6 ; y∈ 1, 2, 3, 4, 5, 6}
That is n(S)= 6*6 = 36,
as any of the 6 possibilities for x, can be combined with of the 6 possibilities for y.
P(E)=n(E)/n(S)=8/36=0.222
Answer: 8/36=0.222
