#### Answer

21

#### Work Step by Step

Losing = (doubles) OR [( green is odd) AND ( red is even)]
The green and red dice are distinct so the set of possible outcomes is
$S=\{(1,1),(1,2),...(1,6),(2,1),(2,2),...(6,6)\} $
(first: green die, second: red die)
$n(S)=36$
Let A = set of outcomes which are doubles
A=$\{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}$
n(A)=6
Let B =set of outcomes where green die is odd and red die is even
B=$\{(1,2), (1,4), (1,6), (3,2), (3,4), (3,6),(5,2), (5,4), (5,6)\}$
$n(B)=9$
$ A\cap B=\emptyset$
Let L = set of losing combinations
$L=A\cup B$
$n(L)=n(A\cup B)=n(A)+n(B)-n(A\cap B)$
$=6+9-0=15$
Since $n(S)=36$, and $L^{\prime}$ is the set of winning combinations,
$n(L^{\prime})=n(S)-n(L)=36-15=21$
(there are 21 winning combinations)