## Finite Math and Applied Calculus (6th Edition)

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)