Answer
$E^{c}=$ {1, 8, 9, 10, 11, 12}
$P(E^{c})=0.5$
Work Step by Step
$E^{c}$ is the complement of E. $E^{c}$ has all outcomes in the sample space S that are not outcomes in the event E. So, $E^{c}=$ {1, 8, 9, 10, 11, 12}
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. So, $N(S)=12$.
E = {2, 3, 4, 5, 6, 7}. So, $N(E)=6$.
$P(E)=\frac{N(E)}{N(S)}=\frac{6}{12}=0.5$
$P(E^{c})=1-P(E)=1-0.5=0.5$
