Answer
$F^{c}=$ {1, 2, 3, 4, 10, 11, 12}
$P(F^{c})=\frac{7}{12}\approx0.583$
Work Step by Step
$F^{c}$ is the complement of F. $F^{c}$ has all outcomes in the sample space S that are not outcomes in the event F. So, $F^{c}=$ {1, 2, 3, 4, 10, 11, 12}
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. So, $N(S)=12$.
F = {5, 6, 7, 8, 9}. So, $N(F)=5$.
$P(F)=\frac{N(F)}{N(S)}=\frac{5}{12}$
$P(F^{c})=1-\frac{5}{12}=\frac{7}{12}\approx0.583$
