Answer:
[tex]P(E^c)=\dfrac{1}{3}[/tex]
Step-by-step explanation:
The sample space for the experiment is:
[tex]S=\{9,10,11,12,13,14,15,16,17,18,19,20\}[/tex]
Given an event E such that:
[tex]E=\{10,11,12,13,14,15,16,17\}[/tex]
The outcomes in the complement of E are the outcomes in S that are not in E.
[tex]E^c=\{9,18,19,20\}[/tex]
Therefore:
[tex]P(E^c)=\dfrac{n(E^c)}{n(S)}\\ =\dfrac{4}{12}\\P(E^c)=\dfrac{1}{3}[/tex]