Answer: [tex]\dfrac{1}{3}[/tex]
Step-by-step explanation:
Given: S = {6,7,8,9,10,11,12,13,14,15,16,17}
E={6,7,8,9,10,11,12,13}
[tex]E^c=[/tex] Event having all elements of S other than E.
[tex]={14, 15 , 16, 17}[/tex]
[tex]P(E^c)=\dfrac{\text{Elements in }E^c}{\text{Elements in }S}\\\\=\dfrac{4}{12}\\\\=\dfrac{1}{3}[/tex]
Hence, [tex]P(E^c)=\dfrac{1}{3}[/tex]