[tex]\displaystyle\\|\Omega|=\binom{21}{4}=\dfrac{21!}{4!17!}=\dfrac{18\cdot19\cdot20\cdot21}{2\cdot3\cdot4}=5985\\|A|=\binom{8}{3}\cdot13=\dfrac{8!}{3!5!}\cdot13=\dfrac{6\cdot7\cdot8}{2\cdot3}\cdot13=728\\\\P(A)=\dfrac{728}{5985}=\dfrac{104}{855}[/tex]