A - not all 3 triplets can be in the starting lineup
A' - all 3 triplets can be in the starting lineup
[tex] \displaystyle
|\Omega|=\binom{14}{6}=\dfrac{14!}{6!8!}=\dfrac{9\cdot10\cdot11\cdot12\cdot13
\cdot14}{2\cdot3\cdot4\cdot5\cdot6}=3003\\
|A'|=\binom{11}{3}=\dfrac{11!}{3!8!}=\dfrac{9\cdot10\cdot11}{2\cdot3}=165\\
|A|=3003-165=2838\\\\
P(A)=\dfrac{2838}{3003}=\dfrac{86}{91}\approx95\% [/tex]
