Question

A box contains 11 transistors, 5 of which are defective. If 5 where selected random find the probility of the none defective ones?

Answer 1

$|\Omega|=C(11,5)=\dfrac{11!}{5!6!}=\dfrac{7\cdot8\cdot9\cdot10\cdot11}{2\cdot3\cdot4\cdot5}=462\\ |A|=C(6,5)=6\\\\ P(A)=\dfrac{6}{462}=\dfrac{1}{77}$