<PRE><B>Question 484245</B>
P(no more than 1 defective one)=P(0 defective)+P(1 defective)
11 CD players, 2 defective, {{{11-2=9}}} not defective
P(0 defective)={{{(9C3)/(11C3)}}}
{{{9C3=9!/(3!(9-3)!)=9!/(3!6!)=(6!*7*8*9)/(2*3*6!)=(7*8*9)/(2*3)=84}}}
{{{11C3=11!/(3!(11-3)!)=11!/(3!8!)=(8!*9*10*11)/(2*3*8!)=165}}}
P(0 defective)={{{(9C3)/(11C3)=84/165=28/55}}}
P(1 defective)={{{(9C2*2C1)/(11C3)}}}
{{{9C2=9!/(2!(9-2)!)=9!/(2!7!)=(7!*8*9)/(2*7!)=36}}}
{{{2C1=2}}}
P(1 defective)={{{(36*2)/165=24/55}}}
P(no more than 1 defective one)=P(0 defective)+P(1 defective)={{{28/55+24/55=52/55=0.945}}}</PRE>