Question 835843
There are 8 ways to choose 3 gaskets (N-no defect, D-defect)
NNN
NND
NDN
NDD
DNN
DND
DDN
DDD
Three have 2 defective gaskets (NDD,DND,DDN) and their probabilities would be 
{{{P=(4/10)(4/10)(6/10)=0.096}}}
Then for all three cases,
{{{P=3(0.096)=0.288}}}