Question 1160659
exactly 3 are defective is 10C3*0.9^7*0.1^3=0.0574
at most 2 is p(0,1,2)
p=0.9298 from binomcdf(10,.1,2)
each of those is 0.3486, 0.3874 and 0.1937 respectively 

at least 3 defective items is 1-at most 2, since those are complementary, and that is 1-0.9289=0.0702

E(X)=10*0.1=1 defective part
Var(X)=1*0.9 (np(1-p)0=0.9
sd is sqrt (V)=0.9487 or about 0.95 parts