Question 983415
First is 5C3*(0.1)^3(0.9)^2; number of ways 3 can be defective out of 5 times the probability is 0.0081  (0.01*0.81)

Fewer than 2 means 1 or 0 are defective
P(1)=5C1(0.1)^1(0.9^4)=0.328
P(0)=(.9^5)=0.590
The probability fewer than 2 are defective is 0.918.  That makes sense since it is a low probability to find one defective.

Probability of 2 defective is 5C2(0.1^2)(0.9^3)=0.073
The probability of 2 or fewer defective (using the above) is 0.918+0.073=0.991.
Therefore, the complement is greater than 2 are defective, which is 0.009.