Question 971730
Binomial distribution:
exactly 2 bulbs defective  is 10C2 (0.1)^2 (0.9)^8= 0.194.  
There are 45 ways two bulbs can be defective, and that is 10C2

No more than 3 are defective.  
P (0) are defective = 0.9^10=0.3486 
P(1) is defective = 10 *0.9^9 *0.1 = 0.3874 (the highest, because it is expected value)
P(2) are defective=0.194 from above
P(3) are defective= 10C3*(.9)^7 * (.1)^3= 0.057
This sum is 0.987
We want NO MORE THAN this, so that is the complement.  It is 0.013.

At least 9 of the bulbs are not defective.  That is the probability of (1) being defective and (0) being defective, which we computed above.  That is 0.7360.