Question 346191
Clearly he must buy at least 8. The probability that all 8 will be nondefective is


(8 choose 8) (0.95)^8 (0.05)^0 ≈ 0.66342


So, 8 is not enough.


The probability that 8 of 9 will be nondefective is


(9 choose 8) (0.95)^8 (0.05)^1 ≈ 0.29854


The probability that 9 of 9 will be nondefective is


(9 choose 9) (0.95)^9 (0.05)^0 ≈ 0.63025


The probability that either 8 or 9 will be nondefective is 


0.29854 + 0.63025 = 0.92879 > .90


So, 9 chips should be enough.