SOLUTION: A student needs eight chips of a certain type to build a circuit. It is known that 5% of these chips are defective. How many chips should he buy for there to be a greater than 90

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Question 346191: A student needs eight chips of a certain type to build a circuit. It is known that 5% of these chips are defective. How many chips should he buy for there to be a greater than 90% probability of having enough chips for the circuit?
Answer by nyc_function(2741) About Me  (Show Source):
You can put this solution on YOUR website!
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.