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) (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.
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