Question 1109991: A manufacturer knows that 6% of the 200 computer monitors it has in stock are defective. Three monitors are randomly selected from their stock. Find the probabilities for parts (a) through (e) below.
a. P(first monitor is defective)
(Round to the nearest tenth as needed.)
b. P(second is defective | first is defective)
(Round to the nearest tenth as needed.)
c. P(second is not defective | first is defective)
(Round to the nearest tenth as needed.)
d. P(second is defective | first is not defective)
(Round to the nearest tenth as needed.)
e. P(third is defective | first two are defective)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! a. 0.06
b. given first is defective (12/200), the second's being defective is 11/199 or 0.055)
c. given first is defective, probability second is not defective is 188/199 or 0.945.
d, This would be 188/200*12/199 or 0.0567
e. Given first two are defective, 10 left and 198 choices, so probability is 10/198 or 0.0505
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