Question 1119070: A carton of 18 lightbulbs includes 5 that are defective. If two bulbs are drawn at random, what are the probabilities that
(a) neither bulb will be defective
(b) exactly 1 bulb will be defective
(c ) both bulbs will be defective
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 13 are not defective, so it would be 13/18*12/17, after one non-defective is removed. Product is 0.5098
1 defective is
*first one 5/18*13/17
*second one 13/18*5/17; those two products are the same, so multiply by 2, the number of ways, and it is 0.4276.
=======
both defective is 5/18*4/17=20/304, or 0.0658. Those three add to 1.0032, which is 1 with rounding error.
|
|
|
