Question 1084635: A shipment of 27 computers, 5 are defective. Four computers are randomly selected and tested.What is the probability that all four are defective if the first, second and third ones are not replaced after being tested?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! probability first one drawn is defective is 5/27
assuming the first was defective, probability second is defective is 4/26
assuming the second defective, third is 3/25
ditto fourth 2/24
probability all 4 are defective is 5/27 * 4/26 * 3/25 * 2/24 = 120/421200 = .0002849 rounded to 7 decimal places.
it can also be calculated as the number of ways you can get 4 out of 5 defective * the number of ways you can get 0 out of 22 non-defective, all divided by the number of ways you can get 4 out of 27.
that formula would be c(5,4) * c(22,0) / c(27,4)
that probability would be equal to .0002849 rounded to 7 decimal places.
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