In a shipment of 30 computers, 4 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?
P=430×329×228×127=24657720=127405P = \frac{4}{30} \times \frac{3}{29} \times \frac{2}{28} \times \frac{1}{27} \\ = \frac{24}{657720} \\ = \frac{1}{27405}P=304×293×282×271=65772024=274051
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