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There are 10 USB drives in a box and 3 are defective. Two drives are to be selected, one after the other. What is the probability of selecting a defective USB followed by anothe…

There are 10 USB drives in a box and 3 are defective. Two drives are to be selected, one after the other. What is the probability of selecting a defective USB followed by anothe…
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There are 10 USB drives in a box and 3 are defective. Two drives are to be selected, one after the other. What is the probability of selecting a defective USB followed by another defective USB?

User Sam Creamer
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1 Answer

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|\Omega|=10\cdot9=90\\|A|=3\cdot2=6\\\\P(A)=(6)/(90)=(1)/(15)\approx6.7\%

User Whiteulver
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