# Statistics

In a shipment of 21 parts, four of the parts are defective. If three parts are selected at random without replacement, find the probability that :

a.) All three parts selected are defective.

4/21*3/20*2/19=.003

b.) None of the tree parts selected are defective.

3/17*2/16*1/15=.001

c.) At least one of the tree parts selected is defective.

4/21*20/20*19/19=.190

Question~ Are all these answers correct?
If not, what am I doing wrong?

I'm also questioning c.) I'm not sure I did that one correctly. Can someone go over this problem with me? I would so greatly appreciate it.

Thank you!

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1. If they say at least one, it is easier to do the question by finding NONE and subtracting that number from 1. At least one would mean, 1, 2 or 3.

I agree with a and b.

for c, take 1- .001 = .999

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