Asked by Grace
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!
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!
Answers
Answered by
John
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
I agree with a and b.
for c, take 1- .001 = .999
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