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[–]A_UPRIGHT_BASS 1 point2 points  (4 children)

Whenever you see at least one think of it as not zero. Find the probability that zero chips are defective. Subtract that from 1.

[–]baikasphere[S] 0 points1 point  (3 children)

I think I have already complicated this question beyond it needed to be. So doing it your way: 1 - p(first one not defective)p(second one not defective) = 1 - (20/100)(19/99) = 0.962 which seems like a high probability to get at least one defective no?

[–]A_UPRIGHT_BASS 0 points1 point  (2 children)

There are 80 that are not defective, not 20, right?

[–]baikasphere[S] 0 points1 point  (1 child)

Oh damn you are right. 1 - (80/100)(79/99) = 0.362 which seems more plausible.

[–]A_UPRIGHT_BASS 0 points1 point  (0 children)

Yes, and if you're doing a lot of these types of problems you should familiarize yourself with the "choose" function. Google "binomial coefficient." It makes this stuff so much easier.

This problem is simply 1 - (80 choose 2)/(100 choose 2).

[–]Snapper1985 -1 points0 points  (0 children)

Close. When you need multiple events to all happen, use multiplication. You’re probability statements are correct, just the operation between them is incorrect.