Posted by **Sherri** on Sunday, September 9, 2007 at 9:58pm.

A sample of 4 different calculators is randomly selected from a group containing 17 that are defective and 36 that have no defects. What is the probability that at least one of the calculators is defective?

- Statistics -
**Amanda**, Monday, September 10, 2007 at 12:03am
17/36

- Statistics -
**MathGuru**, Monday, September 10, 2007 at 8:41am
Find P(0), then take 1 - P(0) for your probability (since the problem says "what is the probability that at least one" is defective). Using the binomial probability function (you can use a binomial probability table as well):

P(x) = (nCx)(p^x)[q^(n-x)]

n = 4

x = 0

p = 17/53 = .32 (note: 53 = 17 + 36)

q = 1 - p = .68

Substituting:

P(0) = (4C0)(.32^0)(.68^4) = .2138

Now, take 1 - .2138 -->this will be your probability.

I hope this helps.

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