SOLUTION: Hello Tutors, Glad you are here to help. a sample of 4 different calculators is randomly selected from a group containging 18 that are defective. and 35 with no defects. What

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Question 366980: Hello Tutors,
Glad you are here to help.
a sample of 4 different calculators is randomly selected from a group containging 18 that are defective. and 35 with no defects. What is the probability that at least 1 calculator is defective.
I started with adding all calculators together to get a total of 53. Then I divided 1/52 since we are looking for at least 1 defect out of 4. the answer I got was .0189. Is this correct?
If not, please solve and show your work.
Thank you kindly
Found 2 solutions by stanbon, robertb:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A sample of 4 different calculators is randomly selected from a group containing 18 that are defective. and 35 with no defects. What is the probability that at least 1 calculator is defective.
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Binomial Problem with n = 4 and P(defective) = 18/(18+35) = 18/53
And P(not defective) = 35/53
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P(at least one defective) = 1 - P(none defective)
P(at least one defective) = 1 - (35/53)^4 = 0.8098
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Cheers,
Stan H.

Answer by robertb(5830) (Show Source):
You can put this solution on YOUR website!
. How? The event that at least 1 calculator is defective is complementary with the event that no calculator is defective. Hence
P(at least 1 calculator is defective) = 1 - P(no calculator is defective).
But P(no calculator is defective) = , and hence the answer above. (Note: This is an example of a hypergeometric probability distribution.)