The Professor's Response
I am having trouble solving this problem: A sample of 4 different calculators is randomly selected from a group containing 13 that are defective and 27 that have no defects. What is the probability that at least one of the calculator is defective.
See the professor's answer below.
Hi Gail,
Here is how to solve this one: P(at least one is defective) = 1 - P( none of the four are defective) = 1 - P(1st is not defective)*P(second is not defective|first is not defective)* ... =
1 - 27/40*26/39*25/38*24/37 = 0.808
In this problem, some students assume the calculators would be replaced, but that is not justified in the problem. Unless with replacement is stated, generally you assume without replacement. The phrase four different calculators points to the fact that you do not want to choose the same one more than once.
Professor McGuckian