The Professor's Response
Hi professor, would you mind doing question 61 from the mixed review? I can't seem to find my mistake. thank you !
See the professor's answer below.
Hi Reina,
Here is the question:
61. In a production of phones 0.8% are defective. A sample of ten phones is randomly selected without replacement from a set of 20,000 and tested. What is the probability that at least one of those ten tested phones is defective?
To answer a probability of at least one question, we will set the problem using the formula P(at least one) = 1 - P(none).
P(at least one is defective) = 1 - P(none are defective)
= 1 - ___ * ___ * ___ * ___ * ___ * ___ * ___ * ___ * ___ * ___
Now, we have ten spaces because we are selecting ten phones. We are trying to find the probability that none of the ten are defective because if we subtract that probability from 1, we will find the probability of at least one defective phone. Each space above, represents a phone that is not defective. This means we need the probability that a phone is not defective.
The P(Defective) = 0.8% = 0.008
The P(not Defective) = 1 - P(Defective) = 1 - 0.008 = 0.992
So each of our ten spaces will be filled with 0.992. This will give us: P(at least one defective phone) = 1 - 0.992*0.992*...*0.992
= 1 - 0.992^10 = 0.0772
Hope that makes sense,
Professor McGuckian