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?
2 answers
17/36
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.
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.
2 answers