Question 1166165: The Telektronic Company purchases large shipments of fluorescent bulbs and uses this acceptance-sampling plan: Randomly select and test 24 bulbs, then accept the whole batch if there is only one or none that doesn’t work. If a particular shipment of thousand of bulbs actually has a 5% rate of defects, what is the probability that this whole shipment will be accepted?
(I believe it's a binomial distribution)
Found 2 solutions by solver91311, Theo:
Answer by solver91311(24713)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! you are correct, as far as i know.
there are 24 bulbs in each shipment.
the probability that any one bulb, randomly selected, will be defective is .05.
you want to find the probability that the whole shipment will be accepted.
the formula for binomial probability is:
p(x) = p^x * q^(n-x) * c(n,x)
c(n,x) is the number of ways you can get x elements out of a set of n elements.
c(n,x) is equal to n! / (x! * (n-x)!).
to be acceptable, the number of defects has to be 0 or 1.
you want the sum of p(0) + p(1).
that tells you the probability that 0 or 1 are defective.
that also tells you the probability that the shipment will be accepted.
p(0) = p^0 * q^(24-0) * c(24,0)
p(1) = p^1 * q^(24-1) * c(24,1)
p = .05
q = 1 - .04 = .95
the formulas becomes:
p(0) = .05^0 * .95^(24) * c(24,0) = .2919890243
p(1) = .05^1 * .95^(23) * c(24,1) = .3688282413
the probability of having 0 or 1 defective bulbs is the sum of p(0) and p(1) = .6608172556.
that's also the probability that the shipment will be accepted, since the shipment will be accepted if 0 or 1 bulbs are defective, i.e. less than 2.
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