Question 442996: Binomial probability:
A company takes a sample of 20 boxes of cereal from their production line. They believe that of the 20 boxes, four will have too little cereal in them and sixteen will have the correct amount. The company then randomly selects three of the boxes from the twenty and weighs them. Assuming the company is correct about the number of light boxes, find the probability that the three boxes selected will all have the correct amount.
If n=20 and r=3, p=16/20 = 4/5, what is the probability of failure?
(I'm just having some solving this problem trouble because I know that Q=1-P, which would give me Q=1/5. Therefore, my equation should be set up as: 20C3(4/5)^3(1/5)^17, but (1/5)^17 is coming out way too big of a number on my calculator that isn't giving me an exact number without it being to the 12 power and I don't know if my answer is correct than, which was 7.65%. Please help to figure out what the heck is Q=? because I'm not sure if it is correctly being done. Thank you and I really appreciate it too.)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Binomial probability:
A company takes a sample of 20 boxes of cereal from their production line. They believe that of the 20 boxes, four will have too little cereal in them and sixteen will have the correct amount.
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p-hat = 16/20 = 4/5
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The company then randomly selects three of the boxes from the twenty and weighs them. Assuming the company is correct about the number of light boxes, find the probability that the three boxes selected will all have the correct amount.
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Binomial problem with n = 3 and p = 4/5
P(x = 3) = (4/5)^3 = 0.512
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If n=20 and r=3, p=16/20 = 4/5, what is the probability of failure?
Comment: n is no longer 20, it is 3. The n = 20 information was stated
so you would have an estimate of "p" for the whole batch.
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Cheers,
Stan H.
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