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Vladimir [108]
3 years ago
15

When a new machine functioning property only 4% of the items produced are defective assume that will randomly select two parts p

roduced on the machine and that we are interested in the number of defective parts found Compute the probability associated with finding no defects exactly one defect and two defects p(no defects) , p(1 defects) , p(2 defects)
Mathematics
1 answer:
siniylev [52]3 years ago
5 0

Answer:

p(no defects) = 92.16%

p(1 defects) =7.68%

p(2 defects) =0.16%

Step-by-step explanation:

We are testing 2 parts (n=2) and the probability for each part to be defective is 4% (A=0.04). The probability of not getting defective parts will be 96% or 0.96 (B=0.96). There are no markings on the parts so the order is not important.

p(no defects) = 0C2 * A^0 * B^2=

p(no defects) = 2!/(0!*2!) * 1 * 0.96^2=0.9216

p(no defects) = 92.16%

p(1 defects) = 1C2 * A^1 * B^1=

p(1 defects) = 2!/(1!*1!) * 0.04* 0.96=0.0768

p(1 defects) =7.68%

p(2 defects) = 2C2 * A^2 * B^0=

p(2 defects) = 2!/(2!*0!) *0.04^2*1=0.0016

p(2 defects) =0.16%

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According to a study done by Otago University, the probability a randomly selected individual will not cover his or her mouth wh
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Answer:

47.52% probability that among 10 randomly observed individuals fewer than 3 do not cover their mouth

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Binomial probability distribution

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This means that p = 0.267

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10 individuals, so n = 10.

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

In which

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P(X = 1) = C_{10,1}.(0.267)^{1}.(0.733)^{9} = 0.1631

P(X = 2) = C_{10,2}.(0.267)^{2}.(0.733)^{8} = 0.2673

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0448 + 0.1631 + 0.2673 = 0.4752

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