Question

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

Answer 1

p(no defects) = 92.16%

p(1 defects) =7.68%

p(2 defects) =0.16%

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%