4. If the fraction of defective parts on a production line is known to be 8%; (a

Question

4. If the fraction of defective parts on a production line is known to be 8%; (a) What is the probability that a sample of 10 parts will contain exactly 2 (b) What is the probability that a sample of 10 parts will contain at least 2 (c) What is the probability that the first defective part will be found on the 3rd defectives? defectives? draw? (d) What is the probability the 4th part selected will be defective, given that the 3rd part selected was defective? (e) What is the probability the 2nd defective part will be found on the 4th part sampled?

Explanation / Answer

P(X) = nCx px qn-x

n = 10

p = 0.08

q = 1 - p = 0.92

a) P(X = 2) = 10x0.082x0.928

= 0.0328

b) P(at least 2) = 1 - P(less than 2 defectives)

= 1 - 0.9210 - 10x0.929x0.08

= 0.188

c) P(first defective part will be found on 3rd) = 0.92x0.92x0.08

= 0.068

d) P(4th defective given that 3rd is defective) = 0.08

e) P(2nd defective on 4th sample) = 3x0.922x0.08 x 0.08

= 0.016

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