Question

A quality control engineer inspects a random sample of 30 calculators from an incoming batch of size 150 and accepts the lot if at most 4 are not in working condition; otherwise the entire lot must be inspected with the cost charged to the vendor. Suppose the lot contains 5 defective calculators.

a.) What is the probability that such a lot will be accepted without further inspection?

b.) Given the third calculator tested is the first calculator to be defective, what is the probability the lot fails initial inspection?

c.) Use a Poisson distribution to approximate the probability a lot will be accepted without further inspection?

Answer 1

a) probability that such a lot will be accepted without further inspection =P(st most 4 are not in wroking condition)

=1-P(5 are not in working condition)=1-⁵C₅*¹⁴⁵C₂₅/¹⁵⁰C₃₀ =1-0.000241 =0.999759

b)

P( lot fails initial inspection given 1st defective is on 3rd trail)=P(rest of 4 defective in next 27)=⁴C₄*¹⁴³C₂₃/¹⁵⁰C₂₇ =0.000940

c)

here expected number of defective in 30 calculator =30*5/150=1

therefore P( lot will be accepted )=P(X<=4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)

=e^-1*1⁰/0!+e^-1*1¹/1!+e^-1*1²/2!+e^-1*1³/3!+e^-1*1⁴/4!=0.996340