Badiuddin askIITians.ismu Expert
Last Activity: 15 Years ago
Dear Sudip
let
A=testing procedure ends at 12th testing
A1 =lot contain 2 defective article
A2 =lot contain 3 defective article
required probability=P(A)
=P(A∩ A1)∩ (A∩A2)
=P(A∩ A1)+(A∩A2)
=P(A1)P(A/A1) + P(A2)P(A/A2)
now P(A/A1)=probability that first 11 draws contain 10 non defective and 1 defective and 12th draw contain a defective article
=18C10 X2C1/20C11 X 1/9
and P(A/A1)=probability that first 11 draws contain 9 non defective and 2 defective and 12th draw contain a defective article
=17C9 X3C2/20C11 X 1/9
Hence required probability
= 0.4 X 18C10 X2C1/20C11 X 1/9 +0.6 X 17C9 X3C2/20C11 X 1/9
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Badiuddin