about to only mathematics

Answer

A

Explanation

The testing proceduce may terminate at the twelfth testing in two mutually exclusive ways.
I: When lot contains 2 defective articles.
II. When lot contains 3 defective articles.
Let A=testing procedure ends at twelthtesting
`A_(1)`=lot contains 2 defective articles
`A_(2)`=lot contains 3 defective articles
`:.` Required probability
`=P(A_(1)).P(A//A_(1))+P(A_(2)).P(A//A_(2))`
Here, `P(A//A_(1))`=probability that first 11 draws contain10 non-defective and one-defective and twelfth draw contains a defective article.
`(overset(18)""C_(10)xxoverset(2)C_(1))/(overset(20)""C_(11))xx(1)/(9)" "`....(i)`
`P(A//A_(2))`=probability that first 11 draws contains9 non-defective and 2-defective articles and twelfth draw contains defective`=`(overset(17)""C_(9)xxoverset(3)C_(2))/(overset(20)""C_(11))xx(1)/(9)" "`....(ii)`
`:. ` Required probability
`=(0.4)P(A//A_(1))+0.6P(A//A_(2))`
`=(0.4xxoverset(18)""C_(10)xxoverset(2)""C_(1))/(overset(20)""C_(11))xx(1)/(9)+(0.6xxoverset(17)""C_(9)xxoverset(3)""C_(2))/(overset(20)""C_(11))xx(1)/(9)=(99)/(1900)`