Correct Answer - A
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
A1=lot contains 2 defective articles
A2=lot contains 3 defective articles
∴ Required probability
=P(A1).P(A/A1)+P(A2).P(A/A2)
Here, P(A/A1)=probability that first 11 draws contain10 non-defective and one-defective and twelfth draw contains a defective article.
18
C10×
2
C
1
20
C11
×
1
9
....(i)P(A//A_(2))=probabilityt
ˆ
f
irst11drawsconta∈s9non-defectiveand2-defectiveartic≤sandtwelfthdrawconta∈sdefective=
17
C9×
3
C
2
20
C11
×
1
9
....(ii):. Requiredprobability=(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)`
