A lot contains 20 articles. The probability that the lot contains exactly 2 defective articles is 0.4 and the probability thatthe lot contains exactly 3 defective articles is 0.6. Articles are drawn in random one by one without replacement andtested till all the defective articles are found. What is the probability that the testing procedure ends at the twelfth testing ?

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)`