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

$\frac{99}{1900}$

The testing procedure 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.

Consider the following events.

A = Testing procedure ends at the twelfth testing.

$A_1$ = Lot contains 2 defective articles.

$A_2$ = Lot contains 3 defective articles.

Required probability

$= P(A) = P(A ∩ A_1)∪ (A ∩ A_2) = P(A ∩ A_1)+P(A ∩ A_2)= P(A_1)P(\frac{A}{A_1})+P(A_2)P(\frac{A}{A_2})$

Now, $P(\frac{A}{A_1}$= Probability that first 11 draws contain 10 non-defective and one defective and 12th draw contains a defective article.

$= \frac{^{18}C_{10}×{^2C}_1}{^{20}C_{11}}× \frac{1}{9}.$

And $P(\frac{A}{A_1})$= Probability that first 11 draws contain 9 non defective and 2 defective articles and

12th draw contains a defective article $= \frac{^{17}C_{9}×{^3C}_2}{^{20}C_{11}}× \frac{1}{9}.$

Hence, required probability $= 0.4×\frac{^{18}C_{10}×{^2C}_1}{^{20}C_{11}}× \frac{1}{9}+0.6 × \frac{^{17}C_{9}×{^3C}_2}{^{20}C_{11}}× \frac{1}{9}=\frac{99}{1900}.$