Probability that the lot contains defective articles. I tried much in the problem but I didn't get my answer correct.
The question is---
A lot contains 20 articles.the probability that the lot contains exactly 2 defective articles is 0.4 and that the lot contains exactly 3 defective articles is 0.6.articles are drawn from the lot at random one by one without replacement and are tested till all defective articles are found.then the probability that the testing procedure ends at the twelfth testing is------
My attempt -----
let E1 be the event that the lot contains exactly 2 defective articles and E2 be the event that the lot contains exactly 3 defective articles.
I noticed that $P(E1)+P(E2)=1 $and also it is easy to see E1 and E2 ate mutually exclusive. So that implies that the lot contains exactly 3 or 2 defective articles.
Hence my approach was $$\frac{\binom{12}{2} \times 0.4 +\binom{12}{4} \times 0.6}{\binom{20}{12}}$$
But I am not getting the answer.please help me in this regard.
Thanks.

If the lot contains two defective articles and the testing procedure ends at the $12^{th} $ testing, then the first defective article must be chosen among the first $11$ and the second must be the $12^{th} $. The probability that this occurs with the first defective article taken at the first test is
$$  \frac{2}{20}   \cdot   \frac{18}{19}      \cdot  \frac{17}{18}   \cdot    \frac{16}{17}....    \cdot  \frac{9}{10}       \cdot \frac{1}{9}$$
where the first and last fraction express the probability of picking the two defective articles. It is not difficult to show that the same probability (with different fractions but identical resulting factors in the overall product of the numerators and denominators) is obtained is we consider that the first defective article is chosen as the $2^{nd} $, $3^{rd} $... $11^{th} $. So the probability is
$$  11   \cdot   \frac{2}{20}    \cdot  \frac{18}{19}  \cdot    \frac{17}{18}     \cdot  \frac{16}{17}....    \cdot   \frac{9}{10}      \cdot \frac{1}{9}$$ 
$$=11 \cdot 2       \cdot \frac {18!}{8!}    \cdot \frac {8!}{20!} $$
$$=\frac {11}{190} $$
By similar considerations, we get that, if the lot contains three defective articles and the testing procedure ends at the $12^{th} $ testing, the probability that this occurs with the first two defective articles chosen in the first two choices is
$$  \frac{3}{20}    \cdot  \frac {2}{19}    \cdot    \frac{17}{18} \cdot    \frac{16}{17}....      \cdot \frac{9}{10}       \cdot \frac {1}{9} $$
where the first two fractions and the last one express the probability of picking the three detective articles. Again, it is not difficult to show that the same probability, with different fractions but identical resulting factors in the product of the numerators and denominators, is obtained if we consider that the first two defective articles are chosen in any of the possible $\binom {11}{2} $ different combinations. So the probability is
$$   \binom {11}{2}    \cdot \frac{3}{20}    \cdot  \frac{2}{19}  \cdot \frac{17}{18}     \cdot \frac{16}{17}....      \cdot \frac{9}{10}    \cdot    \frac{1}{9}$$ 
$$=55 \cdot 6    \cdot \frac {17!}{8!}     \cdot   \frac {8!}{20!} $$
$$=\frac {11}{228} $$
This, the final probability is 
$$0.4    \cdot  \frac {11}{190} + 0.6      \cdot  \frac {11}{228}$$ 
$$ =\frac {99}{1900} \approx 5.2 \%$$
