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.
 A: 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 
$$0.4    \cdot  \frac {11}{190} + 0.6      \cdot  \frac {11}{228}$$ 
$$ =\frac {99}{1900} \approx 5.2 \%$$
A: Since the chance is considered that the procedure might end before testing all the $20$ items, that means that it is known that 
there are no more than $3$ defective items, otherwise they should be tested all.
But we do not know whether the information that the lot contains $2$ or $3$ defective items is given in advance to the responsible
of the tests.
In the hypothesis  that the number of defects is known, the test-responsible will proceed till finding the 3rd in 60% of the cases, or till finding the 2nd in 40% of the cases.
Instead, in the hypothesis that it is not known, the test-responsible will proceed until finding the 3rd defect (in 60% of the cases),
 and when the lot actually contains only $2$ defects (40% of the cases) he shall proceed till the end.
In the first hypothesis the answer $0.4    \cdot  \frac {11}{190} + 0.6      \cdot  \frac {11}{228}$ by Anatoly holds,
in the second it will be just $0.6      \cdot  \frac {11}{228}$ 
A: There are two cases

*

*2 defective items

*3 defective items

Let us see one by one
Case-1: 2 defective items
Total sample space = $_{}^{20}\textrm{P}_{12}$
Because we can select any 12 items and arrange them in 12! ways.
Now the number of ways the 12$^{th}$ item turns to be our last defective item = ( ways to select 1 of the 2 defective objects ) * ( ways to select 10 of the 18 non-defective objects ) * ( ways to arrange these 11 objects ) * ( selecting the last left defective item and only one way to arrange this, i.e. at the 12$^{th}$ position ) = $\binom{2}{1} * \binom{18}{10} * 11! * \binom{1}{1}$
Probabililty(case-1) = $\frac{\binom{2}{1} * \binom{18}{10} * 11! * \binom{1}{1}}{_{}^{20}\textrm{P}_{12}} = \frac{11}{190}$
Case-2: 3 defective items
Similarly here,
Probabililty(case-2) = $\frac{\binom{3}{2} * \binom{17}{9} * 11! * \binom{1}{1}}{_{}^{20}\textrm{P}_{12}} = \frac{11}{228}$
Final Answer:
$0.4 * Probabililty\:(case-1)  +  0.6 * Probabililty\:(case-2) = 0.052$
