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

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$\frac{345}{1574}$

$\frac{99}{1900}$

$\frac{97}{2538}$

$\frac{11}{25}$

answer is B.

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Detailed Solution

(2) E₁ = lot contains 2 defective items
E₂ = lot contains 3 defective items
A = testing ends with the 12^th trial
P(E₁) = 0.4, P(E₂) = 0.6
$P (A | E_{1}) = \frac{2_{C_{1}} \times 18_{C_{10}}}{20_{C_{11}}} \times \frac{1}{9} = \frac{11}{190}$
$P (A | E_{2}) = \frac{3_{C_{1}} \times 17_{C_{9}}}{20_{C_{11}}} \times \frac{1}{9} = \frac{11}{228}$
$\begin{array}{l} P (A) = P (E_{1}) . P (A | E_{1}) + P (E_{2}) . P (A | E_{2}) \\ = \frac{99}{1900} \end{array}$