Let A₁ be the event that the lot contains 2 defective articles and A₂ the event that the lot contains 3 defective articles. Also let A be the event that the testing procedure ends at the twelfth testing. Then according to the question : 

P (A₁) = 0.4 and P (A₂) = 0.6

Since 0 < P (A₁) <1, 0 < P (A₂) < 1, and P (A₁) + P (A₂) = 1 

∴ The events A₁, A₂ form a partition of the sample space. Hence by the theorem of total probability for compound events, we have

NOTE THIS STEP:

 P(A) = P (A₁) P (AI A₁) + P (A₂) P (AI A₂) . . . . . . . . . . . . . . . . . . . . (1) 

Here P (AI A₁) is the probability of the event the testing procedure ends at the twelfth testing when the lot contains 2 defective articles. This is possible when out of 20 articles; first 11 draws must contain 10 non defective and 1 defective articles and 12^th draw must give a defective article. 

∴ P (AI A₁) = ¹⁸C₁₀ x ²C₁/ ²⁰C₁₁ x 1/9 = 11/190 

Similarly, P (AI A₂) = ¹⁷C₉ x ³C₁/ ²⁰C₁₁ x 1/9 = 11/228 

Now substituting the values of P (AI A₁) and P (AI A₂) in eq. (1), we get 

P(A) = 0.4 x 11/190 + 0.6 x 11/228 = 11/475 + 11/380 = 99/1900