A lot contains 20 articles. The probability that the lot contains exactly 2 defective articles is 0.4 and the probability thatthe lot contains exactly 3 defective articles is 0.6. Articles are drawn in random one by one without replacement andtested till all the defective articles are found. What is the probability that the testing procedure ends at the twelfth testing ?
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Let `E_(i)` denote the event that the lot contains 2 defective articles and `E_(2)` the event that the lot contains 3 defective articles. Suppose that A denotes the event that all the defective articles are found by the twelfth testing, then we have `P(A//E_(1))=((""^(2)C_(2))(""^(17)C_(9)))/((""^(20)C_(11)))xx1/9=11/228` (Here up to 11th draw, 2 defective and 9 non-defective articles are drawn and their (i.e., last) defective articles is drawn at the twelfth draw. ] Using total probability theorem, we have `P(A)=P(E_(1))P(A//E_(1))+P(E_(2))P(A//E_(2))` `=0.4xx11/9+0.6xx11/228=99/1990`
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