Factorise the following: `9x^2y^2 – 16 `

Explanation

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`