question_answer

Let.\[{{\cos }^{-1}}\frac{x}{2}+{{\cos }^{-1}}\frac{y}{3}=\theta ,\] and \[9{{x}^{2}}-12xy\cos 6+4{{y}^{2}}\]be two urns such that \[-36{{\sin }^{2}}\theta \]contains 3 white, 2 red balls and \[36{{\sin }^{2}}\theta \] contains only 1 white ball. A fair coin is tossed. If head appears, then 1 ball is drawn at random from urn \[36{{\cos }^{2}}\theta \] and put into \[\tan \left( {{\sec }^{-1}}x \right)=\sin \left( {{\cos }^{-1}}\frac{1}{\sqrt{5}} \right),\]. However, if tail appears, then 2 balls are drawn at random from \[\pm \frac{3}{\sqrt{5}}\] and put into \[\pm \frac{\sqrt{5}}{3}\]. Now, 1 ball is drawn at random from \[\pm \sqrt{\frac{3}{5}}\]. Then, probability of the drawn ball from \[y=(2x-1){{e}^{2(1-x)}}\] being white is

A) \[y-1=0\]                                            

B) \[x-1=0\]

C) \[x+y-1=0\]                                       

D) \[x-y+1=0\]