91Ó°ÊÓ

A point X,Y is chosen at random from disk s

\(S = \left\{ {\left( {x,y} \right) :{{\left( {x - 1} \right)}^2} + {{\left( {y + 2} \right)}^2} \le 9} \right\}.\)

Joint pdf of X and Y is positive for all points inside the circle S.

Area of circle is\(9\pi \)

Joint pdf of X and Y is constant over S

Pdf of x, y will be:

\(f\left( {x,y} \right) = \left\{ \begin{aligned}{l}\frac{1}{{9\pi }}\,\,\,\,\,\,\,for\left( {x,y} \right) \in S\\0\,\,\,\,\,\,\,\,\,\,\,otherwise\end{aligned} \right.\)

Hence, X can lie between -2 and 4.

In short, for-2<x<4

\(\begin{aligned}{f_1}\left( x \right) = \int\limits_{ - 2 - {{\left( {{{\left( {x - 1} \right)}^2}} \right)}^{\frac{1}{2}}}}^{ - 2 + {{\left( {{{\left( {x - 1} \right)}^2}} \right)}^{\frac{1}{2}}}} {\frac{1}{{9\pi }}} dy\\ = \frac{2}{{9\pi }}{\left( {9 - {{\left( {x - 1} \right)}^2}} \right)^{{\raise0.7ex\hbox{$1$} \!\mathord{\left/

{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$2$}}}}\end{aligned}\)

For\( - 2 < x < 4\)and\( - 2 - {\left( {9 - {{\left( {x - 1} \right)}^2}} \right)^{\frac{1}{2}}} < y < - 2 + {\left( {9 - {{\left( {x - 1} \right)}^2}} \right)^{\frac{1}{2}}}\)

Thus,

\(\begin{aligned}{g_2}\left( {y|x} \right) = \frac{{f\left( {x,y} \right)}}{{{f_1}\left( x \right)}}\\ = \frac{1}{9}{\left( {9 - {{\left( {x - 1} \right)}^2}} \right)^{ - {\raise0.7ex\hbox{$1$} \!\mathord{\left/

{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$2$}}}}\end{aligned}\)

b)

When x=2 it follows from part a.

\({g_2}\left( {y|x = 2} \right) = \left\{ \begin{aligned}{l}\frac{1}{{2\sqrt 8 }}\,\,\,{\rm{for}} - 2 - \sqrt 8 < y < - 2 + \sqrt 8 \\0\,\,\,\,\,\,\,\,\,\,{\rm{otherwise}}\end{aligned} \right.\)

Therefore

\(\begin{aligned}{\rm P}\left( {Y > 0|x = 2} \right) = \int\limits_0^{ - 2 + \sqrt 8 } {{g_2}\left( {y|x = 2} \right)dy} \\ = \frac{{ - 2 + \sqrt 8 }}{{2\sqrt 8 }}\\ = \frac{{2 - \sqrt 2 }}{4}\end{aligned}\)