91Ó°ÊÓ

In a certain drug, a particular chemical concentration is a random variable and follows a continuous distribution.

The probability density function is,

\(g\left( x \right) = \left\{ \begin{array}{l}\frac{3}{8}{x^2}\;for\;0 \le x \le 2\\0\;otherwise\end{array} \right.\)

In two separate batches of the chemical, two concentrations X and Y are independent random variables.

As X and Y are independent, so, the joint p.d.f is,

\(\begin{array}{c}g\left( {x,y} \right) = g\left( x \right)g\left( y \right)\\ = \left( {\frac{3}{8}{x^2}} \right) \times \left( {\frac{3}{8}{y^2}} \right)\\ = \frac{9}{{64}}{x^2}{y^2}{I_{\left\{ {\left( {r,s} \right)\left| {0 \le x \le 2,0 \le y \le 2} \right.} \right\}}}\end{array}\)

b.

As we know from the given information, Xand Y have a continuous joint distribution, so, we can say that,

\(\Pr \left( {X = Y} \right) = 0\)

c.

Since XAnd Y have the same probability distribution and are independent, so, we can say that\(\Pr \left( {X > Y} \right) = \Pr \left( {Y < X} \right)\).

Now, referring to part b, we know that\(\Pr \left( {X = Y} \right) = 0\).

So, it follows that \(\Pr \left( {X > Y} \right) = 0.5\).

d.

\(\Pr \left( {X + Y \le 1} \right) = \Pr \left( {X \le 1 - Y} \right)\)

Now,

\(\begin{array}{c}\Pr \left( {X + Y \le 1} \right) = \int\limits_0^1 {\int\limits_0^{1 - y} {f\left( {x,y} \right)dxdy} } \\ = \int\limits_0^1 {\int\limits_0^{1 - y} {\frac{9}{{64}}{x^2}{y^2}dxdy} } \\ = \frac{1}{{1280}}\end{array}\)

Thus, the probability is \(\frac{1}{{1280}}\) .