Step 1

If two random variables X and Y, are stochastically independent then the joint probability distribution of X and Y can be expressed as \(f(x,y)=f_{X}(x)f_{X}(y)\).

Step 2

The probability density function of X is \(f_{X}=2x\), for \(0 \leq x \leq 1\).

The probability density function of Y is, \(f_{X}=\frac{1}{y^{2}}\), for \(1\leq y\).

Thus,

\(P(X\leq 0.5, Y\leq 2)=(\int_{0}^{0.5}2xdx)(\int_{1}^{2}\frac{1}{y^{2}}dy)\)

\(=(2[\frac{x^{2}}{2}]_{0}^{0.5})([\frac{y^{-2+1}}{-2+1}]_{1}^{2})\)

\(=(0.5^{2}-0)(\frac{1}{1}-\frac{1}{2})\)

=0.125

Thus, the required probability is 0.125.

If two random variables X and Y, are stochastically independent then the joint probability distribution of X and Y can be expressed as \(f(x,y)=f_{X}(x)f_{X}(y)\).

Step 2

The probability density function of X is \(f_{X}=2x\), for \(0 \leq x \leq 1\).

The probability density function of Y is, \(f_{X}=\frac{1}{y^{2}}\), for \(1\leq y\).

Thus,

\(P(X\leq 0.5, Y\leq 2)=(\int_{0}^{0.5}2xdx)(\int_{1}^{2}\frac{1}{y^{2}}dy)\)

\(=(2[\frac{x^{2}}{2}]_{0}^{0.5})([\frac{y^{-2+1}}{-2+1}]_{1}^{2})\)

\(=(0.5^{2}-0)(\frac{1}{1}-\frac{1}{2})\)

=0.125

Thus, the required probability is 0.125.