# Probability Distribution and Transformation

Let X have the following distribution function

F(x)= 0 for x<0; x/3 for 0<=x<=2; 1 for x>2.

Let Y=X^(2)

To find P(Y<(X)) Do we simply substitute X^2 for Y and divide to get P(X<1) and then find F(1)?

If not, how would one go about solving this question?

It is exactly as you said, as $P(X^2<X)=P(X<1)= F(1)$ as your distribution function is continous in $1$ (note that $P(X<2)\neq F(2)$ as $P(X=2)= \frac{1}{3}$