Using different pdfs to calculate the same thing I have a question that is asking me to evaluate the following: $P(|X-\frac{1}{2}|<\frac{1}{8})$.  Before I get into any other details in this particular question $Y = X(1-X)=-(X-\frac{1}{2})^2+\frac{1}{4}$, this will come in handy when computing using the pdf of $Y$.
I have two pdfs, one of $X$ which is $1$ and is from $0<x<1$ and $0$ otherwise.
The other one is of $Y$, $\frac{1}{\sqrt{\frac{1}{4}-y}}$ from $0<y<\frac{1}{4}$ and $0$ otherwise.
The question is asking me to evaluate the expression above, using both the pdfs of $X$ and $Y$.  I first did the pdf of $X$ and this is what I did:
$P(|X-\frac{1}{2}|<\frac{1}{8})=P(\frac{3}{8}<x<\frac{5}{8})=\int_\frac{3}{8}^\frac{5}{8}1\space dx = \frac{1}{4}$.
Meanwhile using the pdf of $Y$ I did:
$P(|X-\frac{1}{2}|<\frac{1}{8})=P(|X-\frac{1}{2}|^2<(\frac{1}{8})^2)=P((X-\frac{1}{2})^2+\frac{1}{4}<\frac{1}{64}+\frac{1}{4})=P(Y<\frac{17}{64})=\int_0^\frac{1}{4}1\space dy + \int_\frac{1}{4}^\frac{17}{64}0\space dy = 1$.
My answers are different so I was wondering can see if my logic is correct and where I went wrong...

Why did you use $1$ in the integral for $P(Y< \frac{17}{64})$? You should be using the pdf for $Y$. There is another small error: you say that $P((X- \frac12)^2 + \frac14 < \frac 1{64} + \frac14) = P(Y< \frac {17}{64})$ but this is not true. $Y= -(X-\frac12)^2 + \frac14$. Notice the extra minus sign out front.
You were definitely extremely close with your attempt. Start over from $P(|X-\frac12|^2 < (\frac18)^2)$, as that was your last correct step, and then multiply both sides of the inequality by $-1$ and see where that takes you. And remember to use the right pdf for $Y$ !
