Question about the Solution for Problem 3.7b, Introduction to Probability (Bertsekas, 2nd Edition) I am currently working on the problems in Introduction to Probability (Bertsekas and Tsitsiklis, 2nd Edition) and one of the problems is as follows:

For Problem (b), the final answer is as follows:

My question:
Why do we have cumulative density for 0 <= s <= 1/t? Shouldn't the cumulative density be 0 on the 0 <= s <= 1/t interval? Based on the problem statement, my assumption is that s can only be 0 or [1/t, +inf) so anything outside of that shouldn't be possible (hence, 0 probability). Am i missing something here?
Thanks in advance.
P.S The final answer is taken from the book's Problem Solutions supplement.
 A: 
my approach was to start with the values of Alvin's score (S) then map it back to the corresponding Xs (X = distance from the center) to achieve that score.

This is the right approach, with one small modification: you should map back to the corresponding values of $X$ that achieve that score or less. That is the quantity that the CDF is referring to. Remember that $F_S(s) = P(S \leq s) = \int_{-\infty}^s f(s)ds$.
For example, under the rules in (b) there is no way to get a score of $S = \frac{1}{2t}$. But there is a way to get a score of $S \leq \frac{1}{2t}$. Any dart outside the inner circle will result in $S = 0 \leq \frac{1}{2t}$.
So score values of $0 \leq s < \frac{1}{t}$ all come from missing the inner circle, which is why the CDF has a long 'flat' section over that range.
A: Let $C_x$ and $C_y$ denote the coordinates of Alvin's throw on the dart board with radius $r$.  By assumption, each point is equally likely, hence the joint density function of $C_x$ and $C_y$ is $f_{C_x,C_y}(c_x,c_y) = 1/(\pi r^2)$ such that $c_x^2+c_y^2 \le r^2$.
Let $X$ denote the distance of Alvin's hit from the center.  Clearly, $X^2 = C_x^2 + C_y^2$.  Therefore, the CDF of $X$ is
\begin{eqnarray*} 
F(X) = \mbox{Pr}\left[X \le x\right] = \mbox{Pr}\left[\sqrt{C_x^2 + C_y^2} \le x\right]=\frac{x^2}{r^2},
\end{eqnarray*}
where the penultimate equality holds since if his throw is less than $x$ units away from the center, than $C_x$ and $C_y$ must lie in this smaller circle.
Next let $S = \frac{1}{X}I[X \le t]$, where $I[\cdot]$ denotes the indicator function.  Clearly given $X \le t$, then we have a simple univariate transformation, and the PDF of $S$, given $S \ge 1/t$, is $f_S(s) = \frac{2}{y^3r^2}$.  Given $S$ is 0, it must hold that $X\ge t$.  Evaluating the probability of this event and coupling these we have,
\begin{eqnarray*}
f_S(s) = \begin{cases} 1-\frac{t^2}{r^2} & \mbox{if} \quad s = 0  \\
\frac{2}{s^3r^2} & \mbox{if} \quad s \ge 1/t
\end{cases}.
\end{eqnarray*}
Now we are ready to find the CDF of $S$.  Note, $\mbox{Pr}[S\le s] = \mbox{Pr}[S = 0] + \mbox{Pr}[1/t \le S \le s]$.  Clearly, if $S<0$, then the CDF is $0$, and if $S<1/t$ then only the first term contributes.  If $S\ge 1/t$, then we need to evaluate the second term which turns out to be
\begin{eqnarray*}
\mbox{Pr}[1/t \le S \le s] = \int_{1/t}^s \frac{2}{x^3r^2} \mbox{d}x = \frac{t^2}{r^2}-\frac{1}{s^2r^2}.
\end{eqnarray*}
The result follows by combining these 3 cases as in the solution manual.
To answer your question, "Shouldn't the cumulative density be 0 on the 0 <= s <= 1/t interval?"  Recall that in the name it is CUMULATIVE.  Hence, the probability at 0, extends indefinitely.  Since there is no other change, it remains constant on this interval.
