With the following question :
"Alice and Bob show up at the post office between 3 and 4 p.m. Their arrival times
respectively $T_A$ (Alice) and $T_B$ (Bob) are independent and uniformly distributed
in this interval. We call $T_{min}$ the time of the first arrival (therefore $T_{min} =
min(T_A, T_B)$), and $T_{max}$ the time of the second arrival (thus $T_{max} = max(T_A, T_B)$).
To simplify the notations on the representative measurement of time in hours:
time $0$ is 3 p.m., time $1$ is 4 p.m., time $1/2$ is 3:30 p.m., etc.
Find the distribution function $F_{min}$ of $T_{min}$.
Hint: Alice and Bob both arrive strictly after 3:15 p.m. if and only if $T_{min} > 1/4$."
I don't understand some things in the solution, which is:
"The hint tells us that $T_{min} > 1/4$ iff $T_A > 1/4$ AND $T_B > 1/4$. This observation is true for any time $t$: $T_{min} > t$ iff $T_A > t$ AND $T_B > t$. We remember that $F_{min}(t) = P(T_{min}\leq t)$. So $1 - F_{min}(t) = P(T_{min}> t) = P(T_A > t)P(T_B > t)$ by independence of $T_A$ and $T_B$. So $F_{min}(t) = 1-(1-t)^2 = 2 t - t^2$ for $t \in [0,1]$. For $t\leq0$, $F_{min}(t) = 0$ and for $t\geq1$, $F_{min}(t) = 1$."
What I don't understand:
Why do we do $1 - F_{min}(t)$ to switch the direction of the inequality in $P(T_{min}> t)$? (in short, why don't we directly try to find $P(T_{min}\leq t)$ ?)
How do we get $P(T_A>t) = (1-t)$? I can't find the law or the formula giving this. Also, if it's pure logic, I always come back to my previous question.
By the way $P(T_A>t) = P(T_B>t)$ hence the square of $(1-t)^2$, that I understand.
For your first question: you need to use the observation that "$T_{\min} > t$ if and only if $T_A > t$ and $T_B>t$" which gives a direct route to computing $P(T_{\min} > t)$. There isn't a "corresponding" observation of the form "$T_{\min} \le t$ if and only if $T_A \le t$ and $T_B \le t$" (or something similar), so the provided approach for finding $P(T_{\min} \le t)$ is as direct as you can get, even if it goes via finding $1 - P(T_{\min} > t)$.
$P(T_A > t) = 1-t$ comes from the definition of $T_A$ being uniform on $[0, 1]$.