Not sure if this problem even makes sense, but anyway:
Lets say you have a button which switches on a light. The light lights green with probability $p$ and red with probability $1-p$. If you push the button and it lights green, can we say anything about the value of $p$? In particular, can one calculate the probability of $p$ lying in the interval $[0.5, 1]$?
It seems like a Baysian problem, but my calculations don't make much sense:
$$ U := \lbrace p\text{ lies in upper half of interval [0, 1]} \rbrace\\ G := \lbrace \text{ light is green} \rbrace\\ P(U|G) = \frac{P(G | U)P(U)}{P(G)} $$ So here $P(G|U)$ is some value $\alpha \in [0.5, 1]$, $P(U)=0.5$ (given that all intervals are equally likely), $P(G)=p$. Needless to say, this looks like the wrong approach.
The (a-priori) pdf of $p$ is $f(p)=1,\; 0\leq p \leq 1$. Let $f(p|G)$ denote the (a-posteriori) pdf of $p$ conditioned on the event $G$. It holds \begin{align} f(p|G) &= \frac{\mathbb{P}(G|p) f(p)}{\int_0^1 \mathbb{P}(G|p) f(p) dp} \\ &= \frac{p}{\int_0^1 p dp} \\ &=2p, \; 0\leq p \leq 1. \end{align} Therefore, $\mathbb{P}(p\geq \frac{1}{2}|G) = \int_{1/2}^1f(p|G)dp = 3/4$.