Let $n \in \mathbb{N}$, $k \in \mathbb{N}, k \leq n$. Define $$ P(x) = {n\choose{k}} \cdot x^k \cdot (1-x)^{n-k} \quad \textrm{for} \quad x \in [0,1] $$
The proposed identity is: $$ 1 - \frac{\int_{0}^{1}\int_{0}^{p} P(x)dx \ dp}{\int_{0}^{1}P(x)dx} = \frac{k+1}{n+2} $$
How I met it: I was thinking of a measure of quantifying the underlying probability given a binomial distribution of size $n$ and $k$ successes, which emphasizes the importance of sample size, e.g. the measure for $n=100, k=10$ should be higher than for $n=10, k=1$, even though $\frac{10}{100} = \frac{1}{10}$. This is not a homework question or something similar, just something I thought of out of fun.
- What I know: $P$ is continuous and therefore integratable. The function S with $$ S(p) = \int_{0}^{p} P(x) dx \quad \textrm{for} \quad p \in [0,1]$$ is monotone increasing, as $P(x) \geq 0 \quad \forall x \in [0,1]$ and is therefore integratable. I tried the above identity with a calculator and it seems to be correct, which obviously does not replace a proof. I also know from a video of 3blue1brown that you can arrive at the $\frac{k+1}{n+2}$ measure from a different viewpoint, which I don't know yet.
4. I really have no clue on where to start with this integral, I already fail integrating $P$. On proof techniques I would guess proof by induction, but as already said, I don't know where to start. I am also interested in more background knowledge.
Exchanging integrations, $$\int_0^1\int_0^p P(x)\,dx\,dp=\int_0^1\int_x^1 P(x)\,dp\,dx=\int_0^1(1-x)P(x)\,dx.$$ Now use $\int_0^1 x^m(1-x)^n\,dx=\frac{m!n!}{(m+n+1)!}$ (proven either by induction or using this) to find $$\int_0^1 P(x)\,dx=\frac{1}{n+1},\quad\int_0^1(1-x)P(x)\,dx=\frac{n+1-k}{(n+1)(n+2)}.$$