10 points are uniformly taken within the interval $(0,T)$. Find the probability that m out of the 10 points lie within an interval $(0,X)$ Q: 10 points are uniformly taken within the interval $(0,T)$. Find the probability that m out of the 10 points lie within an interval $(0,X)$, where X is a uniform random variable over the interval $(0,T)$.
My attempt:
Let $A=\left\{\text{the point is in (0,X)} \right\}$, then
$$
\begin{aligned}
&A=\left\{\text{the point is in (0,X)} \right\}\qquad \text{then,}\\
&P(A)=\int_0^{T} P(A|X=x)\cdot P(X=x)dx\\
&P(A|X=x)=\frac{x}{T}\\
&P(X=x)=\frac{1}{T}\qquad X\sim\text{uniform}\\
&P(A)=\frac{1}{T^2}\int_0^{T}xdx\\
&=\frac{T^2}{2T^2}=\frac{1}{2}
\end{aligned}
$$
$$P\left\{\text{ m out of 10 fall in (0,X)} \right\}=\left(\begin{array}{c}10\\ m\end{array}\right)\left(\frac{1}{2} \right)^m\left( \frac{1}{2}\right)^{10-m}$$
$$=\left(\begin{array}{c}10\\ m\end{array}\right)\left(\frac{1}{2} \right)^{10}$$
But this answer is incorrect ! The probability is 0.09
Can I fix my strategy to get the right answer?
 A: There are $11$ intervals among the $10$ points:  $(0, x_1), (x_1, x_2), \ldots, (x_{10}, T)$.  The point $X$ is equally likely to fall in any of those intervals.  That's because if you arrange the points $\{x_i \mid i= 1, \ldots 10 \}$ and $X$ in order, $X$ is equally likely to fall anywhere in the sequence of $11$ points.
You will find $m$ points within the interval $(0, X)$ precisely when $X \in (x_m, x_{m+1})$ (where by convention we say $x_0=0, x_{11}=T$).  Thus, the probability is precisely $\frac{1}{11}$.
A: For a particular value of $X$, it is clear that the probability that a single point lies in $(0,X)$ is
$$\frac{X}{T}$$
Using binomial coefficients, the probability that $m$ points out of $10$ lie within the interval $(0,X)$ for a particular $X$ is
$$\binom{10}{m}\left(\frac{X}{T}\right)^m\left(\frac{T-X}{T}\right)^{10-m}$$
$$=\binom{10}{m}\frac{(X)^m(T-X)^{10-m}}{T^{10}}$$
Since $X$ is uniformly chosen from the interval $(0,T)$, we have that the probability that $m$ points out of $10$ lie within the interval $(0,X)$ for a random variable $X$ is
$$\frac{\int_0^T \binom{10}{m}\frac{(X)^m(T-X)^{10-m}}{T^{10}}\,\mathrm{d}X}{\int_0^T 1\,\mathrm{d}X}$$
$$=\binom{10}{m}T^{-11}\int_0^T (X)^m(T-X)^{10-m}\, \mathrm{d}X$$
Make a u substitution of $u=\frac{X}{T}$,
$$=\binom{10}{m}\int_0^1 u^m(1-u)^{10-m}\,\mathrm{d}u$$
Using the definition of the beta function, we have that this equation simplifies to
$$=\binom{10}{m}B(m+1,11-m)$$
$$=\frac{10!}{m!(10-m)!}\cdot\frac{\Gamma(m+1)\Gamma(11-m)}{\Gamma(12)}$$
$$=\frac{10!}{m!(10-m)!}\cdot\frac{m!(10-m)!}{11!}$$
$$=\frac{10!}{11!}$$
$$=\boxed{\frac{1}{11}}$$
