10 points are uniformly taken within the interval $(0,T)$. Find the probability that m out of the 10 points lie within an interval $(0,X)$

Question

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?

Answer 1

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 $$=\binom{10}{m}B(m+1,11-m)$$ $$=\frac{10!}{m!(10-m)!}\cdot\frac{\Gamma(m+1)\Gamma(11-m)}{\Gamma(12)}$$ $$=\frac{10!}{m!(10-m)!}\cdot\frac{m!(10-m)!}{11!}$$ $$=\frac{10!}{11!}$$ $$=\boxed{\frac{1}{11}}$$

Answer 2

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 $\frac{1}{11}$.