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?
Best Answer
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 $\frac{1}{11}$.