Probability that length of Randomly chosen chord of a circle Find  Probability that length of Randomly chosen chord  of a circle lies between $\frac{2}{3}$ and $\frac{5}{6}$ of its diameter.
My try:  I assumed unit circle with center origin. Let two randomly chosen distinct points be $A(\cos \alpha, \sin \alpha)$ and $B(\cos \beta, \sin \beta)$
Length of the chord is $$p=2\sin \left(\frac{\alpha-\beta}{2}\right)$$  
Now we have to find Probability that
$$\frac{4}{3} \le  2\sin \left(\frac{\alpha-\beta}{2}\right) \le \frac{5}{3}$$  
can i have any clue here?
 A: HINT:-
Take a circle, let its radius be $r$.
Take any point in the circle. It doesn't matter what point you take for the first point.
Now as soon as you choose the second point, you have a chord. Let the two points of the chord make angle $2\theta$ at centre. So the length of the chord is $2\cdot r\cdot \sin\theta$.
Now the given constraint is $\frac{2}{3}\le2\sin\theta \le \frac{5}{6}$.
Now this can be solved easily.
A: Let $r$ be the constant radius of the circle, let $L$ be the continuous random variable that represents the chord’s length, and let $\Theta$ be the continuous random variable that represents the counterclockwise angle measured from the first point chosen to the second point chosen.
As mentioned by others, per symmetry arguments, the location of the first point does not affect the probability.
I would approach this problem by writing $L$ in terms of $\Theta$ by using the law of cosines.
$$\begin{align}
c^2 &= a^2 + b^2 -2ab\cos\gamma \\
L^2 &= r^2 + r^2 -2rr\cos\Theta \\
L^2 &= 2r^2(1-\cos\Theta) \\
\end{align}$$
Let us investigate the values of $\Theta$ that correspond the values of $L$ you specified: $\frac{4r}{3}\le L\le\frac{5r}{3}$.
$$\begin{align}
\frac{L^2}{2r^2} &= 1-\cos\Theta \\
\cos\Theta &= 1 - \frac{L^2}{2r^2} \\
\end{align}$$
Plugging in the range of $L$ gives us that 
$$\begin{align}
\cos\Theta &\in \left[ -\frac{7}{18} , \frac{1}{9} \right] \\
\Theta &\in \left[\arccos\left(\frac19\right) , \arccos\left(-\frac{7}{18}\right)\right] \cup \left[2\pi-\arccos\left(-\frac{7}{18}\right) , 2\pi-\arccos\left(\frac19\right)\right] \\
\end{align}$$
For this problem, solving the other side of the interval isn’t important, because only the length of the interval matters, and the two sides are equal in length—but nevertheless, there it is for reference. Call this final interval $J$ for ease.
The probability of selecting one of the desired values of $\Theta$ boils down to the length of $J$ divided by $2\pi$. This value simplifies to  $$\frac{ \arccos\left(-\frac{7}{18}\right) - \arccos\left(\frac19\right) }{\pi} \approx 16.258\,194\,85 \ \%$$
This was a pretty inventive problem, so I humbly disclaim that my math might not be correct, although I feel confident that it is.
A: The distance between $\alpha$ and $\beta$ cannot be negative.
$p = 2|\sin \frac {\beta - \alpha}{2}| = p = 2\sin \frac {|\beta - \alpha|}{2}$
Now, $\beta - \alpha$ is uniformly distributed.  Let $\theta = \beta - \alpha.$ (In fact, we can fix $\alpha = 0$ and get the same result.) And rather than dealing with the absolute value, put theta on the interval $[0,\pi)$
$\frac 43 < 2 |\sin\frac {\theta}{2}| < \frac 53$
$2\arcsin \frac {4}{6} < \theta < 2\arcsin \frac {5}{6}$
This gives us a range of $\theta$ equal to $2(\arcsin \frac 56 - \arcsin \frac 46)$
And we must divide this by possible values of $\theta.$
$\frac {2}{\pi}(\arcsin \frac 56 - \arcsin \frac 46)$
