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?

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 $\theta.$
$\frac {2}{\pi}(\arcsin \frac 56 - \arcsin \frac 46)$
