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?

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.
