Probability question involving infinite number of vertical chords in a 1 inch circle. 
Infinite number of vertical chords drawn on a circle with a 1 inch radius. What is the probability that a randomly picked chord is shorter than the radius?

The answer should be 
$1 - .5√ 3$ or $.134$ 
but I'm not sure how to approach this problem. 
 A: Hints: Draw a circle of unit radius. Mark on it a vertical chord of
length $1$ and also a horizontal diameter.  Draw lines between the center of the circle and the ends
of the vertical chord.
What is the distance of the vertical chord that you have drawn
from the center of the circle? What is the length of the diameter
that you have drawn? Now suppose
that the midpoint of the vertical chord "chosen at random"
is uniformly distributed on the horizontal diameter. Can you finish from here? 
A: Put the circle in the standard plane with center at the origin.  I'll assume that "random" means "pick a point randomly on the horizontal diameter and draw the perpendicular chord through that point".  WLOG I will just work with that half of the diameter between x = 0 and x = 1.  We first ask "for which x does the perpendicular chord through x have length 1"?  That is a simple question for Pythagorus and he tells us it is x = $\frac{1}{2}$ $\sqrt{3}$.  Thus the "good" chords will come on the interval between x = $\frac{1}{2}$ $\sqrt{3}$ and x = 1.  It is easy to see that the good interval has length 1 - $\frac{1}{2}$ $\sqrt{3}$.  As the total length was 1, that value is the desired probability.
Just to emphasize:  if you have a different notion of "random" in mind (random angle, random point in the circle interior, random endpoint on the curve, whatever) it is quite likely that the answer will be different.  
A: $\qquad\qquad$ 
The total area of the unit circle is $\pi$. Now take a chord the size of the radius. Notice how all other chords parallel to it cut the semi-disc into two regions: the region with chord-length smaller than the radius, and the one with chord-length greater than the radius, but smaller than the diameter. Can you compute the area of each, knowing that the regular hexagon inscribed in the circle has side-length equal to the radius of the circle ?
A: Draw your circle in a Cartesian plane, with center in the origin.
Fix a point on the circle, let's say $(x_0, y_0)$. Then pick another point $(x_1, y_1)$. This is a way to randomly pick a chord in your circle. In particular, the chord joins $(x_0, y_0)$ to $(x_1, y_1)$. 
Without loss of generalities, let's say that $(x_0, y_0) = (1, 0)$ and consider that $(x_1, y_1) = (\cos(\theta), \sin(\theta))$, for some $\theta$.
In this case, we can say that $\theta$ is randomly drawn in the set $[0, 2\pi)$ with equal probability.
The length of the chord is:
$$L = \sqrt{(\cos(\theta) - 1)^2 + \sin^2(\theta)} = \sqrt{\cos^2(\theta) +1 - 2\cos(\theta) + \sin^2(\theta)} = \\
= \sqrt{2(1 - \cos(\theta))}.$$
Now, we want to compute the probability that $L < 1$. Namely:
$$P(L < 1) = P\left(\sqrt{2(1 - \cos(\theta))}<1\right) = P\left(2(1 - \cos(\theta))<1\right) = \\
= P\left(\cos(\theta)>\frac{1}{2}\right) = P\left(\theta\in\left[0, \frac{\pi}{3}\right] \vee \theta\in\left[\frac{5\pi}{3}, 2\pi\right) \right) = \\
= P\left(\theta\in\left[0, \frac{\pi}{3}\right] \right) + P\left(\theta\in\left[\frac{5\pi}{3}, 2\pi\right) \right).$$
The probability density function of $\theta$ is uniform on $[0, 2\pi)$, and hence:
$$P(L < 1) = \int_0^{\frac{\pi}{3}} \frac{1}{2\pi} \text{d}\theta + \int_{\frac{5\pi}{3}}^{2\pi} \frac{1}{2\pi} \text{d}\theta = \frac{1}{3}.$$
The probability you are looking for is $\frac{1}{3}$.
