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Let the interval $[−r, r]$ be the base of a semicircle. If a point is selected at random from this interval, assign a probability to the event that the length of the perpendicular segment from the point to the semicircle is less than $\frac{r}{2}$.

My approach to this problem was that after the point $x$ was past $r\cdot \cos(30)$ on the right side of the semi-circle then the perpendicular would be less than $\frac{r}{2}$. Therefore you need $P(r \cdot \cos(30)\leq x \leq r)$. From here I am a little lost.

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    $\begingroup$ It is not enough that the point is selected "at random" from the interval -- what is probably meant that it is selected uniformly from that interval. Knowing that the probability distribution is uniform should tell you what the probability of landing in the subinterval $[r\cos 30, r]$ is. $\endgroup$ – hmakholm left over Monica Jun 26 '17 at 15:50
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    $\begingroup$ Note, by the way, that $\cos 30^\circ = \frac12\sqrt3$. $\endgroup$ – hmakholm left over Monica Jun 26 '17 at 15:51
  • $\begingroup$ It would be 1 - cos(30) correct? But that would be for only one side as I have to also take into account points that would be made one the other side of the semi circle $\endgroup$ – Dmitriy Jun 26 '17 at 16:05
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The above commenters did a great job pointing me in the right direction but what I didn't understand was the last part of the answer. After you have the regions for which you "want" the x to be in, you have to divide that over the entire region where the x can end up $(2r)$. In other words I missed the basic rule of probability which is $Probability = \frac{\text{specific outcomes that you want}}{\text{all possible outcomes}}$. Since I understand it now my original method and the method gt6989b provided can both be used

Also please note, as Henning Makholm's comment pointed out, that this problem is using a uniform distribution.

For the first method we do $\frac{2P(r\cdot cos(30)\leq x\leq r)}{2r} = \frac{2\cdot (r-r\cdot cos(30))}{2r}=\frac{2\cdot r\cdot (1- cos(30))}{2r}=1-cos(30)$

For the second method we set up the equation as $\frac{r}{2}\geq \sqrt{r^{2}-x^{2}}$ and solve for x which yields $\pm\frac{\sqrt{3}r}{2}\leq x$ since there's a positive and a negative we can just work with one and we have to multiply it by 2 to account for the other half of the semi circle. Now we do $\frac{2\cdot P\left(\frac{\sqrt{3}r}{2}\leq x\right)}{2r}$ and since we can only choose a random point to $r$ the equation now becomes $\frac{2\cdot \left(r-\frac{\sqrt{3}r}{2}\right)}{2r}=1-\frac{\sqrt{3}}{2}=1-cos(30)$

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Picking $x \in [-r,r]$ at random, you want to find the probability of the distance between the $x$-axis and the semi-circle to be at most $r/2$. In other words, think of the upper semicircle $C$ with the equation $y = \sqrt{r^2-x^2}$ and the horizontal line $L$ given by $y=r/2$. If the point you pick is in a region where $C > L$ you are good. If $C<L$, not so good.

Can you find points of intersection (there are 2 by symmetry) and evaluate the length of the respective intervals of the $x$-axis?

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  • $\begingroup$ $\frac{r}{2}\geq \sqrt{r^{2}-x^2}$ and solve for the $x$ to get the $P(\pm\frac{\sqrt{3}r}{2}\leq x)$ $\endgroup$ – Dmitriy Jun 26 '17 at 16:17
  • $\begingroup$ @DYR yes, your full answer to the problem is good $\endgroup$ – gt6989b Jun 26 '17 at 17:55

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