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Three points are selected randomly on the circumference of the circle. We need to find the probability that there will be a semicircle on which all the three points lie.

I tried as follows :

We first fix a point on the circle on the positive X-axis. (Circle is assumed to be centered at the origin).

Let $X$ denotes the position of the second point. Now two cases arise :

(i) $\:0<X<\pi$

(ii) $\pi\leq X\leq2\pi$

Let $A$ denotes the event that the three points are on the same half circle. So we need to find $P(A)$.

$P(A) = \int P(A|X=x)f_X(x)dx$ , where $X$ is uniformly distributed over $(0,2\pi)$.

When (i) is true , $A$ is possible if and only if the third point belongs to the interval $(\pi -x , \pi)$.

Thus , $P(A|X=x) = \dfrac{(\pi - (\pi-x))}{2\pi}=\dfrac{x}{2\pi} , $ ( when (i) is true ).

Similarly when (ii) is true the third point must lie in the interval $(x-\pi,\pi)$.

Thus , $P(A|X=x) = \dfrac{(\pi-(x-\pi))}{2\pi}=\dfrac{2\pi-x}{2\pi}$, ( when (ii) is true ).

So , $P(A) = \int_0^{2\pi}P(A|X=x)f_X(x)dx=\int_0^{\pi} P(A|X=x)f_X(x)dx + \int_{\pi}^{2\pi} P(A|X=x)f_X(x)dx $ which comes out to be $\frac{1}{4}$.

But the solution says its $\frac{3}{4}$. What am i doing wrong ?

I've checked the calculations twice , but still the same. Can anyone help ?

enter image description here

User9523
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  • the first two points are always on a semi circle, the smaller angle between them is random from zero to pi - call it x - the angle the third point can hit to be on a semi circle is pi + x - so for angle x between the first two points, the provability is (pi + x) / 2pi - so if you integrate that over 0 to pi (remembereing we have the smaller angle), along with the probability density which is 1/pi – Cato Oct 18 '16 at 09:44
  • if you think about the first two points, let's say 10 degrees apart - then can you visualise the region where the third one has to land to be on a semi circle with the first 3? It's most of the circle, can you see that? – Cato Oct 18 '16 at 09:51

1 Answers1

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If $X_{1},X_{2}$ are points on the unit circle and a third point $X_{3}$ is chosen then it will not be located on any semicircle that contains $X_{1}$ and $X_{2}$ if it falls on the smallest arc connecting the two mirrorpoints of $X_{1}$ and $X_{2}$ with respect to the origin.

Denoting this arc by $A\left(X_{1},X_{2}\right)$ and its length by $\left|A\left(X_{1},X_{2}\right)\right|$ we find that $\left|A\left(X_{1},X_{2}\right)\right|$ has uniform distribution on $\left[0,\pi\right]$.

The probability that for $X_{1},X_{2},X_{3}$ there is no semicircle that contains them equals:

$$\int_{0}^{\pi}\Pr\left(X_{3}\in A\left(X_{1},X_{2}\right)\mid\left|A\left(X_{1},X_{2}\right)\right|=u\right)\frac{1}{\pi}du=\int_{0}^{\pi}\frac{u}{2\pi}\frac{1}{\pi}du=\left[\frac{u^{2}}{4\pi^{2}}\right]_{0}^{\pi}=\frac{1}{4}$$

drhab
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  • I've added an image in the above question , referring to that , are you suggesting that the three points would not lie on the same circle if the third point lies in the arc between $X1$ and $X2$ (Red ones) ? – User9523 Oct 19 '16 at 07:37
  • Yes. That's exactly what I meant to say. Only the red $X_1$ should carry the name $X_2$ (as mirror of the black $X_2$ with respect to the origin) and vice versa. – drhab Oct 19 '16 at 07:44