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Consider a circle with a circumference of one unit length. Let $d< \dfrac{1}{6}$. Suppose that we independently throw two arcs, each of length $d$, randomly on this circumference so that each arc is uniformly distributed along the circle circumference. The arc attaches itself exactly to the circumference so that arc of length $d$ exactly covers length $d$ of the circumference. What can be said about the probability that the two arcs do not intersect each other?

  1. It equals $(1 - d)$
  2. It equals $(1 - 3d)$
  3. It equals $(1 - 2d)$
  4. It equals $1$
  5. It equals $(1 - d)$ $(1 - d)$
in Numerical Ability by Boss (30.7k points)
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$(1 - 2d)$ will be the correct answer.


Two points on the circumference of any circle divides the circle in two arcs, the length of smaller arc must be less than or equal to half of the circumference, & length of the larger arc must be greater than or equal to half the circumference.

but since here given length of the arc under consideration is strictly less than $\frac{1}{2}$, so henceforth in this answer, whenever I will use the term "arc", I'll be referring to the smaller of those two arcs.

Process of Arc drawing: I am going to follow a specific procedure for drawing any arc of length $d$, which is as follows:

  1. Pick any point on the circumference of the circle, this will be the starting point.
  2. Move $d$ units CLOCKWISE on the circumference of the circle & mark that point as the finishing point.

Suppose we choose our first arc $AB$ of length $d$, randomly anywhere on the circumference on the circle.

Here $A$ is the starting point & $B$ is the end point.

After drawing the arc $AB$, we have to draw another arc $PQ$ on the circle of length $d$, where $P$ will be the starting point & $Q$ will be the end point.

Now if we have to make sure that arc $PQ$ does not intersects with arc $AB$, we have to keep following things in mind hile choosing our starting point $P$:

  1. $P$ can not lie within arc $AB$, otherwise $AB$ and $PQ$ will intersect each other.
  2. $P$ can not lie anywhere within the anticlockwise distance $d$ from the point $A$ other wise, end part of arc $PQ$ will intersect with starting part of arc $AB$.

So, we can conclude that "If $P$ lies anywhere on the circumference of the circle within a distance $d$ from $A$ then the arc $PQ$ & $AB$ will intersect."

So $Probability\left (\text{Arc PQ does not intersects with Arc AB} \right ) = \frac{\text{P lies at least d distance away from A}}{\text{P lies anywhere in the circumference}}$

$\Rightarrow Probability\left (\text{Arc PQ does not intersects with Arc AB} \right ) = \frac{1-2d}{1} = \left ( 1 - 2d \right ).$

by Boss (14.3k points)
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Here 1-2d because 1 is the circumference and 2d is the sum of the two arcs?If wrong,please correct.
+1

P can not lie anywhere within the anticlockwise distance d from the point A other wise, end part of arc PQ will intersect with starting part of arc AB

I understand what you wanted to say but then if arc PQ lies d distance from point A, then also P will be intersecting with the ending point of arc AB i.e. with B. How to take care of that? In short, leaving how much distance from B should PQ start? 

The distance is so negligible that we can safely avoid it. PQ starting even after 1mm or less from B will prevent it from intersecting.

This seems very similar to calculating the vulnerable time in Pure Aloha :P 

0

 YES :) just now solved this and it reminded me of pure aloha!

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