Q.6.16 - A First Course in Probability

Question

Suppose that n points are independently chosen at random on the circumference of a circle, and we want the probability that they all lie in some semicircle. That is, we want the probability that there is a line passing through the center of the circle such that all the points are on one side of that line, as shown in the following diagram: 

Let P1, ... ,Pn denote the n points. Let A denote the event that all the points are contained in some semicircle, and let Ai be the event that all the points lie in the semicircle beginning at the point Pi and going clockwise for 180◦, i = 1, ... , n.

 (a) Express A in terms of the Ai.

 (b) Are the Ai mutually exclusive?

 (c) Find P(A). 

Step-by-Step Solution

Answer

(a) $A={\cup }_{i=1}^n{A}_i$

(b) $A_i$ are almost mutually exclusive, but not quite.

(c) Probability : $P\left(A\right)=n(1/2{)}^{n-1}$     

Step 1: Introduction

The circumference of a circle or ellipse is its perimeter. That is, if the circle were expanded out and straightened out to a line segment, the circumference would be the arc length of the circle. The perimeter, in general, is the length of any closed figure's curve.

Step 2: Given

Shown that line passing through the center of the circle such that all the points are on one side of that line.

Step 3: Explanation (a)

Let

${\mathit{P}}_1,\dots ,{\mathit{P}}_n$denote the n points.

A: event contains all the points in some semicircle.

$A_i$ : event for all the points lie in the semicircle beginning at the point $P_i$

For each i,

$A_i\subset A$

Thus,

${\cup }_{i=1}^n{A}_i\subseteq A$

Now,

For reverse inclusion:

If A occurs,

Then

The semicircle can be rotated clockwise until it begins at $P_i$.

That shows

$A_i$ and ${\cup }_{i=1}^n{A}_i$  occurs.

Such that

$A\subseteq {\cup }_{i=1}^n{A}_i$

Therefore,

$A={\cup }_{i=1}^n{A}_i$

Step 4: Explanation (b)

Let

$P_1,\dots ,P_n$ denote the n points.

A : event contains all the points in some semicircle.

$A_i$ : event for all the points lie in the semicircle beginning at the point $P_i$ If $P_i=P_j$ ,

Then

$A_i$ and $A_j$ cannot be mutually exclusive.

And

$P_k$ lies in the semicircle going clockwise for $180°$ and starts at $P_i=P_j$.

In such case,

Both $A_i$ and $A_j$ occur.

On the other hand,

If $P_i\ne P_j$,

Then

Both $A_i$ and $A_j$ cannot occur.

Also,

If $P_j$ is in the semicircle going clockwise for $180°$ and starts at$P_i$.

Then

$P_i$ is in the semicircle going clockwise for $180°$ and starts at $P_j$.

Thus,

$A_i{A}_j\subseteq \left(P_i=P_j\right)$

Furthermore,

$P\left(P_i=P_j\right)=0$

Such that

$P\left(A_i{A}_j\right)=0$

Thus,

We can say that

$A_i$ are almost mutually exclusive, but not quite.

Step 5: Explanation (c)

Let

$P_1,\dots ,P_n$ denote the n points.

A : event contains all the points in some semicircle.

$A_i$ : event for all the points lie in the semicircle beginning at the point $P_i$

From Part (b),

We have

$P\left(A_i{A}_j\right)=0$

That follows

$P\left(A_i{A}_j{A}_k\right)=0$

And

It is also true for all higher order intersections.

Then

We have

$P\left(A\right)\subseteq P\left(\underset{i=1}{\overset{n}{\cup }}{A}_i\right)$

By the identity of inclusion-exclusion:

$P\left(A\right)=\sum _{i=1}^{n}P\left(A_i\right)-\sum _{i<j}P\left(A_i{A}_j\right)+\sum _{ixj<k}P\left(A_i{A}_j{A}_k\right)-\dots$

Since all other terms are zero,

$P\left(A\right)=\sum _{i=1}^{n}P\left(A_i\right)$

Now,

Fix

$i\in \{1,\dots ,n\}$

And

A randomly chosen$P_i$.

For $j\ne i$,

The probability that the randomly chosen point $P_j$ is in the semicircle going clockwise for $180°$ and starts at

$P_i$ is $1 / 2$ .

$i\in \{1,\dots ,n\}$

And

A randomly chosen $P_i$.

For $j\ne i$,

The probability that the randomly chosen point $P_j$ is in the semicircle going clockwise for $180°$ and starts at

$P_i is 1 / 2$.

Thus,

$P\left(A\right)=\sum _{i=1}^{n}P\left(A_i\right)=n(1/2{)}^{n-1}$

Thus,

$P\left(A\right)=\sum _{i=1}^{n}P\left(A_i\right)=n(1/2{)}^{n-1}$