Q. 7.9 - A First Course in Probability

Question

Nineteen items on the rim of a circle of radius $1$ are to be chosen. Show that for any choice of these points, there will be an arc of (arc) length $1$ that contains at least $4$ of them.

Step-by-Step Solution

Answer

An arc of (arc) length $1$ that contains at least $4$ of them is $E\left[X\right]=\frac{19}{2\pi }>3$.

Step 1: Given Information

 Nineteen items on the rim of a circle of radius $1$.

Step 2: Explanation

Let the neighbourhood of any point on the rim be the arc starting at the point and extending for a length $1$ and let random variable $X$ denote the number of points that lie in neighbourhood of chosen point.

Further, let's define indicator variables $I_j$ as:

$I_j=\left\{\begin{array}{l}1\\ 0\end{array}\right.$if occurs and does not occur.

Step 3: Explanation

whereby $E_{j,}j=1,2,\dots ,19$, denotes the event

$E_j="j$ th item is the neighbourhood of random chosen point ".

$X=\sum _{j=1}^{19}{I}_j$

and therefore the expected number of $X$ is,

$E\left[X\right]=E\left[\sum _{j=1}^{19}{I}_j\right]=\sum _{j=1}^{19}E\left[I_j\right]=\sum _{j=1}^{19}P\left\{E_j\right\}=\sum _{j=1}^{19}\frac{1}{2\pi }=\frac{19}{2\pi }>3$.

Step 4: Final answer

An arc of (arc) length $1$ that contains at least $4$ of them is $E\left[X\right]=\frac{19}{2\pi }>3$.