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# ISI MStat PSB 2004 Problem 4 | Calculating probability using Uniform Distribution This is a very beautiful sample problem from ISI MStat PSB 2004 Problem 4 based on finding the probability using Uniform distribution . Let's give it a try !!

## Problem- ISI MStat PSB 2004 Problem 4

Two policemen are sent to watch a road that is $1 \mathrm{km}$ long. Each of the two policemen is assigned a position on the road which is chosen according to a uniform distribution along the length of the road and independent of the other's position. Find the probability that the
policemen will be less than 1/4 kilometer apart when they reach their assigned posts.

### Prerequisites

Uniform Distribution

Basic geometry

## Solution :

Let X be the position of a policeman and Y be the position of another policeman on the road of 1km length .

As it is given that chosen according to a uniform distribution along the length of the road and independent of the other's position hence we can say that $X \sim U(0,1)$ and $Y \sim U(0,1)$ and X,Y are independent .

Now we have to find the probability that the policemen will be less than 1/4 kilometer apart when they reach their assigned posts , which is

nothing but $P(|X-Y|< \frac{1}{4} )$ .

So , let's calculate the probability $P(|X-Y|< \frac{1}{4} )$ here some sort of geometry will help to calculate it easily !

In general we have 0<X<1 and 0<Y<1 and hence the total probability is the area of the square $1 \times 1$

And in favourable case we have $|X-Y|<1/4 , 0<X<1 , 0<Y<1$ . so, it's basically the area covered by ACBDEF = Area covered by square - area of the triangles BGD and AFH = $1 \times 1$ - $2 \times$ $\frac{1}{2}$ $\times \frac{3}{4} (1- \frac{1}{4} )$ = $1-9/16$ .

Therefore $P(|X-Y|<1/4)= \frac{1-9/16}{1} = \frac{7}{16}$

## Food For Thought

Calculate the same under the condition that road is of length (b-a) , b>a and both are positive real number .

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This is a very beautiful sample problem from ISI MStat PSB 2004 Problem 4 based on finding the probability using Uniform distribution . Let's give it a try !!

## Problem- ISI MStat PSB 2004 Problem 4

Two policemen are sent to watch a road that is $1 \mathrm{km}$ long. Each of the two policemen is assigned a position on the road which is chosen according to a uniform distribution along the length of the road and independent of the other's position. Find the probability that the
policemen will be less than 1/4 kilometer apart when they reach their assigned posts.

### Prerequisites

Uniform Distribution

Basic geometry

## Solution :

Let X be the position of a policeman and Y be the position of another policeman on the road of 1km length .

As it is given that chosen according to a uniform distribution along the length of the road and independent of the other's position hence we can say that $X \sim U(0,1)$ and $Y \sim U(0,1)$ and X,Y are independent .

Now we have to find the probability that the policemen will be less than 1/4 kilometer apart when they reach their assigned posts , which is

nothing but $P(|X-Y|< \frac{1}{4} )$ .

So , let's calculate the probability $P(|X-Y|< \frac{1}{4} )$ here some sort of geometry will help to calculate it easily !

In general we have 0<X<1 and 0<Y<1 and hence the total probability is the area of the square $1 \times 1$

And in favourable case we have $|X-Y|<1/4 , 0<X<1 , 0<Y<1$ . so, it's basically the area covered by ACBDEF = Area covered by square - area of the triangles BGD and AFH = $1 \times 1$ - $2 \times$ $\frac{1}{2}$ $\times \frac{3}{4} (1- \frac{1}{4} )$ = $1-9/16$ .

Therefore $P(|X-Y|<1/4)= \frac{1-9/16}{1} = \frac{7}{16}$

## Food For Thought

Calculate the same under the condition that road is of length (b-a) , b>a and both are positive real number .

## Subscribe to Cheenta at Youtube

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