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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 !!

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.

Uniform Distribution

Basic geometry

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

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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