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 .

Similar Problems and Solutions

Related Program

ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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ISI MStat PSB 2005 Problem 2 | Calculating probability using Binomial Distribution

This is a very beautiful sample problem from ISI MStat PSB 2005 Problem 2 based on finding probability using the binomial distribution. Let's give it a try !!

Problem- ISI MStat PSB 2005 Problem 2

Let \(X\) and \(Y\) be independent random variables with X having a binomial distribution with parameters 5 and \(1 / 2\) and \(Y\) having a binomial distribution with parameters 7 and \(1 / 2 .\) Find the probability that \(|X-Y|\) is even.

Prerequisites

Binomial Distribution

Binomial Expansion

Parity Check

Solution :

Given \( X \sim \) Bin(5,1/2) and \( Y \sim \) Bin(7,1/2) , and they are independent .

Now , we have to find , \( P(|X-Y|=even ) \)

\( |X-Y| \)= even if both X and Y are even or both X and Y are odd .

Therefore \( P(|X-Y|=even )=P(X=even,Y=even) + P(X=odd , Y=odd) \)

P(X=even , Y= even ) =\( ( {5 \choose 0} {(\frac{1}{2})}^5 + {5 \choose 2} {(\frac{1}{2})}^5 + \cdots + {5 \choose 4} {(\frac{1}{2})}^5 )( {7 \choose 0} {(\frac{1}{2})}^7 + {7 \choose 2} {(\frac{1}{2})}^7 + \cdots + {7 \choose 6} {(\frac{1}{2})}^7)\)

=\( ({(\frac{1}{2})}^5 \times \frac{2^5}{2})({(\frac{1}{2})}^7 \times \frac{2^7}{2}) \)

= \(\frac{1}{4} \)

Similarly , one can find P(X=odd , Y=odd ) which is coming out to be \( \frac{1}{4} \) .

Hence , P(|X-Y|) = 14+1/4 = 1/2 .

Food For Thought

Try to find P(X-Y=odd) under the same condition as given in the above problem .

Similar Problems and Solutions

Related Program

ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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Test of Mathematics Solution Subjective 35 - Divisibility by 16

Test of Mathematics at the 10+2 Level

Test of Mathematics Solution Subjective 35 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

Problem

(a) Prove that, for any odd integer n, $ n^4 $ when divided by $16$ always leaves remainder $1$.

(b) Hence or otherwise show that we cannot find integers $n_1 , n_2 , ... , n_8 $ such that $n_1^4 + n_2^4 + ... + n_8^4 = 1993 $.

Solution

For part (a) we consider $n = 2k +1$ and expand it's fourth power binomially to get $ (2k+1)^4 = {{4} \choose {0}} (2k)^4 +{{4} \choose {1}} (2k)^3 + {{4} \choose {2}} (2k)^2 + {{4} \choose {3}} (2k)^1 + {{4} \choose {4}} (2k)^0 = 16k^4 + 32k^3 + 24k^2 + 8k + 1 $

Now $24k^2 + 8k = 8k(3k+1) $ ; if $k$ is even then $8k$ is divisible by $16$ and if $k$ is odd $3k+1$ is even and product of $8k$ and $3k+1$ is divisible by $16$. Since $16k^4 + 32 k^3 $ is already divisible by $16$ we conclude $ (2k+1)^4 $ when divided by $16$ gives $1$ as remainder.

For part (b) we note that $1993$ when divided by $16$, produces $9$ as the remainder. Each of the eight of fourth powers when divided by $16$ produces either $0$ (when $n_i $ is even) or $1$ (when $n_i $ is odd using part (a)) as remainder. Thus they can add up to at most $8$ (modulo $16$) hence can never be equal to $9$ (which $1993$ is modulo $16$).