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# ISI MStat PSB 2007 Problem 6 | Counting Principle & Expectations

This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 6 based on counting principle . Let's give it a try !!

## Problem- ISI MStat PSB 2007 Problem 6

18 boys and 2 girls are made to stand in a line in a random order. Let $X$ be the number of boys standing in between the girls. Find
(a) P(X=5)
(b) $E(X)$

### Prerequisites

Basic Counting Principle

Probability

Discrete random variable

## Solution :

If there are j boys in between 2 girls then first we have to choose j boys out of 18 boys in ${18 \choose j}$ ways now this j boys can arrange among themselves in j! ways and 2 girls can arrange among themselves in 2! ways and now consider these j boys and 2 girls as a single person then this single person along with remaining (18-j) boys can arrange among themselves in (18-j+1)! ways .

Giving all total $2! {18 \choose j} j! (18-j+1)!$ possible arrangements .

Again without any restrictions there are (18+2)!=20! arrangements .

Hence $P(X=j)=\frac{ 2! {18 \choose j} j! (18-j+1)! }{20!} = \frac{19-j}{190}$

(a) $P(X=5)=\frac{19-5}{190}=\frac{7}{95}$

(b) $E(X)= \sum_{j=0}^{18} \frac{j (19-j )}{190} = 9$ just computing the sum of first 18 natural number and sum of squares of first 18 natural numbes.

## Food For Thought

Find the same under the condition that 18 boys and 2 girls sit in a circular table in a random order .

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This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 6 based on counting principle . Let's give it a try !!

## Problem- ISI MStat PSB 2007 Problem 6

18 boys and 2 girls are made to stand in a line in a random order. Let $X$ be the number of boys standing in between the girls. Find
(a) P(X=5)
(b) $E(X)$

### Prerequisites

Basic Counting Principle

Probability

Discrete random variable

## Solution :

If there are j boys in between 2 girls then first we have to choose j boys out of 18 boys in ${18 \choose j}$ ways now this j boys can arrange among themselves in j! ways and 2 girls can arrange among themselves in 2! ways and now consider these j boys and 2 girls as a single person then this single person along with remaining (18-j) boys can arrange among themselves in (18-j+1)! ways .

Giving all total $2! {18 \choose j} j! (18-j+1)!$ possible arrangements .

Again without any restrictions there are (18+2)!=20! arrangements .

Hence $P(X=j)=\frac{ 2! {18 \choose j} j! (18-j+1)! }{20!} = \frac{19-j}{190}$

(a) $P(X=5)=\frac{19-5}{190}=\frac{7}{95}$

(b) $E(X)= \sum_{j=0}^{18} \frac{j (19-j )}{190} = 9$ just computing the sum of first 18 natural number and sum of squares of first 18 natural numbes.

## Food For Thought

Find the same under the condition that 18 boys and 2 girls sit in a circular table in a random order .

## Subscribe to Cheenta at Youtube

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