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 .


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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