This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 6 based on counting principle . Let's give it a try !!
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) \( E(X) \)
Basic Counting Principle
Probability
Discrete random variable
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.
Find the same under the condition that 18 boys and 2 girls sit in a circular table in a random order .
This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 6 based on counting principle . Let's give it a try !!
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) \( E(X) \)
Basic Counting Principle
Probability
Discrete random variable
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.
Find the same under the condition that 18 boys and 2 girls sit in a circular table in a random order .
