Get inspired by the success stories of our students in IIT JAM MS, ISI MStat, CMI MSc Data Science. Learn More

Content

[hide]

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 .

Content

[hide]

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 .

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIALAcademic Programs

Free Resources

Why Cheenta?

Google