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

ISI MStat PSB 2013 Problem 10 | Balls-go-round

This is a very beautiful sample problem from ISI MStat PSB 2013 Problem 10. It's based mainly on counting and following the norms stated in the problem itself. Be careful while thinking !

Problem- ISI MStat PSB 2013 Problem 10


There are 10 empty boxes numbered 1,2,......,10 placed sequentially on a circle as shown in the figure,

We perform 100 independent trials. At each trial one box is selected with probability \frac{1}{10} and a ball is placed in each of the two neighboring boxes of the selected one.

Define X_k be the number of balls in the k^{th} box at the end of 100 trials.

(a) Find E(X_k) for 1 \le k \le 10.

(b) Find Cov(X_k, X_5) for 1 \le k \le 10.

Prerequisites


Counting principles

Binomial Distribution

Independence of Events

Solution :

At first this problem may seem a bit of complex, but when one get to see the pattern it starts unfolding. For his types of problem, I find a useful technique is to follow the picture, in this case the picture is being provided, (if not draw it yourself !!)

Given, X_k : # balls in the k^{th} box at the end of 100 trials.

So, the possible values of X_k are 0,1,2,.....,100. and the probability that at the jth trial a ball will added to the kth box is \frac{1}{10} , (why??) .

Now, P(X_k=x)= {100 \choose x}(\frac{1}{5})^{x} (\frac{4}{5})^{100-x} x=0,1,......,100

Clearly, X_k \sim binomial( 100, \frac{1}{5}), from here one can easily find out the the expectation of X_k. But have you thought like this ??

Now, notice that after every slelection of box, 2 balls are added in the system, so at the end of the 10th trial, there will be 200 balls distributed in the system.

So, X_1 +X_2 +.......+ X_{100} =200, which implies \sum_{k} E(X_k)=200, due to symmetry E(X_k)=E(X_l)  \forall k \neq l, So, E(X_k)=20.

(b) Now this part is the cream of this problem, first notice that number of balls in kth box, is dependent on the number of balls in the (k-2)th and (k+2)th box, and vice versa, So, Cov( X_k, X_l)=0 if |k-l|\neq 2.

So, Cov(X_k, X_5)= 0 \forall  k \neq 3,5 \&\ 7, so we just need to find the Cov(X_7,X_5) and Cov(X_3,X_5), and Cov(X_5,X_5)=Var(X_5) .

Now it is sufficient to find the covariance of any of the the above mentioned covariances, as both are symmetric and identical to each other. But for the finding say Cov(X_3,X_5), lets look whats happening in each trial more closely,

let, X_k= i_{k_1} +i_{k_2}+......+i{k_{100}} where , i_{k_j} = \begin{cases} 1  &  if\ a\ ball\ added\ to\ the\ kth\ box\ at\ the\ jth\ trial\ \\ 0 & otherwise \end{cases}

So, clearly, P(i_{k_j}=1)=\frac{1}{5} ; j=1,2,....,100.

So, Cov(X_3,X_5)=Cov( i_{3_1}+i_{3_2}+.....+i_{3_{100}},i_{5_1}+i_{5_2}+....+i_{5_{100}})=\sum_{j=1}^{100} Cov(i_{3_j},i_{5_j}), [Cov(i_{3_j},i_{5_j*})=0 \forall j\neq j*, why ?? ].

So, Cov(X_3,X_5)= 100 Cov(i_{3_1},i_{5_1})=100( E(i_{3_1}i_{5_1})-E(i_{3_1})E(i_{5_1}))=100(P(i_{3_1}=1, i_{5_1}=1)-P(i_{3_1}=1)P(i_{5_1}=1))=100(\frac{1}{10}- \frac{1}{5}\frac{1}{5})=6.

similarly, Cov(X_7,X_5)=6 also, and its easy to find Var(X_5) , so I leave it as an exercise. So, Cov(X_k,X_5)= \begin{cases} 6 & k=3,7 \\ Var(X_5) & k=5 \\ 0 & k\neq 3,5 \or\ 7\end{cases}. Hence we are done !!


Food For Thought

Wait, lets imagine, these boxes are interchanged in such a way that the hth box is replaced with the kth (\neq h) box, this has been done for all possible pairs of (h,k), Now can you show that all there are precisely 10!\sum_{i+2j=10}(i!j!2^j)^{-1} number of arrangements possible ?

Now imagine, in a game there are 1 balls each in every box( boxes are arranged identically as shown in the question), you pick up the ball from the first box and put it into the 2nd one, now you can't take out any ball from a box in which you just put a ball, so you pick a ball from the 3rd box and put it into the 4th and you go on like this, taking a ball frim the ith box and put it into the (i+1)th box, but cant empty the box you just filled. Now, again after the first round you remove the empty boxes and do the same thing again and again, till all the balls are not accumulated in a single box. Which box you think will contain all the balls after you run this process finitely many time ?? If you are in this lottery and you are to choose a bix before this game begin, which box you must choose ??

if the coordinator of the game, starts with any ith box how should your strategy change ?? Give it a thought !!

For help, look for Josephus Problem, you may be moved by it's beauty !


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


This is a very beautiful sample problem from ISI MStat PSB 2013 Problem 10. It's based mainly on counting and following the norms stated in the problem itself. Be careful while thinking !

Problem- ISI MStat PSB 2013 Problem 10


There are 10 empty boxes numbered 1,2,......,10 placed sequentially on a circle as shown in the figure,

We perform 100 independent trials. At each trial one box is selected with probability \frac{1}{10} and a ball is placed in each of the two neighboring boxes of the selected one.

Define X_k be the number of balls in the k^{th} box at the end of 100 trials.

(a) Find E(X_k) for 1 \le k \le 10.

(b) Find Cov(X_k, X_5) for 1 \le k \le 10.

Prerequisites


Counting principles

Binomial Distribution

Independence of Events

Solution :

At first this problem may seem a bit of complex, but when one get to see the pattern it starts unfolding. For his types of problem, I find a useful technique is to follow the picture, in this case the picture is being provided, (if not draw it yourself !!)

Given, X_k : # balls in the k^{th} box at the end of 100 trials.

So, the possible values of X_k are 0,1,2,.....,100. and the probability that at the jth trial a ball will added to the kth box is \frac{1}{10} , (why??) .

Now, P(X_k=x)= {100 \choose x}(\frac{1}{5})^{x} (\frac{4}{5})^{100-x} x=0,1,......,100

Clearly, X_k \sim binomial( 100, \frac{1}{5}), from here one can easily find out the the expectation of X_k. But have you thought like this ??

Now, notice that after every slelection of box, 2 balls are added in the system, so at the end of the 10th trial, there will be 200 balls distributed in the system.

So, X_1 +X_2 +.......+ X_{100} =200, which implies \sum_{k} E(X_k)=200, due to symmetry E(X_k)=E(X_l)  \forall k \neq l, So, E(X_k)=20.

(b) Now this part is the cream of this problem, first notice that number of balls in kth box, is dependent on the number of balls in the (k-2)th and (k+2)th box, and vice versa, So, Cov( X_k, X_l)=0 if |k-l|\neq 2.

So, Cov(X_k, X_5)= 0 \forall  k \neq 3,5 \&\ 7, so we just need to find the Cov(X_7,X_5) and Cov(X_3,X_5), and Cov(X_5,X_5)=Var(X_5) .

Now it is sufficient to find the covariance of any of the the above mentioned covariances, as both are symmetric and identical to each other. But for the finding say Cov(X_3,X_5), lets look whats happening in each trial more closely,

let, X_k= i_{k_1} +i_{k_2}+......+i{k_{100}} where , i_{k_j} = \begin{cases} 1  &  if\ a\ ball\ added\ to\ the\ kth\ box\ at\ the\ jth\ trial\ \\ 0 & otherwise \end{cases}

So, clearly, P(i_{k_j}=1)=\frac{1}{5} ; j=1,2,....,100.

So, Cov(X_3,X_5)=Cov( i_{3_1}+i_{3_2}+.....+i_{3_{100}},i_{5_1}+i_{5_2}+....+i_{5_{100}})=\sum_{j=1}^{100} Cov(i_{3_j},i_{5_j}), [Cov(i_{3_j},i_{5_j*})=0 \forall j\neq j*, why ?? ].

So, Cov(X_3,X_5)= 100 Cov(i_{3_1},i_{5_1})=100( E(i_{3_1}i_{5_1})-E(i_{3_1})E(i_{5_1}))=100(P(i_{3_1}=1, i_{5_1}=1)-P(i_{3_1}=1)P(i_{5_1}=1))=100(\frac{1}{10}- \frac{1}{5}\frac{1}{5})=6.

similarly, Cov(X_7,X_5)=6 also, and its easy to find Var(X_5) , so I leave it as an exercise. So, Cov(X_k,X_5)= \begin{cases} 6 & k=3,7 \\ Var(X_5) & k=5 \\ 0 & k\neq 3,5 \or\ 7\end{cases}. Hence we are done !!


Food For Thought

Wait, lets imagine, these boxes are interchanged in such a way that the hth box is replaced with the kth (\neq h) box, this has been done for all possible pairs of (h,k), Now can you show that all there are precisely 10!\sum_{i+2j=10}(i!j!2^j)^{-1} number of arrangements possible ?

Now imagine, in a game there are 1 balls each in every box( boxes are arranged identically as shown in the question), you pick up the ball from the first box and put it into the 2nd one, now you can't take out any ball from a box in which you just put a ball, so you pick a ball from the 3rd box and put it into the 4th and you go on like this, taking a ball frim the ith box and put it into the (i+1)th box, but cant empty the box you just filled. Now, again after the first round you remove the empty boxes and do the same thing again and again, till all the balls are not accumulated in a single box. Which box you think will contain all the balls after you run this process finitely many time ?? If you are in this lottery and you are to choose a bix before this game begin, which box you must choose ??

if the coordinator of the game, starts with any ith box how should your strategy change ?? Give it a thought !!

For help, look for Josephus Problem, you may be moved by it's beauty !


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com
Menu
Trial
Whatsapp
rockethighlight